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ELEMENTARY
COMMUTATIVE ALGEBRA
LECTURE NOTES
H.A. NIELSEN
DEPARTMENT OF MATHEMATICAL SCIENCES
UNIVERSITY OF AARHUS
2005

Elementary Commutative Algebra
H.A. Nielsen

Contents
Prerequisites 7
1. A dictionary on rings and ideals 9
1.1. Rings 9
1.2. Ideals 11
1.3. Prime ideals 13
1.4. Chinese remainders 14
1.5. Unique factorization 15
1.6. Polynomials 16
1.7. Roots 18
1.8. Fields 19
1.9. Power series 20
2. Modules 21
2.1. Modules and homomorphisms 21
2.2. Submodules and factor modules 23
2.3. Kernel and cokernel 25
2.4. Sum and product 28
2.5. Homomorphism modules 30
2.6. Tensor product modules 33
2.7. Change of rings 36
3. Exact sequences of modules 39
3.1. Exact sequences 39
3.2. The snake lemma 43
3.3. Exactness of Hom 48
3.4. Exactness of Tensor 49
3.5. Projective modules 50
3.6. Injective modules 52
3.7. Flat modules 54
4. Fraction constructions 57
4.1. Rings of fractions 57
4.2. Modules of fractions 58
4.3. Exactness of fractions 60
4.4. Tensor modules of fractions 62
4.5. Homomorphism modules of fractions 63
4.6. The polynomial ring is factorial 64
5. Localization 65
5.1. Prime ideals 65
5.2. Localization of rings 67
5

6 CONTENTS
5.3. Localization of modules 68
5.4. Exactness and localization 70
5.5. Flat ring homomorphisms 71
6. Finite modules 73
6.1. Finite Modules 73
6.2. Free Modules 75
6.3. Cayley-Hamilton’s theorem 77
6.4. Nakayama’s Lemma 78
6.5. Finite Presented Modules 80
6.6. Finite ring homomorphisms 83
7. Modules of finite length 85
7.1. Simple Modules 85
7.2. The Length 85
7.3. Artinian Rings 87
7.4. Localization 90
7.5. Local artinian ring 91
8. Noetherian modules 93
8.1. Modules and submodules 93
8.2. Noetherian rings 94
8.3. Finite type rings 95
8.4. Power series rings 97
8.5. Fractions and localization 98
8.6. Prime filtrations of modules 98
9. Primary decomposition 101
9.1. Support of modules 101
9.2. Ass of modules 103
9.3. Primary modules 106
9.4. Decomposition of modules 106
9.5. Decomposition of ideals 108
10. Dedekind rings 111
10.1. Principal ideal domains 111
10.2. Discrete valuation rings 112
10.3. Dedekind domains 113
Bibliography 117
Index 119

Prerequisites
The basic notions from algebra, such as groups, rings, fields and their homomorphisms
together with some linear algebra, bilinear forms, matrices and determinants.
Linear algebra: Fraleigh & Beauregard, Linear algebra, New York 1995.
Algebra: Niels Lauritzen, Concrete abstract algebra, Cambridge 2003.
Also recommended: Jens Carsten Jantzen, Algebra 2, Aarhus 2004.
The propositions are stated complete and precise, while the proofs are quite short.
No specific references to the literature are given. But lacking details may all be
found at appropriate places in the books listed in the bibliography.
Nielsen, University of Aarhus, Winter 2004
7

1
A dictionary on rings and ideals
1.1. Rings
1.1.1. Definition. An abelian group is a set A with an addition A×A → A, (a, b) ↦→
a + b and a zero 0 ∈ A satisfying
(1) associative: (a + b) + c = a + (b + c)
(2) zero: a + 0 = a = 0 + a
(3) negative: a + (−a) = 0
(4) commutative: a + b = b + a
for all a, b, c ∈ A. A subset B ⊂ A is a subgroup if 0 ∈ B and a − b ∈ B for
all a, b ∈ B. The factor group A/B is the abelian group whose elements are the
cosets a + B = {a + b|b ∈ B} with addition (a + B) + (b + B) = (a + b) + B. A
homomorphism of groups φ : A → C respects addition φ(a + b) = φ(a) + φ(b).
The projection π : A → A/B, a ↦→ a + B is a homomorphism. If φ(b) = 0 for all
b ∈ B, then there is a unique homomorphism φ ′ : A/B → C such that φ = φ ′ ◦ π.
1.1.2. Definition. A ring is an abelian group R, addition (a, b) ↦→ a + b and zero
0, together with a multiplication R × R → R, (a, b) ↦→ ab and an identity 1 ∈ R
satisfying
(1) associative: (ab)c = a(bc)
(2) distributive: a(b + c) = ab + ac, (a + b)c = ac + bc
(3) identity : 1a = a = a1
(4) commutative : ab = ba
for all a, b, c ∈ R. If (4) is not satisfied then R is a noncommutative ring. A
subring R ′ ⊂ R is an additive subgroup such that 1 ∈ R ′ and ab ∈ R ′ for all
a, b ∈ R ′ . The inclusion R ′ ⊂ R is a ring extension. A homomorphism of rings
φ : R → S is an additive group homomorphism respecting multiplication and
identity
φ(a + b) = φ(a) + φ(b), φ(ab) = φ(a)φ(b), φ(1) = 1
An isomorphism is a homomorphism φ : R → S having an inverse map φ −1 :
S → R which is also a homomorphism. The identity isomorphism is denoted
1R : R → R.
1.1.3. Remark. (1) A bijective ring homomorphism is an isomorphism.
(2) Recall the usual formulas: a + (−b) = a − b, 0a = 0, (−1)a = −a.
(3) The identity 1 is unique.
(4) A ring R is nonzero if and only if the elements 0 = 1.
(5) If φ : R → S is a ring homomorphism, then φ(0) = 0 and φ(R) is a subring
of S.
(6) The unique additive group homomorphism Z → R, 1 ↦→ 1 is a ring homomorphism.
9

10 1. A DICTIONARY ON RINGS AND IDEALS
1.1.4. Proposition. Let R1, R2 be rings. The product ring is the product of additive
groups R1×R2, ((a1, a2), (b1, b2)) ↦→ (a1+b1, a2+b2), with coordinate multiplication
((a1, a2), (b1, b2)) ↦→ (a1b1, a2b2). The element (1, 1) is the identity. The
projections R1 × R2 → R1, (a1, a2) ↦→ a1 and R1 × R2 → R2, (a1, a2) ↦→ a2 are
ring homomorphisms.
1.1.5. Lemma. In a ring R the binomial formula is true
(a + b) n n
n
= a
k
n−k b k
a, b ∈ R and n a positive integer.
k=0
Proof. The multiplication is commutative, so the usual proof for numbers works.
Use the binomial identity
n
+
k − 1
together with induction on n.
n
=
k
n + 1
1.1.6. Definition. a ∈ R is a nonzero divisor if ab = 0 for all b = 0 otherwise a
zero divisor. a is a unit if there is a b such that ab = 1.
1.1.7. Remark. (1) A unit is a nonzero divisor.
(2) If ab = 1 then b is uniquely determined by a and denoted b = a −1 .
1.1.8. Definition. A nonzero ring R is a domain if every nonzero element is a
nonzero divisor and a field if every nonzero element is a unit. Clearly a field is a
domain.
1.1.9. Example. The integers Z is a domain. The units in Z are {±1}. The rational
numbers Q, the real numbers R and the complex numbers C are fields. The natural
numbers N is not a ring.
1.1.10. Example. The set of n × n-matrices with entries from a commutative ring
is an important normally noncommutative ring.
1.1.11. Exercise. (1) Show that the product of two domains is never a domain.
(2) Let R be a ring. Show that the set of matrices
a
U2 =
0
a,
b
b ∈ R
a
with matrix addition and matrix multiplication is a ring.
(3) Show that the set of matrices with real number entries
a,
a −b
b ∈ R
b a
with matrix addition and multiplication is a field isomorphic to C.
(4) Show that the composition of two ring homomorphisms is again a ring homomorphism.
(5) Show the claim 1.1.3 that a bijective ring homomorphism is a ring isomorphism.
(6) Let φ : 0 → R be a ring homomorphism from the zero ring. Show that R is itself the
zero ring.
k

1.2. IDEALS 11
1.2. Ideals
1.2.1. Definition. Let R be a ring. An ideal I is an additive subgroup of R such
that ab ∈ I for all a ∈ R, b ∈ I. A proper ideal is an ideal I = R.
1.2.2.
Lemma. Let {Iα} be a family of ideals in R. Then the additive subgroups
α Iα and
α Iα are ideals.
Proof. The claim for the intersection is clear. Use the formulas bα + cα =
(bα + cα) and a bα = abα to conclude the claim for the sum.
1.2.3. Definition. The intersection, 1.2.2, of all ideals containing a subset B ⊂ R
is the ideal generated by B and denoted (B) = RB = BR. It is the smallest ideal
containing B. A principal ideal (b) = Rb is an ideal generated by one element. A
finite ideal (b1, . . . , bn) is an ideal generated by finitely many elements. The zero
ideal is (0) = {0}. The ring itself is a principal ideal, (1) = R. The ideal product
of two ideals I, J is denoted IJ and is the ideal generated by all ab, a ∈ I, b ∈ J.
This generalizes to the product of finitely many ideals. The power of an ideal is
denoted I n . The colon ideal I : J is the ideal of elements a ∈ R such that aJ ⊂ I.
1.2.4. Example. (1) Every ideal in Z is principal.
(2) In a field (0), (1) are the only ideals.
(3) A subring is normally not an ideal.
(4) Let K be a field. In K × K there are four ideals (0), (1), ((1, 0)), ((0, 1)).
1.2.5. Proposition. Let R be a ring and B a subset, then RB =
b∈B Rb
A principal ideal is
A finite ideal is
RB = {a1b1 + · · · + anbn|ai ∈ R, bi ∈ B}
Rb = {ab|a ∈ R}
(b1, . . . , bn) = Rb1 + · · · + Rbn
Proof. The righthand side is contained in the ideal RB. Moreover the righthand
side is an ideal containing B, so equality.
1.2.6. Definition. Let φ : R → S be a ring homomorphism. For an ideal J ⊂ S
the contracted ideal is φ −1 (J) ⊂ R and denoted J ∩ R. The kernel is the ideal
Ker φ = φ −1 (0). For an ideal I ⊂ R the extended ideal is the ideal φ(I)S ⊂ S
and denoted IS. Note that (J ∩ R)S ⊂ J and I ⊂ (IS) ∩ R.
1.2.7. Lemma. Let I ⊂ R be an ideal and let R/I be the additive factor group.
The multiplication
R/I × R/I → R/I, (a + I, b + I) ↦→ ab + I
is well defined. Together with the addition the conditions of 1.1.2 are satisfied.
Proof. If a + I = a ′ + I and b + I = b ′ + I then a − a ′ , b − b ′ ∈ I and so
ab − a ′ b ′ = a(b − b ′ ) + b ′ (a − a ′ ) ∈ I. Therefore ab + I = a ′ b ′ + I and the
multiplication is well defined. Clearly 1.1.2 are satisfied.
1.2.8. Definition. Let R be a ring and I an ideal, then the factor ring is the additive
factor group R/I with addition (a + I, b + I) ↦→ (a + b) + I and multiplication,
1.2.7, (a + I, b + I) ↦→ ab + I. The projection π : R → R/I, a ↦→ a + I is a ring
homomorphism.

12 1. A DICTIONARY ON RINGS AND IDEALS
1.2.9. Proposition. Let φ : R → S be a ring homomorphism.
(1) Let I ⊂ Ker φ be an ideal. Then there is a unique ring homomorphism
φ ′ : R/I → S such that φ = φ ′ ◦ π.
R
π
R/I
φ
(2) The homomorphism φ ′ : R/ Ker φ → S is a ring isomorphism onto the
subring φ(R) of S.
R
π
R/ Ker φ
φ
φ ′
φ ′
S
φ(R)
(3) For any ideal J ⊂ S, I = φ −1 (J) ⊂ R is an ideal and the map φ ′ : R/I →
S/J is an injective ring homomorphism.
Proof. The statements are clear for the addition and the factor map φ ′ (a + I) =
φ(a) is clearly a ring homomorphism.
1.2.10. Corollary. Let π : R → R/I be the projection. The map I ′ ↦→ J =
π −1 (I ′ ) gives a bijective correspondence between ideals I ′ in R/I and ideals J in
R containing I. Also I ′ = π(J) = J/I. This correspondence preserves inclusions,
sums and intersections of ideals.
1.2.11. Corollary. Let I ⊂ J ⊂ R be ideals. Then there is a canonical isomorphism
R/J → (R/I)/(J/I)
Proof. The kernel of the surjective east-south composite
R
R/J
R/I
(R/I)/(J/I)
is J. By 1.2.9 the horizontal lower factor map gives the isomorphism.
1.2.12. Example. For any integer n the ideals in the factor ring Z/(n) correspond
to ideals (m) ⊂ Z where m divides n.
1.2.13. Definition. Let R be a ring. The additive kernel of the unique ring homomorphism
Z → R is a principal ideal generated by a natural number char(R), the
characteristic of R. Z/(char(R)) is isomorphic to the smallest subring of R.
1.2.14. Proposition. If the characteristic char(R) = p is a prime number, then
the Frobenius homomorphism R → R, a ↦→ a p is a ring homomorphism.
Proof. By the binomial formula 1.1.4
(a + b) p p
p
=
k
k=0
a p−k b k = a p + b p

1.3. PRIME IDEALS 13
since a prime number p divides p
k , 0 < k < p. Clearly (ab) p = apbp .
1.2.15. Exercise. (1) Let I, J be ideals in R. Show that the ideal product
IJ = {a1b1 + · · · + anbn|ai ∈ I, bi ∈ J}
(2) Let I ⊂ R be an ideal. Show that I = I : R.
(3) Show that a ∈ R is a unit if and only if (a) = R.
(4) Show that a ring is a field if and only if (0) = (1) are the only two ideals.
(5) Show that a nonzero ring K is a field if and only if any nonzero ring homomorphism
φ : K → R is injective.
(6) Let m, n be natural numbers. Determine the ideals in Z
(m, n), (m) + (n), (m) ∩ (n), (m)(n)
as principal ideals.
(7) Show that a additive cyclic group has a unique ring structure.
(8) Let p be a prime number. What is the Frobenius homomorphism on the ring Z/(p)?
1.3. Prime ideals
1.3.1. Definition. Let R be a ring and P = R a proper ideal.
(1) P is a prime ideal if for any product ab ∈ P either a ∈ P or b ∈ P . This
amounts to: if a, b /∈ P then ab /∈ P .
(2) P is a maximal ideal if no proper ideal = P contains P .
1.3.2. Proposition. Let P be a prime ideal and I1, . . . In ideals such that I1 . . . In ⊂
P , then some Ik ⊂ P .
Proof. If there exist ak ∈ Ik\P for all k, then since P is prime a1 . . . an ∈
I1 . . . In\P contradicting the inclusion I1 . . . In ⊂ P .
1.3.3. Proposition. Let R be a ring and P an ideal.
(1) P is a prime ideal if and only if R/P is a domain.
(2) P is a maximal ideal if and only if R/P is a field.
Proof. Remark P = R ⇔ R/P = 0. (1) Assume a + P, b + P are nonzero in
R/P . Then a, b /∈ P . So if P is prime then by 1.3.1 ab /∈ P and ab + P is nonzero
in R/P . It follows that R/P is a domain. The converse is similar.
(2) Assume R/P is a field and a /∈ P . Then a + P is a unit in R/P and there is
b such that ab − 1 ∈ P . It follows that the ideal (a) + P = R and therefore P is
maximal. The converse is similar.
1.3.4. Corollary. (1) A maximal ideal is a prime ideal.
(2) A ring is an domain if and only if the zero ideal is a prime ideal.
(3) A ring a field if and only if the zero ideal is a maximal ideal.
1.3.5. Corollary. (1) If φ : R → S is a ring homomorphism and Q ⊂ S a prime
ideal then φ −1 (Q) is a prime ideal of R.
(2) Let I ⊂ R be an ideal. An ideal I ⊂ P is a prime ideal in R if and only if
P/I is a prime ideal in R/I.
(3) Let I ⊂ R be an ideal. An ideal I ⊂ P is a maximal ideal in R if and only if
P/I is a maximal ideal in R/I.
Proof. (1) By 1.2.9 R/φ −1 (Q) is a subring of the domain S/Q. (2) (3) By 1.2.11
R/P (R/I)/(P/I).

14 1. A DICTIONARY ON RINGS AND IDEALS
1.3.6. Example. An ideal in Z is a prime ideal if it is generated by 0 or a prime
number. Any nonzero prime ideal is a maximal ideal.
1.3.7. Definition. For an ideal I in a ring R the radical is
√ I = {a ∈ R|a n ∈ I for some n}
a is nilpotent is a n = 0 for some positive integer n. A ring is reduced if the
nilradical √ 0 = 0, that is if 0 is the only nilpotent element.
1.3.8. Proposition. (1) The radical of an ideal is an ideal.
(2) The nilradical is contained in any prime ideal.
(3) A domain is reduced.
Proof. (1) If am , bn ∈ I then by the binomial formula
(a + b) m+n m+n
m + n
=
a
k
m+n−k b k ∈ I
k=0
and the radical is an ideal. (2) (3) Clearly a nilpotent element is contained in any
prime ideal.
1.3.9. Exercise. (1) Show that the characteristic of a domain is either 0 or a prime
number.
(2) Let m, n be a natural numbers. Show that n + (m) ∈ Z/(m) is a unit if and only if
m, n are relative prime.
(3) Let m be a natural number. Show that Z/(m) is reduced if m is square free.
(4) Show that a product of reduced rings is reduced.
(5) Let a be nilpotent. Show that 1 − a is a unit.
(6) Let I, J be ideals. Show that √ IJ = √ I ∩ J = √ I ∩ √ J.
(7) Assume an ideal I is contained in a prime ideal P . Show that √ I ⊂ P .
1.4. Chinese remainders
1.4.1. Definition. Ideals I, J ⊂ R are comaximal ideals if I + J = R.
1.4.2. Proposition (Chinese remainder theorem). Let I1, . . . , Ik be pairwise comaximal
ideals in a ring R. Then
(1) For a1, . . . , ak ∈ R there is a a ∈ R, such that a−am ∈ Im for m = 1, . . . , k
(2)
(3) The product of projections
is an isomorphism.
Proof. (1) For each m
I1 · · · Ik = I1 ∩ · · · ∩ Ik
R/I1 · · · Ik → R/I1 × · · · × R/Ik
R =
(Im + In) = Im +
n=m
n=m
So choose um ∈ Im and vm ∈
n=m In with um + vm = 1. Put a = a1v1 + · · · +
akvk. Then a − am = · · · + amum + · · · ∈ Im. (2) For a in the intersection assume
by induction that a ∈ I2 · · · Ik. From the proof of (1) a = u1a + av1 ∈ I1 · · · Ik.
(3) Subjectivity follows from (1). The kernel is the intersection which by (2) is the
product. 1.2.9 gives the isomorphism.
In

1.5. UNIQUE FACTORIZATION 15
1.4.3. Corollary. Let P1, . . . , Pk be pairwise different maximal ideals in a ring R.
Then
and
P n1
1 · · · P nk n1
nk
k = P1 ∩ · · · ∩ Pk R/P n1
1 · · · P nk
n1
nk
k → R/P1 × · · · × R/Pk is an isomorphism.
1.4.4. Definition. An element e in a ring R is idempotent if e = e 2 . A nontrivial
idempotent is an idempotent e = 0, 1.
1.4.5. Proposition. A ring R is a product of two nonzero rings if and only if it
contains a nontrivial idempotent e.
Proof. Use that the ideals Re and R(1 − e) are proper and comaximal.
1.4.6. Exercise. (1) Show that for a prime number p the rings Z/(p2 ) and Z/(p) ×
Z/(p) are not isomorphic.
(2) Let n = p n1
1 be a factorization into different primes. Show that
. . . pnk
k
Z/(p n1
1
· · · pnk
k
) → Z/(pn1 1 ) × · · · × Z/(pnk k )
is an isomorphism.
(3) Let elements e1 + e2 = 1 with e1e2 = 0 be given in a ring R. Show that R
R/(e1) × R/(e2).
(4) Let I, J be ideals such that √ I, √ J are comaximal. Show that I, J are comaximal.
1.5. Unique factorization
1.5.1. Lemma. Let R be a domain and (a) = (b) principal ideals. Then there is a
unit u ∈ R such that b = ua.
Proof. b = ua and a = vb giving b = uvb. If b = 0 then uv = 1 showing that u is
a unit.
1.5.2. Definition. Let R be a domain and P the set of principal ideals different
from (0) and R. By 1.5.1 multiplication of generators gives a well defined multiplication
of principal ideals on P. An element in P maximal for inclusion is an
irreducible principal ideal. A generator of an irreducible element is an irreducible
element in R.
1.5.3. Definition. A domain R is a unique factorization domain if
(1) Every irreducible element in P is a prime ideal.
(2) Every element in P is a product of irreducible elements.
1.5.4. Proposition. In a unique factorization domain the factorization of elements
in P into irreducibles is unique up to order.
Proof. Proceed by induction on the shortest factorization of an element in P. Let
(a1) . . . (am) = (b1) . . . (bn) be factorizations into irreducibles. (a1) is a prime
ideal, so by 1.3.2 and reordering (b1) = (a1). By cancellation m − 1 = n − 1 and
(ai) = (bi) after a reordering.
1.5.5. Definition. A domain R is a principal ideal domain if every ideal is principal.
1.5.6. Proposition. A principal ideal domain R is a unique factorization domain.

16 1. A DICTIONARY ON RINGS AND IDEALS
Proof. Let (a) be irreducible and x, y /∈ (a). Then (a, x), (a, y) are principal ideals
properly containing (a) giving (a, x) = (a, y) = R. Let ba + cx = da + ey = 1
and look at (ba+cx)(da+ey) = 1 to see that xy /∈ (a). It follows that (a) is prime,
part (1) of 1.5.3. If (b) is not irreducible, then (b) = (a1)(b1) for some irreducible
(a1) and (b) ⊂ (b1). Continue this process to get (b) = (a1) . . . (an)(bn) for some
irreducibles (a1), . . . (an). The chain of ideals (b) ⊂ (b1) ⊂ · · · ⊂ (bn) has a
union
n (bn) which is a principal ideal. The generator of the union must be in
some (bn). Therefore (bn) = (bn+1) giving that (bn) is irreducible. This gives a
factorization required in 1.5.3 part (2).
1.5.7. Example. The integers Z is a principal ideal domain and therefore a unique
factorization domain.
1.5.8. Definition. The supremum of a set of elements in P is the greatest common
divisor and an infimum is the least common multiple.
1.5.9. Corollary. In a unique factorization domain the greatest common divisor
and the least common multiple of a finite set of elements exist.
If (a) = (p m1
1 ) . . . (pmk k ) and (b) = (pn1 1 ) . . . (pnk k ) with mi, ni ≥ 0 then greatest
common divisor is (p min(m1,n1)
1 ) . . . (p min(mk,nk)
k ) and least common multiple is
).
(p max(m1,n1)
1
) . . . (p max(mk,nk)
k
1.5.10. Exercise. (1) Show that an irreducible element in a principal ideal domain generates
a maximal ideal.
(2) Show that there are infinitely many prime numbers.
(3) Let Z[ √ −1] be the smallest subring of C containing √ −1. Show that Z[ √ −1] is a
principal ideal domain.
(4) Let Z[ √ −5] be the smallest subring of C containing √ −5. Show that Z[ √ −5] is not
a unique factorization domain.
1.6. Polynomials
1.6.1. Definition. Let R be a ring. The polynomial ring R[X] is the additive
group given by the direct sum
n RXn , n = 0, 1, 2, . . . consisting of all finite
sums f = a0 + a1X + . . . amX m , a polynomial with an ∈ R being the n ′ th
coefficient. Multiplication is given by XiX j = Xi+j extended by linearity. If
g = b0 + b1X + . . . bnX n is an other polynomial, then
f + g = (a0 + b0) + (a1 + b1)X + · · · + (ak + bk)X k + . . .
fg = a0b0 + (a0b1 + a1b0)X + · · · + (a0bk + a1bk−1 + · · · + akb0)X k + . . .
A monomial is polynomial of form aX n . The construction may be repeated to
give the polynomial ring in n-variables R[X1, . . . , Xn] or even in infinitely many
variables.
1.6.2. Definition. The degree, deg(f), of a polynomial 0 = f ∈ R[X] is the
index of the highest nonzero coefficient, the leading coefficient. A polynomial
with leading coefficient the identity is a monic polynomial.
1.6.3. Remark. (1) R is identified with the subring of constants in the polynomial
ring R[X1, . . . , Xn].
(2) The nonzero constants are the polynomials of degree 0.
(3) The constant polynomial 1 is the unique monic polynomial of degree 0 and
the identity in the polynomial ring.

1.6. POLYNOMIALS 17
1.6.4. Proposition. Let 0 = f, g ∈ R[X].
(1) If fg = 0 then deg(fg) ≤ deg(f) + deg(g).
(2) If the leading coefficient of f or g is a nonzero divisor in R, then fg = 0 and
deg(fg) = deg(f) + deg(g)
Proof. (1) This is clear. (2)Clearly the leading coefficient of the product is the
product of the leading coefficients.
1.6.5. Corollary. Let R be a domain.
(1) The polynomial ring R[X] is a domain.
(2) The units in R[X] are the constants, which are units in R.
1.6.6. Proposition. Let R be a domain, 0 = f, d ∈ R[X] polynomials with d
monic. Then there are a unique q, r ∈ R[X] such that
f = qd + r, r = 0 or deg(r) < deg(d)
Proof. Induction on deg(f). If deg(f) < deg(d) then q = 0, r = f. Otherwise
if a is the leading coefficient of f, then f − adX deg(f)−deg(d) has degree less than
deg(f). By induction f − adX deg(f)−deg(d) = qd + r giving the claim.
1.6.7. Proposition. Let φ : R → S be a ring homomorphism. For any element
b ∈ S there is a unique ring homomorphism R[X] → S extending φ and mapping
X ↦→ b.
Proof.
a0 + a1X + . . . amX m ↦→ φ(a0) + φ(a1)b + . . . φ(am)b m
is clearly the one and only choice.
1.6.8. Definition. The homomorphism in 1.6.7 is the evaluation map at b in S.
The image of a polynomial f ∈ R[X] is denoted f(b) ∈ S.
1.6.9. Proposition. Let I ⊂ R be an ideal. Then there is a canonical isomorphism.
(R/I)[X] R[X]/IR[X]
Proof. There is an obvious pair of inverse homomorphisms constructed by 1.2.9
and 1.6.7.
1.6.10. Corollary. If P ⊂ R is a prime ideal, then P R[X] ⊂ R[X] is a prime
ideal.
1.6.11. Definition. Let φ : R → S be a ring homomorphism and B ⊂ S a subset.
The ring generated over R by B is
R[B] = φ(R)[B] ⊂ S
the smallest subring of S containing φ(R) ∪ B. If there is a finite subset B such
that R[B] = S then S is a finite type ring or a finitely generated ring over R.
1.6.12. Corollary. Let φ : R → S be a ring homomorphism.
(1) If bα ∈ S and Xα is a family of variables, then there is a surjective ring
homomorphism
R[Xα] → R[bα], Xα ↦→ bα
making R[bα] a factor ring of the polynomial ring R[Xα].

18 1. A DICTIONARY ON RINGS AND IDEALS
(2) If S is a finite type ring over R, then S is a factor ring of a polynomial ring
in finitely many variables over R.
1.6.13. Exercise. (1) Let K be a field. Show that there are infinitely many prime ideals
in K[X].
(2) What are the units in the ring Z[X]/(1 − 2X)?
(3) Determine the prime ideals in Q[X]/(X − X 2 ).
(4) Show that the ring Z[X] is not a principal ideal domain.
(5) Show that the ring Q[X, Y ] is not a principal ideal domain.
1.7. Roots
1.7.1. Definition. Let φ : R → S be a ring homomorphism and f ∈ R[X] a
polynomial. An element b ∈ S is a root of f (in S) if the evaluation f(b) = 0.
1.7.2. Proposition. Let R be a domain. An element a ∈ R is a root of the polynomial
f ∈ R[X] if and only if there is a q ∈ R[X] such that
f = q(X − a)
Proof. By 1.6.6 f = q(X −a)+r. It follows that a is a root if and only if r = 0.
1.7.3. Corollary. Let R is a domain. There are at most deg(f) roots in a nonzero
polynomial f ∈ R[X].
1.7.4. Definition. The multiplicity of a root a of a nonzero polynomial f ∈ R[X]
is highest m such that
f = q(X − a) m
A root of multiplicity m = 1 is a simple root.
1.7.5. Corollary. Let R is a domain. If m1, . . . , mk are the multiplicities of the
roots of a nonzero polynomial f ∈ R[X], then m1 + · · · + mk ≤ deg(f).
1.7.6. Definition. The derivative of a polynomial f = anX n ∈ R[X] is
1.7.7. Lemma. The derivative satisfies
(1) (f + g) ′ = f ′ + g ′ .
(2) (fg) ′ = f ′ g + fg ′
(3) If f is constant, then f ′ = 0.
f ′ = nanX n−1
1.7.8. Proposition. Let R is a domain. An element a ∈ R is a root of multiplicity
m > 1 of a nonzero f ∈ R[X] if and only if a is a root of f and f ′ .
Proof. By 1.6.6 f = q(X − a) 2 + cX + d and by 1.7.7 f ′ = q ′ (X − a) 2 + 2q(X −
a) + c. I follows that a is a root of multiplicity m > 1 if and only if c = d = 0.
1.7.9. Exercise. (1) Let a1, . . . ak be roots with multiplicities m1, . . . , mk in a polynomial
f. Show that m1 + · · · + mk ≤ deg(f).
(2) Let K be a field and let a1, . . . , an ∈ K. Show that the ideal (X1 −a1, . . . , Xn −an)
is maximal in K[X1, . . . , Xn].
(3) Let the characteristic char(R) = n > 0. What is (X n ) ′ in R[X].

1.8. FIELDS 19
1.8. Fields
1.8.1. Definition. Let p ∈ Z be a prime number. The factor ring Fp = Z/(p) is a
field with p elements. Together with Q they constitute the prime fields.
1.8.2. Proposition. Let K be a field then the polynomial ring K[X] is a principal
ideal domain.
Proof. Let d = 0 be a polynomial of lowest degree in an ideal I. Given f ∈ I then
by 1.6.6 f = qd + r with r = f − qd ∈ I. By degree considerations r = 0 and
I = (d).
1.8.3. Corollary. Let K be a field then the polynomial ring K[X] is a unique
factorization domain.
Proof. Follows from 1.5.6.
1.8.4. Definition. A subfield is a subring, which is a field. A field extension is the
inclusion of a subfield K ⊂ L in a field. A finite field extension K ⊂ L is an
extension, where L is finite dimensional as a vector space over K.
1.8.5. Example. (1) Let K be a field and f an irreducible polynomial in K[X].
Then K ⊂ K[X]/(f) is a finite field extension.
(2) Let K ⊂ R ⊂ L be a subring in a finite field extension. Then R is a field.
Namely multiplication on R with a nonzero element of R is a K-linear map
on the finite dimensional K-vector space R and therefore an isomorphism.
1.8.6. Proposition. (1) Let K be a field and f a polynomial in K[X]. Then there
is a finite field extension K ⊂ L such that f factors in linear factors in L[X].
(2) If K ⊂ L1 and K ⊂ L2 are finite field extensions then there is a finite field
extension K ⊂ L such that L1 ∪ L2 ⊂ L.
Proof. (1) Assume f irreducible. In L = K[X]/(f) the class X + (f) is a root of
f. In general proceed adjoining roots of irreducible factors of f. (2) An element x
in a finite field extension K ⊂ K ′ is the root of the irreducible monic polynomial
f generating the kernel of the evaluation homomorphism K[X] → K ′ , X ↦→ x.
Now proceed by (1) adjoining elements in L2 to L1.
1.8.7. Proposition. Let p ∈ Z be a prime number. For any power q = p n there is
a field Fq with q elements, unique up to isomorphism.
Proof. Let Fp ⊂ K be a field extension, where X q − X factors into linear factors,
1.8.6 (1). The subset of roots is the set of elements fixed under n-times the
Frobenius and therefore a subring being a subfield by 1.8.5 (2). The derivative
(X q − X) ′ = −1 so by 1.7.8 there are q elements in this subfield. Uniqueness
follows from 1.8.6 (2).
1.8.8. Exercise. (1) Show that the ring R[X]/(X 2 + 1) is isomorphic to the field of
complex numbers.
(2) Show that the ring F2[X]/(X 2 + X + 1) is a field with 4 elements.
(3) Show that the ring F2[X]/(X 3 + X + 1) is a field with 8 elements.
(4) Let K ⊂ L be a field extension of fields of characteristic 0 and let a ∈ L be a root of
an irreducible polynomial f ∈ K[X]. Show that a is a simple root.
(5) Let p be a prime number. Show that Fp is the only ring with p elements.
(6) Let p be a prime number. Show that a ring with p 2 elements is isomorphic to one of
four non isomorphic Z/(p 2 ), Fp × Fp, Fp[X]/(X 2 ), F p 2.

20 1. A DICTIONARY ON RINGS AND IDEALS
1.9. Power series
1.9.1. Definition. Let R be a ring. The power series ring R[[X]] is the additive
group
n RXn , n = 0, 1, 2, . . . of all power series anX n with n ′ th coefficient
an ∈ R. Multiplication is given by X i X j = X i+j extended by linearity. For
another power series bnX n the sum and product are
anX n + bnX n = (an + bn)X n
anX n · bnX n = (
k
an−kbk)X n
The construction may be repeated to give the power series ring in n-variables
R[[X1, . . . , Xn]] or even in infinitely many variables.
1.9.2. Remark. The polynomial ring is identified as a subring R[X] ⊂ R[[X]] of
power series with only finitely many nonzero terms.
1.9.3. Definition. The order, o(f), of a power series 0 = f ∈ R[[X]] is the index
of the least nonzero coefficient.
1.9.4. Proposition. If R is a domain, then R[[X]] is a domain and for 0 = f, g ∈
R[X]
o(fg) = o(f) + o(g)
Proof. Clearly the lowest nonzero coefficient in the product is the product of the
two lowest nonzero coefficients.
1.9.5. Proposition. A power series f = anX n is a unit if and only if a0 is a
unit.
Proof. It suffices to look at a power series f = 1 − gX. Then the power series
1/f = 1 + gX + g 2 X 2 + · · · + g n X n + . . . is well defined and f · 1/f = 1.
1.9.6. Proposition. If K is a field, then K[[X]] is a principal ideal domain. and
(X) is the only nonzero prime ideal.
Proof. If the lowest order of an element in an ideal I is n. Then clearly I =
(X n ).
1.9.7. Corollary. If K is a field, then K[[X]] is a unique factorization domain.
1.9.8. Proposition. Let I ⊂ R be an ideal. Then there is a canonical surjective
homomorphism
R[[X]]/IR[[X]] → R/I[[X]]
1.9.9. Corollary. If Q ⊂ R[[X]] is a maximal ideal, then P = Q ∩ R ⊂ R is a
maximal ideal and Q = (P, X).
Proof. X ∈ Q so R/Q ∩ R R[[X]]/Q.
1.9.10. Exercise. (1) Show that the ring Z[[X]] is not a principal ideal domain.
(2) Show that the ring Q[[X, Y ]] is not a principal ideal domain.
(3) Let K be a field. Show that (X1, . . . , Xn) is the unique maximal ideal in the power
series ring K[[X1, . . . , Xn]].
(4) Let a ∈ R be nilpotent. Show that the ring R[[X]]/(X − a) is isomorphic to R.
(5) What is R[[X]]/(X − a) if a ∈ R is a unit?
(6) Let I ⊂ R be an ideal. Show that IR[[X]] ⊂ R[[X]] is not a maximal ideal.

2
Modules
2.1. Modules and homomorphisms
2.1.1. Definition. Let R be a ring. A module (R-module) is an abelian group
M, addition (x, y) ↦→ x + y and zero 0, together with a scalar multiplication
R × M → M, (a, x) ↦→ ax satisfying
(1) associative : (ab)x = a(bx)
(2) bilinear : a(x + y) = ax + ay, (a + b)x = ax + bx
(3) identity: 1x = x
for all a, b ∈ R, x, y ∈ M. A submodule M ′ ⊂ M is an additive subgroup such
that ax ∈ M ′ for all a ∈ R, x ∈ M ′ . A homomorphism is an additive group
homomorphism f : M → N respecting scalar multiplication
f(x + y) = f(x) + f(y), f(ax) = af(y)
for all a ∈ R, x, y ∈ M. An isomorphism is a homomorphism f : M → N
having an inverse map f −1 : N → M which is also a homomorphism. The
identity isomorphism is denoted 1M : M → M.
2.1.2. Lemma. Let R be a ring and M a module.
(1) a0 = 0 = 0x
(2) (−1)x = −x
(3) (−a)x = −(ax) = a(−x)
for all a ∈ R, x ∈ M.
Proof. (1) Calculate a0 = a(0 + 0) = a0 + a0 and cancel to get a0 = 0. Similarly
0x = 0. (2) By (1) 0 = 0x = (1 + (−1))x = x + (−1)x, so conclude −x =
(−1)x. (3) Calculate (−a)x = ((−1)a)x = (−1)(ax) = −(ax). Similarly
a(−x) = −(ax).
2.1.3. Lemma. Let R be a ring and f : M → N a homomorphism of modules.
(1) f(0) = 0.
(2) f(ax + by) = af(x) + bf(y).
(3) f(−x) = −f(x) for all x ∈ M.
for all a, b ∈ R, x, y ∈ M.
Proof. (1) Calculate f(0) = f(0 + 0) = f(0) + f(0) and conclude f(0) = 0.
(2) Calculate f(ax + by) = f(ax) + f(by) = af(x) + bf(y). (3) By 2.1.2
f(−x) = f((−1)x) = (−1)f(x) = −f(x).
2.1.4. Example. (1) The zero group is the zero module.
(2) Over the zero ring the zero module is the only module.
(3) The zero subgroup of a module is the zero submodule.
(4) The ring R is a module under multiplication. An ideal is a submodule.
21

22 2. MODULES
(5) If R is a field, a module is a vector space and a homomorphism is a linear
map.
(6) A module over Z is an abelian group and an additive map of abelian groups
is a homomorphism.
2.1.5. Proposition. A bijective homomorphism is an isomorphism.
Proof. Let f : M → N be a bijective homomorphism of R-modules and let g :
N → M be the inverse map. For x, y ∈ N write x = f(g(x)), y = f(g(y)) and
get additivity of g, g(x + y) = g(f(g(x)) + f(g(y))) = g(f(g(x) + g(y))) =
g(x) + g(y). Similarly for a ∈ R g(ax) = g(af(g(x))) = g(f(ag(x))) = ag(x),
so g respects scalar multiplication and is a homomorphism.
2.1.6. Lemma. Let a ∈ R and M be a module. The map M → M, x ↦→ ax is a
homomorphism.
Proof. Let f(x) = ax and calculate f(x+y) = a(x+y) = ax+ay = f(x)+f(y)
and f(bx) = a(bx) = (ab)x = (ba)x = b(ax) = bf(x) to get that f is a
homomorphism. Remark that the last calculation uses that R is commutative 1.1.2
(4).
2.1.7. Definition. Let a ∈ R and M be a module.
(1) The scalar multiplication with a is the homomorphism, 2.1.6,
aM : M → M, x ↦→ ax
(2) a ∈ R is a nonzero divisor on M if scalar multiplication aM is injective, i.e.
ax = 0 for all 0 = x ∈ M. Otherwise a is a zero divisor.
2.1.8. Remark. The two notions of nonzero divisor on R : 1.1.6 as element in the
ring and 2.1.4 as scalar multiplication on the ring coincide.
2.1.9. Example. If R is a field, then scalar multiplication on a vector space is either
zero or an isomorphism.
2.1.10. Lemma. Let φ : R → S be a ring homomorphism and N an S-module.
The map
R × N → N, (a, x) ↦→ φ(a)x
is an R-scalar multiplication on N, 2.1.1.
Proof. Let a, b ∈ R, x, y ∈ N and µ(a, x) = φ(a)x. Calculate µ(a + b, x) =
φ(a + b)x = φ(a)x + φ(b)x = µ(a, x) + µ(b, x), µ(a, x + y) = φ(a)(x + y) =
φ(a)x + φ(a)y = µ(a, x) + µ(a, y), µ(1, x) = φ(1)x = 1x = x and µ(ab, x) =
φ(ab)x = φ(a)φ(b)x = µ(a, µ(bx)) showing the conditions 2.1.1.
2.1.11. Definition. Let φ : R → S be a ring homomorphism. The restriction of
scalars of an S-module N is the same additive group N viewed as an R-module
through φ. The scalar multiplication is 2.1.10
R × N → N, (a, x) ↦→ ax = φ(a)x
An S-module homomorphism g : N → N ′ is also an R-module homomorphism.
2.1.12. Example. (1) The scalar multiplication with a Restriction of scalars for
the unique ring homomorphism Z → R give just the underlying abelian group
of a module, 2.1.4 (6).

2.2. SUBMODULES AND FACTOR MODULES 23
(2) Let I ⊂ R be an ideal. Restriction of scalars along the projection R → R/I
gives any R/I-module as an R-module.
2.1.13. Proposition. Let R be a ring. There is a dictionary:
(1) To an R[X]-module N associate the pair (N, f) consisting of N as R-module
through restriction of scalars and f = XN : N → N, f(y) = Xy scalar
multiplication with X as an R-module homomorphism. An R[X]-homomorphism
g : N → N ′ gives an R-homomorphism such that g ◦ f = f ′ ◦ g.
(2) To a pair (N, f) of an R-module and a homomorphism f : N → N associate
the R[X]-module with abelian group N and scalar multiplication determined
by Xy = f(y) for y ∈ N. Note
( anX n )y = anf ◦n (y)
An R-homomorphism g : N → N ′ such that g ◦ f = f ′ ◦ g is an R[X]homomorphism.
Proof. The statement is an algorithm to follow.
2.1.14. Proposition. Let R be a ring and M a module. The abelian group R ⊕ M
with multiplication
(a + x)(b + y) = ab + (ay + bx)
is a ring. R is a subring and M is an ideal.
Proof. Simple calculations show that the conditions 1.1.2 are satisfied.
2.1.15. Exercise. (1) Show that a composition of homomorphisms is a homomorphism.
(2) Show that composition of scalar multiplications with a, b ∈ R on a module M is a
scalar multiplication with the product, aM ◦ bM = (ab)M .
(3) Let φ : R → S be a ring homomorphism. Show that φ is an R-module homomorphism,
when S is viewed as R-module through restriction of scalars 2.1.11.
(4) Fill out the dictionary 2.1.13.
2.2. Submodules and factor modules
2.2.1. Lemma. Let R be a ring and M a module. Let Nα be a family of submodules.
Then the additive subgroups
α Nα and
α Nα are submodules.
Proof. Use the formulas xα + yα = (xα + yα) and a xα = axα to
conclude that Nα is a submodule. If x, y ∈ Nα for all α, then x + y, ax ∈ Nα
for all α, so Nα is a submodule.
2.2.2. Definition. Let R be a ring and M a module. The intersection of all submodules
containing a subset Y ⊂ M is the submodule generated by Y and denoted
RY . This is the smallest submodule, 2.2.1, of M containing Y . The module M is
generated by Y if RY = M. Let I be an ideal. The submodule generated by all
products ax, a ∈ I, x ∈ M is denoted IM.
2.2.3. Proposition. Let R be a ring and M a module. If Y ⊂ M, then RY =
y∈Y Ry,
RY = {a1y1 + · · · + anyn|ai ∈ R, yi ∈ Y }
Proof. The righthand side is contained in the submodule RY . Moreover the righthand
side is a submodule containing Y , so equality.
2.2.4. Corollary. Let I ⊂ R be an ideal and M a module.

24 2. MODULES
(1)
IM = {a1y1 + · · · + anyn|ai ∈ I, yi ∈ M}
(2) If I = (a) is principal, then
aM = (a)M = {ay|y ∈ M}
Proof. (1) is clear. (2) By (1) an element in aM is biayi = a biyi = ay for
y = biyi ∈ M as claimed.
2.2.5. Lemma. Let R be a ring, M a module and N ⊂ M a submodule. Let M/N
be the abelian factor group, then the map
R × M/N → M/N, (a, x + N) ↦→ ax + N
is well defined and a scalar multiplication, 2.1.1.
Proof. If x+N = y+N then x−y ∈ N and so a(x−y) = ax−ay ∈ N. Therefore
ax + N = ay + N and the multiplication is well defined. Since representatives
may be chosen such that (x + N) + (y + N) = x + y + N, a(x + N) = ax + N,
the laws for scalar multiplication are satisfied.
2.2.6. Definition. Let R be a ring, M a module and N ⊂ M a submodule, then the
factor module is the additive factor group M/N with, 2.2.5, scalar multiplication
a(x + N) = ax + N. The projection p : M → M/N, x ↦→ x + N is a surjective
homomorphism.
2.2.7. Lemma. Let R be a ring, N ⊂ M a submodule and p : M → M/N
the projection. p is surjective and if Y ⊂ M generates M, then p(Y ) ⊂ M/N
generates the factor module.
Proof. Clearly if RY = M then Rp(Y ) = p(RY ) = M/N.
2.2.8. Example. (1) A submodule of R is the same as an ideal.
(2) Both an ideal I ⊂ R and a factor ring R/I are modules.
(3) The module structure on R/I as a factor module and the structure by restriction
of scalars through the projection R → R/I are identical.
2.2.9. Proposition. Let R = R1×R2 be the product ring 1.1.4. There is a bijective
(up to natural isomorphism) correspondence.
(1) If M1 is an R1-module and M2 is an R2-module, then M = M1 × M2 is an
R-module with coordinate scalar multiplication. A pair of homomorphisms
induce a homomorphism on the product.
(2) If M is an R-module then M1 = (1, 0)M is an R1-module and M2 =
(0, 1)M is an R2-module. A homomorphism induces a pair of homomorphisms.
2.2.10. Remark. The correspondence 2.2.9 indicates that the structure of modules
and homomorphisms over a product ring is identified with the structure of pairs of
modules and homomorphisms over each component ring in the product.
2.2.11. Exercise. (1) Give an example of two submodules N, L ⊂ M such that the
union N ∪ L is not a submodule.
(2) Let R be a ring and a ∈ R. Show that the R-module R[X]/(X − a) is isomorphic
to R.
(3) Show that the projection p 2.2.6 is a homomorphism.
(4) Fill in the details in the dictionary 2.2.9

2.3. KERNEL AND COKERNEL 25
2.3. Kernel and cokernel
2.3.1. Lemma. Let R be a ring and f : M → N a homomorphism of modules.
Given submodules M ′ ⊂ M, N ′ ⊂ N, then f −1 (N ′ ) ⊂ M and f(M ′ ) ⊂ N are
submodules.
Proof. If x, y ∈ f −1 (N ′ ) then f(x + y) = f(x) + f(y) ∈ N ′ and for a ∈ R
f(ax) = af(x) ∈ N ′ so x+y, ax ∈ f −1 (N ′ ) proving f −1 (N ′ ) to be a submodule.
The same equations prove that f(M ′ ) is a submodule.
2.3.2. Definition. Let f : M → N be a homomorphism of modules. Then there
are submodules, 2.3.1.
(1) The kernel Ker f = f −1 (0).
(2) The image Im f = f(M).
(3) The cokernel Cok f = N/ Im f.
2.3.3. Proposition. Let f : M → N be a homomorphism of modules.
(1) f is injective if and only if Ker f = 0.
(2) f is surjective if and only if Cok f = 0.
(3) f is an isomorphism if and only if Ker f = 0 and Cok f = 0.
Proof. (1) If f is injective and x ∈ Ker f then f(x) = 0 = f(0) so x = 0.
Conversely if Ker f = 0 and f(x) = f(y) then f(x − y) = 0 so x = y. (2) The
factor module N/ Im f = 0 if and only if Im f = N. (3) This follows from (1)
and (2).
2.3.4. Example. Let a ∈ R give scalar multiplication aM : M → M, x ↦→ ax.
(1) Im aM = aM = {ax ∈ M|x ∈ M}.
(2) Ker aM = {x ∈ M|ax = 0}.
(3) Cok aM = M/aM.
2.3.5. Proposition. Let f : M → N be a homomorphism of modules.
(1) Let L ⊂ Ker f be a submodule. Then there is a unique homomorphism f ′ :
M/L → N such that f = f ′ ◦ p.
M
p
M/L
f
(2) The homomorphism f ′ : M/ Ker f → N is a module isomorphism onto the
submodule Im f of N.
M
p
M/ Ker f
f
f ′
f ′
N
Im f
Proof. (1) If x+L = x ′ +L then x−x ′ ∈ L giving f(x) = f(x ′ ). The factor map
f ′ : M/L → N, x + L ↦→ f(x) is therefore well defined and f = f ′ ◦ p. Since
f, p are homomorphisms and p is surjective it follows that f ′ is a homomorphism.
(2) The kernel of f ′ is Ker f/ Ker f = 0 so by 2.3.3 it is an isomorphism.

26 2. MODULES
2.3.6. Corollary. Let p : M → M/N be the projection onto a factor module. The
map L ′ ↦→ L = p −1 (L ′ ) gives a bijective correspondence between submodules
in M/N and submodules in M containing N. Also L ′ = p(L) = L/N. This
correspondence preserves inclusions, additions and intersections of submodules.
Proof. If L ′ is a submodule of M/N then clearly p(p−1 (L ′ )) = L ′ . If N ⊂ L
is a submodule of M then clearly L ⊂ p−1 (p(L)). Moreover if x ∈ p−1 (p(L))
then p(x) = p(y) for some y ∈ L and therefore y − x ∈ N. It follows that
L = p−1 (p(L)) and the correspondence is bijective. Inclusions are easily seen to
be preserved. Also easily p(L1 + L2) = p(L1) + p(L2) and p−1 (L ′ 1 ∩ L′ 2 ) =
p−1 (L ′ 1 ) ∩ p−1 (L ′ 2 ) hold. The two resting equalities are consequences of this and
bijectivity of the correspondence.
2.3.7. Corollary. Let L ⊂ N ⊂ M be submodules. Then there is a canonical
isomorphism
M/N → (M/L)/(N/L)
Proof. The kernel of the surjective east-south composite
M
M/N
M/L
(M/L)/(N/L)
is N. By 2.3.5 the horizontal lower factor map gives the isomorphism.
2.3.8. Corollary. Let L, N ⊂ M be submodules. Then there is a canonical isomorphism
N/N ∩ L → N + L/L
given by x + N ∩ L ↦→ x + L.
Proof. The kernel of the east-south composite
N
N/N ∩ L
N + L
N + L/L
is N ∩ L. Since x + y + L = x + L for x ∈ N, y ∈ L this composite is also
surjective. By 2.3.5 the horizontal lower factor map gives the isomorphism.
2.3.9. Proposition. Let f : M → N and g : N → L be homomorphisms such that
Im f ⊂ Ker g. Then there is a unique homomorphism g ′ : Cok f → L such that
g = g ′ ◦ p.
Proof. This follows from 2.3.5.
M f
N
g
p
L
g ′
Cok f
2.3.10. Lemma. Let R be a ring and M a module. For x ∈ M the map R →
M, a ↦→ ax is the unique homomorphism such that 1 ↦→ x.

2.3. KERNEL AND COKERNEL 27
Proof. Let f(a) = ax. f(ab) = (ab)x = a(bx) = af(b) shows that f is a
homomorphism. This argument does not use the commutativity 1.1.2 of R.
2.3.11. Definition. Denote the homomorphism 2.3.10
The annihilator of x is the ideal
1x : R → M, a ↦→ ax
Ann(x) = Ker 1x = {a ∈ R|ax = 0}
For a subset Y ⊂ M the annihilator Ann(Y ) =
y∈Y Ann(y) is the ideal of
elements
Ann(Y ) = {a ∈ R|ay = 0, for all y ∈ Y }
2.3.12. Proposition. Let R be a ring and Y a subset of a module M.
(1) Ann(Y ) = Ann(RY ).
(2) a ∈ Ann(M) if and only if aM = 0.
(3) If modules M M ′ then Ann(M) = Ann(M ′ ).
(4) The induced homomorphism
is an isomorphism.
1 ′ x : R/ Ann(x) → Rx, a + Ann(x) ↦→ ax
Proof. (1) If a ∈ Ann(Y ) then a biyi = biayi = 0 giving the not so obvious
Ann(Y ) ⊂ Ann(RY ). (2) Clear since aM(x) = ax, 2.1.7. (3) Let f : M → M ′
be an isomorphism and a ∈ R. Then af(x) = f(ax) expresses that f ◦ aM =
aM ′ ◦ f. Since f is bijective, aM = 0 if and only if aM ′ = 0. By (2) Ann(M) =
Ann(M ′ ). (4) This follows from 2.3.5.
2.3.13. Corollary. Let I, J ⊂ R be ideals.
(1) Ann(R/I) = I
(2) If R/I R/J then I = J.
2.3.14. Lemma. Let I ⊂ R be an ideal and M an R-module. If I ⊂ Ann(M)
then M is an R/I-module with the scalar multiplication
That is, M is an R/ Ann(M)-module.
R/I × M → M, (a + I, x) ↦→ ax
2.3.15. Example. Let a ∈ R give scalar multiplication aM : M → M, x ↦→ ax.
(1) a ∈ Ann(Ker aM).
(2) a ∈ Ann(Cok aM).
(3) Ker aM and Cok aM are modules over the factor ring R/(a), 2.3.14.
2.3.16. Definition. Let R be a ring and L, N ⊂ M submodules. The colon ideal
N : L is the ideal of elements a ∈ R such that aL ⊂ N.
2.3.17. Proposition. The colon ideal of submodules L, N ⊂ M is
N : L = Ann(N + L/N)
Proof. If a ∈ N : L then aL ⊂ N. Therefore a(N + L) ⊂ N and a ∈ Ann(N +
L/N). Conversely if a ∈ Ann(N + L/N) then aL ⊂ N and therefore a ∈ N :
L.
2.3.18. Exercise. (1) Give an example of a homomorphism f : M → N submodules
M1, M2 ⊂ M such that f(M1 ∩ M2) = f(M1) ∩ f(M2).

28 2. MODULES
(2) Give an example of a homomorphism f : M → N submodules N1, N2 ⊂ N such
that f −1 (N1 + N2) = f −1 (M1) + f −1 (M2).
(3) Let R be a ring and M a module. Show that M may be regarded as an R/ Ann(M)module
in a natural way.
(4) Let L, N ⊂ M be submodules. Show that Ann(L + N) = Ann(L) ∩ Ann(N).
(5) Let f : M → N be a surjective homomorphism. Show that Ann(M) ⊂ Ann(N).
(6) Let f : M → N be an injective homomorphism. Show that Ann(N) ⊂ Ann(M).
2.4. Sum and product
2.4.1.
Lemma. Let R be a ring and (Mα) a family of modules. The product
α Mα is the abelian group of all families (xα), xα ∈ Mα with term wise addition.
The setting
r(xα) = (rxα)
is a scalar multiplication on
α Mα. The direct sum
α Mα is the subgroup
of
α Mα consisting of families with only finitely many nonzero terms. This is a
submodule.
Proof. The laws in 2.1.1 are true for each factor and therefore trivially verified for
the product and sum.
2.4.2. Definition. Let R be a ring and Mα a family of modules. By 2.4.1 there are
modules and homomorphisms
(1) The direct product is
α Mα.
(2) The projections pβ :
α Mα → Mβ are the homomorphisms pβ((xα)) =
xβ.
(3) The direct sum is
α Mα. Elements in
α Mα
are written as finite sums
xα.
(4) The injections iβ : Mβ →
α Mα are the homomorphisms given by iβ(x) =
(xα), where xβ = x and xα = 0, α = β.
2.4.3. Proposition. Let R be a ring and Mα a family of modules.
(1) Given a family of homomorphisms fα : L → Mα, then there exists a unique
homomorphism f : L → Mα such that fα = pα ◦ f.
L
fα
f
Mα
pα
α Mα
(2) Given a family of homomorphisms gα : Mα → L, then there exists a unique
homomorphism g : Mα → L such that gα = g ◦ iα.
α Mα
g
iα
gα
Mα
Proof. (1) f(y) = (fα(y)) is the unique homomorphism. (2) g( xα) = gα(xα)
is well defined since only finitely many xα = 0 and a homomorphism.
L

2.4. SUM AND PRODUCT 29
2.4.4. Definition. A family of submodules Mα ⊂ M constitutes a direct sum if
any element x ∈
α Mα has a unique finite representation
x =
xα, xα ∈ Mα
2.4.5. Proposition. The following conditions are equivalent
(1) The family Mα ⊂ M constitutes a direct sum.
(2) The natural homomorphism
Mα →
is an isomorphism.
(3) For all β
α
α
α
Mα
Mβ ∩
Mα = 0
α=β
Proof. (1) and (2) are equivalent. If (1) is true and xβ =
α=β xα ∈ Mβ ∩
α=β Mα, then xβ −
α=β xα = 0. Therefore by uniqueness xβ = 0 and (3)
is true. Conversely if (3) is true and
α xα = 0, then xβ = −
α=β xα ∈
Mβ ∩
α=β Mα = 0. for any β. This shows uniqueness and therefore (1) is
true.
2.4.6. Definition. Let R be a ring. A module isomorphic to a direct sum
α
R of
copies of the ring R is a free module.
A basis of a module, is a subset Y such that any element admits a unique finite
representation
α aαyα, where aα ∈ R, yα ∈ Y .
The standard basis of
α R consists of the elements eα, where each is a family
with 0 for β = α and exactly 1 at index α.
2.4.7. Proposition. A module is free if and only if it admits a basis. If yα is a basis
of F then there is an isomorphism
f :
R → F
where f( aα) = aαyα.
α
Proof. Given a free module f :
α R → F then yα = f(eα) is a basis. Conversely
given a basis yα ∈ F then 1yα : R → F 2.3.11 is a family of homomorphisms
giving a homomorphism f :
α R → F by 2.4.3. As f( aα) = aαyα
it follows that f is bijective and therefore by 2.1.5 an isomorphism.
2.4.8. Remark. The polynomial ring R[X1, . . . , Xn] is free as R-module.
2.4.9. Example. (1) A nonzero ideal is a free module if and only if it has a basis
consisting of a nonzero divisor. Namely if x1 = x2 where in a basis then the
product x1x2 has two different representations.
(2) Let I ⊂ R be an ideal. The module R/I is free if and only if I = 0 or I = R.
2.4.10. Proposition. Any module over a field is free. Conversely if any module
over a nonzero ring is free, then the ring is a field
Proof. By Zorn’s lemma any vector space admits a basis. If I ⊂ R is an ideal and
R/I is free, then I = Ann(R/I) is either 0 or R. So R is a field.

30 2. MODULES
2.4.11. Proposition. Let F be a free module with basis yα. For a module M and
a family of elements xα ∈ M there is a unique homomorphism g : F → M such
that g(yα) = xα given by g( aαyα) = aαxα.
Proof. The basis yα ∈ F gives the isomorphism 2.4.7 f :
α R → F . The
family 1xα : R → M gives a homomorphism g ′ :
R → M by 2.4.3. Then
g = g ′ ◦ f −1 .
2.4.12. Corollary. Let M be an R-module and
M R the free module with basis
ex indexed by x ∈ M. The homomorphism
R → M, axex ↦→ ax x
M
is surjective identifying M as a factor module of a free module in a natural way.
2.4.13. Definition. A module is indecomposable if it is not isomorphic to a direct
sum of two nonzero submodules, otherwise decomposable.
2.4.14. Example. Q is an indecomposable Z-module. Namely if m1
n1
nonzero numbers in two submodules, then n1m2 m1
n1
ber in the intersection.
α
= n2m1 m2
n2
, m2
n2 are
is a nonzero num-
2.4.15. Exercise. (1) Show that if a ring is decomposable as a module, then it is the
product of two nonzero rings.
(2) Let Mα be a finite family of modules. Show that Mα = Mα,
(3) Let Nα ⊂ Mα be a family of submodules modules. Show that
Mα/ Nα Mα/Nα
and that Mα/ Nα Mα/Nα
2.5. Homomorphism modules
2.5.1. Lemma. Let R be a ring and f, g : M → N homomorphisms.
(1) (f + g)(x) = f(x) + g(x) is a homomorphism.
(2) If a ∈ R, then (af)(x) = af(x) is a homomorphism.
Proof. Calculate according to 2.1.1. (1) (f + g)(x + y) = f(x + y) + g(x + y) =
f(x) + f(y) + g(x) + g(y) = (f + g)(x) + (f + g)(y) and (f + g)(ax) =
f(ax) + g(ax) = a(f(x) + g(x)) = a(f + g)(x). (2) (af)(x + y) = af(x + y) =
af(x) + af(y) = (af)(x) + (af)(y) and (af)(bx) = af(bx) = abf(x) =
baf(x) = b(af)(x). The last calculation uses that R is commutative 1.1.2 (4).
2.5.2. Definition. Let R be a ring and M, N modules. By 2.5.1, the homomorphism
module HomR(M, N) is the additive group of all homomorphism with
scalar multiplication
R × HomR(M, N) → HomR(M, N), (a, f) ↦→ af = [x ↦→ af(x)]
2.5.3. Definition. Let a ∈ R be a ring and f : M → M ′ , g : N → N ′ , h, k :
M ′ → N homomorphisms of modules. By 2.5.1
(1) (h + k) ◦ f = h ◦ f + k ◦ f.
(2) (ah) ◦ f = a(h ◦ f).
(3) g ◦ (h + k) = g ◦ h + g ◦ k.
(4) g ◦ (ah) = a(g ◦ h).

There is induced a homomorphism
of R-modules.
2.5. HOMOMORPHISM MODULES 31
Hom(f, g) : HomR(M ′ , N) → HomR(M, N ′ )
(h : M ′ → N) ↦→ (g ◦ h ◦ f : M → N ′ )
2.5.4. Definition. Let R, S be a rings. A functor is a construction T , which to
R-modules M, N associates S-modules T (M), T (N) and a homomorphism
such that
HomR(M, N) → HomS(T (M), T (N)), f ↦→ T (f)
(1) T (1M) = 1 T (M)
(2) T (g ◦ f) = T (g) ◦ T (f)
In case the homomorphism goes
and
HomR(M, N) → HomS(T (N), T (M)), f ↦→ T (f)
(1) T (1M) = 1 T (M)
(2) T (g ◦ f) = T (f) ◦ T (g)
the functor is contravariant. Clearly functors transform isomorphisms into isomorphism.
Given functors T, T ′ a natural homomorphism is a family νM : T (M) → T ′ (M)
of homomorphisms, such that for each f : M → N the following diagram commutes
T (M)
T (f)
νN
T (N)
In the contravariant case the diagram is
T (M)
T (f)
T (N)
νM
T ′ (M)
νM
T ′ (f)
T ′ (N)
T ′ (M)
T ′ (f)
νN
T ′ (N)
A natural isomorphism is a natural homomorphism such that each νM is an isomorphism.
2.5.5. Proposition. Let R be a ring.
(1) The construction
is a functor.
(2) The construction
is a contravariant functor
N ↦→ HomR(M, N), g ↦→ Hom(1M, g)
M ↦→ HomR(M, N), f ↦→ Hom(f, 1N)

32 2. MODULES
Proof. By 2.5.3 the construction on homomorphisms are homomorphisms. Given
also homomorphisms f ′ : M ′ → M ′′ and g ′ : N ′ → N ′′ . Then
and
Hom(1M, g ′ ◦ g) = Hom(1M, g ′ ) ◦ Hom(1M, g)
Hom(f ′ ◦ f, 1N) = Hom(f, 1N) ◦ Hom(f ′ , 1N)
showing the conditions on compositions.
2.5.6. Corollary. Let a ∈ R give scalar multiplications aM, aN.
(1)
Hom(aM, 1N) = Hom(1M, aN) : HomR(M, N) → HomR(M, N) f ↦→ af
is scalar multiplication a HomR(M,N).
(2) The map R → HomR(M, M), a ↦→ aM is a homomorphism.
2.5.7. Example. Let R = R1 × R2 be the product ring. The constructions in 2.2.9
is: (1) A functor which to a pair of an R1-module and an R2-module associates an
R-module.(2) A functor which to an R-module associates a pair of an R1-module
and an R2-module.
2.5.8. Proposition. Let R be a ring and Mα a family of modules. For any module
N there are natural isomorphisms
(1) HomR(
α Mα, N)
α HomR(Mα, N)
(2) HomR(N,
α Mα)
α HomR(N, Mα)
Proof. This is 2.4.3 reformulated. (1) A homomorphism g :
α Mα → N is
uniquely determined by the family gα = g ◦ iα : Mα → N. (2) A homomorphism
f : N →
α Mα is uniquely determined by the family fα = pα ◦f : N → Mα.
2.5.9. Lemma. Let R be a ring and M, N modules. For x ∈ M there is a homomorphism
HomR(M, N) → N, f ↦→ f(x).
Proof. Calculate according to 2.1.1 (f + g) ↦→ (f + g)(x) = f(x) + g(x) and
(af) ↦→ (af)(x) = af(x).
2.5.10. Definition. The natural homomorphism 2.5.9
is the evaluation at x.
evx : HomR(M, N) → N, f ↦→ f(x)
2.5.11. Lemma. There is a natural homomorphism
M → HomR(HomR(M, N), N), x ↦→ evx
Proof. Calculate according to 2.1.1 evx+y(f) = f(x + y) = f(x) + f(y) =
evx(f) + evy(f). and evax(f) = f(ax) = af(x) = aevx(f) to see that the map
is a homomorphism.
2.5.12. Proposition. Let R be a ring and M a module. The evaluation
ev1 : HomR(R, M) M, f ↦→ f(1)
is a natural isomorphism. x ↦→ 1x 2.3.11 is the inverse.
Proof. Calculate the composite ev1(x ↦→ 1x) = 1x(1) = x and 1 f(1)(a) =
af(1) = f(a) proving the claims.

2.6. TENSOR PRODUCT MODULES 33
2.5.13. Definition. Let R be a ring and M a module. The dual module is
M ∨ = HomR(M, R)
If f : M → N is a homomorphism, then the dual homomorphism is
f ∨ = Hom(f, 1R) : N ∨ → M ∨
This construction is a contravariant functor.
2.5.14. Lemma. There is a natural homomorphism
where evx(f) = f(x) 2.5.10.
M → M ∨∨ = HomR(HomR(M, R), R), x ↦→ evx
Proof. This is a special case of 2.5.11
2.5.15. Definition. A module M is a reflexive module if the homomorphism 2.1.14,
is an isomorphism.
2.5.16. Example. Let R be a ring.
M → M ∨∨
(1) The module R is reflexive.
(2) If (a) = R, (0) in a domain, then R/(a) is not reflexive.
2.5.17. Exercise. (1) Show that HomZ(Q, Z) = 0.
(2) Calculate HomZ(Z/(m), Z/(n)) = 0 for integers m, n.
(3) Let I ⊂ R be an ideal and M a module. Show that HomR(R/I, M) = {x ∈ M|I ⊂
Ann(x)}.
(4) Let R be a ring. Show that a free module with a finite basis is a reflexive module.
(5) If (n) ⊂ (m) ⊂ Z, then show that (m)/(n) is a reflexive Z/(n)-module.
2.6. Tensor product modules
2.6.1. Definition. Let R be a ring and M, N modules. The tensor product module
M ⊗R N = F/F ′
is the factor module F/F ′ , where F = ⊕M×NR is the free module with basis
(x, y) = e (x,y) 2.4.6 and F ′ is the submodule generated by all elements of form
(x1 + x2, y) − (x1, y) − (x2, y), (x, y1 + y2) − (x, y1) − (x, y2)
(ax, y) − a(x, y), (x, ay) − a(x, y)
The projection of the basis element (x, y) onto M ⊗R N is x ⊗ y = (x, y) + F ′ .
2.6.2. Remark. The relations are interpreted.
(1) There are identities in M ⊗R N
(x1 + x2) ⊗ y = x1 ⊗ y + x2 ⊗ y, x ⊗ (y1 + y2) = x ⊗ y1 + x ⊗ y2
(2) The map
ax ⊗ y = a(x ⊗ y) = x ⊗ ay
⊗ : M × N → M ⊗R N, (x, y) ↦→ x ⊗ y
has partial maps x ↦→ x⊗y : M → M ⊗R N and y ↦→ x⊗y : N → M ⊗R N
that are all homomorphisms.

34 2. MODULES
(3) The formation of partial homomorphism are again homomorphisms.
and
N → HomR(M, M ⊗R N)), y ↦→ (x ↦→ x ⊗ y)
M → HomR(N, M ⊗R N)), x ↦→ (y ↦→ x ⊗ y)
2.6.3. Proposition. Given a map µ : M × N → L such that the partial maps x ↦→
µ(x, y) : M → L and y ↦→ µ(x, y) : N → L are homomorphisms. Then there
exists a unique homomorphism u : M ⊗R N → L such that u(x ⊗ y) = µ(x, y).
⊗
M × N
M ⊗R N
µ u
L
Proof. By 2.6.1 M ⊗R N = F/F ′ . The homomorphism 2.4.11 F → K, (x, y) ↦→
µ(x, y) has F ′ in the kernel. 2.3.5 gives the homomorphism u.
2.6.4. Remark. Two homomorphisms u, v : M ⊗RN → L are equal if u(x⊗y) =
v(x ⊗ y) for all x ∈ M, y ∈ N.
2.6.5. Proposition. Let R be a ring and f : M → M ′ , g : N → N ′ homomorphisms
of modules. Then there is induced a homomorphism
Proof. The map south-east
M ⊗R N → M ′ ⊗R N ′ , x ⊗ y ↦→ f(x) ⊗ g(y)
M × N
f×g
M ′ × N ′
⊗
M ⊗R N
⊗
M ′
⊗R N ′
satisfies the assumptions in 2.6.3 to induce the right vertical map x ⊗ y ↦→ f(x) ⊗
g(y).
2.6.6. Definition. f ⊗ g : M ⊗R N → M ′ ⊗R N ′ is the induced homomorphism
2.6.5.
2.6.7. Proposition. Let R be a ring. The constructions
(1)
(2)
are functors.
M ↦→ M ⊗R N, f ↦→ f ⊗ 1N
N ↦→ M ⊗R N, g ↦→ 1M ⊗ g
Proof. Given also homomorphisms f ′ : M ′ → M ′′ and g ′ : N ′ → N ′′ . Then by
2.6.4
f ′ ◦ f ⊗ g ′ ◦ g = f ′ ⊗ g ′ ◦ f ⊗ g
The rest follows directly from 2.6.5.

2.6. TENSOR PRODUCT MODULES 35
2.6.8. Corollary. Let a ∈ R give scalar multiplications aM, aN. The homomorphisms
aM ⊗ 1N = 1M ⊗ aN : M ⊗R N → M ⊗R N, x ⊗ y ↦→ a(x ⊗ y)
is scalar multiplication aM⊗RN.
2.6.9. Example. Let R be a ring.
(1) Then there is an isomorphism
R ⊗R R R, a ⊗ b ↦→ ab
(2) Let M, N, L be modules. Composition of maps gives a homomorphism
HomR(N, L) ⊗R HomR(M, N) → HomR(M, L), g ⊗ f ↦→ g ◦ f
by 2.5.3. This is a natural homomorphism in each variable.
(3) For a module M composition
HomR(M, M) ⊗R HomR(M, M) → HomR(M, M), g ⊗ f ↦→ g ◦ f
gives HomR(M, M) a structure of a normally noncommutative ring. The
map R → HomR(M, M), a ↦→ aM is a ring homomorphism.
2.6.10. Proposition. Let R be a ring and M, N, L modules. Then there are natural
isomorphisms
(1) M ⊗R R M, x ⊗ a ↦→ ax
(2) (M ⊗R N) N ⊗R M, x ⊗ y ↦→ y ⊗ x
(3) (M ⊗R N) ⊗R L M ⊗R (N ⊗R L), (x ⊗ y) ⊗ z ↦→ x ⊗ (y ⊗ z)
Proof. (1) M × R → M, (x, a) ↦→ ax induces the homomorphism M ⊗R R →
M, x ⊗ a ↦→ ax by 2.6.3. The map M → R ⊗R M, x ↦→ 1 ⊗ x is the inverse. (2)
M × N → N ⊗R M, (x, y) ↦→ y ⊗ x induces the homomorphism (M ⊗R N)
N ⊗R M, x ⊗ y ↦→ y ⊗ x by 2.6.3. The inverse is constructed similarly and the
composites are the identities by the uniqueness statement in 2.6.3. (3) For a fixed
z ∈ L the map M × N → M ⊗R (N ⊗R L), (x, y) ↦→ x ⊗ (y ⊗ z) induces the
homomorphism µz : M ⊗R N → M ⊗R (N ⊗R L), x ⊗ y ↦→ x ⊗ (y ⊗ z) by 2.6.3.
Finally the map (M⊗RN)×L → M⊗R(N⊗RL), (x⊗y, z) ↦→ µz(x⊗y). induces
the homomorphism (M ⊗RN)⊗RL M ⊗R(N ⊗RL), (x⊗y)⊗z ↦→ x⊗(y⊗z)
by 2.6.3. The inverse is constructed similarly and the composites are the identities
by the uniqueness statement in 2.6.3.
2.6.11. Proposition. Let R be a ring and Mα a family of modules. For any module
N there is a natural isomorphism
(
Mα) ⊗R N
(Mα ⊗R N)
α
giving the identification ( xα) ⊗ y = (xα ⊗ y).
Proof. 2.4.3 and 2.6.3 give a pair of inverse homomorphisms. Fix y ∈ N. The
family gα : Mα → (Mα ⊗R N), xα ↦→ xα ⊗ y induces by 2.4.3 a homomorphism
gy : Mα → (Mα ⊗R N). The map ( Mα) × N → (Mα ⊗R
N), ( xα, y) ↦→ gy( xα) induces by 2.6.3 the homomorphism ( Mα) ⊗R
N → (Mα⊗RN), ( xα)⊗y ↦→ xα⊗y. The family iα⊗1N : Mα⊗RN →
( Mα) ⊗R N induces by 2.4.3 the inverse.
α

36 2. MODULES
2.6.12. Example. Let R be a ring, F a free module with basis yα and G a free
module with basis zβ. Then F ⊗R G is a free module with basis yα ⊗ zβ.
2.6.13. Proposition. Let R be a ring and M, N, L modules. Then there is a natural
isomorphism
HomR(M ⊗R N, L) HomR(M, HomR(N, L))
(x ⊗ y ↦→ g(x)(y)) ← g
f ↦→ (x ↦→ [y ↦→ f(x ⊗ y)])
Proof. A given f : M ⊗R N → L is mapped to the composite homomorphism
M → HomR(N, M ⊗R N) → HomR(N, L), 2.5.4 and 2.6.2. This is a homomorphism
as map of f by 2.5.4. Given g : M → HomR(N, L) the map M × N →
L, (x, y) ↦→ g(x)(y) induces a homomorphism M ⊗R N → L, x ⊗ y ↦→ g(x)(y)
by 2.6.3. Clearly the maps are inverse to each other and therefore giving an isomorphism
by 2.1.5.
2.6.14. Exercise. (1) Show that Q ⊗Z Q/Z 0.
(2) Show that Z/(m) ⊗Z Z/(n) = 0 if (m, n) = Z.
(3) Let P, Q ⊂ R be different maximal ideals and M a module. Show that M/P M ⊗R
M/QM = 0.
2.7. Change of rings
2.7.1. Proposition. (1) Let φ : R → S be a ring homomorphism and N an S
module. The restriction scalars 2.1.11 viewing N as an R-module through
φ, R × N → N, (a, x) ↦→ ax = φ(a)x, is a functor from S-modules to
R-modules.
(2) Let I ⊂ R be an ideal. Restriction of scalars along R → R/I identifies
R/I-modules M with R-modules such that I ⊂ Ann(M). For R/I-modules
M, N there is a natural isomorphism
HomR(M, N) Hom R/I(M, N)
Proof. This is a restatement of 2.1.11 using 2.5.4.
2.7.2. Lemma. Let R → S be a ring homomorphism, M an R-module and N an
S-module. Then
is an S-scalar multiplication.
S × M ⊗R N → M ⊗R N, (b, x ⊗ y) ↦→ x ⊗ by
Proof. For fixed b ∈ S the map M × N → M ⊗R N, (x, y) ↦→ x ⊗ by induces
the homomorphism µb : M ⊗R N → M ⊗R N, x ⊗ y ↦→ x ⊗ by by 2.6.3. This
gives a well defined scalar multiplication S × M ⊗R N → M ⊗R N, (b, x ⊗ y) ↦→
µb(x ⊗ y).
2.7.3. Definition. Let R → S be a ring homomorphism and M an R-module. The
change of ring S-module is M ⊗R S with S-scalar multiplication 2.7.2
S × M ⊗R S → M ⊗R S, (b, x ⊗ c) ↦→ x ⊗ bc

2.7.4. Proposition. The construction
2.7. CHANGE OF RINGS 37
M ↦→ M ⊗R S
and
f : M → M ′ ↦→ f ⊗ 1S : M ⊗R S → M ′ ⊗R S
is a functor from R-modules to S-modules.
Proof. This is clear from 2.7.2 and 2.6.7.
2.7.5. Proposition. Let R → S be a ring homomorphism, M an R-module and N
an S-modules. Then there is a natural isomorphism of S-modules.
M ⊗R S ⊗S N M ⊗R N, x ⊗ b ⊗ y ↦→ x ⊗ by
Proof. The homomorphism v : S ⊗S N → N, b ⊗ y ↦→ by is an isomorphism,
2.6.10. The homomorphism 1M ⊗ v : M ⊗R S ⊗S N M ⊗R N is an R-module
isomorphism, 2.6.7. The identity x ⊗ bc ⊗ y = x ⊗ b ⊗ cy proves this to be an
S-module homomorphism.
2.7.6. Proposition. Let R → S be a ring homomorphism, M an R-module and N
an S-modules. Then there is a natural isomorphism
HomR(M, N) HomS(M ⊗R S, N), f ↦→ (x ⊗ b ↦→ bf(x))
Proof. A given f is mapped to the composite M⊗R → N ⊗R S → N which is an
S-homomorphism. Given a homomorphism g : M ⊗R S → N then the composite
M → M ⊗R S → N is an R-homomorphism and an inverse to the first given
map.
2.7.7. Lemma. Let R → S be a ring homomorphism, M an R-module and N an
S-module. Then
S × HomR(N, M) → HomR(N, M), (b, f : N → M) ↦→ (y ↦→ f(by))
is an S-scalar multiplication.
Proof. The map (b, f) ↦→ f ◦ bN satisfies the laws 2.1.1.
2.7.8. Definition. Let R → S be a ring homomorphism and M an R-module. The
induced module is the S-module HomR(S, M) with S-scalar multiplication 2.7.7
S × HomR(S, M) → HomR(S, M), (b, f : S → M) ↦→ (c ↦→ f(bc))
2.7.9. Proposition. The induced module
M ↦→ HomR(S, M)
and
f : M → M ′ ↦→ Hom(1S, f) : HomR(S, M) → HomR(S, M ′ )
is a functor from R-modules to S-modules.
Proof. This is clear from 2.7.7 using 2.5.4.
2.7.10. Proposition. Let R → S be a ring homomorphism and M an R-module
and N an S-modules. Then there is a natural isomorphism
HomR(N, M) HomS(N, HomR(S, M)), f ↦→ (y ↦→ [b ↦→ f(by)])
Proof. g ↦→ (y ↦→ g(y)(1)) is an inverse.
2.7.11. Example. Let I ⊂ R be an ideal and R → R/I the projection.

38 2. MODULES
(1) The change of ring functor maps an R-module M to the R/I-module M/IM.
The natural isomorphism 2.7.6 is
HomR(M, N) Hom R/I(M/IM, N)
for any R/I-module N.
(2) The induced module functor maps an R-module M to the R/I-module {x ∈
M|I ⊂ Ann(x)}. The natural isomorphism 2.7.10 is
HomR(N, M) Hom R/I(N, HomR(R/I, M))
for any R/I-module N.
2.7.12. Definition. Let R → S, S ′ be ring homomorphisms. The tensor product
ring over R is S ⊗R S ′ with multiplication given by (b ⊗ b ′ )(c ⊗ c ′ ) = bc ⊗ b ′ c ′
extended by linearity. R → S ⊗R S ′ , r ↦→ r ⊗ 1 = 1 ⊗ r is the natural ring
homomorphism.
2.7.13. Proposition. Let φ, φ ′ : R → S, S ′ and ψ, ψ ′ : S, S ′ → T give a commutative
diagram of ring homomorphisms, ψ ◦ φ = ψ ′ ◦ φ ′ . Then b ⊗ b ′ ↦→ ψ(b)ψ ′ (b ′ )
is the unique homomorphism making the following diagram commutative.
Proof. This is clear by 2.6.3.
R
S ′
S
S ⊗R S
′
T
2.7.14. Example. Let R → S be a ring homomorphism. Then
is an isomorphism.
R[X] ⊗R S S[X]
2.7.15. Exercise. (1) Show that the change of rings of a free R-module is a free Smodule.
(2) Let φ : R → S be a ring homomorphism. Show that the change of rings of a
scalar multiplication a : M → M on an R-module is a scalar multiplication φ(a) :
M ⊗R S → M ⊗R S.
(3) Show that the change of rings of the composition of two homomorphisms is the
composition of the change of rings of each homomorphism.
(4) Show the isomorphism
R[X] ⊗R R[Y ] R[X, Y ]

3
Exact sequences of modules
3.1. Exact sequences
3.1.1. Definition. Let f : M → N and g : N → L be homomorphisms of
modules. The sequence
of homomorphisms is a
M f
g
N
L
(1) 0-sequence: g ◦ f = 0 or equivalently Im f ⊂ Ker g
(2) exact sequence: Im f = Ker g
For a sequence of more homomorphisms the conditions should be satisfied for
every consecutive composition. E.g. The sequence
M f
g
N
h
L
K
is a 0-sequence if g ◦ f = 0 and h ◦ g = 0. The sequence is exact if Im f = Ker g
and Im g = Ker h.
3.1.2. Remark. An interpretation of 2.3.3 gives:
(1) The sequence
is exact if and only if f is injective.
(2) The sequence
0
M f
f
M
N
is exact if and only if f is surjective.
(3) The sequence
0
f
M
N
is exact if and only if f is an isomorphism.
3.1.3. Proposition. (1) For a homomorphism f : M → N the sequence
0
Ker f
f
M
N
N
0
0
Cok f
is exact.
(2) For scalar multiplication with a ∈ R on M the sequence
0
is exact.
Ker aM
aM
M
M
39
M/aM
0
0

40 3. EXACT SEQUENCES OF MODULES
3.1.4. Proposition. The 0-sequence
0
f
M
g
N
is exact if and only if the following equivalent statements are satisfied.
(1) f is an isomorphism onto Ker g.
(2) Given a homomorphism h : K → N such that g ◦ h = 0 then there is a
unique h ′ : K → M such that h = f ◦ h ′ .
0
f g
M
N
h
K
′
h
Proof. (1) This is clearly equivalent with exactness. (2) Assume the sequence exact.
Im h ⊂ Ker g = Im f, so by (1) put h ′ = f −1 ◦ h. Assume (2) satisfied and
apply it to Ker g → M to see that (1) is satisfied.
3.1.5. Proposition. The 0-sequence
M f
g
N
is exact if and only if the following equivalent statements are satisfied.
(1) The factor homomorphism 2.3.9 g ′ : Cok f → L induced by g is an isomorphism.
(2) Given a homomorphism k : N → K such that k ◦ f = 0 then there is a
unique k ′ : L → K such that k = k ′ ◦ g.
M f
L
g
N
L
k
k
′
K
Proof. (1) The equivalence follows from 2.3.9. (2) Assume the sequence exact. By
2.3.5 there is k ′′ : Cok f → K such that k ′′ ◦ p = k. By (1) put k ′ = k ′′ ◦ g −1 .
Assume (2) satisfied and apply it to N → Cok f to see that (1) is satisfied.
3.1.6. Proposition. Let
Mα
Nα
Lα
L
L
be a family of exact sequences. Then there are exact sequences:
(1) The sum
Mα
Nα
Lα
(2) The product
Mα
Nα
0
0
Lα
Proof. Clear since kernel and image is calculated componentwise.
3.1.7. Definition. An exact sequence
0
f
M
g
N
is a short exact sequence. That is f is injective, Im f = Ker g and g is surjective.
L
0

3.1. EXACT SEQUENCES 41
3.1.8. Proposition.
quence
(1) Let I ⊂ R be an ideal, then there is a short exact se-
0 I R R/I 0
(2) Let M ⊂ N be a submodule, then there is a short exact sequence
0
M
N
N/M
(3) For scalar multiplication with nonzero divisor a ∈ R on M the sequence
0
aM
M
M
M/aM
is a short exact sequence.
(4) Given a homomorphism f : M → N there are associated two short exact
sequences.
and
0
0
Ker f
Im f
f
M
Im f
N
Cok f
(5) For scalar multiplication with any a ∈ R on M there are associated two short
exact sequences.
and
0
0
Ker aM
aM
M
aM
M
aM
M/aM
3.1.9. Definition. Let f : M → N be a homomorphism.
(1) f has a retraction if there is a homomorphism u : N → M such that u ◦ f =
1M.
(2) f has a section if there is a homomorphism v : N → M such that f ◦v = 1N.
3.1.10. Proposition. Let f : M → N be a homomorphism.
(1) If f has a retraction u : N → M then f is injective, u is surjective and
N = Im f ⊕ Ker u
(2) If f has a section v : N → M then f is surjective, v is injective and
M = Ker f ⊕ Im v
Proof. (1) u(f(x)) = x so f is injective and u is surjective. If y ∈ N then
y = f(u(y)) + (y − f(u(y)) and u(y − f(u(y))) = 0, so N = Im f + Ker u. Let
y ∈ Im f ∩ Ker u. Then y = f(x) gives x = u(f(x)) = u(y) = 0, so y = 0.
Conclude by 2.4.5 that the sum is direct. (2) y = f(v(y)) so f is a retraction of v.
Finish by (1).
3.1.11. Lemma. For a short exact sequence
the following are equivalent
0
(1) f has a retraction.
(2) g has a section.
f
M
g
N
L
0
0
0
0
0
0
0

42 3. EXACT SEQUENCES OF MODULES
For any retraction u there is a unique section v and wise-verse such that
1N = f ◦ u + v ◦ g
Proof. If u is a retraction of f, then Ker g = Im f ⊂ Ker(1N − f ◦ u). By 3.1.5
there is a homomorphism v : L → N such that v ◦ g = 1N − f ◦ u. This is a
section of g. Conversely if v is a section of g then Im(1N − v ◦ g) ⊂ Ker g, so
there is a homomorphism u : N → M such that f ◦ u = 1N − v ◦ g, 3.1.4. u is a
retraction of f. The equation is clearly satisfied.
3.1.12. Definition. Let R be a ring and f : M → N, g : N → L homomorphisms.
A short exact sequence
0
f
M
g
N
is a split exact sequence if equivalently 3.1.11 f has a retraction or g has a section.
3.1.13. Proposition. A sequence
0
f
M
g
N
is a split exact sequence if and only if there are homomorphism u : N → M, v :
L → N satisfying
L
L
0
0
g ◦ f = 0, u ◦ f = 1M, g ◦ v = 1L, f ◦ u + v ◦ g = 1N
If the sequence is split exact then
0
v
L
u
N
N
is split exact and (x, y) ↦→ f(x)+v(y) and z ↦→ u(z)+g(z) gives the isomorphism
M ⊕ L N
Proof. The sequence is a 0-sequence f is injective and g is surjective. From f ◦
u + v ◦ g = 1N follows that z ∈ Ker g ⊂ Im f, so the sequence is short exact. The
rest is contained in 3.1.10.
3.1.14. Corollary. A (contravariant) functor preserves split exact sequences. If
0
f
M
is split exact and T a functor, then
is split exact.
0
g
N
L
T (M) T (f)
T (N) T (g)
T (L)
Proof. By 3.1.13 a split exact sequence is characterized by a set of equations.
These are preserved by the functor, 2.5.4.
3.1.15. Example. A short exact sequence
0
f
M
g
N
where L is a free module is a split exact sequence. Namely let xα ∈ L be a
basis and choose yα ∈ N with g(yα) = xα. The define v : L → N by setting
v(xα) = yα, 2.4.11.
L
0
0
0
0

3.2. THE SNAKE LEMMA 43
3.1.16. Example. Let Zi is the family of modules each a copy of Z indexed by the
natural numbers. Then the short exact sequence
0
i Zi
i Zi
is not split exact.
The element f = (1, 2, 22 , . . . , 2n , . . . ) +
fk = (0, . . . , 0, 2 n−k , . . . ) +
But in
i Zi/ i Zi
0
i Zi is divisible by 2 k for all k. If
i Zi for n ≥ k, then 2 k fk = f in
i Zi the only element divisible with all 2 k is 0, so no section exists.
3.1.17. Exercise. (1) Show that the sequence
0
Z
is short exact, but not split exact.
(2) Show that the sequence
0
Q
n
Z
Z
is exact, but not split exact for n = 0, 1.
(3) Show that the sequence
is exact, but not split exact.
(4) Show that the sequence
is split exact.
0
0
Z/(2) 1↦→2
Z/(4)
Z/(2) 1↦→3
Z/(6)
Q/Z
Z/(n)
3.2. The snake lemma
Z/(2)
Z/(3)
3.2.1. Example. Given a commutative diagram of homomorphisms
M
u
M ′
there is induced a commutative diagram
0
0
Ker f
u
Ker f ′
M
u
M ′
where the rows are exact sequences.
The diagram splits into two diagrams
0
0
Ker f
u
Ker f ′
f
f ′
f
f ′
M
u
M ′
N
v
N ′
N
0
0
Cok f
0
0
v
v
N ′ Cok f ′
f
Im f
f
v
′
Im f ′
0
0
i
0
0
Zi/
i Zi.

44 3. EXACT SEQUENCES OF MODULES
and
0
0
Im f
N
where the rows are short exact sequences.
Cok f
v
v
v
Im f ′
N ′ Cok f ′
3.2.2. Lemma. Given a commutative diagram of homomorphisms
0
M
u
M ′
f
f ′
N
v
N ′
g
g ′
L
w
L ′
where the rows exact sequences. The snake homomorphism δ : Ker w → Cok u
is well defined by: For z ∈ Ker w choose y ∈ N such that g(y) = z. The
element v(y) ∈ Ker g ′ so there is x ′ ∈ M ′ such that f ′ (x ′ ) = v(y). Then δ(z) =
x ′ + Im u ∈ Cok u.
Proof. Assume g(y ′ ) = z and f ′ (x ′′ ) = v(y ′ ). There is x ∈ M with f(x) = y−y ′ .
Now f ′ (u(x)) = v(f(x)) = v(y − y ′ ) = f ′ (x ′ − x ′′ ) so u(x) = x ′ − x ′′ since f ′
is injective. Then x ′ + Im u = x ′′ + Im u as wanted. The choices made respect
addition and scalar multiplication showing that δ is a homomorphism.
3.2.3. Remark. The snake is
Ker u
f
Ker v
0
0
f
g
M N L 0
u
v
w
0
M ′
f ′
N ′
g ′
L ′
Cok u
The construction of δ is schematically
M ′
Cok u
f ′
N
v
N ′
g
f ′
Ker w
L
Cok v
g
g ′
x ′
δ(z)
Ker w
Cok w
0
v(y)
z
y
z

3.2. THE SNAKE LEMMA 45
3.2.4. Theorem (snake lemma). Given a commutative diagram of homomorphisms
0
M
u
M ′
f
f ′
N
v
N ′
g
g ′
L
w
L ′
where the rows exact sequences. There is induced a six term long exact sequence
Ker u
f
Ker v
δ
f ′
Cok v
Cok u
g
g ′
Ker w
Cok w
Proof. By construction of δ it is clear that the sequence is a 0-sequence: If y ∈
Ker v then to calculate δ(g(y)) the choice v(y) = 0 gives δ ◦ g = 0. Also
f ′ (δ(z)) = v(y) + Im v shows that f ′ ◦ δ = 0. Exactness at Ker v and Cok v
are clear. Given z ∈ Ker w such that δ(z) = 0. By 3.2.2 choose y, g(y) = z
and x ′ , f ′ (x ′ ) = v(y). Then δ(z) = x ′ + Im u = 0, so choose x, u(x) = x ′ .
Now v(f(x)) = f ′ (u(x)) = v(y) so y − f(x) ∈ Ker v and g(y − f(x)) =
g(y) = z. Therefore exactness at Ker w. Given x ′ + Im u ∈ Cok u such that
f ′ (x ′ ) + Im v = 0 ∈ Cok v. Choose y, v(y) = f ′ (x ′ ) and put z = g(y). Then
w(g(y)) = g ′ (v(y)) = g ′ (f ′ (x ′ )) = 0. Now z ∈ Ker w and δ(z) = x ′ + Im u.
Therefore exactness at Cok u.
3.2.5. Corollary. If f is injective then the f : Ker u → Ker v is injective and the
long exact sequence is
0
Ker u
f
Ker v
δ
f ′
Cok v
Cok u
g
g ′
0
Ker w
Cok w
If g ′ is surjective then g ′ : Cok v → Cok w is surjective and the long exact sequence
is
Ker u
f
Ker v
δ
f ′
Cok v
Cok u
g
g ′
Ker w
Cok w
3.2.6. Corollary. (1) If v is injective and u is surjective, then w is injective.
(2) If v is surjective and w is injective, than u is surjective.
(3) If v is an isomorphism, then w is injective if and only if u is surjective.
3.2.7. Proposition. Given submodules N, L ⊂ M, then there is a short exact
sequence
0
0
M/N ∩ L x↦→(x,x)
M/N ⊕ M/L
(x,y)↦→x−y
M/N + L
0

46 3. EXACT SEQUENCES OF MODULES
Proof. The commutative diagram
0
0
N ∩ L
N ⊕ L
N + L
x↦→(x,x)
(x,y)↦→x−y
M
M ⊕ M
M
where the rows are short exact sequences, gives by 3.2.4 a five term long exact
sequence
0
M/N ∩ L
M/N ⊕ M/L
M/N + L
3.2.8. Proposition (five lemma). Given a commutative diagram of homomorphisms
M1
u1
M ′ 1
M2
u2
M ′
2
M3
u3
M ′
3
M4
u4
M ′
4
M5
u5
M ′
5
where the rows are exact sequences. If u1 is surjective, u2, u4 are isomorphism
and u5 is injective, then u3 is an isomorphism.
Proof. Let fi : Mi → Mi+1, f ′ i : M ′ i → M ′ i+1 . Split the given diagram in three as
follows
0
0
0
0
0
M1
u1
Ker f ′
2
Cok f2
u ′′
3
Ker f ′
4
Im f2
u ′ 3
Im f ′
2
M2
u2
M ′
2
M4
u4
M ′
4
M3
u3
M ′
3
Cok f1
u ′ 3
Im f ′
2
Im f4
u5
M ′
5
Cok f2
u ′′
3
Cok f ′
2
Note that Cok f1 Im f2 and Cok f ′ 2 Ker f ′ 4 . Now use 2.3.3 and the snake
lemma to conclude that Ker u3 = 0 and Cok u3 = 0 and u3 is therefore an isomorphism.
3.2.9. Proposition (windmill lemma). Given homomorphism M f
There is induced an eight term long exact sequence
0
Ker f
Ker g ◦ f
Cok f
f
g
Ker g
Cok g ◦ f
0
0
0
0
0
0
0
Cok g
0
g
N
0
L .

Proof. Look at the two diagrams
0
0
3.2. THE SNAKE LEMMA 47
0
Ker g
M
f
g◦f
L
1
By the snake lemma the sequences
0
N
M
g
L
Ker f
δ=f
Cok f
Ker g
Ker g ◦ f
f
g
N
Ker g
δ=g
Cok g
Cok g ◦ f
1
M
g◦f
L
Cok f
0
are exact and overlap to give the windmill sequence.
3.2.10. Remark. The windmill is
Ker g ◦ f
Cok g ◦ f
Cok f
Ker g ◦ f
Ker g
f
Ker f
M
N
g◦f
g
Cok f
0
L
Cok g
Cok g ◦ f
3.2.11. Exercise. (1) Given a short exact sequence
0
M
N
Show that Ann(N) ⊂ Ann(M) ∩ Ann(L).
(2) Give a short exact sequence
0
M
N
such that Ann(N) = Ann(M) ∩ Ann(L).
(3) Given ideals I, J ⊂ R. Show that there is a short exact sequence.
0
R/I ∩ J
R/I ⊕ R/J
L
L
0
0
0
0
0
R/I + J
0

48 3. EXACT SEQUENCES OF MODULES
3.3. Exactness of Hom
3.3.1. Proposition. Given an exact sequence
0
f
M
g
N
L
and an R-module K. Then the following sequence is exact
0
HomR(K, M)
HomR(K, N)
HomR(K, L)
Proof. Given h : K → M such that f ◦ h = 0 then h = 0 since f is injective.
Given k : K → N such that g ◦ k = 0 then by 3.1.4 there is h : K → M such that
f ◦ h = k. So the sequence is exact.
3.3.2. Proposition. Given a sequence
0
f
M
g
N
L
Such that for any module K, the following sequence is exact
0
HomR(K, M)
Then the original sequence is exact.
HomR(K, N)
HomR(K, L)
Proof. For K = M the identity 1M is mapped to g ◦ f ◦ 1M = 0 so it is a 0sequence.
Assume f(x) = 0. Take K = R then 1x is mapped to f ◦ 1x = 1 f(x) =
0. Therefore 1x = 0 and so x = 0. That insures that f is injective. Assume
g(y) = 0. Take K = R then 1y is mapped to g ◦ 1y = 1 g(y) = 0. There exists a
homomorphism h : R → M such that f ◦ h = 1y. By 2.5.12 h = 1x and therefore
f(x) = y. The original sequence is now proven exact.
3.3.3. Proposition. Given an exact sequence
M f
g
N
L
and a module K. Then the following sequence is exact
0
HomR(L, K)
HomR(N, K)
0
HomR(M, K)
Proof. Given h : L → K such that h ◦ g = 0 then h = 0 since g is surjective.
Given k : N → K such that k ◦ f = 0 then by 3.1.5 there is h : L → K such that
h ◦ g = k. So the sequence is exact.
3.3.4. Proposition. Given a sequence
M f
g
N
L
such that for any module K, the following sequence is exact
0
HomR(L, K)
Then the original sequence is exact.
HomR(N, K)
0
HomR(M, K)
Proof. For K = L the identity 1L is mapped to 1L ◦g ◦f = 0 so it is a 0-sequence.
Take K = Cok g then pg : L → Cok g has pg ◦ g = 0, but by exactness 0 is the
unique homomorphism satisfying this, so pg = 0. Therefore Cok g = 0 and g is
surjective. Take K = Cok f, p : N → Cok f the projection. p ◦ f = 0 so by
exactness there exists a unique q : L → Cok f such that q ◦ g = p. It follows that
Ker g ⊂ Ker p = Im f. All together the original sequence is exact.

3.3.5. Proposition. Given a sequence
The following are equivalent.
0
3.4. EXACTNESS OF TENSOR 49
f
M
g
N
(1) The sequence is split exact.
(2) For any K the following sequence is exact
0
HomR(K, M)
HomR(K, N)
(3) For any K the following sequence is exact
0
HomR(L, K)
HomR(N, K)
L
0
HomR(K, L)
HomR(M, K)
If the conditions are true, then the sequences (2) and (3) are split exact.
Proof. (1) ⇒ (2), (1) ⇒ (3) are clear by 3.1.14 giving that the sequences (2), (3)
are split exact. (2) ⇒ (1): Let K = L, then there is a section to g. By 3.3.2 and
3.1.11 the original sequence is split exact. (3) ⇒ (1): Let K = M, then there is a
retraction to f. By 3.3.4 and 3.1.11 the original sequence is split exact.
3.3.6. Exercise. (1) Show that the sequence
0
HomZ(Q/Z, Z)
HomZ(Q/Z, Q)
is exact, but the rightmost map is not surjective.
(2) Show that the sequence
0
HomZ(Z/(n), Z)
n
HomZ(Z/(n), Z)
is exact, but the rightmost map is not surjective.
3.4. Exactness of Tensor
3.4.1. Proposition. Given an exact sequence
M f
g
N
L
0
and an R-module K. Then the following sequence is exact
M ⊗R K
N ⊗R K
L ⊗R K
0
0
HomZ(Q/Z, Q/Z)
HomZ(Z/(n), Z/(n))
Proof. Let K ′ be any module. By 3.3.4 it is enough to see that the sequence
0
HomR(L ⊗R K, K ′ )
is exact. By 2.6.13 it amounts to see that the sequence
0
HomR(L, HomR(K, K ′ ))
is exact. This follows from 3.3.3.
0
HomR(N ⊗R K, K ′ )
HomR(M ⊗R K, K ′ )
HomR(M, HomR(K, K ′ ))
HomR(N, HomR(K, K ′ ))

50 3. EXACT SEQUENCES OF MODULES
3.4.2. Proposition. Given a split exact sequence
0
f
M
g
N
L
and a module K. Then the following sequence is split exact
0
K ⊗R M
K ⊗R N
Proof. This follows from the functor properties 3.1.14.
0
K ⊗R L
3.4.3. Proposition. Let I ⊂ R be an ideal. For any module M, the homomorphism
is an isomorphism.
M ⊗R R/I → M/IM, x ⊗ a + I ↦→ ax + IM
Proof. x + IM ↦→ x ⊗ 1 is an inverse.
3.4.4. Corollary. Let I, J ⊂ R be ideals. Then
is an isomorphism.
R/I ⊗R R/J → R/(I + J), a + I ⊗ b + J ↦→ ab + I + J
3.4.5. Proposition. Let I1, . . . , Ik ⊂ R be pairwise comaximal ideals and M a
module. Then the product of projections
is an isomorphism.
Proof. By Chinese remainders 1.4.2
M/I1 · · · IkM → M/I1M × · · · × M/IkM
R/I1 · · · Ik → R/I1 × · · · × R/Ik
is an isomorphism. Tensor with M and use 3.4.3 to get the isomorphism.
3.4.6. Exercise. (1) Calculate Z/(m) ⊗Z Z/(n) for all integers m, n.
(2) Let I ⊂ R be an ideal. Show that R/I ⊗R R/I R/I.
(3) Let I ⊂ R be an ideal. Show that I ⊗R R/I I/I 2 .
(4) Let 2Z be scalar multiplication. Show that 2Z ⊗ 1 Z/(2) : Z ⊗Z Z/(2) → Z ⊗Z Z/(2)
is not injective.
3.5. Projective modules
3.5.1. Definition. An R-module F is a projective module if for any exact sequence
N → L → 0 the sequence
is exact.
HomR(F, N) → HomR(F, L) → 0
3.5.2. Proposition. A module F is projective if and only if any surjective homomorphism
M → F → 0 has a section.
Proof. Assume F projective and g : M → F surjective. Then HomR(F, M) →
HomR(F, F ) → 0 is exact. So there exists a v : F → M such that g ◦ v = 1F .
v is then a section. Conversely given g : N → L surjective and h : F → L. Let
M = Ker N ⊕F → L, (y, z) ↦→ g(y)−h(z) and pN : M → N, pF : M → F the
0

3.5. PROJECTIVE MODULES 51
projections, then g ◦ pN = h ◦ pF . Now pF is surjective since g is. Let v : F → M
be a section of pF , then h ′ = pN ◦ v satisfies h = g ◦ h ′ .
pN
N g
L
h
M
F
′
h
v
3.5.3. Corollary. A short exact sequence
0
f
M
pF
g
N
0
F
where F is a projective module is a split exact sequence.
3.5.4. Corollary. A free module is projective. Over a field every module is projective.
3.5.5. Example. Let I ⊂ R be an ideal. If R/I is projective, then there is a ring
decomposition R/I × R ′ R.
3.5.6. Proposition. A direct summand in a projective module is projective.
Proof. Let F ⊕ F ′ be projective and g : M → F surjective. By 3.5.2 there is a
section v ′ : F ⊕ F ′ → M ⊕ F ′ to (g, 1F ′). Then v(y) = pM ◦ v ′ (y, 0) is a section
to g and F is projective.
3.5.7. Proposition. A module is projective if and only if it is a direct summand in
a free module.
Proof. By 2.4.12 any module is a factor module of a free module. By 3.5.2 a
projective factor module has a section, and is therefore by 3.1.10 a direct summand.
3.5.8. Proposition. Let Fα be a family of projective modules, then the direct sum
α Fα is a projective module.
Proof. Let N → L be surjective. Then by 2.5.8
is the product
0
HomR( Fα, N) → HomR( Fα, L)
HomR(Fα, N) → HomR(Fα, L)
which is surjective by 3.1.6. So Fα is projective.
3.5.9. Proposition. Let F, F ′ be projective modules. Then F ⊗R F ′ is projective.
Proof. F ⊗R F ′ is clearly a direct summand in a free module.
3.5.10. Proposition. Let R → S be a ring homomorphism and F a projective
module. The change of ring module F ⊗R S is a projective S-module.
Proof. A direct summand of a free R-module is changed to a direct summand of a
free S-module.

52 3. EXACT SEQUENCES OF MODULES
3.5.11. Proposition. Any module M admits an exact sequence
F → M → 0
where F is a projective module. That is, any module is a factor module of a projective
module.
Proof. Take F free, 2.4.12.
3.5.12. Exercise. (1) Let R = R1 × R2. Show that R1, R2 are projective ideals in R.
(2) Show that the ideal (2)/(6) in the ring Z/(6) is projective.
(3) Show that the ideal (2)/(4) in the ring Z/(4) is not projective.
3.6. Injective modules
3.6.1. Definition. An R-module E is an injective module if for any exact sequence
0 → M → N the sequence
is exact.
HomR(N, E) → HomR(M, E) → 0
3.6.2. Proposition. A module E is an injective module if and only if any injective
homomorphism 0 → E → L has a retraction.
Proof. Assume E injective and f : E → L injective. Then HomR(L, E) →
HomR(E, E) → 0 is exact. So there exists a u : L → E such that u ◦ f = 1E.
Then u is a retraction. Conversely given f : M → N injective and h : M → E.
Let L = Cok M → E ⊕ N, x ↦→ h(x) − f(x) and iE : E → L, iN : N → L the
injections, then iN ◦ f = iE ◦ h. Now iE is injective since f is. Let u : L → E be
a retraction of iE, then h ′ = u ◦ iN satisfies h = h ′ ◦ f.
0
M
3.6.3. Corollary. A short exact sequence
0
f
E
f
N
h
h
′
E iE
L
u
g
N
iN
L
where E is an injective module is a split exact sequence.
3.6.4. Example. Let I ⊂ R be an ideal. If I is injective, then there is a ring
decomposition R/I × R ′ R.
3.6.5. Proposition. A direct summand in an injective module is injective.
Proof. Let E ⊕ E ′ be an injective module and f : E → L an injective homomorphism.
By 3.6.2 there is a retraction u ′ : L ⊕ E ′ → E ⊕ E ′ to (f, 1E ′). Then
u(y) = pE ◦ u ′ (y, 0) is a retraction to f and E is injective.
3.6.6. Proposition. Let Eα be a family of injective modules, then the direct product
α Eα is an injective module
0

Proof. Let M → N be injective. Then by 2.5.8
is the product
3.6. INJECTIVE MODULES 53
HomR(N, Eα) → HomR(M, Eα)
HomR(N, Eα) → HomR(M, Eα)
which is surjective by 3.1.6. So Eα is injective.
3.6.7. Proposition. A module E is injective, if for any ideal I ⊂ R
is exact.
HomR(R, E) → HomR(I, E) → 0
Proof. Let f : E → L be an injective homomorphism. The set of submodules
f(E) ⊂ L ′ ⊂ L and retractions u ′ : L ′ → E is nonempty and inductively ordered.
By Zorn’s lemma a maximal L ′ , u ′ exists. If L/L ′ = 0 choose a y ∈ L\L ′ . The
homomorphism Ann(y + L ′ ) → E, a ↦→ u ′ (ay) extends to u ′′ : R → E by
hypothesis. The setting L ′ + Ry → E, x + ay ↦→ u ′ (x) + u ′′ (a) is a well defined
retraction. This contradicts maximality.
3.6.8. Definition. Let R be a domain. M is a divisible module if scalar multiplication
with a nonzero a ∈ R is surjective.
3.6.9. Proposition. (1) An injective module over a domain is divisible.
(2) A divisible module over a principal ideal domain is injective.
(3) Over a field any module is injective.
Proof. (1) Let E be an injective module and a = 0. Choose x ∈ E and look
at 1x : R → E. Scalar multiplication aR is injective, so there is an extension
1y : R → E such that 1x = 1y ◦ aR. Then ay = x. (2) Let E be divisible. To
extend f : (a) → E, choose x ∈ E such that ax = f(a), then 1x : R → E, 1 → x
extends f. (3) This is clear.
3.6.10. Proposition. Let R → S be a ring homomorphism and E an injective
R-module. The induced module HomR(S, E) is an injective S-module.
Proof. Let M → N be an injective homomorphism of S-modules. Then by
2.7.10 HomS(N, HomR(S, E)) → HomS(M, HomR(S, E)) is HomR(N, E) →
HomR(M, E) being surjective since E is an injective R-module.
3.6.11. Lemma. (1) Q and Q/Z are divisible and therefore injective Z modules.
(2) The homomorphism 2.5.11
is injective.
x ↦→ evx : M → HomZ(HomZ(M, Q/Z), Q/Z)
Proof. (1) Clear by 3.6.9. (2) Let x = 0 and choose h : Z/ Ann(x) → Q/Z
nonzero. Now Z/ Ann(x) Zx ⊂ M so extend h to h ′ : M → Q/Z. Then
evx(h ′ ) = h(1) = 0.
3.6.12. Proposition. Any module M admits an exact sequence
0 → M → E
where E is an injective module. That is, any module is a submodule of an injective
module.

54 3. EXACT SEQUENCES OF MODULES
Proof. By 2.4.12 choose a surjection F → HomZ(M, Q/Z) where F is a free
R-module. Then
0 → M → HomZ(F, Q/Z)
is exact by 3.6.11. Since F
α R the module
HomZ(F, Q/Z)
HomZ(R, Q/Z)
is injective, 3.6.6 and 3.6.10.
3.6.13. Proposition. A homomorphism f : M → N is injective if and only if
is surjective for any injective module E.
HomR(N, E) → HomR(M, E) → 0
Proof. Assume that Ker f = 0 and choose 0 → Ker f → E, 3.6.12. The sequence
HomR(N, E) → HomR(M, E) → HomR(Ker f, E) → 0
is exact. So HomR(N, E) → HomR(M, E) is not surjective.
3.6.14. Definition. A submodule N ⊂ M is an essential extension if any nonzero
submodule L ⊂ M has nonzero intersection N ∩ L = 0. An essential extension
M ⊂ E with E injective is an injective envelope of M.
3.6.15. Proposition. Any module M has an injective envelope. If M ⊂ E, E ′ are
two injective envelopes, then there is an isomorphism f : E → E ′ fixing M.
Proof. By 3.6.12 choose M ⊂ E ′ with E ′ injective. By Zorn’s lemma choose
M ⊂ E ⊂ E ′ maximal among the essential extensions of M. If E = E ′ then the
set of modules N ′ ⊂ E ′ such that E ∩ N ′ is nonempty by maximality of E. Let
N be maximal among these by Zorn’s lemma. It follows that E ⊕ N = E ′ and E
is injective by 3.6.6. Given two envelopes, let f : E → E ′ be any homomorphism
fixing M. Then f is injective, since M ⊂ E is essential. If f is not surjective, then
E ′ f(E) ⊕ E ′′ contradicting that M ⊂ E ′ is essential.
3.6.16. Exercise. (1) Let R be a domain. Show that the fraction field is an injective
module.
(2) Let R be a domain. Show that the torsion free divisible module injective.
(3) Show that for a ring that all modules are projective if and only all modules are injective.
α
3.7. Flat modules
3.7.1. Definition. An R-module F is a flat module if for any exact sequence 0 →
M → N the sequence
0 → M ⊗R F → N ⊗ F
is exact.
3.7.2. Example. If a ∈ R is a nonzero divisor and M is a flat modules, then aM
is injective and a is a nonzero divisor on M.
3.7.3. Proposition. A direct summand in a flat module is flat.
Proof. Let F ⊕ F ′ be flat and M → N injective. Then M ⊗R (F ⊕ F ′ ) →
N ⊗R (F ⊕ F ′ ) is injective. Conclusion by 2.6.11.

3.7. FLAT MODULES 55
3.7.4. Proposition. Let Fα be family of flat modules, then the direct sum
α Fα
is a flat module.
Proof. Let M → N be injective. Then M ⊗R ( Fα) → N ⊗R ( Fα) is
injective by 2.6.11 and 3.1.6, so the product is flat.
3.7.5. Example. A free module is flat.
3.7.6. Corollary. A projective module is flat.
Proof. By 3.5.7 a projective module is a direct summand in a free. Conclusion by
3.7.4.
3.7.7. Proposition. Let F, F ′ be flat modules. Then F ⊗R F ′ is flat.
Proof. Let M → N be injective. Then by 2.6.10 M ⊗R (F ⊗R F ′ ) → N ⊗R
(F ⊗R F ′ ) is (M ⊗R F ) ⊗R F ′ → (N ⊗R F ) ⊗R F ′ being injective by using the
definition twice.
3.7.8. Proposition. Let R → S be a ring homomorphism and F a flat R-module.
The change of ring module F ⊗R S is a flat S-module.
Proof. Let M → N be an injective homomorphism of S-modules. Then by 2.7.5
M ⊗S (F ⊗R S ′ ) → N ⊗S (F ⊗R S) is M ⊗R F → N ⊗R F being injective since
F is a flat R-module.
3.7.9. Proposition. Let R be a ring and F a module. The following are equivalent.
(1) F is a flat module.
(2) HomR(F, E) is an injective module for any injective module E.
Proof. Let M → N be injective. By 3.6.13 M ⊗R F → N ⊗R F is injective if and
only if HomR(N ⊗R F, E) → HomR(M ⊗R F, E) is surjective for any injective
module E. By 2.6.13 this is HomR(N, HomR(F, E)) → HomR(M, HomR(F, E)).
3.7.10. Corollary. Given a short exact sequence
0
M
N
F
where F is a flat module. Then M is flat if and only if N is flat.
Proof. Let E be an injective module. By 3.7.9 and 3.6.3 the sequence
0
HomR(F, E)
HomR(N, E)
0
HomR(M, E)
is split exact. By 3.6.5 and 3.6.6 HomR(N, E) is injective if and only if HomR(M, E)
is so. Conclusion by 3.7.9.
3.7.11. Corollary. Given a short exact sequence
0
M
where F is a flat module. For any module L there is a short exact sequence
0
L ⊗R M
N
L ⊗R N
F
0
L ⊗R F
0
0

56 3. EXACT SEQUENCES OF MODULES
Proof. Let E be an injective module. By 3.7.9 and 3.6.3 the sequence
0
HomR(F, E)
is split exact. So also the sequence
0
HomR(L, HomR(F, E))
HomR(N, E)
HomR(M, E)
HomR(L, HomR(N, E))
HomR(L, HomR(M, E))
is split exact. By 2.6.13 this is natural isomorphic to the sequence
0
Conclusion by 3.6.13.
HomR(L ⊗R F, E)
HomR(L ⊗R N, E)
HomR(L ⊗R M, E)
3.7.12. Proposition. A module F is flat, if for any ideal I ⊂ R
is exact.
0 → I ⊗R F → R ⊗R F
Proof. By 3.7.9 it suffices to see that HomR(F, E) is injective for any injective E.
By 3.6.7 this amounts to HomR(R, HomR(F, E)) → HomR(I, HomR(F, E)) being
surjective. By 2.6.13 this homomorphism is HomR(R⊗RF, E) → HomR(I⊗R
F, E) which is surjective since E is injective.
3.7.13. Exercise. (1) Show that Z/(n) is not a flat Z-module for n = 0, 1.
(2) Show that Z/(2) is a flat Z/(6)-module.
(3) Show that Z/(2) is not a flat Z/(4)-module.
0
0
0

4
Fraction constructions
4.1. Rings of fractions
4.1.1. Lemma. Let R be a ring and U ⊂ R such that 1 ∈ U and for u, v ∈ U the
product uv ∈ U. On U × R is defined a relation
(u, a) ∼ (u ′ , a ′ ) ⇔ there is v ∈ U such that vu ′ a = vua ′
(1) The relation is an equivalence relation.
(2) If (u, a) ∼ (u ′ , a ′ ) and (v, b) ∼ (v ′ , b ′ ) then (uv, va + ub) ∼ (u ′ v ′ , v ′ a ′ +
u ′ b ′ ).
(3) If (u, a) ∼ (u ′ , a ′ ) and (v, b) ∼ (v ′ , b ′ ) then (uv, ab) ∼ (u ′ v ′ , a ′ b ′ ).
Proof. The claims are proved by simple calculations. (1) Symmetry is clear. Reflexive
follows as 1 ∈ U. Transitive: if (u, a) ∼ (u ′ , a ′ ), (u ′ , a ′ ) ∼ (u ′′ , a ′′ ) then
vu ′ a = vua ′ , v ′ u ′′ a ′ = v ′ u ′ a ′′ . Since multiplication is commutative (v ′ vu ′ )u ′′ a =
v ′ u ′′ vua ′ = (v ′ vu ′ )ua ′′ give (u, a) ∼ (u ′′ , a ′′ ). (2) From wu ′ a = wua ′ , w ′ v ′ b =
w ′ vb ′ follow that w ′ vv ′ wu ′ a = w ′ vv ′ wua ′ , wuu ′ w ′ v ′ b = wuu ′ w ′ vb ′ . So now
ww ′ (u ′ v ′ )(va + ub) = ww ′ (uv)(v ′ a ′ + u ′ b ′ ) as needed. (3) Is similar.
4.1.2. Definition. Let R be a ring. U ⊂ R is a multiplicative subset if 1 ∈ U
and for u, v ∈ U the product uv ∈ U. The ring of fractions U −1R is given by
on U × R under the relation 4.1.1
equivalence classes a
u
(u, a) ∼ (u ′ , a ′ ) ⇔ there is v ∈ U such that vu ′ a = vua ′
The addition is
a b
+
u v
and the multiplication is
a b
·
u v
The canonical ring homomorphism is
= va + ub
uv
= ab
uv
ι : R → U −1 R, a ↦→ a
1
4.1.3. Proposition. Let φ : R → S be a ring homomorphism and U ⊂ R a
multiplicative subset. If all elements in φ(U) ⊂ S are units, then there exists a
unique ring homomorphism φ ′ : U −1 R → S such that φ = φ ′ ◦ ι.
φ
R
ι
U −1R Proof. φ ′ ( a
u ) = φ(a)φ(u)−1 is the well defined unique ring homomorphism. Observe
that the elements of form bv −1 satisfy the rules for fractions.
57
φ ′
S

58 4. FRACTION CONSTRUCTIONS
4.1.4. Proposition. Let U ⊂ R be a multiplicative subset.
(1) a
1 = 0 in U −1R if and only if Ann(a) ∩ U = ∅.
(2) The canonical ring homomorphism R → U −1R is injective if and only if U
consists only of nonzero divisors.
Proof. a
1 = 0 if and only if va = 0 for some v ∈ U.
4.1.5. Corollary. If R is a domain and U is the nonzero elements then K = U −1 R
is a field and the canonical homomorphism identifies R as a subring.
4.1.6. Definition. The field K in 4.1.5 is the fraction field of R.
4.1.7. Corollary. If R is a domain with fraction field K and φ : R → L is an
injective ring homomorphism into a field, then there is a unique homomorphism
K → L extending φ.
4.1.8. Definition. The total ring of fractions of R is U −1 R, where U is the set of
nonzero divisors of R.
4.1.9. Corollary. A ring is a subring of its total ring of fractions.
4.1.10. Proposition. Let U = {u n } the powers of an element u ∈ R. There is an
isomorphism
R[X]/(uX − 1) → U −1 R, X ↦→ 1
u
The ring of fractions with only one denominator is of finite type.
Proof. A pair of inverse homomorphism are constructed by 1.6.7 and 4.1.3.
4.1.11. Exercise. (1) Show that if U contains a nilpotent element, then U −1 R = 0.
(2) Show that
Ker R → U −1 R = {a ∈ R|ua = 0, for some u ∈ U}
(3) Let U = {u1, . . . , um} and u = u1 · · · um. Then show that
U −1 R = {u n } −1 R
(4) Let R1, R2 be domains with fraction fields K1, K2. Show that the total ring of fractions
of R1 × R2 is K1 × K2.
(5) Show that the ring
{ a
∈ Q|a ∈ Z, n ∈ N}
2n is of finite type over Z.
4.2. Modules of fractions
4.2.1. Lemma. Let R be a ring and U ⊂ R multiplicative. On U × M is defined
a relation
(u, x) ∼ (u ′ , x ′ ) ⇔ there is v ∈ U such that vu ′ x = vux ′
(1) The relation is an equivalence relation.
(2) If (u, x) ∼ (u ′ , x ′ ) and (v, y) ∼ (v ′ , y ′ ) then (uv, vx + uy) ∼ (u ′ v ′ , v ′ x ′ +
u ′ y ′ ).
(3) If (u, a) ∼ (u ′ , a ′ ) in U × R 4.1.1 and (v, x) ∼ (v ′ , x ′ ) then (uv, ax) ∼
(u ′ v ′ , a ′ x ′ ).
Proof. The claims are proved by simple calculations. See the proof 4.1.1.

4.2. MODULES OF FRACTIONS 59
4.2.2. Definition. Let R be a ring and U a multiplicative subset. The module of
fractions U −1M is given by equivalence classes x
u on U × M under the relation
4.2.1
(u, x) ∼ (u ′ , x ′ ) ⇔ there is v ∈ U such that vu ′ x = vux ′
The addition is
x y
+
u v
and the U −1R-scalar multiplication is
a y
·
u v
The canonical homomorphism is
= vx + uy
uv
= ay
uv
M → U −1 M, x ↦→ x
1
4.2.3. Lemma. Let R be a ring and f : M → N a homomorphism of R-modules.
Then there is a homomorphism U −1f : U −1M → U −1N, x f(x)
u ↦→ u of U −1R modules.
Proof. The claims are proved by simple calculations. If (u, x) ∼ (u ′ , x ′ ) then
vu ′ x = vux ′ and therefore vu ′ f(x) = vuf(x ′ ) showing (u, f(x)) ∼ (u ′ , f(x ′ )).
So the map is well defined. The rest is similar.
4.2.4. Proposition. The construction 4.2.2
and 4.2.3
M ↦→ U −1 M
f : M → N ↦→ U −1 f : U −1 M → U −1 N
is a functor from R-modules to U −1 R-modules.
Proof. Follows from the definitions by simple calculations as in the proof of 4.2.3.
For example, U −1 (f + g) = U −1f + U −1g, follows from f(x)+g(x)
u
g(x)
u .
= f(x)
u +
4.2.5. Remark. The induced homomorphism relates to the canonical homomorphism
such that the diagram is commutative.
M
f
N
U −1M U −1f
U −1N That is, the canonical homomorphism is a natural homomorphism.
4.2.6. Proposition. Let U ⊂ R be a multiplicative subset and M a module.
(1) x
1 = 0 in U −1M if and only if Ann(x) ∩ U = ∅.
(2) The canonical homomorphism M → U −1M is injective if and only if U
consists only of nonzero divisors on M.
Proof. x
1 = 0 if and only if vx = 0 for some v ∈ U.

60 4. FRACTION CONSTRUCTIONS
4.2.7. Proposition. If Mα is a family of modules, then the homomorphism
U −1 (
Mα) →
is a natural isomorphism of U −1 R-modules.
α
α
U −1 Mα
Proof. This is the method of common denominators in a finite sum.
xi
=
ui
1
(Πj=iuj)xi
Πiui
i
4.2.8. Exercise. (1) Show that if U contains a nilpotent element, then U −1 M = 0.
(2) Show that
Ker M → U −1 M = {x ∈ M|ux = 0, for some u ∈ U}
(3) Let U = {u1, . . . , um} and u = u1 · · · um. Then show that
U −1 M = {u n } −1 M
(4) Show that U −1 M = 0 if and only if U ∩ Ann(x) = ∅ for all x ∈ M.
(5) Show that the fraction homomorphism of a composition is the composition of the
respective fraction homomorphisms.
(6) Let M be a free R-module. Show that U −1 M is a free U −1 R-module
4.3. Exactness of fractions
4.3.1. Proposition. Let R be a ring and U a multiplicative subset. Given an exact
sequence of R-modules
M f
Then the following sequence is exact
U −1 M
i
g
N
U −1N
L
U −1L Proof. If y
u ∈ U −1N maps to g(y)
u = 0 then there is v ∈ U such that 0 = vg(y) =
g(vy). Choose x ∈ M such that f(x) = vy. Then x
exactness.
4.3.2. Corollary. Given a short exact sequence
0
f
M
Then the following sequence is exact
0
U −1M g
N
U −1N
L
vu
maps to f(x)
vu
0
U −1L If the first sequence is split exact, also the second sequence is split exact.
0
= y
u proving
4.3.3. Corollary. For a homomorphism f : M → N there are natural isomorphisms
of U −1 R-modules.
(1) U −1 Ker f Ker U −1 f.
(2) U −1 Im f Im U −1 f.
(3) U −1 Cok f Cok U −1 f.

4.3. EXACTNESS OF FRACTIONS 61
Proof. Represent the statements using short exact sequences. (1) The kernel is
determined by the exact sequence 0 → Ker f → M → N, 3.1.4. (3) The cokernel
is determined by the exact sequence M → N → Cok f → 0, 3.1.5. (2) The
image is determined by the exact sequence 0 → Ker f → M → Im f → 0, 2.3.5,
3.1.4.
4.3.4. Corollary. For submodules N, L ⊂ M there are natural identifications of
U −1 R-submodules and factor modules.
(1) U −1 (M/N) = U −1 M/U −1 N.
(2) U −1 (N + L) = U −1 N + U −1 L.
(3) U −1 (N ∩ L) = U −1 N ∩ U −1 L.
Proof. Represent the statements using short exact sequences. (1) 0 → N → M →
M/N → 0 is short exact giving 0 → U −1 N → U −1 M → U −1 (M/N) → 0. The
wanted isomorphism follows form 3.1.5. (2) N + L is the image of N ⊕ L → M
so conclude by 4.3.3. (3) N ∩ L is the kernel of N ⊕ L → M so conclude by
4.3.3.
4.3.5. Corollary. For ideals I, J ⊂ R there are natural identifications in U −1 R.
(1) U −1 (R/I) = U −1 R/U −1 I.
(2) U −1 (I + J) = U −1 I + U −1 J.
(3) U −1 (I ∩ J) = U −1 I ∩ U −1 J.
(4) U −1 (IJ) = U −1 IU −1 J.
Proof. (1) (2) (3) These are special cases of 4.3.4. (4) Both sides have the same
, a ∈ I, b ∈ J, u ∈ U.
generators ab
u
4.3.6. Proposition. Let R → U −1 R be the canonical homomorphism.
(1) For an ideal I ⊂ R the extended ideal
IU −1 R = U −1 I
(2) For an ideal J ⊂ U −1 R the extended contracted ideal
U −1 (J ∩ R) = J
(3) For an ideal I ⊂ R the contracted extended ideal
I ⊂ IU −1 R ∩ R
Proof. (1) This is clear. (2) U −1 (J ∩ R) ⊂ J is true for any ring homomorphism
1.2.6. If b
b
1 b
u ∈ J then 1 ∈ J giving b ∈ J ∩ R and ub = u ∈ U −1 (J ∩ R). (3) This
is true for any ring homomorphism 1.2.6.
4.3.7. Proposition. Let R → U −1 R be the canonical homomorphism. For an
ideal I ⊂ R the contracted extended ideal
I = IU −1 R ∩ R
if and only if each u ∈ U is a nonzero divisor on R/I.
Proof. Apply the snake lemma 3.2.4 to
0
0
I
0
R
U −1R/U −1I U −1R/U −1I
0
R/I
0

62 4. FRACTION CONSTRUCTIONS
and get the exact sequence
Conclusion from 4.2.6.
0 → I → IU −1 R ∩ R → Ker(R/I → U −1 (R/I)) → 0
4.3.8. Corollary. Let P ⊂ R be a prime ideal. Then U −1 P ⊂ U −1 R is either a
prime ideal or the whole ring.
Proof. By 4.1.5 and 4.3.5 U −1 R/U −1 P = U −1 (R/P ) is either 0 or a domain.
4.3.9. Corollary. Let R be a principal ideal domain. Then U −1 R is a principal
ideal domain.
Proof. A restricted ideal is principal by hypothesis and the extension of a principal
ideal is principal. Conclude by 4.3.6.
4.3.10. Proposition. Let R be a unique factorization domain. Then U −1 R is a
unique factorization domain.
Proof. By 4.3.7 the extension of an irreducible element is either a unit or an irreducible
element. Since a principal ideal ( a a
u ) = ( 1 ), a factorization into irreducibles
in R gives a factorization in U −1R. Now if (a) = (p1) . . . (pn) is a factorization
in R and ( a
1 ) is irreducible in U −1R. Then all but one pi
1 is a unit in U −1R so
( a pi
1 ) = ( 1 ) is a prime ideal 4.3.9. The conditions 1.5.3 are satisfied.
4.3.11. Exercise. (1) Let U ⊂ R be multiplicative. Show that
U −1 (R[X]) = (U −1 R)[X]
(2) Let φ : R → S be a ring homomorphism and U ⊂ R a multiplicative subset. Show
that
U −1 S = φ(U) −1 S
4.4. Tensor modules of fractions
4.4.1. Proposition. Let R be a ring and U a multiplicative subset. For any module
M, the homomorphism
M ⊗R U −1 R → U −1 M, x ⊗ a ax
↦→
u u
is a natural isomorphism of U −1R-modules. Proof. By 2.6.3 there is an R-module homomorphism x ⊗ a
u
b a
ba bax b ax
(x ⊗ ) = x ⊗ ↦→ =
v
u
vu
vu
map U −1 M → M ⊗R U −1 R, x
u
v
ax → u . By definition
u , so the this is a U −1 R-homomorphism. The
↦→ x ⊗ 1
u
is an inverse.
4.4.2. Remark. The two constructions, module change of ring to a fraction ring
and fraction module are natural isomorphic functors from modules to modules over
the fraction ring.
4.4.3. Corollary. Let R be a ring and U a multiplicative subset. Then U −1 R is a
flat R-module.
Proof. This follows from 4.4.1 and 4.3.1.
4.4.4. Corollary. Let R be a ring and U a multiplicative subset and M, N modules.
Then there is a natural isomorphism
U −1 (M ⊗R N) U −1 M ⊗ U −1 R U −1 N

4.5. HOMOMORPHISM MODULES OF FRACTIONS 63
Proof. This follows from 2.7.5 and 4.4.1.
4.4.5. Corollary. Let I ⊂ R be an ideal and U a multiplicative subset For a
module M module
U −1 (IM) U −1 IU −1 M
Proof. Use that IM = Im(I ⊗R M → M) and 4.4.4.
4.4.6. Exercise. (1) Let M be a flat R-module. Show that U −1 M is a flat U −1 Rmodule.
(2) Let M be a projective R-module. Show that U −1 M is a projective U −1 R-module.
4.5. Homomorphism modules of fractions
4.5.1. Proposition. Let R be a ring and U a multiplicative subset. For any modules
M, N there is a natural homomorphism
of U −1 R-modules.
U −1 HomR(M, N) → Hom U −1 R(U −1 M, U −1 N)
Proof. Given f : M → N and u ∈ U the setting x
homomorphism U −1 M → U −1 N.
v ↦→ f(x)
uv is a U −1R 4.5.2. Proposition. Let R be a ring and U a multiplicative subset. For any Rmodule
M and any U −1 R-modules N there is a natural isomorphism
of U −1 R-modules.
HomR(M, N) → Hom U −1 R(U −1 M, N)
Proof. This is the change of rings isomorphism 2.7.6 interpreted according to
4.4.2.
4.5.3. Example. Let R be a ring and U a multiplicative subset. R → U −1 R the
canonical homomorphism. The induced module functor maps an R-module M to
the U −1 R-module HomR(U −1 R, M). The natural isomorphism 2.7.10 is
for any U −1 R-module N.
HomR(N, M) Hom U −1 R(N, HomR(U −1 R, M))
4.5.4. Example. The homomorphism 4.5.1 is in general neither injective nor surjective.
(1) Not surjective:
(2) Not injective:
0 = HomZ(Q, Z) → HomQ(Q, Q) = Q
0 = HomZ(Q, Q/Z) ⊗Z Q → HomQ(Q, Q/Z ⊗Z Q) = 0
4.5.5. Exercise. (1) Let M be a Z-module and N a Q-module. Show that there is an
isomorphism HomZ(M, N) HomQ(M ⊗Z Q, N).

64 4. FRACTION CONSTRUCTIONS
4.6. The polynomial ring is factorial
4.6.1. Definition. Let R be a unique factorization domain and let f = anX n +
· · · + a0 be a polynomial over R. Then the content of polynomial f, c(f), is the
greatest common divisor of the coefficients a0, . . . , an.
4.6.2. Proposition (Gauss’ lemma). Let R be a unique factorization domain. For
polynomials f, g ∈ R[X]
c(fg) = c(f)c(g)
Proof. Assume by cancellation that c(f), c(g) are units in R. For any irreducible
p ∈ R the projections of f, g in R/(p)[X] are nonzero. Since R has unique factorization
the ideal (p) is a prime ideal. It follows that the projection of the product
fg in R/(p)[X] is also nonzero and therefore p is not a common divisor of the
coefficients of the product fg.
4.6.3. Proposition. Let R be a unique factorization domain. Then the ring of
polynomials R[X] is a unique factorization domain.
Proof. Let K be the fraction field of R, then the polynomial ring K[X] is a principal
ideal domain. Let f ∈ R[X] and use unique factorization in K[X] to get
0 = a ∈ R and p1, . . . , pn ∈ R[X], irreducible in K[X], such that
af = p1 . . . pn
Assume by 4.6.2 that a = 1 and c(p1), . . . , c(pn) are units in R. Apply 4.6.2 and
1.6.5 to see that p1, . . . , pn are irreducible in R[X]. An irreducible p ∈ R generates
a prime ideal (p) ⊂ R[X]. A non constant irreducible p ∈ R[X] generates a prime
ideal (p) ⊂ K[X] and therefore also a prime ideal (p) ⊂ R[X]. So conditions
1.5.3 are satisfied.
4.6.4. Theorem. Let K be a field. Then the polynomial ring K[X1, . . . , Xn] is a
unique factorization domain.
Proof. Follows by induction from 4.6.3.
4.6.5. Exercise. (1) Let f ∈ Z[X] be monic and assume f = gh where g, h ∈ Q[X]
are monic. Show that g, h ∈ Z[X].
(2) Let f ∈ Z[X] be monic and irreducible in Z/(n)[X]. Show that f is irreducible
Q[X].
(3) Let K be a field. Show that the polynomial ring K[X1, X2, . . . ] in countable many
variables is a unique factorization domain

5
Localization
5.1. Prime ideals
5.1.1. Theorem (Krull). A nonzero ring contains a maximal ideal.
Proof. The nonempty set of ideals different from R is ordered by inclusion. Given
an increasing chain Iα then ∪Iα is an ideal different from R which is a maximum
for the chain. Conclusion by Zorn’s lemma.
5.1.2. Corollary. Any proper ideal in a ring is contained in a maximal ideal.
Proof. The factor ring is nonzero and contains a maximal ideal. Conclusion by
1.2.10.
5.1.3. Proposition. Let P1, . . . , Pn ⊂ R be ideals with at most 2 not being prime
ideals. If an ideal
I ⊂ P1 ∪ · · · ∪ Pn
then I ⊆ Pi for some i.
Proof. Assume n > 1 and I is not contained in any sub union. Moreover assume
the numbering such that P3, . . . , Pn are prime ideals. Then for each i there is
The element
ai ∈ (I ∩ Pi)\ ∪j=i Pj
an + a1 . . . an−1
is in I but not in any Pi, giving a contradiction. So n = 1.
5.1.4. Proposition. Let P1, . . . , Pn ⊂ R be prime ideals and I any ideal. If for
some a ∈ R
a + I ⊂ P1 ∪ · · · ∪ Pn
then I ⊆ Pi for some i.
Proof. If a ∈ ∩iPi then conclusion by 5.1.3. On the contrary after renumbering
there exists j with 1 ≤ j < n such that
a ∈ P1 ∩ · · · ∩ Pj\Pj+1 ∪ · · · ∪ Pn
Assume no inclusions between the prime ideals and choose by 5.1.3
b ∈ I ∩ Pj+1 ∩ · · · ∩ Pn\P1 ∪ · · · ∪ Pj
Then a + b /∈ ∪iPi contradicts the hypothesis.
5.1.5. Proposition. Let R → U −1 R be the canonical homomorphism. Extension
and contraction gives a bijective correspondence between prime ideals in R disjoint
from U and all prime ideals in U −1 R.
(1) For a prime ideal P ⊂ R\U the extended ideal P U −1 R is a prime ideal in
U −1 R and the contracted P U −1 R ∩ R = P .
65

66 5. LOCALIZATION
(2) For a prime ideal Q ⊂ U −1 R the contracted ideal Q ∩ R is a prime ideal
and the extended (Q ∩ R)U −1 R = Q
Proof. Conclusion by 4.3.6, 4.3.7, 4.3.8.
5.1.6. Corollary. Let R be a ring and U a multiplicative subset.
(1) An ideal maximal among the ideals disjoint from U is a prime ideal.
(2) Any ideal disjoint from U is contained in a prime ideal disjoint form U.
Proof. The prime ideals disjoint from U are the prime ideals in U −1 R.
5.1.7. Proposition. The nilradical of a ring R is the intersection of all prime ideals
P .
√
0 = P
Proof. By 1.3.8 the nilradical is contained in any prime ideal. Suppose u ∈ R is
not nilpotent. Then {u n } −1 R is nonzero. Then contraction of a maximal ideal,
5.1.1, is a prime ideal in R not containing u.
5.1.8. Corollary. Let R be a ring.
(1) The radical of an ideal I is the intersection of all prime idealsP containing I
√
I = P
P
I⊂P
(2) For ideals I, J ⊂ R, √ I ∩ J = √ I ∩ √ J.
(3) If U is a multiplicative subset, then U −1√ 0 = √ 0 in U −1 R.
If R is reduced, then U −1 R is reduced.
Proof. (1) Use 5.1.7 on the factor ring R/I. (2) This follows from (1). (3) Use the
correspondence 5.1.5.
5.1.9. Definition. A prime ideal minimal for inclusion among prime ideals is a
minimal prime ideal.
5.1.10. Proposition. Any prime ideal of Q ⊂ R contains a minimal prime ideal
P ⊂ Q.
Proof. The set of prime ideals in R is ordered by inclusion. Given a decreasing
chain Pα then ∩Pα is a prime ideal. Conclusion by Zorn’s lemma.
5.1.11. Corollary. Let R ⊂ S be a subring and P ⊂ R a minimal prime ideal.
Then there is a minimal prime ideal Q ⊂ S contracting to P = Q ∩ R.
5.1.12. Exercise. (1) Let K ⊂ R be an infinite subfield and I, P1, . . . , Pn any ideals.
Show that if I ⊂ P1 ∪ · · · ∪ Pn then I ⊂ Pi for some i.
(2) Let P, P1, P2 be proper ideals. Show that if P is a maximal ideal and P n ⊂ P1 ∪ P2
then P = P1 of P = P2.

5.2. LOCALIZATION OF RINGS 67
5.2. Localization of rings
5.2.1. Definition. A ring R which contains precisely one maximal ideal P is a
local ring and denoted (R, P ). The residue field of R is R/P denoted by k(P ).
A ring homomorphism φ : R → S of local rings (R, P ), (S, Q) is a local ring
homomorphism if φ(P ) ⊂ Q.
5.2.2. Lemma. Let R be a ring and Q = ∅ a subset. Then R is a local ring with
maximal ideal Q if and only if R\Q is the set of units in R.
Proof. Use that an ideal I = R if and only if it contains a unit.
5.2.3. Lemma. A ring homomorphism φ : R → S of local rings (R, P ), (S, Q) is
a local ring homomorphism if the extended ideal P S ⊂ Q or the contracted ideal
Q ∩ R = P . The residue homomorphism k(P ) → k(Q) is a field extension.
Proof. The contraction Q ∩ R is a prime ideal containing P . The rest is clear.
5.2.4. Lemma. Let R be a ring and P a prime ideal. Then U = R\P is a multiplicative
subset. The ring of fractions U −1 R is a local ring. The maximal ideal is
the extended ideal P U −1 R. The residue field is U −1 R/P U −1 R which is canonical
isomorphic to the fraction field of R/P .
5.2.5. Definition. Let R be a ring, P a prime ideal and U = R\P . The localized
ring at P is the local ring RP = U −1 R. The residue field is denoted k(P ) =
RP /P RP .
5.2.6. Example. Let the ring be Z.
(1) The local ring at (0) is the fraction field Q = Z (0).
(2) The local ring Z (p) for a prime number p is identified with a subring of Q
Z (p) = { m
|p not dividing n}
n
The residue field Fp = Z (p)/(p).
(3) Any nonzero ideal in Z (p) is principal of the form (pn ) for some n.
5.2.7. Proposition. Let (R, P ) be a local ring. One of the following conditions is
satisfied:
(1) The characteristic char(R) = 0. P ∩ Z = (0) and Q ⊂ R is a subfield.
Q → k(P ) is a field extension.
(2) The characteristic char(R) = 0. P ∩ Z = (p), p a prime number. Z (p) ⊂ R
is a local subring. Fp → k(P ) is a field extension.
(3) The characteristic char(R) = p n , a power of a prime number. Z/(p n ) ⊂ R
is a local subring. Fp → k(P ) is a field extension.
Proof. (1) (2) are clear by 5.2.3 and 5.2.6. (3) If the characteristic is nonzero then
any prime ideal contracts Q ∩ Z = (p). So a prime number q = p gives a unit in
R. There is a local homomorphism Z (p) → R, 4.1.3. The nontrivial kernel is (p n )
by 5.2.6.
5.2.8. Example. A field K is a local ring with maximal ideal (0). The power series
ring K[[X]] is a local ring with maximal ideal (X) and residue field K.
5.2.9. Remark. Let R be a domain.
(1) The local ring at (0) is the fraction field K = R (0).

68 5. LOCALIZATION
(2) Any local ring RP is identified with a subring of the fraction field K.
(3) The intersection
R =
RP , P a maximal ideal
P
5.2.10. Remark. Let R × S be a product of rings.
(1) A prime ideal is of the form P × S or R × Q for uniquely determined prime
ideals P ⊂ R or Q ⊂ S.
(2) The local ring at P ×S is identified with RP through the projection R ×S →
R.
(3) The local ring at R × Q is identified with SQ through the projection R × S →
S.
5.2.11. Proposition. Let P be a prime ideal and R → RP the canonical homomorphism.
Extension and contraction gives a bijective correspondence between
prime ideals in R contained in P and all prime ideals in RP .
(1) For a prime ideal Q ⊂ P the extended ideal QRP is a prime ideal in RP and
the contracted QRP ∩ R = Q.
(2) For a prime ideal Q ′ ⊂ RP the contracted ideal Q ′ ∩ R ⊂ P is a prime ideal
and the extended (Q ′ ∩ R)RP = Q ′
Proof. This is a special case of 5.1.5.
5.2.12. Lemma. Let Q ⊂ P ⊂ R be prime ideals. Then QRP is a prime ideal in
RP and canonically
Proof. By fraction rules a w av
u / v = uw .
RQ = (RP )QRP
5.2.13. Definition. The intersection of all maximal ideals in a ring is the Jacobson
radical.
5.2.14. Remark. The Jacobson radical contains the nilradical. In a local ring the
Jacobson radical is the maximal ideal.
5.2.15. Exercise. (1) Show that a local ring is never a product of two nonzero rings.
(2) Show that a ∈ R is in the Jacobson radical if and only if 1+ab is a unit for all b ∈ R.
(3) Let p be a prime number. Describe the prime ideals in the ring Z (p).
(4) Let P be a prime ideal. Show that k(P ) is the fraction field of R/P .
(5) Let (R, P ) be a local ring. Show that (R[[X]], (P, X) is a local ring and the canonical
homomorphism R → R[[X]] is a local homomorphism.
5.3. Localization of modules
5.3.1. Definition. Let R be a ring, P a prime ideal and U = R\P . For a module
M, the localized module at P is the module MP = U −1 M over the local ring RP .
For a homomorphism f : M → N the localized homomorphism is fP : MP →
NP and the residue homomorphism is f(P ) : M ⊗R k(P ) → N ⊗R k(P ).
The constructions are functors, 2.7.4, 4.2.4.
5.3.2. Lemma. Let Q ⊂ P ⊂ R be prime ideals and M an R-module Then QRP
is a prime ideal in RP and canonically
Proof. See proof of 5.2.12.
MQ = (MP )QRP

5.3. LOCALIZATION OF MODULES 69
5.3.3. Proposition. Let R be a ring and P a prime ideal. If Mα is a family of
modules, then the homomorphism
(
Mα)P →
(Mα)P
is an isomorphism of RP -modules.
Proof. See 4.2.7.
5.3.4. Corollary. For a homomorphism f : M → N
(1) (Ker f)P Ker fP .
(2) (Im f)P Im fP .
(3) (Cok f)P Cok fP .
Proof. See 4.3.3.
α
5.3.5. Corollary. Let R be a ring and P a prime ideal. For submodules N, L ⊂ M
(1) (M/N)P MP /NP .
(2) (N + L)P NP + LP .
(3) (N ∩ L)P NP ∩ LP .
Proof. See 4.3.4.
5.3.6. Proposition. Let R be a ring, P a prime ideal and M a module.
(1) MP M ⊗R RP .
(2) MP /P RP MP M ⊗R k(P ).
Proof. See 4.4.1.
5.3.7. Proposition. Let R be a ring, P a prime ideal.
(1) For an R module M and an RP -module N there is a natural isomorphism
M ⊗R RP ⊗RP N M ⊗R N
(2) For an R module M, L there is a natural isomorphism
Proof. See 4.2.7 and 2.7.4.
(M ⊗R L)P MP ⊗RP LP
5.3.8. Definition. Let R be a ring. F is a locally free module is FP is a free
RP -module for all prime ideals P .
5.3.9. Lemma. Let R be a ring. F is a locally free module if FQ is a free RQmodule
for all maximal ideals Q.
Proof. A prime ideal P ⊂ Q is contained in a maximal ideal. By 5.3.6 FP
(FQ)PQ is free.
5.3.10. Example. A free module is a locally free module.
5.3.11. Exercise. (1) Let P ⊂ R be a prime ideal and R → S a ring homomorphism.
Show that RP → SP is a ring homomorphism.
(2) Let Q ⊂ S be a prime ideal and R → S a ring homomorphism. Show that RQ∩R →
SQ is a local ring homomorphism.
(3) Let R = K × L be a product of fields. Show that ideal K × {0} is locally free but
not free.
α

70 5. LOCALIZATION
5.4. Exactness and localization
5.4.1. Proposition. Let R be a ring and M a module. The following conditions
are equivalent.
(1) M = 0.
(2) MP = 0 for all prime ideals P .
(3) MP = 0 for all maximal ideals P .
Proof. (1) ⇒ (2) ⇒ (3) is clear. (3) ⇒ (1): Let 0 = x ∈ M be given. Then
Ann(x) ⊂ P is contained in a maximal ideal, 5.1.2. Clearly 0 = x
1 ∈ MP
contradicts (3).
5.4.2. Corollary. Let R be a ring and f : M → N a homomorphism. The following
conditions are equivalent.
(1) f is injective.
(2) fP is injective for all prime ideals P .
(3) fP is injective for all maximal ideals P .
Proof. Use 5.4.1 on Ker f.
5.4.3. Corollary. Let R be a ring and f : M → N a homomorphism. The following
conditions are equivalent.
(1) f is surjective.
(2) fP is surjective for all prime ideals P .
(3) fP is surjective for all maximal ideals P .
Proof. Use 5.4.1 on Cok f.
5.4.4. Corollary. Let R be a ring and f : M → N a homomorphism. The following
conditions are equivalent.
(1) f is an isomorphism.
(2) fP is an isomorphism for all prime ideals P .
(3) fP is an isomorphism for all maximal ideals P .
5.4.5. Corollary. Let R be a ring and
0
f
M
g
N
a sequence of homomorphisms. The following conditions are equivalent.
(1) The sequence is short exact.
(2) The sequence
0
MP
fP
NP
is short exact for all prime ideals P .
(3) The sequence
0
MP
fP
NP
is short exact for all maximal ideals P .
L
gP
LP
gP
LP
5.4.6. Corollary. Let R be a ring and F a module. The following conditions are
equivalent.
(1) F is flat.
(2) FP is flat for all prime ideals P .
0
0
0

(3) FP is flat for all maximal ideals P .
5.5. FLAT RING HOMOMORPHISMS 71
Proof. Let 0 → M → N. Use 5.3.7 and 5.4.2 on M ⊗R F → N ⊗R F .
5.4.7. Proposition. Let R be a ring and M a module. Then there is an exact
sequence
0 → M →
P maximal
Proof. Let 0 = x ∈ M be given. Then Ann(x) ⊂ P is contained in a maximal
ideal, 5.1.2. Clearly 0 = x
1 ∈ MP .
5.4.8. Corollary. Let R be a ring. Then there is an injective ring homomorphism
R →
P maximal
5.4.9. Corollary. Let R be a ring. The following conditions are equivalent.
(1) R is reduced.
(2) RP is reduced for all prime ideals P .
(3) RP is reduced for all maximal ideals P .
Proof. Use 5.1.8 and 5.4.10.
5.4.10. Exercise. (1) Let R be a ring and
0
f
M
g
N
a split exact sequence. Show that the localized sequence is split exact for all prime
ideals P .
RP
MP
L
5.5. Flat ring homomorphisms
5.5.1. Definition. A flat R-module F is faithfully flat if for any M = 0 the tensor
product M ⊗R F = 0.
5.5.2. Example. A nonzero free module is faithfully flat.
5.5.3. Lemma. Let F be a faithfully flat R-module and M → N a homomorphism.
If M ⊗R F → N ⊗R F is injective, then M → N is injective.
Proof. Let f : M → N. Then 0 → Ker f ⊗R F → M ⊗R F → N ⊗R F is exact.
It follows, that Ker f = 0.
5.5.4. Definition. A ring homomorphism R → S is a flat ring homomorphism if
S is a flat R-module and a faithfully flat ring homomorphism if S is faithfully flat.
5.5.5. Proposition. Let R → S be a faithfully flat ring homomorphism. for any
ideal I ⊂ R the extended contracted returns I, i.e.
I = IS ∩ R
Proof. Tensor the homomorphism R/I → R/IS∩R with S. The induced R/I⊗R
S → R/IS ∩ R ⊗R S is canonically isomorphic to the identity S/IS → S/IS.
By 5.5.3 R/I → R/IS ∩ R is injective giving I = IS ∩ R.
5.5.6. Corollary. A faithfully flat ring homomorphism R → S is injective.
5.5.7. Proposition. Let φ : R → S be a ring homomorphism. The following
conditions are equivalent
0

72 5. LOCALIZATION
(1) φ is flat.
(2) φP : RP → SP is flat for all prime ideals P ⊂ R.
(3) φP : RP → SP is flat for all maximal ideals P ⊂ R.
Proof. Use 5.4.6.
5.5.8. Proposition. A flat local homomorphism (R, P ) → (S, Q) is faithfully flat.
Proof. Let 0 = x ∈ M be an R-module. R/ Ann(x) Rx and I = Ann(x) ⊂ P ,
so IS ⊂ Q. Then R/I ⊗R S S/IS = 0 and therefore M ⊗R S = 0 giving the
claim.
5.5.9. Proposition (going-down). Let φ : R → S be a flat ring homomorphism
and Q ⊂ S a prime ideal. For any prime ideal P ′ ⊂ P = Q ∩ R there is a prime
ideal Q ′ ⊂ Q contracting to P ′ = Q ′ ∩ R.
Proof. The local homomorphism RP → SQ is faithfully flat. The ring k(P ′ ) ⊗R
SQ is nonzero and therefor contains a maximal ideal Q ′′ . The contraction to S,
Q ′ = Q ′′ ∩ S contracts to P ′ = R ∩ Q ′ .
5.5.10. Proposition. Let R → S be a flat homomorphism. The following conditions
are equivalent.
(1) R → S is faithfully flat.
(2) Any prime ideal P ⊂ R is the contraction P = Q ∩ R of a prime ideal
Q ⊂ S.
Proof. Assume (2). Let M = 0 be an R-module. Then MP = 0 for some P . Let
P = Q ∩ R, then (M ⊗R S)Q MP ⊗RP SQ = 0 as RP → SQ is faithfully flat.
So (1) is true.
5.5.11. Proposition. The inclusion R → R[X1, . . . , Xn] is a faithfully flat homomorphism
Proof. The R-module R[X1, . . . , Xn] is free.
5.5.12. Exercise. (1) Show that a free module is faithfully flat.
(2) Let R → S and S → T be flat homomorphisms. Show that the composite R → S is
flat.
(3) Show that Q is a flat but not faithfully flat Z-module.
(4) Let R be a ring and I = √ 0 the nilradical. Show that IR[X] is the nilradical of
R[X].

6
Finite modules
6.1. Finite Modules
6.1.1. Definition. Let R be a ring. A finite module is generated by finitely many
elements. The finite free module with standard basis e1, . . . , en is denoted R n .
6.1.2. Lemma. Let R be a ring and M a module. The following are equivalent.
(1) M is generated by n elements x1, . . . , xn.
(2) There is a surjective homomorphism R n → M → 0, ei ↦→ xi.
Proof. See 2.4.12.
6.1.3. Proposition. Let R → S be a ring homomorphism. If an R-module M is
generated by n elements x1, . . . , xn. Then the change of rings S-module M ⊗R S
is generated by x1 ⊗ 1, . . . , xn ⊗ 1 over S.
Proof. Follows from 6.1.2 and 3.4.1
6.1.4. Corollary. Let R be a ring and U a multiplicative subset. If M is a finite
R-module, then U −1 M is a finite U −1 R-module.
6.1.5. Proposition. For a short exact sequence
0
f
M
the following hold
(1) If N is finite, then L is finite.
(2) If M, L are finite, then N is finite.
g
N
Proof. (1) If y1, . . . , yn generates N, then g(y1), . . . , g(yn) generates L. (2) Choose
u : R n → M → 0 and v : R m → L → 0 exact. By 3.5.4 there is w : R m → N
such that g ◦ w = v. There is a diagram
0
0
Rn
u
M
Conclusion by the snake lemma 3.2.4.
6.1.6. Corollary. Let
0
f
f
M
L
Rn ⊕ Rm
f◦u+w
g
N
g
N
0
Rm
v
L
be a split exact sequence. Then N is finite if and only if M, L are finite.
Proof. Let u be a retraction of f. By 6.1.5 Im u = M is finite. The rest is contained
in 6.1.5.
73
L
0
0
0

74 6. FINITE MODULES
6.1.7. Corollary. Let f : M → N be a homomorphism.
(1) If M is finite, then Im f is finite.
(2) If Ker f, Im f are finite, then M is finite.
(3) If N is finite, then Cok f is finite.
(4) If Im f, Cok f are finite, then N is finite.
Proof. Use 6.1.5 on the exact sequences 3.1.8.
6.1.8. Corollary. Let M, N be modules. Then M ⊕ N is finite if and only if M
and N are finite.
Proof. Use 6.1.6.
6.1.9. Proposition. Let R be a ring and M, N finite modules. Then M ⊗R N is
finite.
Proof. Use 6.1.2 and 6.1.5. Let R m → M and R n → N be surjective. Then
R m ⊗R R n → M ⊗R N is surjective, 3.4.1.
6.1.10. Proposition. Let R be a ring and M a module. The following are equivalent.
(1) M is finite and projective.
(2) M is a direct summand in a finite free module.
Proof. Use 6.1.2 and 6.1.8.
6.1.11. Proposition. Let R be a ring and M, N finite modules.
(1) If M is projective, then HomR(M, N) is finite.
(2) If M, N are projective, then HomR(M, N) is projective.
Proof. Use 6.1.8.
6.1.12. Proposition. Let F be a finite projective module and E an injective module.
(1) F ⊗R E is injective.
(2) HomR(F, E) is injective.
Proof. (1) (2) Both modules become summands in injective modules.
6.1.13. Proposition. Let R be a ring and U a multiplicative subset. For a finite
module M the following hold:
(1) U −1 M = 0 if and only if there is a u ∈ U such that uM = 0.
(2) Ann(U −1 M) = U −1 Ann(M) in U −1 R.
Proof. Let x1, . . . , xn generate M. (1) U −1 M = 0 if and only if u1x1 = · · · =
unxn. Put u = u1 . . . un. (2) Ann(M) = Ann(x1) ∩ · · · ∩ Ann(xn). Now use (1)
and 4.3.4.
6.1.14. Proposition. Let R be a ring and Mα a family of modules. For any finite
module N there is a natural isomorphism
HomR(N,
Mα)
HomR(N, Mα)
α
Proof. A homomorphism f : N → Mα has an image in a finite sum.
6.1.15. Exercise. (1) Show that
n Z/(n) is not a finite Z-module.
α

6.2. FREE MODULES 75
(2) Let K be a field and R = K[X1, X2, . . . ] the polynomial ring in countable many
variables. Show that R is a finite module, but the ideal (X1, X2, . . . ) is not a finite
module.
6.2. Free Modules
6.2.1. Definition. Let R be a ring and let R n be the free module with standard
basis e1, . . . , en.
(1) Let A = (aij) be a m × n-matrix with m rows and n columns, where the
entry aij ∈ R. Identify Rn with n-columns. Then matrix multiplication
x = (xj) ↦→ y = Ax, yi =
j
aijxj
defines a homomorphism R n → R m .
(2) Let f : R n → R m be a homomorphism. Then define a m × n-matrix A =
(aij) by
f(ej) =
i
aijei
6.2.2. Proposition. (1) The dictionary defined in 6.2.1 gives a canonical isomorphism
between the module of m × n-matrices and HomR(R n , R m ).
(2) Matrix multiplication corresponds to composition of homomorphisms and the
identity matrix corresponds to the identity homomorphism.
(3) Invertible matrices correspond to isomorphisms.
Proof. Do linear algebra homework.
6.2.3. Definition. Let R be a ring.
(1) Let A = (aij) be a m × n-matrix. The (m − 1) × (n − 1) matrix derived
from A be deleting i-row and j-column is Aij.
(2) For a square matrix A the determinant is defined by row expansion and induction:
det(a11) = a11
det A =
i
(−1) 1+j a1j det A1j
(3) The determinant of a k × k-matrix derived from A by choosing entries from
k rows and columns is a k-minor of A.
(4) If A is a n × n-matrix, then the cofactor matrix A ′ = (a ′ ij ) has entries
given by (n − 1)-minors.
a ′ ij = (−1) i+j det Aji
6.2.4. Proposition. (1) The determinant of the identity matrix is 1.
(2) The determinant is calculated by any expansion
det A =
(−1) i+j aij det Aij =
i
(3) If A, B are n × n-matrices then the product rule holds
det AB = det A det B
j
(−1) i+j aij det Aij

76 6. FINITE MODULES
(4) Let A be an n × n-matrix with cofactor matrix A ′ . Then the matrix Cramer’s
rule holds
AA ′ = A ′ A = det A(1n)
Where (1n) is the n × n identity matrix.
(5) A square matrix A is invertible if and only if det A is a unit in R.
Proof. More linear algebra homework.
6.2.5. Proposition. Let f : R n → R n be a homomorphism represented by an
n × n-matrix A.
(1) f is a surjective if and only if det A is a unit.
(2) f is injective if and only if det A is a nonzero divisor.
Proof. (1) If f is surjective, then a section is represented by a matrix B such that
AB = (1n). Then det A is a unit. Conversely by 6.2.4. (2) If det A is a nonzero
divisor, then by 6.2.4 A ′ A = det A(1n) gives an injective homomorphism, so f is
injective. If det A is a zero divisor, then there is number k < n such that Ann(1 −
minors) = · · · = Ann(k − minors) = 0 and 0 = b ∈ Ann((k + 1) − minors).
Assume that the k−minor from first k rows and columns ck+1 has ck+1b = 0. Let
cj(−1) k+1+j be the k-minor from first k rows and first k + 1 columns excluding
number j and put cj = 0, j > k + 1. Then A(cj) is a column with entries being
(k + 1) − minors so f((bcj)) = A(bcj) = 0 and f is not injective.
6.2.6. Proposition. Let R be a ring and f : R n → R m homomorphism.
(1) If f is surjective, then n ≥ m.
(2) If f is injective, then n ≤ m.
Proof. (1) If n < m let p : R m → R n be the projection onto first n coordinates.
Then f ◦p is surjective and represented by an m×m-matrix A with m-column zero.
A section to f ◦ p is represented by an m × m-matrix B such that BA = (1m). But
the product AB must have a zero m-column, so the contradiction gives n ≥ m.
(2) If n > m let i : R m → R n be injection onto first m coordinates. Then i ◦ f is
injective and represented by an n × n-matrix A with n-row zero. Then det A = 0
contradicting 6.2.5. So n ≤ m.
6.2.7. Proposition. (1) A finite free module has a finite basis.
(2) The number of elements in a basis for a finite free module is independent of
the basis.
Proof. Let F be finite free generated by n elements. If y1, . . . , ym is part of a
basis, then by projection F → R m there is a surjective homomorphism R n → R m .
Conclusion by 6.2.6.
6.2.8. Definition. The number of elements 6.2.7 in a basis for a finite free module
F is the rank, rankR F .
6.2.9. Proposition. If x1, . . . , xn generates a free module F of rank n, then they
constitutes a basis.
Proof. Choose a basis and an isomorphism f : R n → F . The homomorphism
g : R n → F, ei ↦→ xi is surjective. The composite f −1 ◦ g : R n → R n is
surjective and therefore by 6.2.5 an isomorphism. Then g is an isomorphism and
xi a basis.

6.3. CAYLEY-HAMILTON’S THEOREM 77
6.2.10. Proposition. Let F, F ′ be finite free modules. Then
(1) F ⊕ F ′ is free and rankR F ⊕ F ′ = rankR F + rankR F ′ .
(2) F ⊗R F ′ is free and rankR F ⊗R F ′ = rankR F · rankR F ′ .
(3) HomR(F, F ′ ) is free and rankR HomR(F, F ′ ) = rankR F · rankR F ′ .
6.2.11. Exercise. (1) Let R n → R n be a surjective homomorphism. Show that it is an
isomorphism.
6.3. Cayley-Hamilton’s theorem
6.3.1. Remark. Let R be a ring and f : M → M a homomorphism. By 2.1.13
view M as an R[X]-module, where Xx = f(x) for x ∈ M. The homomorphism
1.6.7, 2.6.9, R[X] → HomR(M, M), a ↦→ aM, X ↦→ f is a ring homomorphism.
The image is R[f] the smallest subring containing 1M, f. M is naturally a R[f]module
and the R[X]-module above is the restriction of scalars.
6.3.2. Proposition. Let A be an n × n-matrix and I the ideal generated by the
entries aij. The polynomial
det(X1n − A) = a0 + a1X + . . . an−1X n−1 + X n
has a0, . . . , an−1 ∈ I and gives the relation
as n × n-matrix.
a0(1n) + a1A + . . . an−1A n−1 + A n = 0
Proof. View R n as a module over the ring R[X], 6.3.1, with scalar multiplication
Xx = Ax, x ∈ R n
Let the n × n-matrix U = X(1n) − A over R[X] have cofactor matrix U ′ . The
relations above give U ′ Uej = 0, written out by 6.2.4
det U ej = 0
for all j. That is det U M = 0. By calculation
in R[X], with ai ∈ I.
det U = a0 + a1X + · · · + an−1X n−1 + X n
6.3.3. Proposition. Let I ⊂ R be an ideal and f : M → M a homomorphism
on a finite module generated by n elements. Suppose Im f ⊂ IM, then there exist
a0, . . . , an−1 ∈ I such that
in HomR(M, M).
a01M + a1f + . . . an−1f n−1 + f n = 0
Proof. Let x1, . . . , xn generate M and write
f(xj) =
i
aijxi
for an n × n-matrix A with entries aij ∈ I. View this over the ring R[X], 6.3.1.
Then
Xxj =
i
aijxi

78 6. FINITE MODULES
Let the n × n-matrix U = X(1n) − A over R[X] have cofactor matrix U ′ . The
relations above give U ′ Uxj = 0, written out by 6.2.4
det U xj = 0
for all j. That is det U M = 0. By calculation
in R[X], with ai ∈ I.
det U = a0 + a1X + · · · + an−1X n−1 + X n
6.3.4. Proposition. Let I ⊂ R be an ideal and M a finite module. If IM = M
then I + Ann(M) = R. That is there is a ∈ I such that (1 + a)M = 0.
Proof. By 6.3.3
Put a = a0 + · · · + an−1.
a01M + · · · + 1M = (a0 + · · · + an−1)1M + 1M = 0
6.3.5. Corollary. Let I ⊂ R be an ideal and M a finite module. If IM = M and
all elements 1 + I are nonzero divisors on M, then M = 0.
6.3.6. Corollary. Let I ⊂ R be an ideal and N ⊂ M a submodule. Suppose M/N
is a finite module and M = N + IM. Then I + (N : M) = R.
6.3.7. Proposition. Let R be a ring and M a finite module. If a homomorphism
f : M → M is surjective, then it is an isomorphism.
Proof. Regard M, f as a module over R[X] 6.3.1. Then (X)M = M, so by 6.3.4
there is p ∈ R[X] such that 1 + pX ∈ Ann(M). For any u ∈ Ker f, calculate
u = u + p(f) ◦ f(u) = (1 + pX)u = 0. So f is an isomorphism.
6.3.8. Exercise. (1) Let √ 0M = M. Show that M = 0.
6.4. Nakayama’s Lemma
6.4.1. Proposition. Let (R, P ) be a local ring and M a finite module. The following
conditions are equivalent.
(1) M = 0.
(2) P M = M.
(3) M ⊗R k(P ) = 0.
Proof. (1) ⇒ (2) ⇔ (3) is clear. (2) ⇒ (1): Elements in 1 + P are units in R, so
by 6.3.5 M = 0.
6.4.2. Corollary. Let (R, P ) be a local ring and N ⊂ M a submodule. Suppose
M/N is a finite module and M = N + P M. Then N = M.
6.4.3. Corollary. Let (R, P ) be a local ring and M, N finite modules. If M ⊗R
N = 0, then M = 0 or N = 0.
Proof. If M, N = 0 then M ⊗R k(P ), N ⊗R k(P ) = 0 are vector spaces over
k(P ). Now M ⊗R N ⊗R k(P ) M ⊗R k(P )⊗ k(P ), N ⊗R k(P ) = 0, giving the
statement.
6.4.4. Corollary. Let R be a ring and M a finite module. The following conditions
are equivalent.
(1) M = 0.

(2) P MP = MP for all prime ideals P .
(3) P MP = MP for all maximal ideals P .
(4) M ⊗R k(P ) = 0 for all prime ideals P .
(5) M ⊗R k(P ) = 0 for all maximal ideals P .
Proof. Combine 6.4.1 with 5.4.1.
6.4. NAKAYAMA’S LEMMA 79
6.4.5. Corollary. Let (R, P ) be a local ring and f : M → N a homomorphism.
Assume N is finite. The following conditions are equivalent.
(1) f is surjective.
(2) f(P ) is surjective.
Proof. f is surjective if and only if Cok f = 0. Cok f is finite, so it is zero if and
only if Cok f ⊗R k(P ) = Cok(f(P )) = 0, 6.4.1.
6.4.6. Corollary. Let R be a ring and f : M → N a homomorphism. Assume N
is finite. The following conditions are equivalent.
(1) f is surjective.
(2) f(P ) is surjective for all prime ideals P .
(3) f(P ) is surjective for all maximal ideals P .
6.4.7. Corollary. Let (R, P ) be a local ring and M a finite module. Let x1, . . . , xn ∈
M. The following conditions are equivalent.
(1) x1, . . . , xn generate M.
(2) x1, . . . , xn generate M ⊗R k(P ).
6.4.8. Corollary. Let R be a ring and M a finite module. Let x1, . . . , xn ∈ M.
The following conditions are equivalent.
(1) x1, . . . , xn generate M.
(2) x1, . . . , xn generate M ⊗R k(P ) for all prime ideals P .
(3) x1, . . . , xn generate M ⊗R k(P ) for all maximal ideals P .
6.4.9. Proposition. Let f : R n → R m be a homomorphism represented by an
m × n-matrix A. The following are equivalent.
(1) f is surjective.
(2) n ≥ m and the ideal (m − minors) = R.
Proof. (1) ⇒ (2): If f is surjective, then for any maximal ideal f(P ) is surjective
linear map. So n ≥ m and some m-minor is nonzero in k(P ). Therefore (m −
minors) is not contained in P , so (m − minors) = R. (2) ⇒ (1): f(P ) is
surjective for all maximal ideals, so f surjective by 6.4.6.
6.4.10. Proposition. Let f : R n → R m be a homomorphism represented by an
m × n-matrix A. The following are equivalent.
(1) f is injective.
(2) n ≤ m and the ideal Ann(n − minors) = 0.
Proof. See the proof 6.2.5 (2).
6.4.11. Exercise. (1) Let R be a domain and f : R n → R m a homomorphism represented
by an m × n-matrix. Show that f is injective if and only if n ≤ m and some
n − minor = 0.

80 6. FINITE MODULES
6.5. Finite Presented Modules
6.5.1. Definition. Let R be a ring. A finite presented module is a module M
having an exact sequence
R n → R m → M → 0
or equivalently there is a short exact sequence
with N finite.
0 → N → R m → M → 0
6.5.2. Example. A finite projective module is finite presented, 6.1.10.
6.5.3. Lemma. For a short exact sequence
0
f
M
g
N
the following hold:
(1) If M, L are finite presented, then N is finite presented.
(2) If L is finite presented and N is finite , then M is finite.
(3) If N is finite presented and M is finite , then L is finite presented.
Proof. (1) Choose u : R n → M → 0 and v : R m → L → 0 exact with finite
kernels. By 3.5.4 there is w : R m → N such that g ◦ w = v. There is a diagram
0
0
Rn
u
M
f
L
Rn ⊕ Rm
f◦u+w
g
N
0
Rm
v
L
By the snake lemma 3.2.4 the sequence 0 → Ker u → Ker f ◦u+w → Ker v → 0
is exact. By 6.1.5 Ker f ◦ u + w is finite. (2) Choose v : R m → L → 0 exact with
finite kernel and w : R m → N such that g ◦ w = v. There is a diagram
0
0
0
f
M
R m
w
N
g
Rm
v
L
By the snake lemma 3.2.4 the sequence 0 → Ker w → Ker v → M → Cok w →
0 is exact. By 6.1.5 M is finite. (3) Choose w : R m → N → 0 exact with finite
kernel. Then v = g ◦ w : R m → L → 0 is exact and there is a diagram
0
0
0
f
M
R m
w
N
g
Rm
v
L
By the snake lemma 3.2.4 the sequence 0 → Ker w → Ker v → M → 0 is exact.
By 6.1.5 Ker v is finite and L is finite presented.
6.5.4. Corollary. Let
0
f
M
g
N
be a split exact sequence. Then N is finite presented if and only if M, L are finite
presented.
L
0
0
0
0
0
0
0

Proof. By 3.1.13 there is a split exact sequence
so the statement follows from 6.5.3.
0
6.5. FINITE PRESENTED MODULES 81
v
L
u
N
M
6.5.5. Corollary. Let f : M → N be a homomorphism.
(1) If M is finite and Im f finite presented, then Ker f is finite.
(2) If Ker f, Im f are finite presented, then M is finite presented.
(3) If N is finite presented and Im f finite, then Cok f is finite presented.
(4) If Im f, Cok f are finite presented, then N is finite presented.
Proof. Use the sequences 3.1.8.
6.5.6. Proposition. Let R be a ring and M, N finite presented modules.
(1) M ⊕ N is finite presented.
(2) M ⊗R N is finite presented.
Proof. (1) This is clear from 6.5.4. (2) If M = R n then M ⊗R N is finite presented
by (1). In general chose u : R n → M → 0 exact with finite kernel. The sequence
Ker u ⊗R N → R n ⊗R N → M ⊗R N → 0 is exact. So Ker u ⊗ 1N is finite.
Conclusion by 6.5.5.
6.5.7. Proposition. Given submodules N, L ⊂ M. Then
(1) If M/N, M/L are finite and M/N + L is finite presented, then M/N ∩ L is
finite.
(2) If M/N, M/L are finite presented and M/N ∩ L is finite, then M/N + L is
finite presented.
Proof. Use the sequence 3.2.7.
6.5.8. Proposition. Let R be a ring and U a multiplicative subset.
(1) For a finite module M and any module N the natural homomorphism
0
U −1 HomR(M, N) → Hom U −1 R(U −1 M, U −1 N)
is injective.
(2) For a finite presented module M and any module N the homomorphism
U −1 HomR(M, N) → Hom U −1 R(U −1 M, U −1 N)
is a natural isomorphism.
Proof. (1) If M = R this is an isomorphism. Then this is also an isomorphism
for M = R n since both functors respect finite direct sums. In general choose
0 → K → R n → M → 0 exact. There is a diagram
0
0
U −1 HomR(M, N)
HomU −1R(U −1M, U −1N)
U −1 HomR(R n , N)
HomU −1R(U −1Rn , U −1N) giving injectivity. (2) In (1) K is finite, so the last vertical map is injective. Conclusion
by the five lemma 3.2.8.
U −1 HomR(K, N)
HomU −1R(U −1K, U −1N)

82 6. FINITE MODULES
6.5.9. Corollary. Let P ⊂ R be a prime ideal and M a finite presented module.
For any module N there is a natural isomorphism
HomR(M, N)P HomRP (MP , NP )
6.5.10. Proposition. Let R be a ring and F a module. The following conditions
are equivalent.
(1) F is flat.
(2) For any module N and a relation 0 =
i yi ⊗ xi ∈ N ⊗R F , there exist
zj ∈ F and aij ∈ R such that 0 =
i aijyi ∈ N and xi =
j aijzj ∈ F .
(3) For any relation 0 =
i bixi ∈ F , there exist zj ∈ F and aij ∈ R such that
0 =
i aijbi ∈ R and xi =
j aijzj ∈ F .
Proof. (1) ⇒ (2): Let f : Rn → N, ei ↦→ yj, then 0 → Ker f ⊗R F → F n →
N ⊗R F is exact. By assumption (xi) ∈ Ker f ⊗R F , so (xi) =
j aij ⊗ zj
with (aij) ∈ Ker f. (2) ⇒ (3) is clear. (3) ⇒ (1): Let I ⊂ R be an ideal and
bi ⊗ xi ∈ Ker(I ⊗R F → F ). Then 0 =
i aijbi and xi =
j aijzj. Now
calculate bi ⊗ xi =
j i aijbi ⊗ xi = 0. By 3.7.12 F is flat.
6.5.11. Proposition. Let (R, P ) be a local ring and F a finite presented module.
The following conditions are equivalent.
(1) F is free.
(2) F is projective.
(3) F is flat.
(4) P ⊗R F → F is injective.
Proof. (1) ⇒ (2) ⇒ (3) ⇒ (4) are clear. (4) ⇒ (1): Choose xi ∈ F such that
xi ⊗ 1 give a basis for F ⊗R k(P ). The homomorphism f : R n → F, ei ↦→ xi is
surjective by 6.4.5. P ⊗R Ker f → P n → P ⊗R F → 0 is exact, so P Ker f =
Ker(P n → P ⊗R F ) = Ker(P n → P ⊗R F → F ) = Ker f. by the hypothesis.
Ker f is finite 6.5.5 and therefore Ker f = 0 by 6.4.1, so F is free.
6.5.12. Corollary. Let (R, P ) be a local ring and f : F → F ′ a homomorphism
of finite free modules. The following conditions are equivalent.
(1) f has a retraction u : F ′ → F .
(2) f is injective and Cok f is free.
(3) f(P ) is injective.
Proof. (1) ⇒ (2): Cok f is projective, so free by 6.5.11. (2) ⇒ (3) is clear. (3) ⇒
(1): Let f ∨ : F ′∨ → F ∨ be the dual homomorphism. f ∨ (P ) is surjective, so f ∨ is
surjective, 6.4.5. A section v of f ∨ gives a retraction u = v ∨ .
6.5.13. Corollary. Let R be a ring and F a finite presented module. The following
conditions are equivalent.
(1) F is projective.
(2) F is flat.
(3) F is locally free.
(4) FP is free for all maximal ideals P .
Proof. (1) ⇒ (2) ⇒ (3) ⇒ (4) are clear by 6.5.11. (4) ⇒ (1): Let N → L → 0
be exact. By hypothesis HomRP (FP , NP ) → HomRP (FP , LP ) → 0 is exact for
all maximal ideals. By 6.5.9 HomR(F, N)P → HomR(F, L)P → 0 is exact for

6.6. FINITE RING HOMOMORPHISMS 83
all maximal ideals. By 5.4.3 HomR(F, N) → HomR(F, L) → 0 is exact and F is
projective.
6.5.14. Exercise. (1) Let I ⊂ R be an ideal. Show that R/I is a finite presented
R-module if and only if I is a finite ideal.
(2) Show that Q is a flat, but not projective Z-module.
6.6. Finite ring homomorphisms
6.6.1. Definition. A ring homomorphism φ : R → S is a finite ring homomorphism
if S is a finite R-module. If R ⊂ S is a subring, then a finite ring homomorphism
is a finite ring extension.
6.6.2. Proposition. Let R be a ring.
(1) Let f ∈ R[X] be a monic polynomial. Then the homomorphism R →
R[X]/(f) is finite.
(2) Let f : M → M be a homomorphism of a finite R-module. Then the homomorphism
6.3.1, R → R[f] is finite.
Proof. (2) Follows from (1) and 6.3.3.
6.6.3. Lemma. Let φ : R → S be a finite ring homomorphism. If N is a finite
S-module, then by restriction along φ the R-module N is finite.
6.6.4. Proposition. Let R ⊂ S be a finite ring extension of domains. Then R is a
field if and only if S is a field.
Proof. Let R be a field, a minimal equation 6.3.3 for scalar multiplication by a
nonzero b ∈ S, bS as R-module homomorphism
gives
b n + · · · + a0 = 0
b −1 = −a −1
0 (an−1b n−2 + · · · + a1) ∈ S
Let S be a field and 0 = a ∈ R. An equation 6.3.3 for scalar multiplication a −1
S as
R-homomorphism
a −n + · · · + a0 = 0
gives
a −1 = −(a0a n−1 + · · · + an−1) ∈ R
6.6.5. Corollary. Let R → S be a finite ring homomorphism. A prime ideal Q ⊂ S
is maximal if and only if the contraction Q ∩ R is maximal.
6.6.6. Proposition (going-up). Let R ⊂ S be a finite ring extension and P ⊂ R a
prime ideal. Then there is a prime ideal Q ⊂ S contracting
.
P = Q ∩ R
Proof. RP ⊂ SP is a finite ring extension. Since SP = 0 there is a maximal
ideal in SP contracting to P RP by 6.6.5. The corresponding prime ideal Q ⊂ S
contracts to P .
6.6.7. Proposition. Let R ⊂ S be a finite ring extension and E an R-module. The
following are equivalent.

84 6. FINITE MODULES
(1) E is an injective R-module.
(2) HomR(S, E) is an injective S-module.
Proof. (1) ⇒ (2): This is 3.6.10. (2) ⇒ (1): Let E ⊂ E ′ be an injective envelope,
3.6.15. HomR(S, E) → HomR(S, E ′ ) is injective and identifies HomR(S, E)
as an essential submodule, since S is a finite R-module. E ′ is injective, so also
HomR(S, E ′ ). It follows that
HomR(S, E)
HomR(R, E)
E
HomR(S, E ′ )
HomR(R, E ′ )
E ′
with the right down map being surjective. So E = E ′ .
6.6.8. Exercise. (1) Show that Z → Z[ √ −5] is finite.
(2) Let p be a prime number. Show that Z → Z (p) is not finite.

7
Modules of finite length
7.1. Simple Modules
7.1.1. Definition. A nonzero module M is a simple module if 0 and M are the
only submodules.
7.1.2. Proposition. Let f : M → M ′ be a homomorphism
(1) If M is simple, then f is either zero or injective.
(2) If M ′ is simple, then f is either zero or surjective.
(3) If M, M ′ are both simple, then f is either zero or an isomorphism.
Proof. This follows from 2.3.3.
7.1.3. Proposition. A simple module is isomorphic to a factor module R/P for
some uniquely determined maximal ideal P . A simple module is finite, in fact
generated by one element.
Proof. A nonzero f : R → M must be surjective and R/P is simple exactly when
P is maximal. Clearly P = Ann(M).
7.1.4. Corollary. Any nonzero finite module has a simple factor module.
Proof. If M = 0 then MP = 0 for some maximal ideal. By 6.4.1 M ⊗R k(P ) =
M/P M = 0. Choose a nonzero linear projection M/P M → k(P ), giving M →
k(P ) → 0 exact.
7.1.5. Example. (1) If K is a field, a one dimensional vector space is simple.
(2) If R is a principal ideal domain, then R/(p) is a simple module for all irreducible
(p).
(3) Z/(p) are simple for all prime numbers p.
(4) Let K be a field. K[X]/(X − a) is a simple K[X]-module.
7.1.6. Lemma. Let I ⊂ R be an ideal. A simple R/I-module is a simple Rmodule.
7.1.7. Exercise. (1) Show that Z does not contain any simple modules.
(2) Show that Q does not have any simple factor Z-module.
(3) Let L, L ′ ⊂ M be simple submodules. Show that either L ∩ L ′ = 0 of L = L ′ .
(4) Let L = L ′ ⊂ M be simple submodules. Show that L ⊕ L ′ = L + L ′ .
7.2. The Length
7.2.1. Definition. Let R be a ring. A module M is of finite length if it admits a
composition series by submodules
0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn−1 ⊂ Mn = M
such that each factor Mi/Mi−1 is a simple module.
85

86 7. MODULES OF FINITE LENGTH
7.2.2. Lemma. Any two finite composition series have the same number of submodules.
Proof. Let l(M) be the least length of a composition series. If M is simple then
l(M) = 1. Let 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M be any composition series. For a
submodule N ⊂ M there is a filtration 0 = N ∩M0 ⊂ N ∩M1 ⊂ · · · ⊂ N ∩Mn =
N with factors N ∩Mi/N ∩Mi−1 ⊂ Mi/Mi−1 being either simple or 0. It follows
that l(N) ≤ l(M). If l(N) = l(M) then each factor is nonzero and therefore
N = M. Applying this to the composition series of M gives l(M) ≤ n ≤ l(M)
as wanted.
7.2.3. Definition. The number of nontrivial submodules in a filtration as above
will be denoted ℓR(M) and is the length of M.
7.2.4. Proposition. Given an exact sequence
0 → N → M → L → 0
Then M has a finite length if and only if both N and L have finite length. In that
case
ℓR(M) = ℓR(N) + ℓR(L)
Proof. A finite filtration with simple quotients in M induces filtrations in both N
and L, and vice versa.
The filtrations in N and L give a filtration in M with the same quotients, so the
length formula follows from 7.2.2.
7.2.5. Corollary. Let f : M → N be a homomorphism.
(1) If M has finite length if and only if Ker f, Im f have finite length. If finite
length
ℓR(M) = ℓR(Ker f) + ℓR(Im f)
(2) If N has finite length if and only if Im f, Cok f have finite length. If finite
length
ℓR(N) = ℓR(Im f) + ℓR(Cok f)
Proof. Use the sequences 3.1.8.
7.2.6. Corollary. Let f : M → M be a homomorphism on a module of finite
length. Then
ℓR(Ker f) = ℓR(Cok f)
and the following are equivalent:
(1) f is injective.
(2) f is surjective.
(3) f is an isomorphism.
Proof. Use the sequences 3.1.8.
7.2.7. Corollary. Let N ⊂ M be a submodule and suppose M has finite length.
(1) ℓR(N) ≤ ℓR(M).
(2) ℓR(N) = ℓR(M) if and only if N = M.
7.2.8. Corollary. Let N, L ⊂ M be a submodules and suppose N, L has finite
length. Then N + L, N ∩ L have finite length and
ℓR(N + L) + ℓR(N ∩ L) = ℓR(N) + ℓR(L)

7.3. ARTINIAN RINGS 87
7.2.9. Proposition. Given submodules N, L ⊂ M. Then the following are equivalent.
(1) M/N, M/L have finite length.
(2) M/N ∩ L has finite length.
If finite length
ℓR(M/N + L) + ℓR(M/N ∩ L) = ℓR(M/N) + ℓR(M/L)
Proof. Use the sequences 3.2.8.
7.2.10. Proposition. A R-module M of finite length is finite and generated by
ℓR(M) or less elements.
Proof. There is a sequence 0 → N → M → L → 0 with L simple. Conclusion
by induction.
7.2.11. Proposition. Let I ⊂ R be an ideal. Suppose an R/I-module M has finite
length. Then M has finite length as R-module and
ℓ R/I(M) = ℓR(M)
Proof. This follows from 7.1.6 and 7.2.2.
7.2.12. Example. Let K be a field. A module M is of finite length if it is a finite
dimensional vector space. Then
7.2.13. Exercise. (1) Compute
(2) Compute
ℓZ(Z/(p n1
1
ℓK(M) = rankK M
. . . pnk
k )) = n1 + · · · + nk
ℓ K[X](K[X]/((X − a1) n1 . . . (X − ak) nk )) = n1 + · · · + nk
7.3. Artinian Rings
7.3.1. Lemma. Let M be a module. The following conditions are equivalent.
(1) Any decreasing sequence Mi of submodules is stationary, Mn = Mn+1 for
n >> 0.
(2) Any nonempty subset of submodules of M contains a minimal element.
Proof. (1) ⇒ (2): Suppose a nonempty subset of submodules do not contain a
minimal element. Then choose a non stationary sequence. (2) ⇒ (1): A descending
sequence containing a minimal element is stationary.
7.3.2. Definition. A module M which satisfies the conditions of 7.3.1 is an artinian
module.
7.3.3. Definition. A ring R is an artinian ring if R is an artinian module over
itself.
7.3.4. Proposition. Let 0 → N → M → L → 0 be an exact sequence of modules
over the ring R. Then M is artinian if and only if N and L are artinian.

88 7. MODULES OF FINITE LENGTH
Proof. Let M be artinian. A chain in N gives a chain in M, so N is artinian. A
chain in L gives a chain in M, which becomes stationary, so also the original chain,
so L is artinian. Conversely if N and L are artinian and Mi a chain in M, then
the induced chains in N and L become stationary. By the snake lemma 3.2.4 the
original chain is stationary.
7.3.5. Corollary. Let f : M → N be a homomorphism.
(1) M is artinian if and only if Ker f, Im f are artinian.
(2) N is artinian if and only if Im f, Cok f are artinian
Proof. Use the sequences 3.1.8.
7.3.6. Proposition. Let f : M → M be a homomorphism on an artinian module.
Then the following are equivalent
(1) f is injective
(2) f is an isomorphism
Proof. There is a number n such that Im f ◦n = Im f ◦2n . For x ∈ M there is y
such that f ◦n (x) = f ◦2n (y). Then f ◦n (x − f ◦n (y)) = 0 so x = f ◦n (y) since f is
injective. It follows that f ◦n is surjective and so is f.
7.3.7. Proposition. Given submodules N, L ⊂ M. Then the following are equivalent.
(1) M/N, M/L are artinian
(2) M/N ∩ L is artinian
Proof. use the sequences 3.2.8.
7.3.8. Corollary. Given ideals I, J ⊂ R. Then the following are equivalent.
(1) R/I, R/J are artinian
(2) R/I ∩ J is artinian
7.3.9. Corollary. Let R be an artinian ring and I an ideal. Then the factor ring
R/I is artinian.
The product of two artinian rings is artinian.
7.3.10. Proposition. An artinian domain is a field.
Proof. By 7.3.6 scalar multiplication with a nonzero element is an isomorphism.
7.3.11. Proposition. Let R be an artinian ring. Then all prime ideals are maximal
and there are only finitely many such.
Proof. By 7.3.10 primes are maximal. If Pi are different maximal ideals, then
P1 · · · Pn+1 ⊂ P1 · · · Pn is a strictly decreasing chain. So there are only finitely
many maximal ideals.
7.3.12. Proposition. Let R be an artinian ring.
(1) The factor ring R/ √ 0 is a finite product of fields.
(2) The nilradical √ 0 is nilpotent, √ 0 k = 0 for some k.

7.3. ARTINIAN RINGS 89
Proof. (1) Let P1, . . . , Pn be the prime and maximal ideals 7.3.11. The nilradical
√ 0 = P1 ∩ · · · ∩ Pn by 5.1.7. Conclude by Chinese remainders 1.4.3. (2) √ 0 k =
√ 0 k+1 for some k. If √ 0 k = 0 let (a) be minimal among ideals I such that
I √ 0 k = 0. By minimality (a) = (a) √ 0 k , so a = ab for some b ∈ √ 0 k . But b is
nilpotent 1.3.8, so a = 0 gives a contradiction. It follows, that √ 0 k = 0.
7.3.13. Proposition. A ring R is artinian if and only if it has finite length.
Proof. The factor module √ 0 i / √ 0 i+1 is an artinian module over R/ √ 0 which is
a product of fields, so it has finite length.
7.3.14. Corollary. Let R be an artinian ring and M a module. The following are
equivalent.
(1) M is finite.
(2) M has finite length.
(3) M is finite presented.
7.3.15. Corollary. Let R be artinian and R → S a finite ring homomorphism.
(1) S is artinian.
(2) A finite length S-module N is by restriction of scalars a finite length Rmodule.
(3) A finite length R-module M gives by change of rings M ⊗R S as finite length
S-module.
7.3.16. Proposition. Let M be a R-module of finite length.
(1) The ring R/ Ann(M) is artinian.
(2) There are only finitely many prime ideals Ann(M) ⊂ P .
(3) Any prime ideal Ann(M) ⊂ P is maximal.
(4) M is finite presented.
Proof. (1) Let x ∈ M, then R/ Ann(x) Rx is artinian. If x1, . . . , xn generate
M, then Ann(M) = Ann(x1)∩· · ·∩Ann(Xn). By 7.3.8 R/ Ann(M) is artinian.
7.3.17. Example. Let K be a field. The ring R =
artinian.
(1) The maximal ideals in R are
(2) The simple types are
N
Pi = {a : N → K|a(i) = 0}
R/Pi Ji = {a : N → K|a(j) = 0, i = j}
(3) Ji are the only simple ideals and the sum is an ideal
Ji = J = R
(4) J is an ideal which has no complement in R.
i
K = {a : N → K} is not
7.3.18. Exercise. (1) Show that a vector space is artinian if and only if it is finite dimensional.
(2) Show that Z is not artinian.
(3) Show that R[X] is not artinian for a nonzero ring R.
(4) Show that Q[X]/(X 2 − X) is artinian.

90 7. MODULES OF FINITE LENGTH
(5) Show that a ring with finitely many ideals is artinian.
(6) Let R be a principal ideal domain and a = 0. Show that R/(a) is artinian.
7.4. Localization
7.4.1. Proposition. Any artinian ring is the product of finitely many artinian local
rings. Let R be artinian with maximal ideals P1, . . . , Pk. Then there is n1, . . . , nk
such that P n1
1 . . . P nk
k = 0 and
R R/P n1
nk
1 × · · · × R/Pk Each R/P ni
1 is a local artinian ring.
Proof. This follows from 7.3.11, 7.3.12 and Chinese remainders 1.4.2.
7.4.2. Example. A reduced artinian ring is a finite product of fields.
7.4.3. Corollary. Let R be an artinian ring and U ⊂ R a multiplicative subset.
The ring of fractions U −1 R is an artinian ring.
7.4.4. Example. Let P ⊂ R be a maximal ideal and M a module. Assume s /∈ P
and n ∈ N.
(1) Rs + P n = A.
(2) Scalar multiplication by s : M/P n M → M/P n M, x → sx is an isomorphism.
(3) The canonical map M/P n M → (M/P n M)P is an isomorphism.
(4) If M is finite then M/P n M has finite length.
7.4.5. Proposition. Let R be artinian with maximal ideals P1, . . . , Pk. A finite
module M has a decomposition
M
The length is
Proof. Let R R/P n1
ℓR(M) =
R/P nk
k and MPi M ⊗R R/P ni
i .
1
i
i
MPi
ℓRP i (MPi )
nk
× · · · × R/Pk . Then M M ⊗R R/P n1
1 × · · · × M ⊗R
7.4.6. Proposition. Let M be an R-module of finite length and Ann(M) ⊂ P1, . . . , Pk
the maximal ideals. Then MPi is a finite length RPi-module and
ℓR(M) =
ℓRP (MPi i )
Proof. The ring R/ Ann(M) R/P n1
1
i
× · · · × R/P nk
k
by 7.3.16 and 7.4.1.
7.4.7. Proposition. Let (R, P ) → (S, Q) be a local ring homomorphism and assume
that k(P ) → k(Q) is a finite extension. If N is a finite length S-module, then
N is a finite length R-module and
Proof. Reduce to N = k(Q).
ℓR(N) = rank k(P )(k(Q)) · ℓS(N)
7.4.8. Proposition. Let (R, P ) → (S, Q) be a local ring homomorphism and assume
that S/P S is a finite length S-module. Let M be a finite length R-module.

7.5. LOCAL ARTINIAN RING 91
(1) M ⊗R S is a finite length S-module.
(2) In general
ℓS(M ⊗R S) ≤ ℓS(S/P S) · ℓR(M)
(3) If R → S is flat then
ℓS(M ⊗R S) = ℓS(S/P S) · ℓR(M)
Proof. The case M = k(P ) is clear. Conclude by induction.
7.4.9. Exercise. (1) Let K ⊂ L be a finite field extension and W a finite vector space
over L. Show that
rankK(W ) = rankK(L) · rankL(W )
(2) Let K ⊂ L be a finite field extension and V a finite vector space over K. Show that
rankL(V ⊗K L) = rankK(V )
7.5. Local artinian ring
7.5.1. Lemma. Let (R, P ) be a local artinian ring and M a finite module. The
following are equivalent.
(1) M = 0.
(2) HomR(k(P ), M) = 0.
(3) HomR(M, k(P )) = 0.
Proof. Clear by a filtration.
7.5.2. Proposition. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope.
(1) The are isomorphisms
k(P ) HomR(k(P ), k(P )) HomR(k(P ), E)
(2) For a finite module M, the module HomR(M, E) has finite length and
ℓR(HomR(M, E)) = ℓR(M)
Proof. (1) A nonzero homomorphism f : k(P ) → E has Im f = k(P ) since the
extension is essential. (2) This follows from (1) by use of a filtration.
7.5.3. Corollary. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. Then E has finite length
ℓR(E) = ℓR(R)
7.5.4. Proposition. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. There is a natural isomorphism for any finite module M
x ↦→ evx : M → HomR(HomR(M, E), E)
Proof. The case M = k(P ) is clear by 7.5.2. Let 0 → N → M → L → 0 be a
short exact sequence. Then the following diagram has exact rows.
0
0
N
HomR(HomR(N, E), E)
M
HomR(HomR(M, E), E)
Conclusion by the five lemma 3.2.8 and induction on the length.
L
HomR(HomR(L, E), E)
0
0

92 7. MODULES OF FINITE LENGTH
7.5.5. Corollary. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. There is an isomorphism
11E : R → HomR(E, E)
7.5.6. Proposition. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. Let M → E ′ be an injective envelope of a finite module. Then
(1) HomR(k(P ), M) HomR(k(P ), E ′ )
(2) E ′ E n , where n = ℓR(HomR(k(P ), M).
(3) ℓR(E ′ ) = ℓR(HomR(k(P ), M) ℓR(E).
Proof. (1) A homomorphism f : k(P ) → E ′ has Im f ⊂ M since the extension
is essential. (2) By induction on length of M follows that E ′ E n , where n is
determined by (1) and 7.5.2. (3) This follows from (2).
7.5.7. Proposition. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. The following are equivalent.
(1) R is injective.
(2) R E.
(3) HomR(k(P ), R) HomR(k(P ), k(P )) k(P ).
Proof. (1) ⇒ (2): By 7.5.6 R E n and by 7.5.3 n = 1. (2) ⇒ (3): Immediate
from 7.5.6. (3) ⇒ (1): Let R ⊂ E ′ be an injective envelope. By 7.5.6 R = E ′ .
7.5.8. Definition. A ring satisfying the conditions 7.5.7 is a local artinian Gorenstein
ring.
7.5.9. Example. Let R be a principal ideal domain and p ∈ R an irreducible
element. Then R/(p n ) is a local artinian Gorenstein ring.
7.5.10. Exercise. (1) Let p be a prime number. Show that Z/(p k ) is a local artinian
Gorenstein ring.

8
Noetherian modules
8.1. Modules and submodules
8.1.1. Lemma. Let M be a module. The following conditions are equivalent.
(1) Any increasing sequence Mn of submodules is stationary, Mn = Mn+1 for
n >> 0.
(2) Any nonempty subset of submodules of M contains a maximal element.
(3) Any submodule of M is finite.
Proof. (1) ⇔ (2): See the proof of 7.3.1. (2) ⇒ (3): Let N be a submodule of
M and choose a maximal element N ′ in the set of finite submodules of N. For
y ∈ N, the module N ′ + Ry is finite, so N ′ = N ′ + Ry gives N = N ′ finite.
(3) ⇒ (1): The union ∪Mn is a submodule, generated by x1, . . . , xm ∈ Mk, so
Mn = ∪Mn, n > k.
8.1.2. Definition. A module M which satisfies the conditions of 8.1.1 is a noetherian
module.
8.1.3. Proposition. Let 0 → N → M → L → 0 be an exact sequence of modules
over the ring R. Then M is noetherian if and only if N and L are noetherian.
Proof. See the proof of 7.3.4.
8.1.4. Corollary. Let f : M → N be a homomorphism.
(1) M is noetherian if and only if Ker f, Im f are noetherian.
(2) N is noetherian if and only if Im f, Cok f are noetherian.
Proof. Use the sequences 3.1.8.
8.1.5. Proposition. Let f : M → M be a homomorphism on a noetherian module.
Then the following are equivalent
(1) f is surjective
(2) f is an isomorphism
Proof. Analog of the proof of 7.3.4. There is a number n such that Ker f ◦2n =
Ker f ◦n . For x ∈ Ker f ◦n there is y such that x = f ◦n (y). Then f ◦2n (y) =
f ◦n (x) = 0, so y ∈ Ker f ◦2n = Ker f ◦n . Then x = f ◦n (y) = 0 and f is
injective.
8.1.6. Proposition. A finite direct sum of noetherian modules is noetherian.
8.1.7. Proposition. Given submodules N, L ⊂ M. Then the following are equivalent.
(1) M/N, M/L are noetherian
(2) M/N ∩ L is noetherian
93

94 8. NOETHERIAN MODULES
Proof. Use the sequence 3.2.7.
8.1.8. Example. A module of finite length is noetherian.
8.1.9. Exercise. (1) Show that
N Z is not noetherian.
(2) Show that
N Z is not noetherian.
(3) Show that Q is not a noetherian Z-module.
8.2. Noetherian rings
8.2.1. Definition. A ring R is a noetherian ring if R is a noetherian module.
8.2.2. Proposition. The following conditions are equivalent.
(1) R is noetherian.
(2) Any ideal is finite.
(3) Any increasing sequence of ideals is stationary.
(4) Any nonempty subset of ideals of contains an ideal maximal for inclusion.
Proof. This is follows from 8.1.1.
8.2.3. Proposition. The nilradical in a noetherian ring is nilpotent, i.e. √ 0 n = 0
for some n. Some power of the radical of an ideal is contained in the ideal, i.e.
√ I n ⊂ I for some n
Proof. √ 0 = (b1, . . . , bn) such that b k i = 0. Then ( aibi) nk = 0.
8.2.4. Proposition. Given ideals I, J ⊂ R. Then the following are equivalent.
(1) R/I, R/J are noetherian
(2) R/I ∩ J is noetherian
Proof. This is a special case of 8.1.7.
8.2.5. Lemma. (1) Any factor ring of a noetherian ring is a noetherian ring.
(2) A principal ideal domain is noetherian.
(3) If M is a noetherian R-module, then R/ Ann(M) is a noetherian ring.
Proof. (1), (2) are clear. (3) Analog of the proof of 7.3.16. Let x ∈ M, then
R/ Ann(x) Rx is noetherian. If x1, . . . , xn generate M, then Ann(M) =
Ann(x1) ∩ · · · ∩ Ann(Xn). By 8.2.4 R/ Ann(M) is artinian.
8.2.6. Proposition. Let R be a noetherian ring. Any finite R-module is noetherian.
Proof. Let M be a finite R-module. There is a surjective homomorphism R n →
M → 0, 6.1.2. Conclusion by 8.2.3.
8.2.7. Corollary. Let R be a noetherian ring. Any finite R-module is finite presented.
8.2.8. Lemma. Let R be a noetherian ring and M, N noetherian modules.
(1) M ⊗R N is noetherian.
(2) HomR(M, N) is noetherian.
Proof. Conclusion by 6.1.8, 6.1.9, 8.2.6.

8.3. FINITE TYPE RINGS 95
8.2.9. Proposition. Let R be a noetherian ring and U a multiplicative subset. For
a finite module M and any module N the homomorphism
is an isomorphism.
U −1 HomR(M, N) → Hom U −1 R(U −1 M, U −1 N)
Proof. Conclusion by 6.5.8, 8.2.7.
8.2.10. Proposition. Let R be a noetherian ring. If E is an injective R-module,
then U −1 E is an injective U −1 R-module.
Proof. Let I ⊂ R be an ideal. Then HomR(R, E) → HomR(I, E) → 0 is exact.
By 8.2.9 Hom U −1 R(U −1 R, U −1 E) → Hom U −1 R(U −1 I, U −1 E) → 0 is exact.
So U −1 E is injective 3.6.7, using that any ideal is extended 4.3.6.
8.2.11. Proposition. For a ring R, the following are equivalent.
(1) R is a noetherian ring.
(2) For any family Eα of injective modules, the sum
α Eα is injective.
Proof. (1) ⇒ (2): Let I ⊂ R be an ideal. A homomorphism f : I → Eα
has Im f contained in a finite sum, which is injective 3.6.6. So f extends to R →
Eα, giving injectivity. (2) ⇒ (1): Let In ⊂ R be an increasing chain of ideals.
Put I = ∪In and choose an injective envelope I/In ⊂ En. The homomorphism
f : I → En extends to f ′ : R → En. Since Im f is contained in a finite
sum, I/In = 0 for n >> 0.
8.2.12. Proposition. If R is a ring such that every prime ideal is finite, then it is
noetherian.
Proof. If R is not noetherian, then the set of not finite ideals is nonempty and by
Zorn’s lemma it has a maximal element I. Since I is not prime there is a, b /∈ I
and ab ∈ I. The ideals I + (a) and I : (a) are both greater that I and therefore
finite. If I + (a) = (a1, . . . , am, a), ai ∈ I and I : (a) = (b1, . . . , bn), then
I = (a1, . . . , am, ab1, . . . , abn) is finite. It follows that R must be noetherian.
8.2.13. Exercise. (1) Show that a principal ideal domain is noetherian.
(2) Let I ⊂ R be an ideal in a noetherian ring. Show that R/I is noetherian.
(3) Let K be a field and R = K[X1, X2, . . . ] the polynomial ring in countable many
variables. Show that R is not noetherian.
(4) Let K be a field. Show that the ring R =
N K is not noetherian.
8.3. Finite type rings
8.3.1. Proposition (Hilbert’s basis theorem). Let R be a noetherian ring. Then the
ring of polynomials R[X] is noetherian.
Proof. Assume I ⊂ R[X] to be a not finite ideal. Choose a sequence f1, f2, · · · ∈
I such that
fi = aiX di + terms of lower degree , ai = 0
and fi+1 has lowest degree in I − (f1, . . . , fi). The ideal of leading coefficients is
finitely generated by a1, . . . , an. Then an+1 = b1a1 + · · · + bnan and d1 ≤ · · · ≤
dn+1 = d gives
fn+1 − b1X d−d1 f1 − · · · − bnX d−dn fn
in I − (f1 . . . , fn) of degree less than d. By contradiction the ideal I is finite.

96 8. NOETHERIAN MODULES
8.3.2. Corollary. Let R be a noetherian ring and R → S a finite type ring over R.
Then S is noetherian.
8.3.3. Corollary. Let R be a noetherian ring and M a finite module. Then the ring
R ⊕ M, 2.1.14, is noetherian.
8.3.4. Example. Let I ⊂ R be an ideal in a noetherian ring. Then there are noetherian
rings.
(1) GIR = ⊕n≥0I n /I n+1 .
(2) BIR = ⊕n≥0I n .
8.3.5. Theorem (Krull’s intersection theorem). Let I be an ideal in a noetherian
ring R. Then there is a ∈ I such that
1 + a ∈ Ann(
I n )
Proof. Let I = (u1, . . . , um). If b ∈ I n then b = fn(u1, . . . , um) where fn ∈
R[X1, . . . , Xm] are homogeneous of degree n. By Hilbert’s basis theorem, 8.3.1
there is N such that
fN+1 = f1g1 + · · · + fNgN
where gn is homogeneous of degree N − n + 1 > 0. By substitution b ∈ bI and
I ∩ I n = ∩I n . Conclusion by 6.3.4.
8.3.6. Corollary. Let I be an ideal in a noetherian ring R such that the elements
1 + a, a ∈ I are nonzero divisors. Then
I n = 0
n
8.3.7. Corollary. Let I be an ideal in a noetherian ring R and M a finite module.
Then there is a ∈ I such that
1 + a ∈ Ann(
I n M)
Proof. Use 8.3.5 on the ring R ⊕ M and the ideal I ⊕ M. Clearly (I + M) n =
I n + I n−1 M.
8.3.8. Corollary. Let I be an ideal in a noetherian ring R and M a finite module
such that the elements 1 + a, a ∈ I are nonzero divisors on M. Then
I n M = 0
n
8.3.9. Proposition. If R ⊂ S be a finite extension. Then R is noetherian if and
only if S is noetherian.
Proof. Assume S is noetherian. Let Eα be a family of injective R-modules. Then
HomR(S, Eα) is an injective S-module. Since S is finite over R, HomR(S, Eα)
HomR(S, Eα) 6.1.14. By 6.6.7 Eα is injective and by 8.2.11 R is noetherian.
8.3.10. Example. Let K be a field and R = K[X1, X2, . . . ]/(X1 − x2X3, x2 −
x3X4, . . . ). Put P = (X1, X2, . . . ).
(1) P is maximal
(2) P 2 = P .
(3) P (∩nP n ) = ∩nP n .
n
n

8.4. POWER SERIES RINGS 97
8.3.11. Exercise. (1) Show that if R[X] is noetherian, then R is noetherian.
(2) Show that the subring Z[2X, 2X 2 , . . . ] ⊂ Z[X] is not noetherian. Conclude that the
extension is not finite.
8.4. Power series rings
8.4.1. Proposition. Let R be a noetherian ring. Then the power series ring R[[X]]
is noetherian.
Proof. Let P ⊂ R[[X]] be a prime ideal. Then P +(X)/(X) = (a1, . . . , an) ⊂ R
is a finite ideal. If X ∈ P then P = (a1, . . . , an, X) is finite. Suppose X /∈ P
and choose fi = ai + terms of positive degree ∈ P . If g ∈ P then Xg1 =
g − b11f1 + · · · + bn1fn ∈ P ∩ (X) for some bi1 ∈ R. Since P is prime, g1 ∈ P .
Now Xg2 = g1 − b12f1 + · · · + bn2fn ∈ P ∩ (X) and so on. Put hi = bikX k ,
then g = hifi. P is finite and R[[X]] is noetherian by 8.2.12.
8.4.2. Proposition. Let R be a principal ideal domain. Then the power series ring
R[[X]] is a unique factorization domain.
Proof. Let P ⊂ R[[X]] be a minimal nonzero prime ideal. Then P + (X)/(X) =
(a) ⊂ R Suppose P = (X) and choose f = a + terms of positive degree ∈ P .
If g ∈ P then Xg1 = g − b1f ∈ P ∩ (X) for some b1 ∈ R. Since P is prime,
g1 ∈ P . Now Xg2 = g1 − b2f ∈ P ∩ (X) and so on. Put h = bkX k , then
g = hf. P = (f) is principal. The conditions of 1.5.3 are satisfied since R[[X]] is
noetherian 8.4.1.
8.4.3. Proposition. Let I ⊂ R be an ideal in a noetherian ring. Then there is a
canonical isomorphism
R[[X]]/IR[[X]] R/I[[X]]
Proof. The projection R[[X]] → R[[X]]/IR[[X]] factors over R/I[[X]] since I is
finitely generated. Then there is an inverse to the homomorphism 1.9.8.
8.4.4. Corollary. If P ⊂ R is a prime ideal in a noetherian ring, then P R[[X]] ⊂
R[[X]] is a prime ideal.
8.4.5. Proposition. Let R be a noetherian ring.
(1) The inclusion R[X] ⊂ R[[X]] is a flat homomorphism.
(2) The inclusion R ⊂ R[[X]] is a faithfully flat homomorphism.
Proof. (1) Let I ⊂ R[X] be an ideal and let ai ⊗ fi ∈ K = Ker(I ⊗R[X] R[[X]] → R[[X]]. Then aifi = 0. Write fi = gi + Xnhi and get ai ⊗ fi =
Xn (ai ⊗ hi). It follows that K ⊂
n Xn (I ⊗R[X] R[[X]]). I ⊗R[X] R[[X]]
is a finite R[[X]]-module and 1 + aX is a unit, so conclusion by 8.3.8. (2) The
homomorphism is flat by (1). For any maximal ideal P ⊂ R the homomorphism
RP → RP [[X]] is local and flat.
8.4.6. Exercise. (1) Show that if R[[X]] is noetherian, then R is noetherian.

98 8. NOETHERIAN MODULES
8.5. Fractions and localization
8.5.1. Proposition. Let R be a noetherian ring and U a multiplicative subset. Then
U −1 R is a noetherian ring.
Proof. Any ideal is extended 4.3.6.
8.5.2. Corollary. Let R be a noetherian ring and U a multiplicative subset. If M
is a noetherian R-module, then U −1 M is a noetherian U −1 R-module.
Proof. By 6.1.3 change of ring of a finite module is finite.
8.5.3. Corollary. Let R be a noetherian ring and M a finite module. Let P be a
prime ideal. Then RP is a noetherian ring and MP is a finite RP -module.
8.5.4. Proposition (Krull’s intersection theorem). Let (R, P ) be a noetherian local
ring and M a finite module. Then
P n M = 0
Proof. This follows from 8.3.7.
n
8.5.5. Corollary. Let (R, P ) be a noetherian local ring and I ⊂ P an ideal. Then
I n = 0
n
8.5.6. Proposition. Let (R, P ) be a noetherian local ring and F a finite module.
The following conditions are equivalent.
(1) F is free.
(2) F is projective.
(3) F is flat.
(4) P ⊗R F → F is injective.
Proof. This follows from 6.5.13.
8.5.7. Exercise. (1) Is it true that if U −1 R is noetherian, then R is noetherian?
8.6. Prime filtrations of modules
8.6.1. Proposition. Let R be a ring and M = 0 a nonzero module. An ideal
Ann(x) maximal in the set of ideals {Ann(y)|0 = y ∈ M} is a prime ideal.
Proof. Let Ann(x) be a maximal annihilator. Suppose a, b ∈ R such that ab ∈
Ann(x) and b /∈ Ann(x). Then
Ann(x) ⊆ Ann(bx) = R
Consequently Ann(x) = Ann(bx) in particular a ∈ Ann(x).
8.6.2. Corollary. Let R be a noetherian ring and M = 0 a nonzero module. Then
there is x ∈ M such that Ann(x) is a prime ideal.
8.6.3. Theorem. Let R be a noetherian ring and M = 0 a finite R-module. Then
there exists a finite filtration of M by submodules
0 = M0 ⊂ M1 ⊂ · · · ⊂ Mr−1 ⊂ Mr = M
such that Mi/Mi−1, i = 1, . . . , r is isomorphic to an R-module of the form R/Pi
where Pi is a prime ideal in R.

8.6. PRIME FILTRATIONS OF MODULES 99
Proof. The set of submodules of M for which the theorem is true is nonempty by
8.2.6. Let N ⊂ M be maximal in this set. Suppose N = M. By 8.2.6 applied to
M/N there is a chain N ⊂ N ′ ⊂ M such that N ′ /N is isomorphic to an R-module
of the form R/P ′ where P ′ is a prime ideal. This contradicts the maximality of N.
So N = M.
8.6.4. Corollary. Let R be a nonzero noetherian ring. Then there exists a finite
filtration of ideals
0 = I0 ⊂ I1 ⊂ · · · ⊂ Ir−1 ⊂ Ir = R
such that Ii/Ii−1, i = 1, . . . , r is isomorphic to an R-module of the form R/Pi
where Pi is a prime ideal in R.
8.6.5. Example. In Z there is a filtration
0 =⊂ (p n ) ⊂ (p n−1 ) ⊂ · · · ⊂ (p) ⊂ Z
of any length with factors Z/(p) for any prime number p.
8.6.6. Proposition. A ring R is artinian if and only if it is noetherian and all prime
ideals are maximal.
Proof. By 7.3.13 an artinian ring has finite length and therefore noetherian. Primes
are maximal by 7.3.11. Conversely by 8.6.4 there is a finite composition series.
8.6.7. Corollary. Let R be a ring and M a module. The following are equivalent
(1) M has finite length.
(2) R/ Ann(M) is artinian and M is finite.
8.6.8. Proposition. If R ⊂ S be a finite extension. Then R is artinian if and only
if S is artinian.
Proof. Suppose S is artinian. By 8.3.9 R is noetherian. By 6.6.5 any prime ideal
in R is maximal. Conclusion by 8.6.6.
8.6.9. Proposition. Let R be a noetherian ring. The number of minimal prime
ideals is finite.
Proof. Choose a filtration 8.6.4, 0 = I0 ⊂ · · · ⊂ Ir = R with Ii/Ii−1 R/Pi,
where Pi is a prime ideal. Let P be a minimal prime ideal in R. Then (Ii/Ii−1)P
(R/Pi)P = 0 if and only if Pi ⊂ P . Thus P = Pi for some i since RP = 0.
8.6.10. Proposition. Let R be a noetherian ring such that the local rings RP are
domains for all maximal ideals P . Then R is a finite product of domains.
Proof. Let P1, . . . , Pn be the minimal primes 8.6.9. The intersection is 0 5.4.9 and
they are comaximal since a domain has a unique minimal prime. Conclusion by
Chinese remainders 1.4.2.
8.6.11. Exercise. (1) Compute a filtration 8.6.3 of the Z-module Z/(36).

9
Primary decomposition
9.1. Support of modules
9.1.1. Definition. Let R be a ring and M a module.
(1) The set of prime ideals is the spectrum and denoted Spec(R).
(2) For a ring homomorphism φ : R → S restriction defines a map
(3) For a subset B ⊂ R
is a subset of the spectrum.
(4) The support of M is
φ ∗ : Spec(S) → Spec(R)
Q ↦→ φ −1 (Q)
V (B) = {P ∈ Spec(R)|MP = 0}
Supp(M) = {P ∈ Spec(R)|MP = 0}
(5) A minimal prime ideal in Supp(M) is a minimal prime of M.
9.1.2. Proposition. Let R be a ring.
(1) Let I ⊂ R be an ideal. Then
and is identified with Spec(R/I).
(2) Let U be a multiplicative subset.
Supp(R/I) = V (I)
Supp(U −1 R) = {P ∈ Spec(R)|P ∩ U = ∅}
and is identified with Spec(U −1 R).
Proof. This is a restatement of 1.3.5, 5.1.5.
9.1.3. Proposition. Let 0 → N → M → L → 0 be a short exact sequence of
modules. Then
Supp(M) = Supp(N) ∪ Supp(L)
Proof. This follows from 5.4.5.
9.1.4. Corollary. (1) Let N ⊂ M be a submodule. Then
Supp(M) = Supp(N) ∪ Supp(M/N)
(2) Given submodules N, L ⊂ M. Then
Supp(M/N ∩ L) ∪ Supp(M/N + L) = Supp(M/N) ∪ Supp(M/L)
and
Supp(M/N + L) ⊂ Supp(M/N) ∩ Supp(M/L)
Proof. (2) Use the sequence 3.2.7.
101

102 9. PRIMARY DECOMPOSITION
9.1.5. Proposition. Let R be a ring and M a module.
(1) M = 0 if and only if Supp(M) = ∅.
(2) For any module
Supp(M) ⊂ V (Ann(M))
.
(3) If M is finite, then
Supp(M) = V (Ann(M))
Proof. (1) See 5.4.1. (2) If MP = 0 then for u /∈ P there is x ∈ M such that
ux = 0. So Ann(M) ⊂ P . (3) Let x1, . . . , xn generate M. If Ann(M) =
∩ Ann(xi) ⊂ P then some Ann(xi) ⊂ P , so xi
1 = 0 in MP .
9.1.6. Proposition. Let N1, . . . , Nk ⊂ M be submodules such that Supp(M/Ni)∩
Supp(M/Nj) = ∅, i = j. Then the homomorphism
M/ ∩i Ni →
M/Ni
is an isomorphism.
Proof. By induction on k it is enough to treat the case k = 2. By 9.1.4 the support
of the cokernel is empty.
9.1.7. Proposition. Let R be a ring and M, N modules.
(1)
Supp(M ⊗R N) ⊂ Supp(M) ∩ Supp(N)
(2) If M, N are finite, then
Supp(M ⊗R N) = Supp(M) ∩ Supp(N)
Proof. There is an isomorphism (M ⊗R N) MP ⊗RP NP . (1) This is clear. (2)
This follows from 6.4.3.
9.1.8. Corollary. Let φ : R → S be a ring homomorphism and M an R-module.
(1) For the change of rings module
(2) If M is finite, then
Supp(M ⊗R S) ⊂ φ ∗−1 (Supp(M))
Supp(M ⊗R S) = φ ∗−1 (Supp(M))
9.1.9. Corollary. Let R be a ring, I an ideal in R and M a finite R-module. Then
Supp(M/IM) = Supp(M) ∩ V (I)
9.1.10. Corollary. Let R be a ring, U a multiplicative subset and M a finite Rmodule.
Then
Supp(U −1 M) = Supp(M) ∩ Spec(U −1 R)
9.1.11. Proposition. Let (R, P ) → (S, Q) be a local homomorphism and M a
finite R-module. If Supp(M) = ∅ then Supp(M ⊗R S) = ∅.
Proof. The homomorphism R → S is faithfully flat 5.5.8.
9.1.12. Proposition. Let M be a finite R-module and P ∈ Supp(M). Then there
is a nonzero homomorphism M → R/P .
i

9.2. ASS OF MODULES 103
Proof. The module HomR(M, R/P )P HomRP (MP , k(P )) is nonzero 7.1.4.
9.1.13. Proposition. Let R be a noetherian ring and M a finite module. M has
finite length if and only if Supp(M) consists only of maximal ideals.
Proof. This is 8.6.7.
9.1.14. Proposition. Let R be a noetherian ring and F a finite module. If R has
only finitely many maximal ideals and FP is free of rank n, then F is free of rank
n.
Proof. If R is artinian, this is clear from 7.4.5. If P1, . . . , Pk are the maximal
ideals, then choose x1, . . . , xn ∈ F giving a basis for F/P1 · · · PnF . The homomorphism
R n → F, ei ↦→ xi is an isomorphism 6.4.6 and 8.1.5.
9.1.15. Exercise. (1)
9.2. Ass of modules
9.2.1. Definition. Let M be an R-module. A prime ideal P ⊂ R is an associated
prime ideal of M if P = Ann(x) for some x ∈ M. The set of prime ideals
associated to M is Ass(M).
9.2.2. Proposition. Let P ⊂ R be a prime ideal.
(1) P ∈ Ass(M) if and only if there is an injective homomorphism R/P → M.
(2)
Ass(M) ⊂ Supp(M)
(3) Let 0 = N ⊂ R/P be a nonzero submodule, then
Proof. This is clear from the definition.
Ass(N) = {P }
9.2.3. Proposition. Let 0 → N → M → L → 0 be a short exact sequence of
modules. Then
Ass(N) ⊂ Ass(M) ⊂ Ass(N) ∪ Ass(L)
Proof. The left inclusion is trivial. Next let P ∈ Ass(M) such that P /∈ Ass(N).
Choose a submodule L of M such that L R/P . Then Ass(L ∩ N) ⊂ Ass(L) ∩
Ass(N). It follows that L ∩ N = 0, and therefore M/N contains a submodule
isomorphic to R/P .
9.2.4. Corollary. Let 0 → N → M → L → 0 be a split exact sequence of
modules.
Ass(M) = Ass(N) ∪ Ass(L)
9.2.5. Corollary. (1) Let N ⊂ M be a submodule. Then
(2) Let M, N be modules. Then
Ass(N) ⊂ Ass(M) ⊂ Ass(N) ∪ Ass(M/N)
Ass(M ⊕ N) = Ass(M) ∪ Ass(N)
(3) Given submodules N, L ⊂ M. Then
Ass(M/N ∩ L) ⊂ Ass(M/N) ∪ Ass(M/L) ⊂ Ass(M/N ∩ L) ∪ Ass(M/N + L)
Proof. (3) Use the sequence 3.2.7.

104 9. PRIMARY DECOMPOSITION
9.2.6. Proposition. Let M be a module and P ⊂ Ass(M). Then there is a submodule
N ⊂ M such that
Ass(N) = P , Ass(M/N) = Ass(M)\P
Proof. Choose by Zorn’s lemma N maximal in the set of submodules N ′ ⊂ M
for which Ass(N ′ ) ⊂ P. Let Q ∈ Ass(M/N) and choose N ⊂ L ⊂ M with
L/N R/Q. Then Ass(L) ⊂ P ∪ {Q}. By maximality of N follows that
Ass(L) ⊂ P. So Q /∈ P and Q ∈ Ass(L) ⊂ Ass(M).
9.2.7. Proposition. Let R be a noetherian ring and M a module. Then M = 0 if
and only if Ass(M) = ∅.
Proof. If M = 0, then by 8.6.2 Ass(M) = ∅.
9.2.8. Corollary. Let R be a noetherian ring and f : M → N a homomorphism.
The following conditions are equivalent.
(1) f is injective.
(2) fP is injective for all prime ideals P ∈ Ass(M).
Proof. Use 9.2.7 on Ker f.
9.2.9. Corollary. Let R be a noetherian ring and M a module. Then a ∈ R is a
nonzero divisor on M if and only if
a /∈ ∪ P ∈Ass(M)P
The set of zero divisors on M is ∪ P ∈Ass(M)P .
Proof. aM is injective if and only if aMP
This happens when a /∈ ∪ P ∈Ass(M)P .
is injective for all P ∈ Ass(M), 8.2.8.
9.2.10. Proposition. Let U ⊂ R be a multiplicative subset and M a module.
(1) In general
(2) If R is noetherian
Ass(M) ∩ Spec(U −1 R) ⊂ Ass U −1 R(U −1 M)
Ass(M) ∩ Spec(U −1 R) = Ass U −1 R(U −1 M)
Proof. Let P be a prime ideal. For any homomorphism R/P → M there is a
commutative diagram
R/P
M
U −1R/P
U −1M (1) If P ∈ Ass(M) and P ∩ U = ∅ then U −1 P ∈ Ass(U −1 M). (2) If U −1 P ∈
Ass(U −1 M), then P ∩ U = ∅. By 8.2.9 there is a diagram as above. R/P → M
is injective by 9.2.8. So P ∈ Ass(M).
9.2.11. Proposition. Let R be a noetherian ring and M a module. Then any minimal
prime P ∈ Supp(M) is contained in Ass(M).
Proof. Assume P ∈ Supp(M) is minimal. Then the RP -module MP has support
exactly in the maximal ideal, so {P RP } = Ass(MP ). Conclusion by 9.2.10.

9.2. ASS OF MODULES 105
9.2.12. Definition. A non minimal prime ideal in Ass(M) is an embedded prime
of M.
9.2.13. Proposition. Let R be a noetherian ring and M a finite module. Then
Ass(M) is a finite set.
Proof. Follows immediately from 9.2.4 and 8.1.5.
9.2.14. Corollary. Let R be a noetherian ring and M a finite module. The following
are equivalent
(1) M has finite length.
(2) Supp(M) consists of maximal ideals.
(3) Ass(M) consists of maximal ideals.
Proof. (3) ⇒ (2): The minimal ideals in the support are maximal.
9.2.15. Lemma. Let (R, P ) be a local ring M a module. Then P ∈ Ass(M) if
and only if HomR(k(P ), M) = 0.
9.2.16. Proposition. Let R be a noetherian ring and M a finite module. For any
module N
Ass(HomR(M, N)) = Supp(M) ∩ Ass(N)
Proof. By 8.2.9 HomR(M, N)P HomRP (MP , NP ). So reduce to the case
where (R,P) is local. Now
HomR(k(P ), HomR(M, N)) = HomR(M, HomR(k(P ), N))
Conclusion by Nakayama’s lemma 6.4.1 and 9.2.15.
= Hom k(P )(M ⊗R k(P ), HomR(k(P ), N))
9.2.17. Proposition. Let R be a noetherian ring and F a finite module. Assume
rank F ⊗R k(P ) = n for all primes P . Then F is locally free (projective) if and
only if FP is free for all P ∈ Ass(R).
Proof. Let Q be a maximal ideal and 0 → K → R n Q → FQ → 0 exact. Ass(K) ⊂
Ass(R), so KP = 0 for all P ∈ Ass(K). By 9.2.7 K = 0.
9.2.18. Proposition. Let (R, P ) be a noetherian local ring. If there is a nonzero
finite injective module E, then R is artinian.
Proof. Let Q be a prime and f : R/Q → E. If a ∈ P \Q then a R/Q is injective,
so there is f ′ : R/Q → E such that f = f ′ ◦ a R/Q. That is P HomR(R/Q, E) =
HomR(R/Q, E), so by Nakayama’s lemma 6.4.1 HomR(R/Q, E) = 0 if Q = P .
By 9.2.7 0 = HomR(R/P, E) ⊂ HomR(R/Q, E), so Q = P . R is artinian by
8.6.6.
9.2.19. Exercise. (1) Show that
(2) Show that
Ass(Z/(n)) = {(p)|p prime dividing n}
Ass(K[X, Y ]/(X) ∩ (X 2 , Y 2 )) = {(X), (X, Y )}
and point out an embedded prime.
(3) Let I ⊂ R be an ideal such that √ I = I. Show that R/I has no embedded prime
ideals.

106 9. PRIMARY DECOMPOSITION
(4) Let I, J ⊂ R be a ideals such that JRP ⊂ IRP for all P ∈ Ass(R/I). Show that
J ⊂ I.
9.3. Primary modules
9.3.1. Definition. A submodule N ⊂ M is a primary submodule or more precisely
P -primary if Ass(M/N) = {P }.
9.3.2. Proposition. A prime ideal P ⊂ R is a P -primary submodule.
Proof. This is 9.2.2.
9.3.3. Proposition. Let R be a noetherian ring and M a finite module. For a
submodule N ⊂ M the following are equivalent.
(1) N ⊂ M is primary for some prime
(2) The set of zero divisors on M/N is contained in the radical Ann(M/N).
Proof. (1) ⇒ (2): Ann(M/N) = P the set of zero divisors by 9.2.9. (2) ⇒ (1):
If P1, P2 ∈ Ass(M/N) then P1 ∪ P2 ⊂ Ann(M/N) ⊂ P1 ∩ P2, so P1 = P2.
9.3.4. Corollary. Let R be a noetherian ring and I ⊂ R a proper ideal. The
following are equivalent.
(1) I ⊂ R is primary for some prime.
(2) Any zero divisor in R/I is nilpotent.
9.3.5. Corollary. Let R be a noetherian ring.
(1) If an ideal I ⊂ R is P -primary then √ I = P .
(2) If the radical √ I = P is a maximal ideal, then I ⊂ R is a P -primary
submodule.
(3) A finite power P n ⊂ R of a maximal ideal is a P -primary submodule.
9.3.6. Proposition. Let R be a noetherian ring and M a finite module.
(1) If N ⊂ M is P -primary, then Ann(M/N) ⊂ R is P -primary.
(2) If N, N ′ ⊂ M are P -primary, then N ∩ N ′ is P -primary.
Proof. (1) This follows from 9.3.3. (2) This follows from 9.2.5.
9.3.7. Proposition. Let R be a noetherian ring and M a finite module. Suppose
N ⊂ M is P -primary and U ⊂ R is multiplicative subset.
(1) If U ∩ P = ∅, then U −1 N ⊂ U −1 M is P U −1 R-primary.
(2) If U ∩ P = ∅, then U −1 N = U −1 M.
Proof. This follows from 9.2.10.
9.3.8. Exercise. (1) Let K be a field. Show that (X 2 , Y ) ⊂ K[X, Y ] is (X, Y )primary.
(2) Let p be a prime number. Show that (p k ) ⊂ Z is a primary ideal.
9.4. Decomposition of modules
9.4.1. Definition. A submodule L ⊂ M has a primary decomposition if there
exist a family Ni ⊂ M of Pi-primary submodules, such that
L = N1 ∩ · · · ∩ Nn
A primary decomposition is a reduced primary decomposition if Pi = Pj for i = j
and no Ni can be excluded.

9.4. DECOMPOSITION OF MODULES 107
9.4.2. Lemma. Let R be a noetherian ring and M a finite module. For each Pi ∈
Ass(M) there is a submodule Ni ⊆ M such that Ass(Ni) = Ass(M) − {Pi} and
Ass(M/Ni)) = {Pi}. M injects
0 → M →
M/Ni
Proof. The submodule Ni is given by 9.2.6. Ass(∩Ni) = ∅, so conclusion by
9.2.2.
9.4.3. Proposition. Let R be a noetherian ring and M a finite module. A proper
submodule L ⊂ M has a reduced primary decomposition
and for any such
and
is exact.
Proof. Apply 9.4.2 to M/L.
L = N1 ∩ · · · ∩ Nn
Ass(M/L) = {P1, . . . , Pn}
0 → M/L →
M/Ni
9.4.4. Proposition. Let R be a noetherian ring and M a finite module. If
L = N1 ∩ · · · ∩ Nn
is a reduced primary decomposition of L ⊂ M and Pi is minimal in Ass(M/L),
then
and therefore uniquely determined.
Ni = M ∩ LPi
Proof. Clearly Ni ⊂ M ∩ LPi . By localization Ass(M ∩ LPi /Ni) = ∅. So
equality.
9.4.5. Proposition. Let R be a noetherian ring and M a finite module. Let L ⊂ M
such that M/L = 0 has finite length. If Ass(M/L) = {P1, . . . , Pn}, then there is
a reduced primary decomposition
where
and an isomorphism
L = N1 ∩ · · · ∩ Nn
Ni = M ∩ LPi
M/L
M/Ni
Proof. This follows from 9.4.3, 9.4.4 and 9.1.6.
9.4.6. Proposition. Let R be a noetherian ring and M a finite length module. If
Ass(M) = {P1, . . . , Pn}, then there is a reduced primary decomposition
where
0 = N1 ∩ · · · ∩ Nn
Ni = Pi ni M, M/Ni MPi
i
i
i

108 9. PRIMARY DECOMPOSITION
and isomorphisms
M
Proof. This follows from 7.4.5.
i
MPi
M/Pi niM 9.4.7. Proposition. Let R be a noetherian ring and M a finite module. Let L ⊂ M
have a reduced primary decomposition
L = N1 ∩ · · · ∩ Nn
where Ni is Pi-primary. Assume U to be a multiplicative subset disjoint from
exactly P1, . . . Pk. Then
is a reduced primary decomposition.
U −1 L = U −1 N1 ∩ · · · ∩ U −1 Nk
Proof. This follows from 9.3.7 and 9.4.3.
9.4.8. Exercise. (1) Describe the primary decomposition over a field.
9.5. Decomposition of ideals
9.5.1. Proposition. A proper ideal I ⊂ R has a reduced primary decomposition
and for any such
and
is exact.
Proof. This is a case of 9.4.3.
I = Q1 ∩ · · · ∩ Qn
Ass(R/I) = {P1, . . . , Pn}
0 → R/I →
R/Qi
9.5.2. Proposition. If
I = Q1 ∩ · · · ∩ Qn
is a reduced primary decomposition and Pi is minimal in Ass(R/I), then
and therefore uniquely determined.
Proof. This is a case of 9.4.4.
Qi = R ∩ IRPi
9.5.3. Definition. Let P be a prime ideal. The symbolic power of P is
P (n) = R ∩ P n RP
9.5.4. Proposition. Let
P n = Q1 ∩ · · · ∩ Qn
be a reduced primary decomposition of a power of a prime ideal P . If Ass(R/Q1) =
{P }, then
Q1 = P (n)
Proof. This is a case of 9.4.4.
i
i

9.5. DECOMPOSITION OF IDEALS 109
9.5.5. Proposition. Let I ⊂ R be a proper ideal in a noetherian ring such that R/I
has finite length. If Ass(R/I) = {P1, . . . , Pn}, then there is a reduced primary
decomposition
I = Q1 ∩ · · · ∩ Qn
where
and an isomorphism
Proof. This is a case of 9.4.3 and 9.5.4.
Qi = R ∩ QPi
R/I
R/Qi
9.5.6. Proposition. Let R be an artinian ring. If Ass(M) = {P1, . . . , Pn}, then
there is a reduced primary decomposition
where
and isomorphisms
Proof. This is a case of 9.4.6.
0 = Q1 ∩ · · · ∩ Qn
Qi = Pi ni , R/Qi RPi
R
i
RPi
i
R/Pi ni
9.5.7. Proposition. Let a proper ideal I ⊂ R in a noetherian ring have a reduced
primary decomposition
I = Q1 ∩ · · · ∩ Qn
where Qi is Pi-primary. Assume U to be a multiplicative subset disjoint from
exactly P1, . . . Pk. Then
is a reduced primary decomposition.
Proof. This is a case of 9.4.7.
U −1 I = U −1 Q1 ∩ · · · ∩ U −1 Qk
9.5.8. Example. Let R be a unique factorization domain. A factorization into
powers of different irreducible primes is a reduced primary decomposition of a
principal ideal.
9.5.9. Proposition. Let R be a noetherian ring. The following are equivalent
(1) R is reduced.
(2) RP is a field for all P ∈ Ass(R).
(3) RP is a domain for all P ∈ Ass(R).
Proof. (1) ⇒ (2): The maximal ideal P RP = 0. (3) ⇒ (1): This follows from
√ 0 = ∩P ∈Ass(R)P .
9.5.10. Corollary. Let R be a reduced noetherian ring. Then all elements in
Ass(R) are minimal primes. That is, there are no embedded primes.
9.5.11. Exercise. (1) Let I ⊂ R be an ideal. Show that if P = √ I is a maximal ideal,
then I is a P -primary ideal.
(2) Let I ⊂ R be an ideal. Show that if I contains a power of a maximal ideal P , then I
is a P -primary ideal.
i

110 9. PRIMARY DECOMPOSITION
(3) Let K be a field and I = (X 2 , XY ) ⊂ K[X, Y ]. Show that √ I = (X), but I is not
(X)-primary.

10
Dedekind rings
10.1. Principal ideal domains
10.1.1. Lemma. Let R be a domain. The set of elements x ∈ M in a module with
Ann(x) = 0 is a submodule.
Proof. The product of two nonzero elements is nonzero.
10.1.2. Definition. Let R be a domain. An element x ∈ M in a module is a torsion
element if Ann(x) = 0. By 10.1.1 the set of torsion elements is a submodule
T (M) ⊂ M, the torsion submodule. If T (M) = 0 then M is a torsion free
module.
10.1.3. Lemma. Let R be a domain.
(1) A flat module is torsion free.
(2) M/T (M) is torsion free.
(3) Let K be the fraction field. Then T (M) = Ker(M → M ⊗R K).
(4) If M → N → L is exact, then T (M) → T (N) → T (L) is exact.
(5) If U ⊂ R is multiplicative, then U −1 T (M) = T (U −1 M).
Proof. (1) If 0 = a ∈ R and F flat, then aF is injective. (2), (4), (5) These are
clear. (3) This follows from 4.4.1.
10.1.4. Corollary. Let R be a domain and M a module. The following are equivalent.
(1) M is torsion free.
(2) MP is torsion free for all prime ideals P .
(3) MP is torsion free for all maximal ideals P .
10.1.5. Lemma. Let R be a principal ideal domain. A submodule of a finite free
module is free.
Proof. Let F ⊂ R n be a submodule and p : R n → R the last projection. Then
p(F ) is a principal ideal and free. By induction F ∩ Ker p is free, so F F ∩
Ker p ⊕ p(F ) is free.
10.1.6. Proposition. Let R be a principal ideal domain.
(1) A torsion free module is flat.
(2) A finite torsion free module is free.
Proof. (1) For a nonzero ideal (a) ⊂ R the composite R (a) → R is aM.
For a torsion free F the homomorphism aF : F (a) ⊗R F → F is injective.
So F is flat by 3.7.12. (2) Let K be the fraction field. Let F ⊂ F ⊗R K be a
torsion free submodule and suppose x1, . . . , xn ∈ F give a basis for F ⊗R K.
Then F ′ = Rx1 ⊕ · · · ⊕ Rxn ⊂ F is a free submodule such that F/F ′ is a finite
111

112 10. DEDEKIND RINGS
torsion module. There is 0 = a ∈ Ann(F/F ′ ), so F aF ⊂ F ′ . Conclusion by
10.1.5.
10.1.7. Proposition. Let R be a principal ideal domain. A finite module M decomposes
M = T (M) ⊕ F
as a direct sum of the torsion submodule and a finite free submodule F .
Proof. The sequence 0 → T (M) → M → M/T (M) → 0 is split exact.
10.1.8. Proposition. Let R be a principal ideal domain. A finite torsion module
M has a primary decomposition
Where
M = ⊕ (p)Mp
Mp = {x ∈ M|p n x = 0, for some n}
Proof. This is the primary decomposition 9.4.6.
10.1.9. Proposition. Let R be a principal ideal domain and (p) an irreducible
principal ideal. A finite torsion module M such that M = Mp has decomposition
Where n1 ≥ · · · ≥ nk.
M = R/(p n1 ) ⊕ · · · ⊕ R/(p nk )
Proof. Let n1 = n be such that p n ∈ Ann(M), but p n−1 x = 0 for some x ∈ M.
The short exact sequence 0 → Rx → M → M/Rx → 0 of R/(p n )-modules
is split exact, since R/(p n ) Rx is an injective module 7.5.9. Conclusion by
induction on ℓR(M).
10.1.10. Exercise. (1) Show that a nonzero finite Z-submodule of Q is a free module
of rank 1.
(2) Show that a finite torsion Z-module is a finite group.
(3) Let K be a field. Show that a finite torsion K[X]-module is a finite K vector space.
(4) Let R be a noetherian domain such that every nonzero prime ideal is maximal. Let
M be a finite module. Show that the torsion submodule T (M) has finite length.
10.2. Discrete valuation rings
10.2.1. Definition. A local principal ideal domain, which is not a field, is a discrete
valuation ring. A generator of the maximal ideal is a local parameter or a
uniformizing parameter.
10.2.2. Proposition. Let (R, (p)) be a discrete valuation ring. Then
Proof. See 8.5.5.
∩n(p n ) = 0
10.2.3. Proposition. Let (R, (p)) be a discrete valuation ring. Any nonzero ideal
is of the form (p n ) for a unique n = 0, 1, 2, . . . .
Proof. Let 0 = x R, by 10.2.2 there is n such that x ∈ (p n ) − (p n+1 ). Since any
ideal is finitely generated it follows that a nonzero ideal is of the form (p n ). n is
unique by unique factorization.

10.3. DEDEKIND DOMAINS 113
10.2.4. Corollary. Let (R, (p)) be a discrete valuation ring. Any nonzero element
in the fraction field K of R has a unique representation up n where u is a unit and
n ∈ Z.
10.2.5. Definition. Let K be a field. A surjective map v : K\{0} → Z satisfying
(1) v(xy) = v(x) + v(x).
(2) If x + y = 0, v(x + y) ≥ min(v(x), v(y).
is a valuation on K.
10.2.6. Proposition. Let v be a valuation on a field K. Then
R = {x|v(x) ≥ 0}
is the discrete valuation ring of v. A local parameter is any element p, v(p) = 1.
Proof. Clear from the definition.
10.2.7. Proposition. Let (R, (p)) be a discrete valuation ring with fraction field
K. The map
v : K\{0} → Z, up n ↦→ n
is a valuation and R is the discrete valuation ring of v.
Proof. Clear from the definitions.
10.2.8. Proposition. Let R, (p) be a discrete valuation ring. A finite module M
has decomposition
Where n1 ≥ · · · ≥ nk > 0.
M = R/(p n1 ) ⊕ · · · ⊕ R/(p nk ) ⊕ R n
Proof. This is a case of 10.1.7 and 10.1.9.
10.2.9. Exercise. (1) Let K be a field. Show that the subring K[[X 2 , X 3 ]] ⊂ K[[X]]
is not a discrete valuation ring.
10.3. Dedekind domains
10.3.1. Definition. A noetherian domain R, which is not a field, is a Dedekind
domain if all local rings RP at nonzero prime ideals are discrete valuation rings.
10.3.2. Proposition. Let R be a Dedekind domain.
(1) Any nonzero prime ideal is maximal.
(2) If U ⊂ R is multiplicative, then U −1 R is a field or a Dedekind domain.
Proof. (1) (0), P are the only prime ideals in a prime ideal P . (2) This is clear
from 5.2.11.
10.3.3. Proposition. Let R be a domain which is not a field. The following conditions
are equivalent.
(1) R is a Dedekind domain.
(2) Every nonzero proper ideal in R is a product of finitely many maximal ideals.
Proof. The primary ideals P (n) = P n . Conclusion by 9.5.4 and Chinese remainders
1.4.2.
10.3.4. Proposition. Let R be Dedekind domain.
(1) If R is a unique factorization domain then it is a principal ideal domain.

114 10. DEDEKIND RINGS
(2) If R has only finitely many maximal ideals then it is a principal ideal domain.
Proof. (1) Any nonzero prime ideal is principal. Conclusion by 10.3.3. (2) Let
P, P2 . . . , Pn be the finitely many maximal ideals. Choose a ∈ P \P 2 ∪P2 · · ·∪Pn,
5.1.3. Then (a) is P primary. By 10.3.3 (a) = P k , so (a) = P . As all maximal
ideals are principal, conclusion by 10.3.3.
10.3.5. Proposition. Let R be Dedekind domain. An ideal I is generated by at
most two elements.
Proof. Let Ass(R/I) = {P1, . . . , Pn} and U = R\P1 ∪ · · · ∪ Pn, then by 10.3.4
U −1 R is a principal ideal domain. By 10.1.6 U −1 R U −1 I, so choose by 8.2.9 a
homomorphism f : R → I such that U −1 f is an isomorphism. Then f is injective
9.2.8 and the ideal f(R) = (a) ⊂ R satisfies: Pi /∈ Ass(R/(a)) for any i. Let
Q1, . . . , Qm ∈ Ass(I/(a)) and choose b ∈ I\Q1 ∪ · · · ∪ Qm. b is a nonzero
divisor on I/(a) and therefore b I/(a) is an isomorphism as I/(a) has finite length.
It follows that I = (a, b).
10.3.6. Theorem. Let R be Dedekind domain.
(1) A torsion free module is flat.
(2) A finite torsion free module is projective.
(3) Any ideal is projective.
(4) Let F be a finite torsion free module. Then there is a number n and an ideal
I such that
F R n ⊕ I
Proof. (1), (2), (3) These follow from 10.1.6, 6.5.13. (4) Let R have fraction field
K and assume rankK F ⊗RK = n+1. Choose a nonzero homomorphism F → R
and get by induction on n, F I1⊕· · ·⊕In+1 for nonzero ideals Ij in R. It suffices
to treat the case n = 1. Let Ass(R/I1) = {P1, . . . , Pm} and U = R\P1∪· · ·∪Pm,
then by 10.3.4 U −1 R is a principal ideal domain. By 10.1.6 U −1 I2 U −1 R, so
choose by 8.2.9 a homomorphism f : I2 → R such that U −1 f is an isomorphism.
Then f is injective 9.2.8 and the ideal f(I2) ⊂ R satisfies: Pi /∈ Ass(R/f(I2)) for
any i. It follows that I1 + f(I2) = R. Conclusion by a surjection I1 ⊕ I2 → R.
10.3.7. Proposition. Let R be a Dedekind domain. A finite module M decomposes
M = T (M) ⊕ F
as a direct sum of the torsion submodule and a finite torsion free submodule F .
Proof. By 10.3.6 the projection M → M/T (M) splits.
10.3.8. Proposition. Let R be a Dedekind domain. A finite torsion module M has
decomposition
M = R/P n1
nk
1 ⊕ · · · ⊕ R/Pk Where P1, . . . , Pk are not necessarily distinct maximal ideals.
Proof. This follows from 9.5.1 and 10.2.9.
10.3.9. Corollary. Let R be a Dedekind domain and M a finite module. Then M
has decomposition
M = R/P n1
1
⊕ · · · ⊕ R/P nk
k ⊕ Rn ⊕ Q1 · · · Ql
Where P1, . . . , Pk, Q1, . . . , Ql are not necessarily distinct maximal ideals.

10.3. DEDEKIND DOMAINS 115
10.3.10. Exercise. (1) Show that the ring Z[ √ −5] is a Dedekind domain.
(2) Show that the ring Z[ √ 5] is not a Dedekind domain.

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117

118 BIBLIOGRAPHY

A/B, 9
P -primary, 106
0-sequence, 39
abelian group, 9
addition, 9
annihilator, 27
artinian module, 87
artinian ring, 87
artinian rings, 87
ass of modules, 103
associated prime ideal, 103
associative, 9, 21
basis, 29
bilinear, 21
binomial formula, 10
canonical homomorphism, 59
canonical ring homomorphism, 57
Cayley-Hamilton’s theorem, 77
change of ring, 36
change of rings, 36
characteristic, 12
Chinese remainder theorem, 14
Chinese remainders, 14
coefficient, 16, 20
cofactor matrix, 75
cokernel, 25
colon ideal, 11, 27
comaximal ideals, 14
commutative, 9
composition series, 85
constants, 16
content of polynomial, 64
contracted ideal, 11
contravariant, 31
cosets, 9
Cramer’s rule, 76
decomposable, 30
decomposition of ideals, 108
decomposition of modules, 106
Dedekind domain, 113
Dedekind domains, 113
Index
119
degree, 16
derivative, 18
determinant, 75
direct product, 28
direct sum, 28, 29
discrete valuation ring, 112
discrete valuation ring of v, 113
discrete valuation rings, 112
distributive, 9
divisible module, 53
domain, 10
dual homomorphism, 33
dual module, 33
embedded prime, 105
essential extension, 54
evaluation, 32
evaluation map, 17
exact sequence, 39
exact sequences, 39
exactness and localization, 70
exactness of fractions, 60
exactness of hom, 48
exactness of tensor, 49
extended ideal, 11
factor group, 9
factor module, 24
factor ring, 11
faithfully flat, 71
faithfully flat ring homomorphism, 71
field, 10
field extension, 19
fields, 19
finite field extension, 19
finite ideal, 11
finite length, 85
finite module, 73
finite modules, 73
finite presented module, 80
finite presented modules, 80
finite ring extension, 83
finite ring homomorphism, 83
finite ring homomorphisms, 83
finite type ring, 17

120 INDEX
finite type rings, 95
finitely generated ring, 17
five lemma, 46
flat module, 54
flat modules, 54
flat ring homomorphism, 71
flat ring homomorphisms, 71
fraction field, 58
fractions and localization, 98
free module, 29
free modules, 75
Frobenius homomorphism, 12
functor, 31
Gauss’ lemma, 64
generated, 23
going-down, 72
going-up, 83
Gorenstein ring, 92
greatest common divisor, 16
Hilbert’s basis theorem, 95
homomorphism, 9, 21
homomorphism module, 30
homomorphism modules, 30
homomorphism modules of fractions, 63
ideal, 11
ideal generated by, 11
ideal product, 11
ideals, 11
idempotent, 15
identity, 9, 21
identity isomorphism, 9, 21
image, 25
indecomposable, 30
induced module, 37
injections, 28
injective envelope, 54
injective module, 52
injective modules, 52
irreducible element, 15
irreducible principal ideal, 15
isomorphism, 9, 21
Jacobson radical, 68
kernel, 11, 25
kernel and cokernel, 25
Krull’s intersection theorem, 96, 98
Krull’s theorem, 65
leading coefficient, 16
least common multiple, 16
length, 86
linear map, 22
local artinian ring, 91
local parameter, 112
local ring, 67
local ring homomorphism, 67
localization, 90
localization of modules, 68
localization of rings, 67
localized homomorphism, 68
localized module, 68
localized ring, 67
locally free module, 69
maximal ideal, 13
minimal prime, 101
minimal prime ideal, 66
minor, 75
module, 21
module of fractions, 59
modules and homomorphisms, 21
modules and submodules, 93
modules of fractions, 58
monic polynomial, 16
monomial, 16
multiplication, 9
multiplication of principal ideals, 15
multiplicative subset, 57
multiplicity, 18
Nakayama’s lemma, 78
natural homomorphism, 31
natural isomorphism, 31
negative, 9
nilpotent, 14
nilradical, 14
noetherian module, 93
noetherian ring, 94
noetherian rings, 94
noncommutative ring, 9
nontrivial idempotent, 15
nonzero divisor, 10, 22
order, 20
polynomial, 16
polynomial ring, 16
polynomials, 16
power series, 20
power series ring, 20
power series rings, 97
primary decomposition, 106
primary modules, 106
primary submodule, 106
prime fields, 19
prime filtrations of modules, 98
prime ideal, 13
prime ideals, 13, 65
principal ideal, 11
principal ideal domain, 15
principal ideal domains, 111
product ring, 10
projection, 9, 24

projections, 28
projective module, 50
projective modules, 50
proper ideal, 11
radical, 14
rank, 76
reduced, 14
reduced primary decomposition, 106
reflexive module, 33
residue field, 67
residue homomorphism, 68
restriction of scalars, 22
retraction, 41
ring, 9
ring extension, 9
ring generated, 17
ring of fractions, 57
rings, 9
rings of fractions, 57
root, 18
roots, 18
scalar multiplication, 21, 22
section, 41
short exact sequence, 40
simple module, 85
simple modules, 85
simple root, 18
snake homomorphism, 44
snake lemma, 45
spectrum, 101
split exact sequence, 42
standard basis, 29
subfield, 19
subgroup, 9
submodule, 21
submodule generated, 23
submodules and factor modules, 23
subring, 9
sum and product, 28
support, 101
support of modules, 101
symbolic power, 108
tensor modules of fractions, 62
tensor product, 33
tensor product modules, 33
tensor product ring, 38
the length, 85
the polynomial ring is factorial, 64
the snake lemma, 43
torsion element, 111
torsion free module, 111
torsion submodule, 111
total ring of fractions, 58
uniformizing parameter, 112
INDEX 121
unique factorization, 15
unique factorization domain, 15
unit, 10
valuation, 113
vector space, 22
windmill lemma, 46
zero, 9
zero divisor, 10, 22
zero ideal, 11
zero module, 21
zero submodule, 21