Commutative algebra - Department of Mathematical Sciences - old ...

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).