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

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.