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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.