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

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9.2. ASS OF MODULES 105<br />

9.2.12. Definition. A non minimal prime ideal in Ass(M) is an embedded prime<br />

<strong>of</strong> M.<br />

9.2.13. Proposition. Let R be a noetherian ring and M a finite module. Then<br />

Ass(M) is a finite set.<br />

Pro<strong>of</strong>. Follows immediately from 9.2.4 and 8.1.5.<br />

9.2.14. Corollary. Let R be a noetherian ring and M a finite module. The following<br />

are equivalent<br />

(1) M has finite length.<br />

(2) Supp(M) consists <strong>of</strong> maximal ideals.<br />

(3) Ass(M) consists <strong>of</strong> maximal ideals.<br />

Pro<strong>of</strong>. (3) ⇒ (2): The minimal ideals in the support are maximal.<br />

9.2.15. Lemma. Let (R, P ) be a local ring M a module. Then P ∈ Ass(M) if<br />

and only if HomR(k(P ), M) = 0.<br />

9.2.16. Proposition. Let R be a noetherian ring and M a finite module. For any<br />

module N<br />

Ass(HomR(M, N)) = Supp(M) ∩ Ass(N)<br />

Pro<strong>of</strong>. By 8.2.9 HomR(M, N)P HomRP (MP , NP ). So reduce to the case<br />

where (R,P) is local. Now<br />

HomR(k(P ), HomR(M, N)) = HomR(M, HomR(k(P ), N))<br />

Conclusion by Nakayama’s lemma 6.4.1 and 9.2.15.<br />

= Hom k(P )(M ⊗R k(P ), HomR(k(P ), N))<br />

9.2.17. Proposition. Let R be a noetherian ring and F a finite module. Assume<br />

rank F ⊗R k(P ) = n for all primes P . Then F is locally free (projective) if and<br />

only if FP is free for all P ∈ Ass(R).<br />

Pro<strong>of</strong>. Let Q be a maximal ideal and 0 → K → R n Q → FQ → 0 exact. Ass(K) ⊂<br />

Ass(R), so KP = 0 for all P ∈ Ass(K). By 9.2.7 K = 0.<br />

9.2.18. Proposition. Let (R, P ) be a noetherian local ring. If there is a nonzero<br />

finite injective module E, then R is artinian.<br />

Pro<strong>of</strong>. Let Q be a prime and f : R/Q → E. If a ∈ P \Q then a R/Q is injective,<br />

so there is f ′ : R/Q → E such that f = f ′ ◦ a R/Q. That is P HomR(R/Q, E) =<br />

HomR(R/Q, E), so by Nakayama’s lemma 6.4.1 HomR(R/Q, E) = 0 if Q = P .<br />

By 9.2.7 0 = HomR(R/P, E) ⊂ HomR(R/Q, E), so Q = P . R is artinian by<br />

8.6.6.<br />

9.2.19. Exercise. (1) Show that<br />

(2) Show that<br />

Ass(Z/(n)) = {(p)|p prime dividing n}<br />

Ass(K[X, Y ]/(X) ∩ (X 2 , Y 2 )) = {(X), (X, Y )}<br />

and point out an embedded prime.<br />

(3) Let I ⊂ R be an ideal such that √ I = I. Show that R/I has no embedded prime<br />

ideals.

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