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