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

104 9. PRIMARY DECOMPOSITION
9.2.6. Proposition. Let M be a module and P ⊂ Ass(M). Then there is a submodule
N ⊂ M such that
Ass(N) = P , Ass(M/N) = Ass(M)\P
Proof. Choose by Zorn’s lemma N maximal in the set of submodules N ′ ⊂ M
for which Ass(N ′ ) ⊂ P. Let Q ∈ Ass(M/N) and choose N ⊂ L ⊂ M with
L/N R/Q. Then Ass(L) ⊂ P ∪ {Q}. By maximality of N follows that
Ass(L) ⊂ P. So Q /∈ P and Q ∈ Ass(L) ⊂ Ass(M).
9.2.7. Proposition. Let R be a noetherian ring and M a module. Then M = 0 if
and only if Ass(M) = ∅.
Proof. If M = 0, then by 8.6.2 Ass(M) = ∅.
9.2.8. Corollary. Let R be a noetherian ring and f : M → N a homomorphism.
The following conditions are equivalent.
(1) f is injective.
(2) fP is injective for all prime ideals P ∈ Ass(M).
Proof. Use 9.2.7 on Ker f.
9.2.9. Corollary. Let R be a noetherian ring and M a module. Then a ∈ R is a
nonzero divisor on M if and only if
a /∈ ∪ P ∈Ass(M)P
The set of zero divisors on M is ∪ P ∈Ass(M)P .
Proof. aM is injective if and only if aMP
This happens when a /∈ ∪ P ∈Ass(M)P .
is injective for all P ∈ Ass(M), 8.2.8.
9.2.10. Proposition. Let U ⊂ R be a multiplicative subset and M a module.
(1) In general
(2) If R is noetherian
Ass(M) ∩ Spec(U −1 R) ⊂ Ass U −1 R(U −1 M)
Ass(M) ∩ Spec(U −1 R) = Ass U −1 R(U −1 M)
Proof. Let P be a prime ideal. For any homomorphism R/P → M there is a
commutative diagram
R/P
M
U −1R/P
U −1M (1) If P ∈ Ass(M) and P ∩ U = ∅ then U −1 P ∈ Ass(U −1 M). (2) If U −1 P ∈
Ass(U −1 M), then P ∩ U = ∅. By 8.2.9 there is a diagram as above. R/P → M
is injective by 9.2.8. So P ∈ Ass(M).
9.2.11. Proposition. Let R be a noetherian ring and M a module. Then any minimal
prime P ∈ Supp(M) is contained in Ass(M).
Proof. Assume P ∈ Supp(M) is minimal. Then the RP -module MP has support
exactly in the maximal ideal, so {P RP } = Ass(MP ). Conclusion by 9.2.10.