Proposition 6

Let A be a Noetherian ring, M an A-module, S a multiplicative subset of A and Y the set of elements of $\mathrm{Ass}_A(M)$ which do not meet S. Then the kernel N of the canonical mapping $M \to S^{-1}M$ is the unique submodule of M which satisfies the relations

By Proposition 4 of no. 1, there exists a submodule $N'$ of M which satisfies the relations $\mathrm{Ass}(N') = \mathrm{Ass}(M) - Y$ and $\mathrm{Ass}(M/N') = \Psi$. We need to prove $N = N'$. Consider the commutative diagram

where $p, u, v$ are the canonical homomorphisms. We shall show that $S^{-1}p$ and $v$ are injective, which will prove that $u$ and $p$ have the same kernel and hence $N = N'$.

As $\mathrm{Ass}(N') \cap \Psi = \varnothing$, every element of $\mathrm{Ass}(N')$ meets S. Then $\mathrm{Ass}_{S^{-1}A}(S^{-1}N') = \varnothing$ (Corollary to Proposition 5), whence $S^{-1}N' = \{0\}$ (no. 1, Corollary 1 to Proposition 2), which proves that $S^{-1}p$ is injective (Chapter II, §2, no. 4, Theorem 1). On the other hand, if $x$ belongs to the kernel K of $v$, then $\mathrm{Ann}(x) \cap S \neq \varnothing$ (Chapter II, §2, no. 2, Proposition 4); hence $\mathrm{Ass}(K) = \varnothing$ since $\mathrm{Ass}(K) \subset \mathrm{Ass}(M/N') = \Psi$; we deduce that $K = \{0\}$ (no. 1, Corollary 1 to Proposition 2) and $v$ is injective.