Lemma 3

Let $A$ be a noetherian ring, $x$ an element of $A$, $M$ a finitely generated $A$-module, and $p$ a prime ideal of $A$ associated with $M$. Suppose that the homothety $x_M$ is injective. Let $q$ be a prime ideal of $A$, minimal among those containing $p + xA$. Then $q$ is associated with the $A$-module $M/xM$.

Let $N$ denote the submodule of $M$ formed by the elements $m$ such that $pm = 0$. We have $N \cap xM = xN$; in fact, if an element $m$ of $M$ is such that $pxm = 0$, then $pm = 0$ since $x_M$ is injective, hence $m \in N$. Consequently, the $A$-module $N/xN$ is isomorphic to the submodule $(N + xM)/xM$ of $M/xM$, and it is enough to prove that $q$ is associated with $N/xN$. Since $p$ is associated with $M$, there exists an element $m$ of $M$ such that $p = \mathrm{Ann}(m)$; we have $m \in N$, whence $p = \mathrm{Ann}(N)$ and therefore $\mathrm{Supp}(N/xN) = V(p + xA)$ by II, § 4, No. 4, corollary to prop. 18; consequently, $q$ is associated with $N/xN$ (IV, § 1, No. 4, th. 2).