Associated primes of a localized finite module come from upstairs

By [L1], choose $u\in S^{-1}M$ with $u\mathrm{\ne }0$ and ${\mathrm{Ann}}_{S^{-1}R}(u)=\mathfrak{q}$. Write $u=m\mathrm{/}s$ with $m\in M$ and $s\in S$. Since $1\mathrm{/}s$ is a unit of $S^{-1}R$, the element $m\mathrm{/}1=(s\mathrm{/}1)u$ is also nonzero and has the same annihilator $\mathfrak{q}$. By [L4], there is a prime ideal $\mathfrak{p}\subseteq R$ with $\mathfrak{p}\cap S=\mathrm{\varnothing }$ and $\mathfrak{q}=S^{-1}\mathfrak{p}$. If $am=0$, then $a\mathrm{/}1$ annihilates $m\mathrm{/}1$, so $a\mathrm{/}1\in \mathfrak{q}$ and therefore $a\in \mathfrak{p}$. Thus ${\mathrm{Ann}}_R(m)\subseteq \mathfrak{p}$.

By [L3], write $\mathfrak{p}=(f_1,\dots ,f_n)$. Since each $f_i\mathrm{/}1\in \mathfrak{q}={\mathrm{Ann}}_{S^{-1}R}(m\mathrm{/}1)$, there exists $s_i\in S$ with $s_i{f}_{i}m=0$ in $M$. Put $h=s_1\cdots s_n$. Then $hm\mathrm{\ne }0$, because $h\mathrm{/}1$ is a unit and $(h\mathrm{/}1)(m\mathrm{/}1)=hm\mathrm{/}1$. Also each generator $f_i$ kills $hm$, so $\mathfrak{p}\subseteq {\mathrm{Ann}}_R(hm)$.

If $a\in {\mathrm{Ann}}_R(hm)$, then $a\mathrm{/}1$ annihilates $hm\mathrm{/}1$. Since $h\mathrm{/}1$ is a unit, $a\mathrm{/}1$ also annihilates $m\mathrm{/}1$, so $a\mathrm{/}1\in \mathfrak{q}$ and therefore $a\in \mathfrak{p}$ by step 1.1. Hence ${\mathrm{Ann}}_R(hm)\subseteq \mathfrak{p}$, and step 2.1 gives ${\mathrm{Ann}}_R(hm)=\mathfrak{p}$.

The prime ideal $\mathfrak{p}$ is therefore the annihilator of the nonzero element $hm\in M$, so $\mathfrak{p}\in {\mathrm{Ass}}_R(M)$ by [L1]. Together with step 1.1, this proves the claim.