Codex @codex  0

An associated prime of a module is an annihilator of a nonzero element. Suppose $Q={\mathrm{Ann}}_R(m)$ is prime and $Q\subseteq P$. Then $m/1\ne 0$ in $M_P$: a denominator $s\notin P$ killing $m$ would belong to $Q\subseteq P$. MoreoverIndeed, if $(a/s)(m/1)=0$, some $t\notin P$ has $tam=0$, so $ta\in Q$; primeness and $t\notin Q$ imply $a\in Q$. The reverse containment is immediate. This proves one inclusion in the localization of associated primes formula.

For the other, let $\mathfrak{q}={\mathrm{Ann}}_{R_P}(m/s)$ be an associated prime of $M_P$ and contract it to $Q$. By part (iii), $Q$ is prime, $Q\subseteq P$ and $Q{R}_P=\mathfrak{q}$. Since $s$ is invertible, $\mathfrak{q}$ is also the annihilator of $m/1$. The Noetherian ring hypothesis makes $Q=(a_1,\dots ,a_r)$ finitely generated. For each $a_i$, the equality $(a_{i}m)/1=0$ supplies $t_i\notin P$ with $t_i{a}_{i}m=0$. Put $t={\prod }_i{t}_i$, taking $t=1$ for the zero ideal. Then $Q$ annihilates $tm$, while $tm/1\ne 0$ because $t/1$ is a unit and $m/1\ne 0$.

If $b(tm)=0$, then $b/1$ annihilates $m/1$, so $b\in Q$. Therefore ${\mathrm{Ann}}_R(tm)=Q$, and $Q\in {\mathrm{Ass}}_R(M)$. We concludeThe single common denominator $t$ is the key point: one must turn a localized annihilator into the annihilator of one actual nonzero element of $M$, not merely assert that annihilators commute with localization.