Let $A$ be a commutative ring, $I \lhd A$ (ideal of $A$) and $M$ be a finitely generated $A$-module. Then we have the following identity: $$\sqrt{\operatorname{Ann }_A(M/IM)} = \sqrt{\operatorname{Ann}_A (M) + I},$$ where $\operatorname{Ann }_A(N) = \{a\in A: na = 0 \quad \forall n\in N\}$ denotes the annihilator of $N$ in $R$.
One direction is easy: If $r = a + s \in \operatorname{Ann}_A (M) + I$, then for each $m\in M$ $$(m+ IM)r = ma + ms + IM = ms + IM = 0 + IM,$$ so $\operatorname{Ann}_A (M) + I \subseteq \operatorname{Ann }_A(M/IM)$ and taking radicals yields $"\supseteq "$. However, I am struggling with the converse inclusion:
Let $r \in \sqrt{\operatorname{Ann }_A(M/IM)}$. Then there exists some $n\in \mathbb N$ such that $r^n \in \operatorname{Ann }_A(M/IM)$. But that means $r^n m \in IM$ for each $m\in M$. Now consider the $R$-module homomorphism $$\varphi \colon M \to M; \quad \varphi(m) := r^n m.$$ Then as noted $\varphi(M) \subseteq IM$. By the Cayley-Hamilton theorem there is $k\in \mathbb N$ and coefficients $a_0, \dots, a_{k-1} \in I$ such that $$\varphi^{k} + a_{k-1}\varphi^{k-1} + \dots + a_1 \varphi + a_0 = 0.$$ Does this get me anywhere? Plugging in $m\in M$ in this equation did not give me any insight. Any hints are appreciated!
I came up with the answer: Note that it suffices to show $\operatorname{Ann }_A(M/IM) \subseteq \sqrt{\operatorname{Ann }_A(M) + I}$, since for ideals $\mathfrak a \subseteq \mathfrak b \subseteq A$ one always has $\sqrt{\mathfrak a} \subseteq \sqrt{\mathfrak b} $ and $\sqrt{\sqrt{ \mathfrak b}} = \sqrt{\mathfrak b}$.
Now take $r\in \operatorname{Ann }_A(M/IM)$. Then $rM \subseteq IM$, so by Cayley-Hamilton we have $$(r^k + a_{k-1} r^{k-1} + \dots + a_1 r + a_0)M = 0$$ for some $k\in \mathbb N$ and $a_i \in I$. Putting $a:= a_{k-1}r^{k-1} + \dots + a_1 r + a_0 \in I$ this is equivalent to $$r^k + a \in \operatorname{Ann }_R(M).$$ But this means $r^k \in \operatorname{Ann }_R(M) + I$ and $r \in \sqrt{\operatorname{Ann }_A(M) + I}$, as desired.