This is repeat post from relation of annihilators on exact sequence and the hint seems unclear.
$0 \rightarrow M^{\prime} \rightarrow M \rightarrow M^{\prime \prime} \rightarrow 0$ is an exact sequence of $k[x_0, \dotsc, x_n]$-modules. To prove: ${\mathrm{Ann}}~M = {\mathrm{Ann}}~{M^{\prime}} \cap {\mathrm{Ann}}~{M^{\prime \prime}}$.
The part "$\subseteq$" is obvious. For the other side one require some property of $k[x_0, \dotsc, x_n]$, as even a ring being a PID doesn't help.
The discussion of the above post gear towards the fact ${\mathrm{Supp}}~M = {\mathrm{Supp}}~M^{\prime} \cup {\mathrm{Supp}}~M^{\prime \prime}$, for which a ring being Noetherian is enough (over finitely generated modules). However, the counterexample shows it is not true over ${\mathbb{Z}}$. What am I missing?
The question seems to be a tiny step from Hilbert-Serre in Hartshorne (Theorem 7.5).
Best Answer
It's not true. Let $A=k[x_0]$. One can put $M=A/x^2A$ into the middle of a short exact sequence flanked by two copies of $A/xA$. In this case the annihilator of $M$ is $x^2A$ but those of $M'$ and $M''$ are both $xA$.
(This is really the same example as Matthew Emerton's in the linked post).