relation of annihilators on exact sequence

Question

Let $0\to M' \to M \rightarrow M''\to 0$ be an exact sequence of modules. I want to show that ${\rm Ann}(M)= {\rm Ann}(M')\cap {\rm Ann}(M^{''})$.

The "$\subset$" case I have shown, but I can't show the "$\supset$" case.

Answer 1

This won't be true in general; it is related to the possibility of the short exact sequence being non-split.

If $M = M'\oplus M,$ then your equation is true [easy exercise].

But consider the simplest example of a non-split short exact sequence, such as $0 \to \mathbb Z/p \to \mathbb Z/p^2 \to \mathbb Z/p \to 0.$ In this case your question is false.

In general (i.e. for modules over more general rings), the question of your equation holds in any particular case can be quite delicate.

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Note though that obviously $\mathrm{Ann}(M)\mathrm{Ann}(M'') \subset\mathrm{Ann}(M),$ and so combining this with $\mathrm{Ann}(M) \subset \mathrm{Ann}(M') \cap \mathrm{Ann}(M''),$ we find that $V(\mathrm{Ann}(M)) = V(\mathrm{Ann}(M')) \cup V(\mathrm{Ann}(M'')).$

One could also think about this in terms of localizing at prime ideals: $M_{\mathfrak p}$ is non-zero if and only at least one of $M'_{\mathfrak p}$ or $M''_{\mathfrak p}$ is non-zero, since localization is exact.

One could also think geometrically, in terms of supports and stalks.