# relation of annihilators on exact sequence

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.

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I think you mean the intersection of the two ideals on the right? The union of ideals is not in general an ideal. – John Stalfos Sep 11 '12 at 1:42
sorry missprint then is it true, John?, I wonder the proof. – Sang Cheol Lee Sep 11 '12 at 1:56

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.

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

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.

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thanks Matt. if ring is a polynomial ring, is it true? or is $V({\rm Ann}(M))= V({\rm Ann}(M^')) \cup V({\rm Ann}(M^{''}))$ where V means a zero set. – Sang Cheol Lee Sep 11 '12 at 3:12
Dear Sang, No; just replace $\mathbb Z$ by $k[t]$ and $p$ by $t$. As I wrote, in general, it is a delicate question in any particular case, related to the geometry of the situation. Regards, – Matt E Sep 11 '12 at 3:16
@Sang: Dear Sang, The statement about supports is true, though. I will add this to my answer. Regards, – Matt E Sep 11 '12 at 3:18
thank you Matt! – Sang Cheol Lee Sep 11 '12 at 3:50
@MattE is there a typo in your example of a non-split SES? The only map $\mathbb{Z}/p^2\rightarrow \mathbb{Z}$ is 0, so I can't see the exactness in the middle-nor on the right. – Kevin Carlson Sep 11 '12 at 13:00