1
$\begingroup$

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.

  • 0
    I think you mean the intersection of the two ideals on the right? The union of ideals is not in general an ideal.2012-09-11
  • 0
    sorry missprint then is it true, John?, I wonder the proof.2012-09-11

1 Answers 1

2

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.

  • 0
    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.2012-09-11
  • 0
    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,2012-09-11
  • 0
    @Sang: Dear Sang, The statement about supports is true, though. I will add this to my answer. Regards,2012-09-11
  • 0
    @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.2012-09-11
  • 0
    @KevinCarlson: Dear Kevin, Thanks for pointing this out. Regards,2012-09-12