0
$\begingroup$

This might be a trivial question to experts but not to me whatsoever. Suppose that $(R,m,k)$ is a Noetherian local ring, $M$ is an $R$-finite module whose depth is $n$. One then defines the type of $M$ by the formula (as in the text "Cohen-Macaulay Rings" of Bruns and Herzog): $$ \tau(M) = \mbox{dim}_k\mbox{Ext}_R^n(k,M). $$ This definition only makes sense if $\mbox{Ext}_R^n(k,M)$ is a vector space over $k$. One knows that $\mbox{Ext}_R^n(k,M)$ is an $R$-module but why is it a vector space over $k$ then? I guess one needs to verify that $m \subset ann(\mbox{Ext}_R^n(k,M))$ but this is not evident to me at all. Thanks for any comment and answer!

$\endgroup$
2
  • 3
    $\begingroup$ (Maybe math.stackexchange.com is a better place for some of your questions?) $\endgroup$ Jul 26, 2011 at 5:12
  • 2
    $\begingroup$ Dear Mariano, thanks for your suggestion. I've just recently been exposed to online math forums. This mathoverflow is the only site I knew, now you tell me another one. $\endgroup$ Jul 26, 2011 at 6:29

1 Answer 1

4
$\begingroup$

The action of an element $r\in R$ on $\mbox{Ext}_R^n(k,M)$ is the map $\mbox{Ext}_R^n(k,M)\to \mbox{Ext}_R^n(k,M)$ which is induced by either the map $k\to k$ given by multiplication by $r$, or by the map $M\to M$ given by multiplication by $r$ (the two induced maps are the same) Now, if $r\in\mathfrak m$ then the map $k\to k$ is zero, so...

$\endgroup$
1
  • 1
    $\begingroup$ The same answer can be found on Weibel's "An introduction to Homological Algebra", chapter 3. I should have used this book instead of Hilton-Stambach. Thanks. $\endgroup$ Jul 26, 2011 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.