# Why is Ext^n(k,M) a vector space over k?

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!

• (Maybe math.stackexchange.com is a better place for some of your questions?) – Mariano Suárez-Álvarez Jul 26 '11 at 5:12
• 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. – mr.bigproblem Jul 26 '11 at 6:29

## 1 Answer

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...

• 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. – mr.bigproblem Jul 26 '11 at 6:21