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!



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


