Lemma 10.72.5. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a nonzero finite $R$-module. Then $\text{depth}(M)$ is equal to the smallest integer $i$ such that $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(R/\mathfrak m, M)$ is nonzero.
Proof. Let $\delta (M)$ denote the depth of $M$ and let $i(M)$ denote the smallest integer $i$ such that $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(R/\mathfrak m, M)$ is nonzero. We will see in a moment that $i(M) < \infty $. By Lemma 10.63.18 we have $\delta (M) = 0$ if and only if $i(M) = 0$, because $\mathfrak m \in \text{Ass}(M)$ exactly means that $i(M) = 0$. Hence if $\delta (M)$ or $i(M)$ is $> 0$, then we may choose $x \in \mathfrak m$ such that (a) $x$ is a nonzerodivisor on $M$, and (b) $\text{depth}(M/xM) = \delta (M) - 1$. Consider the long exact sequence of Ext-groups associated to the short exact sequence $0 \to M \to M \to M/xM \to 0$ by Lemma 10.71.6:
Since $x \in \mathfrak m$ all the maps $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(\kappa , M) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(\kappa , M)$ are zero, see Lemma 10.71.8. Thus it is clear that $i(M/xM) = i(M) - 1$. Induction on $\delta (M)$ finishes the proof. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)