The Stacks project

Lemma 47.11.1. Let $A$ be a Noetherian ring, let $I \subset A$ be an ideal, and let $M$ be a finite $A$-module such that $IM \not= M$. Then the following integers are equal:

  1. $\text{depth}_ I(M)$,

  2. the smallest integer $i$ such that $\mathop{\mathrm{Ext}}\nolimits _ A^ i(A/I, M)$ is nonzero, and

  3. the smallest integer $i$ such that $H^ i_ I(M)$ is nonzero.

Moreover, we have $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(N, M) = 0$ for $i < \text{depth}_ I(M)$ for any finite $A$-module $N$ annihilated by a power of $I$.

Proof. We prove the equality of (1) and (2) by induction on $\text{depth}_ I(M)$ which is allowed by Algebra, Lemma 10.71.4.

Base case. If $\text{depth}_ I(M) = 0$, then $I$ is contained in the union of the associated primes of $M$ (Algebra, Lemma 10.62.9). By prime avoidance (Algebra, Lemma 10.14.2) we see that $I \subset \mathfrak p$ for some associated prime $\mathfrak p$. Hence $\mathop{\mathrm{Hom}}\nolimits _ A(A/I, M)$ is nonzero. Thus equality holds in this case.

Assume that $\text{depth}_ I(M) > 0$. Let $f \in I$ be a nonzerodivisor on $M$ such that $\text{depth}_ I(M/fM) = \text{depth}_ I(M) - 1$. Consider the short exact sequence

\[ 0 \to M \to M \to M/fM \to 0 \]

and the associated long exact sequence for $\mathop{\mathrm{Ext}}\nolimits ^*_ A(A/I, -)$. Note that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I, M)$ is a finite $A/I$-module (Algebra, Lemmas 10.70.9 and 10.70.8). Hence we obtain

\[ \mathop{\mathrm{Hom}}\nolimits _ A(A/I, M/fM) = \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, M) \]

and short exact sequences

\[ 0 \to \mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I, M) \to \text{Ext}^ i_ A(A/I, M/fM) \to \mathop{\mathrm{Ext}}\nolimits ^{i + 1}_ A(A/I, M) \to 0 \]

Thus the equality of (1) and (2) by induction.

Observe that $\text{depth}_ I(M) = \text{depth}_{I^ n}(M)$ for all $n \geq 1$ for example by Algebra, Lemma 10.67.9. Hence by the equality of (1) and (2) we see that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I^ n, M) = 0$ for all $n$ and $i < \text{depth}_ I(M)$. Let $N$ be a finite $A$-module annihilated by a power of $I$. Then we can choose a short exact sequence

\[ 0 \to N' \to (A/I^ n)^{\oplus m} \to N \to 0 \]

for some $n, m \geq 0$. Then $\mathop{\mathrm{Hom}}\nolimits _ A(N, M) \subset \mathop{\mathrm{Hom}}\nolimits _ A((A/I^ n)^{\oplus m}, M)$ and $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(N, M) \subset \text{Ext}^{i - 1}_ A(N', M)$ for $i < \text{depth}_ I(M)$. Thus a simply induction argument shows that the final statement of the lemma holds.

Finally, we prove that (3) is equal to (1) and (2). We have $H^ p_ I(M) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ p_ A(A/I^ n, M)$ by Lemma 47.8.2. Thus we see that $H^ i_ I(M) = 0$ for $i < \text{depth}_ I(M)$. For $i = \text{depth}_ I(M)$, using the vanishing of $\mathop{\mathrm{Ext}}\nolimits _ A^{i - 1}(I/I^ n, M)$ we see that the map $\mathop{\mathrm{Ext}}\nolimits _ A^ i(A/I, M) \to H_ I^ i(M)$ is injective which proves nonvanishing in the correct degree. $\square$


Comments (0)


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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AVZ. Beware of the difference between the letter 'O' and the digit '0'.