The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

45.11 Depth

In this section we revisit the notion of depth introduced in Algebra, Section 10.71.

Lemma 45.11.1. 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 10.67.8. 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 45.8.2. 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. $\square$

Lemma 45.11.2. Let $A$ be a Noetherian ring. Let $0 \to N' \to N \to N'' \to 0$ be a short exact sequence of finite $A$-modules. Let $I \subset A$ be an ideal.

  1. $\text{depth}_ I(N) \geq \min \{ \text{depth}_ I(N'), \text{depth}_ I(N'')\} $

  2. $\text{depth}_ I(N'') \geq \min \{ \text{depth}_ I(N), \text{depth}_ I(N') - 1\} $

  3. $\text{depth}_ I(N') \geq \min \{ \text{depth}_ I(N), \text{depth}_ I(N'') + 1\} $

Proof. Assume $IN \not= N$, $IN' \not= N'$, and $IN'' \not= N''$. Then we can use the characterization of depth using the Ext groups $\mathop{\mathrm{Ext}}\nolimits ^ i(A/I, N)$, see Lemma 45.11.1, and use the long exact cohomology sequence

\[ \begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ A(A/I, N') \to \mathop{\mathrm{Hom}}\nolimits _ A(A/I, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(A/I, N'') \\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, N') \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, N'') \to \ldots \end{matrix} \]

from Algebra, Lemma 10.70.6. This argument also works if $IN = N$ because in this case $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I, N) = 0$ for all $i$. Similarly in case $IN' \not= N'$ and/or $IN'' \not= N''$. $\square$

Lemma 45.11.3. Let $A$ be a Noetherian ring, let $I \subset A$ be an ideal, and let $M$ a finite $A$-module with $IM \not= M$.

  1. If $x \in I$ is a nonzerodivisor on $M$, then $\text{depth}_ I(M/xM) = \text{depth}_ I(M) - 1$.

  2. Any $M$-regular sequence $x_1, \ldots , x_ r$ in $I$ can be extended to an $M$-regular sequence in $I$ of length $\text{depth}_ I(M)$.

Proof. Part (2) is a formal consequence of part (1). Let $x \in I$ be as in (1). By the short exact sequence $0 \to M \to M \to M/xM \to 0$ and Lemma 45.11.2 we see that $\text{depth}_ I(M/xM) \geq \text{depth}_ I(M) - 1$. On the other hand, if $x_1, \ldots , x_ r \in I$ is a regular sequence for $M/xM$, then $x, x_1, \ldots , x_ r$ is a regular sequence for $M$. Hence (1) holds. $\square$

Lemma 45.11.4. Let $R$ be a Noetherian local ring. If $M$ is a finite Cohen-Macaulay $R$-module and $I \subset R$ a nontrivial ideal. Then

\[ \text{depth}_ I(M) = \dim (\text{Supp}(M)) - \dim (\text{Supp}(M/IM)). \]

Proof. We will prove this by induction on $\text{depth}_ I(M)$.

If $\text{depth}_ I(M) = 0$, then $I$ is contained in one of the associated primes $\mathfrak p$ of $M$ (Algebra, Lemma 10.62.18). Then $\mathfrak p \in \text{Supp}(M/IM)$, hence $\dim (\text{Supp}(M/IM)) \geq \dim (R/\mathfrak p) = \dim (\text{Supp}(M))$ where equality holds by Algebra, Lemma 10.102.7. Thus the lemma holds in this case.

If $\text{depth}_ I(M) > 0$, we pick $x \in I$ which is a nonzerodivisor on $M$. Note that $(M/xM)/I(M/xM) = M/IM$. On the other hand we have $\text{depth}_ I(M/xM) = \text{depth}_ I(M) - 1$ by Lemma 45.11.3 and $\dim (\text{Supp}(M/xM)) = \dim (\text{Supp}(M)) - 1$ by Algebra, Lemma 10.62.10. Thus the result by induction hypothesis. $\square$

Lemma 45.11.5. Let $R \to S$ be a flat local ring homomorphism of Noetherian local rings. Denote $\mathfrak m \subset R$ the maximal ideal. Let $I \subset S$ be an ideal. If $S/\mathfrak mS$ is Cohen-Macaulay, then

\[ \text{depth}_ I(S) \geq \dim (S/\mathfrak mS) - \dim (S/\mathfrak mS + I) \]

Proof. By Algebra, Lemma 10.98.3 any sequence in $S$ which maps to a regular sequence in $S/\mathfrak mS$ is a regular sequence in $S$. Thus it suffices to prove the lemma in case $R$ is a field. This is a special case of Lemma 45.11.4. $\square$

Lemma 45.11.6. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $M$ be an $A$-module. Let $Z = V(I)$. Then $H^0_ I(M) = H^0_ Z(M)$. Let $N$ be the common value and set $M' = M/N$. Then

  1. $H^0_ I(M') = 0$ and $H^ p_ I(M) = H^ p_ I(M')$ and $H^ p_ I(N) = 0$ for all $p > 0$,

  2. $H^0_ Z(M') = 0$ and $H^ p_ Z(M) = H^ p_ Z(M')$ and $H^ p_ Z(N) = 0$ for all $p > 0$.

Proof. By definition $H^0_ I(M) = M[I^\infty ]$ is $I$-power torsion. By Lemma 45.9.1 we see that

\[ H^0_ Z(M) = \mathop{\mathrm{Ker}}(M \longrightarrow M_{f_1} \times \ldots \times M_{f_ r}) \]

if $I = (f_1, \ldots , f_ r)$. Thus $H^0_ I(M) \subset H^0_ Z(M)$ and conversely, if $x \in H^0_ Z(M)$, then it is annihilated by a $f_ i^{e_ i}$ for some $e_ i \geq 1$ hence annihilated by some power of $I$. This proves the first equality and moreover $N$ is $I$-power torsion. By Lemma 45.8.1 we see that $R\Gamma _ I(N) = N$. By Lemma 45.9.1 we see that $R\Gamma _ Z(N) = N$. This proves the higher vanishing of $H^ p_ I(N)$ and $H^ p_ Z(N)$ in (1) and (2). The vanishing of $H^0_ I(M')$ and $H^0_ Z(M')$ follow from the preceding remarks and the fact that $M'$ is $I$-power torsion free by More on Algebra, Lemma 15.79.4. The equality of higher cohomologies for $M$ and $M'$ follow immediately from the long exact cohomology sequence. $\square$


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 0AVY. Beware of the difference between the letter 'O' and the digit '0'.