
## 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. $\square$

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