
## 10.71 Depth

Here is our definition.

Definition 10.71.1. Let $R$ be a ring, and $I \subset R$ an ideal. Let $M$ be a finite $R$-module. The $I$-depth of $M$, denoted $\text{depth}_ I(M)$, is defined as follows:

1. if $IM \not= M$, then $\text{depth}_ I(M)$ is the supremum in $\{ 0, 1, 2, \ldots , \infty \}$ of the lengths of $M$-regular sequences in $I$,

2. if $IM = M$ we set $\text{depth}_ I(M) = \infty$.

If $(R, \mathfrak m)$ is local we call $\text{depth}_{\mathfrak m}(M)$ simply the depth of $M$.

Explanation. By Definition 10.67.1 the empty sequence is not a regular sequence on the zero module, but for practical purposes it turns out to be convenient to set the depth of the $0$ module equal to $+\infty$. Note that if $I = R$, then $\text{depth}_ I(M) = \infty$ for all finite $R$-modules $M$. If $I$ is contained in the Jacobson radical of $R$ (e.g., if $R$ is local and $I \subset \mathfrak m_ R$), then $M \not= 0 \Rightarrow IM \not= M$ by Nakayama's lemma. A module $M$ has $I$-depth $0$ if and only if $M$ is nonzero and $I$ does not contain a nonzerodivisor on $M$.

Example 10.67.2 shows depth does not behave well even if the ring is Noetherian, and Example 10.67.3 shows that it does not behave well if the ring is local but non-Noetherian. We will see depth behaves well if the ring is local Noetherian.

Lemma 10.71.2. Let $R$ be a ring, $I \subset R$ an ideal, and $M$ a finite $R$-module. Then $\text{depth}_ I(M)$ is equal to the supremum of the lengths of sequences $f_1, \ldots , f_ r \in I$ such that $f_ i$ is a nonzerodivisor on $M/(f_1, \ldots , f_{i - 1})M$.

Proof. Suppose that $IM = M$. Then Lemma 10.19.1 shows there exists an $f \in I$ such that $f : M \to M$ is $\text{id}_ M$. Hence $f, 0, 0, 0, \ldots$ is an infinite sequence of successive nonzerodivisors and we see agreement holds in this case. If $IM \not= M$, then we see that a sequence as in the lemma is an $M$-regular sequence and we conclude that agreement holds as well. $\square$

Lemma 10.71.3. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a nonzero finite $R$-module. Then $\dim (\text{Supp}(M)) \geq \text{depth}(M)$.

Proof. The proof is by induction on $\dim (\text{Supp}(M))$. If $\dim (\text{Supp}(M)) = 0$, then $\text{Supp}(M) = \{ \mathfrak m\}$, whence $\text{Ass}(M) = \{ \mathfrak m\}$ (by Lemmas 10.62.2 and 10.62.7), and hence the depth of $M$ is zero for example by Lemma 10.62.18. For the induction step we assume $\dim (\text{Supp}(M)) > 0$. Let $f_1, \ldots , f_ d$ be a sequence of elements of $\mathfrak m$ such that $f_ i$ is a nonzerodivisor on $M/(f_1, \ldots , f_{i - 1})M$. According to Lemma 10.71.2 it suffices to prove $\dim (\text{Supp}(M)) \geq d$. We may assume $d > 0$ otherwise the lemma holds. By Lemma 10.62.10 we have $\dim (\text{Supp}(M/f_1M)) = \dim (\text{Supp}(M)) - 1$. By induction we conclude $\dim (\text{Supp}(M/f_1M)) \geq d - 1$ as desired. $\square$

Lemma 10.71.4. Let $R$ be a Noetherian ring, $I \subset R$ an ideal, and $M$ a finite nonzero $R$-module such that $IM \not= M$. Then $\text{depth}_ I(M) < \infty$.

Proof. Since $M/IM$ is nonzero we can choose $\mathfrak p \in \text{Supp}(M/IM)$ by Lemma 10.39.2. Then $(M/IM)_\mathfrak p \not= 0$ which implies $I \subset \mathfrak p$ and moreover implies $M_\mathfrak p \not= IM_\mathfrak p$ as localization is exact. Let $f_1, \ldots , f_ r \in I$ be an $M$-regular sequence. Then $M_\mathfrak p/(f_1, \ldots , f_ r)M_\mathfrak p$ is nonzero as $(f_1, \ldots , f_ r) \subset I$. As localization is flat we see that the images of $f_1, \ldots , f_ r$ form a $M_\mathfrak p$-regular sequence in $I_\mathfrak p$. Since this works for every $M$-regular sequence in $I$ we conclude that $\text{depth}_ I(M) \leq \text{depth}_{I_\mathfrak p}(M_\mathfrak p)$. The latter is $\leq \text{depth}(M_\mathfrak p)$ which is $< \infty$ by Lemma 10.71.3. $\square$

Lemma 10.71.5. 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 10.62.18 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 10.70.6:

$\begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(\kappa , M) \to \mathop{\mathrm{Hom}}\nolimits _ R(\kappa , M) \to \mathop{\mathrm{Hom}}\nolimits _ R(\kappa , M/xM) \\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(\kappa , M) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(\kappa , M) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(\kappa , M/xM) \to \ldots \end{matrix}$

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 10.70.8. Thus it is clear that $i(M/xM) = i(M) - 1$. Induction on $\delta (M)$ finishes the proof. $\square$

Lemma 10.71.6. Let $R$ be a local Noetherian ring. Let $0 \to N' \to N \to N'' \to 0$ be a short exact sequence of finite $R$-modules.

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

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

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

Proof. Use the characterization of depth using the Ext groups $\mathop{\mathrm{Ext}}\nolimits ^ i(\kappa , N)$, see Lemma 10.71.5, and use the long exact cohomology sequence

$\begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(\kappa , N') \to \mathop{\mathrm{Hom}}\nolimits _ R(\kappa , N) \to \mathop{\mathrm{Hom}}\nolimits _ R(\kappa , N'') \\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(\kappa , N') \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(\kappa , N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(\kappa , N'') \to \ldots \end{matrix}$

from Lemma 10.70.6. $\square$

Lemma 10.71.7. Let $R$ be a local Noetherian ring and $M$ a nonzero finite $R$-module.

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

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

Proof. Part (2) is a formal consequence of part (1). Let $x \in R$ be as in (1). By the short exact sequence $0 \to M \to M \to M/xM \to 0$ and Lemma 10.71.6 we see that the depth drops by at most 1. On the other hand, if $x_1, \ldots , x_ r \in \mathfrak m$ is a regular sequence for $M/xM$, then $x, x_1, \ldots , x_ r$ is a regular sequence for $M$. Hence we see that the depth drops by at least 1. $\square$

Lemma 10.71.8. Let $(R, \mathfrak m)$ be a local Noetherian ring and $M$ a finite $R$-module. Let $x \in \mathfrak m$, $\mathfrak p \in \text{Ass}(M)$, and $\mathfrak q$ minimal over $\mathfrak p + (x)$. Then $\mathfrak q \in \text{Ass}(M/x^ nM)$ for some $n \geq 1$.

Proof. Pick a submodule $N \subset M$ with $N \cong R/\mathfrak p$. By the Artin-Rees lemma (Lemma 10.50.2) we can pick $n > 0$ such that $N \cap x^ nM \subset xN$. Let $\overline{N} \subset M/x^ nM$ be the image of $N \to M \to M/x^ nM$. By Lemma 10.62.3 it suffices to show $\mathfrak q \in \text{Ass}(\overline{N})$. By our choice of $n$ there is a surjection $\overline{N} \to N/xN = R/\mathfrak p + (x)$ and hence $\mathfrak q$ is in the support of $\overline{N}$. Since $\overline{N}$ is annihilated by $x^ n$ and $\mathfrak p$ we see that $\mathfrak q$ is minimal among the primes in the support of $\overline{N}$. Thus $\mathfrak q$ is an associated prime of $\overline{N}$ by Lemma 10.62.8. $\square$

Lemma 10.71.9. Let $(R, \mathfrak m)$ be a local Noetherian ring and $M$ a finite $R$-module. For $\mathfrak p \in \text{Ass}(M)$ we have $\dim (R/\mathfrak p) \geq \text{depth}(M)$.

Proof. If $\mathfrak m \in \text{Ass}(M)$ then there is a nonzero element $x \in M$ which is annihilated by all elements of $\mathfrak m$. Thus $\text{depth}(M) = 0$. In particular the lemma holds in this case.

If $\text{depth}(M) = 1$, then by the first paragraph we find that $\mathfrak m \not\in \text{Ass}(M)$. Hence $\dim (R/\mathfrak p) \geq 1$ for all $\mathfrak p \in \text{Ass}(M)$ and the lemma is true in this case as well.

We will prove the lemma in general by induction on $\text{depth}(M)$ which we may and do assume to be $> 1$. Pick $x \in \mathfrak m$ which is a nonzerodivisor on $M$. Note $x \not\in \mathfrak p$ (Lemma 10.62.9). By Lemma 10.59.12 we have $\dim (R/\mathfrak p + (x)) = \dim (R/\mathfrak p) - 1$. Thus there exists a prime $\mathfrak q$ minimal over $\mathfrak p + (x)$ with $\dim (R/\mathfrak q) = \dim (R/\mathfrak p) - 1$ (small argument omitted; hint: the dimension of a Noetherian local ring $A$ is the maximum of the dimensions of $A/\mathfrak r$ taken over the minimal primes $\mathfrak r$ of $A$). Pick $n$ as in Lemma 10.71.8 so that $\mathfrak q$ is an associated prime of $M/x^ nM$. We may apply induction hypothesis to $M/x^ nM$ and $\mathfrak q$ because $\text{depth}(M/x^ nM) = \text{depth}(M) - 1$ by Lemma 10.71.7. We find $\dim (R/\mathfrak q) \geq \text{depth}(M/x^ nM)$ and we win. $\square$

Lemma 10.71.10. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $R \to S$ be a finite ring map. Let $\mathfrak m_1, \ldots , \mathfrak m_ n$ be the maximal ideals of $S$. Let $N$ be a finite $S$-module. Then

$\min \nolimits _{i = 1, \ldots , n} \text{depth}(N_{\mathfrak m_ i}) = \text{depth}(N)$

Proof. By Lemmas 10.35.20, 10.35.22, and Lemma 10.35.21 the maximal ideals of $S$ are exactly the primes of $S$ lying over $\mathfrak m$ and there are finitely many of them. Hence the statement of the lemma makes sense. We will prove the lemma by induction on $k = \min \nolimits _{i = 1, \ldots , n} \text{depth}(N_{\mathfrak m_ i})$. If $k = 0$, then $\text{depth}(N_{\mathfrak m_ i}) = 0$ for some $i$. By Lemma 10.71.5 this means $\mathfrak m_ i S_{\mathfrak m_ i}$ is an associated prime of $N_{\mathfrak m_ i}$ and hence $\mathfrak m_ i$ is an associated prime of $N$ (Lemma 10.62.16). By Lemma 10.62.13 we see that $\mathfrak m$ is an associated prime of $N$ as an $R$-module. Whence $\text{depth}(N) = 0$. This proves the base case. If $k > 0$, then we see that $\mathfrak m_ i \not\in \text{Ass}_ S(N)$. Hence $\mathfrak m \not\in \text{Ass}_ R(N)$, again by Lemma 10.62.13. Thus we can find $f \in \mathfrak m$ which is not a zerodivisor on $N$, see Lemma 10.62.18. By Lemma 10.71.7 all the depths drop exactly by $1$ when passing from $N$ to $N/fN$ and the induction hypothesis does the rest. $\square$

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).