The Stacks project

Proof. Suppose that $IM = M$. Then Lemma 10.20.1 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. $\square$

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 10.63.2 and 10.63.7), and hence the depth of $M$ is zero for example by Lemma 10.63.18. 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 10.72.2 it suffices to prove $\dim (\text{Supp}(M)) \geq d$. We may assume $d > 0$ otherwise the lemma holds. By Lemma 10.63.10 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. $\square$

Proof. Since $M/IM$ is nonzero we can choose $\mathfrak p \in \text{Supp}(M/IM)$ by Lemma 10.40.2. 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 10.72.3. $\square$

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. $\square$

Proof. Use the characterization of depth using the Ext groups $\mathop{\mathrm{Ext}}\nolimits ^ i(\kappa , N)$, see Lemma 10.72.5, and use the long exact cohomology sequence

from Lemma 10.71.6. $\square$

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 10.72.6 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. $\square$

Proof. Pick a submodule $N \subset M$ with $N \cong R/\mathfrak p$. By the Artin-Rees lemma (Lemma 10.51.2) 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 10.63.3 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 10.63.8. $\square$

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 10.63.9). By Lemma 10.60.13 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 10.72.8 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 10.72.7. We find $\dim (R/\mathfrak q) \geq \text{depth}(M/x^ nM)$ and we win. $\square$

Proof. If $M_\mathfrak p = 0$, then $\text{depth}(M_\mathfrak p) = \infty $ and the lemma holds. If $\text{depth}(M) \leq \dim (R/\mathfrak p)$, then the lemma is true. If $\text{depth}(M) > \dim (R/\mathfrak p)$, then $\mathfrak p$ is not contained in any associated prime $\mathfrak q$ of $M$ by Lemma 10.72.9. Hence we can find an $x \in \mathfrak p$ not contained in any associated prime of $M$ by Lemma 10.15.2 and Lemma 10.63.5. Then $x$ is a nonzerodivisor on $M$, see Lemma 10.63.9. Hence $\text{depth}(M/xM) = \text{depth}(M) - 1$ and $\text{depth}(M_\mathfrak p / x M_\mathfrak p) = \text{depth}(M_\mathfrak p) - 1$ provided $M_\mathfrak p$ is nonzero, see Lemma 10.72.7. Thus we conclude by induction on $\text{depth}(M)$. $\square$

Proof. By Lemmas 10.36.20, 10.36.22, and Lemma 10.36.21 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 10.72.5 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 10.63.16). By Lemma 10.63.13 we see that $\mathfrak m$ is an associated prime of $N$ as an $R$-module. Whence $\text{depth}_\mathfrak m(N) = 0$. 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 10.63.13. Thus we can find $f \in \mathfrak m$ which is not a zerodivisor on $N$, see Lemma 10.63.18. By Lemma 10.72.7 all the depths drop exactly by $1$ when passing from $N$ to $N/fN$ and the induction hypothesis does the rest. $\square$