The Stacks project

Proof. We prove the lemma by induction on $d$. If $d = 0$, then $M$ is finite and there is no case to which the lemma applies. If $d = 1$, then we have to show that $g : M \to M$ is injective. The kernel $K$ has support $\{ \mathfrak m\}$ because by assumption $\dim \text{Supp}(M) \cap V(g) = 0$. Hence $K$ has finite length. Hence $f_1 : K \to K$ injective implies the length of the image is the length of $K$, and hence $f_1 K = K$, which by Nakayama's Lemma 10.20.1 implies $K = 0$. Also, $\dim \text{Supp}(M/gM) = 0$ and so $M/gM$ is Cohen-Macaulay of depth $0$.

Assume $d > 1$. Observe that $g$ is good for $(M/f_1M, f_2, \ldots , f_ d)$, as is easily seen from the definition. By induction, we have that (a) $g$ is a nonzerodivisor on $M/f_1M$ and (b) $M/(g, f_1)M$ is Cohen-Macaulay with maximal regular sequence $f_2, \ldots , f_{d - 1}$. By Lemma 10.68.4 we see that $g, f_1$ is an $M$-regular sequence. Hence $g$ is a nonzerodivisor on $M$ and $f_1, \ldots , f_{d - 1}$ is an $M/gM$-regular sequence. $\square$

Proof. Choose a $M$-regular sequence $f_1, \ldots , f_ d$ with $d = \dim (\text{Supp}(M))$. If $g$ is good with respect to $(M, f_1, \ldots , f_ d)$ we win by Lemma 10.103.2. In particular the lemma holds if $d = 1$. (The case $d = 0$ does not occur.) Assume $d > 1$. Choose an element $h \in R$ such that (i) $h$ is good with respect to $(M, f_1, \ldots , f_ d)$, and (ii) $\dim (\text{Supp}(M) \cap V(h, g)) = d - 2$. To see $h$ exists, let $\{ \mathfrak q_ j\}$ be the (finite) set of minimal primes of the closed sets $\text{Supp}(M)$, $\text{Supp}(M)\cap V(f_1, \ldots , f_ i)$, $i = 1, \ldots , d - 1$, and $\text{Supp}(M) \cap V(g)$. None of these $\mathfrak q_ j$ is equal to $\mathfrak m$ and hence we may find $h \in \mathfrak m$, $h \not\in \mathfrak q_ j$ by Lemma 10.15.2. It is clear that $h$ satisfies (i) and (ii). From Lemma 10.103.2 we conclude that $M/hM$ is Cohen-Macaulay. By (ii) we see that the pair $(M/hM, g)$ satisfies the induction hypothesis. Hence $M/(h, g)M$ is Cohen-Macaulay and $g : M/hM \to M/hM$ is injective. By Lemma 10.68.4 we see that $g : M \to M$ and $h : M/gM \to M/gM$ are injective. Combined with the fact that $M/(g, h)M$ is Cohen-Macaulay this finishes the proof. $\square$

Proof. Let $Z = \text{Supp}(M) \subset \mathop{\mathrm{Spec}}(R)$. By Lemma 10.60.13 in the chain $Z \supset Z \cap V(g_1) \supset \ldots \supset Z \cap V(g_1, \ldots , g_ c)$ each step decreases the dimension at most by $1$. Hence by assumption each step decreases the dimension by exactly $1$ each time. Thus we may successively apply Lemma 10.103.3 to the modules $M/(g_1, \ldots , g_ i)$ and the element $g_{i + 1}$.

To extend $g_1, \ldots , g_ c$ by one element if $c < d$ we simply choose an element $g_{c + 1} \in \mathfrak m$ which is not in any of the finitely many minimal primes of $Z \cap V(g_1, \ldots , g_ c)$, using Lemma 10.15.2. $\square$

Proof. By Lemma 10.72.7 we have $\text{depth}(M/xM) = \text{depth}(M)-1$. By Lemma 10.63.10 we have $\dim (\text{Supp}(M/xM)) = \dim (\text{Supp}(M)) - 1$. $\square$

Proof. Omitted. $\square$

Proof. By Lemma 10.72.9 we have $\text{depth}(M) \leq \dim (R/\mathfrak p)$. Of course $\dim (R/\mathfrak p) \leq \dim (\text{Supp}(M))$ as $\mathfrak p \in \text{Supp}(M)$ (Lemma 10.63.2). Thus we have equality in both inequalities as $M$ is Cohen-Macaulay. Then $\mathfrak p$ must be minimal in $\text{Supp}(M)$ otherwise we would have $\dim (R/\mathfrak p) < \dim (\text{Supp}(M))$. Finally, minimal primes in the support of $M$ are equal to the minimal elements of $\text{Ass}(M)$ (Proposition 10.63.6) hence $M$ has no embedded associated primes (Definition 10.67.1). $\square$

Proof. We will prove this by induction on $\dim (R)$. If $\dim (R) = 0$, then the statement is clear. Assume $\dim (R) > 0$. Then $n > 0$. Choose an element $x \in \mathfrak p_1$, with $x$ not in any of the minimal primes of $R$, and in particular $x \not\in \mathfrak p_0$. (See Lemma 10.15.2.) Then $\dim (R/xR) = \dim (R) - 1$ by Lemma 10.60.13. The module $M/xM$ is Cohen-Macaulay over $R/xR$ by Proposition 10.103.4 and Lemma 10.103.6. The support of $M/xM$ is $\mathop{\mathrm{Spec}}(R/xR)$ by Lemma 10.40.9. After replacing $x$ by $x^ n$ for some $n$, we may assume that $\mathfrak p_1$ is an associated prime of $M/xM$, see Lemma 10.72.8. By Lemma 10.103.7 we conclude that $\mathfrak p_1/(x)$ is a minimal prime of $R/xR$. It follows that the chain $\mathfrak p_1/(x) \subset \ldots \subset \mathfrak p_ n/(x)$ is a maximal chain of primes in $R/xR$. By induction we find that this chain has length $\dim (R/xR) = \dim (R) - 1$ as desired. $\square$

Proof. Follows immediately from Lemma 10.103.9. $\square$

Proof. We may and do assume $\mathfrak p \not= \mathfrak m$ and $M$ not zero. Choose a maximal chain of primes $\mathfrak p = \mathfrak p_ c \subset \mathfrak p_{c - 1} \subset \ldots \subset \mathfrak p_1 \subset \mathfrak m$. If we prove the result for $M_{\mathfrak p_1}$ over $R_{\mathfrak p_1}$, then the lemma will follow by induction on $c$. Thus we may assume that there is no prime strictly between $\mathfrak p$ and $\mathfrak m$. Note that $\dim (\text{Supp}(M_\mathfrak p)) \leq \dim (\text{Supp}(M)) - 1$ because any chain of primes in the support of $M_\mathfrak p$ can be extended by one more prime (namely $\mathfrak m$) in the support of $M$. On the other hand, we have $\text{depth}(M_\mathfrak p) \geq \text{depth}(M) - \dim (R/\mathfrak p) = \text{depth}(M) - 1$ by Lemma 10.72.10 and our choice of $\mathfrak p$. Thus $\text{depth}(M_\mathfrak p) \geq \dim (\text{Supp}(M_\mathfrak p))$ as desired (the other inequality is Lemma 10.72.3). $\square$

Proof. By induction on the number of variables it suffices to prove this for $M[x] = M \otimes _ R R[x]$ over $R[x]$. Let $\mathfrak m \subset R[x]$ be a maximal ideal, and let $\mathfrak p = R \cap \mathfrak m$. Let $f_1, \ldots , f_ d$ be a $M_\mathfrak p$-regular sequence in the maximal ideal of $R_{\mathfrak p}$ of length $d = \dim (\text{Supp}(M_{\mathfrak p}))$. Note that since $R[x]$ is flat over $R$ the localization $R[x]_{\mathfrak m}$ is flat over $R_{\mathfrak p}$. Hence, by Lemma 10.68.5, the sequence $f_1, \ldots , f_ d$ is a $M[x]_{\mathfrak m}$-regular sequence of length $d$ in $R[x]_{\mathfrak m}$. The quotient

has support equal to the primes lying over $\mathfrak p$ because $R_\mathfrak p \to R[x]_\mathfrak m$ is flat and the support of $M_{\mathfrak p}/(f_1, \ldots , f_ d)M_{\mathfrak p}$ is equal to $\{ \mathfrak p\}$ (details omitted; hint: follows from Lemmas 10.40.4 and 10.40.5). Hence the dimension is $1$. To finish the proof it suffices to find an $f \in \mathfrak m$ which is a nonzerodivisor on $Q$. Since $\mathfrak m$ is a maximal ideal, the field extension $\kappa (\mathfrak m)/\kappa (\mathfrak p)$ is finite (Theorem 10.34.1). Hence we can find $f \in \mathfrak m$ which viewed as a polynomial in $x$ has leading coefficient not in $\mathfrak p$. Such an $f$ acts as a nonzerodivisor on

and hence acts as a nonzerodivisor on $Q$. $\square$