## 10.130 Openness of Cohen-Macaulay loci

In this section we characterize the Cohen-Macaulay property of finite type algebras in terms of flatness. We then use this to prove the set of points where such an algebra is Cohen-Macaulay is open.

Lemma 10.130.1. Let $S$ be a finite type algebra over a field $k$. Let $\varphi : k[y_1, \ldots , y_ d] \to S$ be a quasi-finite ring map. As subsets of $\mathop{\mathrm{Spec}}(S)$ we have

$\{ \mathfrak q \mid S_{\mathfrak q} \text{ flat over }k[y_1, \ldots , y_ d]\} = \{ \mathfrak q \mid S_{\mathfrak q} \text{ CM and }\dim _{\mathfrak q}(S/k) = d\}$

For notation see Definition 10.125.1.

Proof. Let $\mathfrak q \subset S$ be a prime. Denote $\mathfrak p = k[y_1, \ldots , y_ d] \cap \mathfrak q$. Note that always $\dim (S_{\mathfrak q}) \leq \dim (k[y_1, \ldots , y_ d]_{\mathfrak p})$ by Lemma 10.125.4 for example. Moreover, the field extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is finite and hence $\text{trdeg}_ k(\kappa (\mathfrak p)) = \text{trdeg}_ k(\kappa (\mathfrak q))$.

Let $\mathfrak q$ be an element of the left hand side. Then Lemma 10.112.9 applies and we conclude that $S_{\mathfrak q}$ is Cohen-Macaulay and $\dim (S_{\mathfrak q}) = \dim (k[y_1, \ldots , y_ d]_{\mathfrak p})$. Combined with the equality of transcendence degrees above and Lemma 10.116.3 this implies that $\dim _{\mathfrak q}(S/k) = d$. Hence $\mathfrak q$ is an element of the right hand side.

Let $\mathfrak q$ be an element of the right hand side. By the equality of transcendence degrees above, the assumption that $\dim _{\mathfrak q}(S/k) = d$ and Lemma 10.116.3 we conclude that $\dim (S_{\mathfrak q}) = \dim (k[y_1, \ldots , y_ d]_{\mathfrak p})$. Hence Lemma 10.128.1 applies and we see that $\mathfrak q$ is an element of the left hand side. $\square$

Lemma 10.130.2. Let $S$ be a finite type algebra over a field $k$. The set of primes $\mathfrak q$ such that $S_{\mathfrak q}$ is Cohen-Macaulay is open in $S$.

This lemma is a special case of Lemma 10.130.5 below, so you can skip straight to the proof of that lemma if you like.

Proof. Let $\mathfrak q \subset S$ be a prime such that $S_{\mathfrak q}$ is Cohen-Macaulay. We have to show there exists a $g \in S$, $g \not\in \mathfrak q$ such that the ring $S_ g$ is Cohen-Macaulay. For any $g \in S$, $g \not\in \mathfrak q$ we may replace $S$ by $S_ g$ and $\mathfrak q$ by $\mathfrak qS_ g$. Combining this with Lemmas 10.115.5 and 10.116.3 we may assume that there exists a finite injective ring map $k[y_1, \ldots , y_ d] \to S$ with $d = \dim (S_{\mathfrak q}) + \text{trdeg}_ k(\kappa (\mathfrak q))$. Set $\mathfrak p = k[y_1, \ldots , y_ d] \cap \mathfrak q$. By construction we see that $\mathfrak q$ is an element of the right hand side of the displayed equality of Lemma 10.130.1. Hence it is also an element of the left hand side.

By Theorem 10.129.4 we see that for some $g \in S$, $g \not\in \mathfrak q$ the ring $S_ g$ is flat over $k[y_1, \ldots , y_ d]$. Hence by the equality of Lemma 10.130.1 again we conclude that all local rings of $S_ g$ are Cohen-Macaulay as desired. $\square$

Lemma 10.130.3. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. The set of Cohen-Macaulay primes forms a dense open $U \subset \mathop{\mathrm{Spec}}(S)$.

Proof. The set is open by Lemma 10.130.2. It contains all minimal primes $\mathfrak q \subset S$ since the local ring at a minimal prime $S_{\mathfrak q}$ has dimension zero and hence is Cohen-Macaulay. $\square$

Lemma 10.130.4. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. If $\dim (S) > 0$, then there exists an element $f \in S$ which is a nonzerodivisor and a nonunit.

Proof. Let $I \subset S$ be the radical ideal such that $V(I) \subset \mathop{\mathrm{Spec}}(S)$ is the set of primes $\mathfrak q \subset S$ with $S_\mathfrak q$ not Cohen-Macaulay. See Lemma 10.130.3 which also tells us that $V(I)$ is nowhere dense in $\mathop{\mathrm{Spec}}(S)$. Let $\mathfrak m \subset S$ be a maximal ideal such that $\dim (S_\mathfrak m) > 0$ and $\mathfrak m \not\in V(I)$. Such a maximal ideal exists as $\dim (S) > 0$ using the Hilbert Nullstellensatz (Theorem 10.34.1) and Lemma 10.114.5 which implies that any dense open of $\mathop{\mathrm{Spec}}(S)$ has the same dimension as $\mathop{\mathrm{Spec}}(S)$. Finally, let $\mathfrak q_1, \ldots , \mathfrak q_ m$ be the minimal primes of $S$. Choose $f \in S$ with

$f \equiv 1 \bmod I,\quad f \in \mathfrak m,\quad f \not\in \bigcup \mathfrak q_ i$

This is possible by Lemma 10.15.3. Namely, we have $S/(I \cap \mathfrak m) = S/I \times S/\mathfrak m$ by Lemma 10.15.4. Thus we can first choose $g \in S$ such that $g \equiv 1 \bmod I$ and $g \in \mathfrak m$. Then $g + (I \cap \mathfrak m) \not\subset \mathfrak q_ i$ since $V(I \cap \mathfrak m) \not\supset V(\mathfrak q_ i)$. Hence the lemma applies. Clearly $f$ is not a unit. To show that $f$ is a nonzerodivisor, it suffices to prove that $f : S_\mathfrak q \to S_\mathfrak q$ is injective for every prime ideal $\mathfrak q \subset S$. If $S_\mathfrak q$ is not Cohen-Macaulay, then $\mathfrak q \in V(I)$ and $f$ maps to a unit of $S_\mathfrak q$. On the other hand, if $S_\mathfrak q$ is Cohen-Macaulay, then we use that $\dim (S_\mathfrak q/fS_\mathfrak q) < \dim (S_\mathfrak q)$ by the requirement $f \not\in \mathfrak q_ i$ and we conclude that $f$ is a nonzerodivisor in $S_\mathfrak q$ by Lemma 10.104.2. $\square$

Lemma 10.130.5. Let $R$ be a ring. Let $R \to S$ be of finite presentation and flat. For any $d \geq 0$ the set

$\left\{ \begin{matrix} \mathfrak q \in \mathop{\mathrm{Spec}}(S) \text{ such that setting }\mathfrak p = R \cap \mathfrak q \text{ the fibre ring} \\ S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} \text{ is Cohen-Macaulay} \text{ and } \dim _{\mathfrak q}(S/R) = d \end{matrix} \right\}$

is open in $\mathop{\mathrm{Spec}}(S)$.

Proof. Let $\mathfrak q$ be an element of the set indicated, with $\mathfrak p$ the corresponding prime of $R$. We have to find a $g \in S$, $g \not\in \mathfrak q$ such that all fibre rings of $R \to S_ g$ are Cohen-Macaulay. During the course of the proof we may (finitely many times) replace $S$ by $S_ g$ for a $g \in S$, $g \not\in \mathfrak q$. Thus by Lemma 10.125.2 we may assume there is a quasi-finite ring map $R[t_1, \ldots , t_ d] \to S$ with $d = \dim _{\mathfrak q}(S/R)$. Let $\mathfrak q' = R[t_1, \ldots , t_ d] \cap \mathfrak q$. By Lemma 10.130.1 we see that the ring map

$R[t_1, \ldots , t_ d]_{\mathfrak q'} / \mathfrak p R[t_1, \ldots , t_ d]_{\mathfrak q'} \longrightarrow S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}$

is flat. Hence by the critère de platitude par fibres Lemma 10.128.8 we see that $R[t_1, \ldots , t_ d]_{\mathfrak q'} \to S_{\mathfrak q}$ is flat. Hence by Theorem 10.129.4 we see that for some $g \in S$, $g \not\in \mathfrak q$ the ring map $R[t_1, \ldots , t_ d] \to S_ g$ is flat. Replacing $S$ by $S_ g$ we see that for every prime $\mathfrak r \subset S$, setting $\mathfrak r' = R[t_1, \ldots , t_ d] \cap \mathfrak r$ and $\mathfrak p' = R \cap \mathfrak r$ the local ring map $R[t_1, \ldots , t_ d]_{\mathfrak r'} \to S_{\mathfrak r}$ is flat. Hence also the base change

$R[t_1, \ldots , t_ d]_{\mathfrak r'} / \mathfrak p' R[t_1, \ldots , t_ d]_{\mathfrak r'} \longrightarrow S_{\mathfrak r}/\mathfrak p' S_{\mathfrak r}$

is flat. Hence by Lemma 10.130.1 applied with $k = \kappa (\mathfrak p')$ we see $\mathfrak r$ is in the set of the lemma as desired. $\square$

Lemma 10.130.6. Let $R$ be a ring. Let $R \to S$ be flat of finite presentation. The set of primes $\mathfrak q$ such that the fibre ring $S_{\mathfrak q} \otimes _ R \kappa (\mathfrak p)$, with $\mathfrak p = R \cap \mathfrak q$ is Cohen-Macaulay is open and dense in every fibre of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$.

Proof. The set, call it $W$, is open by Lemma 10.130.5. It is dense in the fibres because the intersection of $W$ with a fibre is the corresponding set of the fibre to which Lemma 10.130.3 applies. $\square$

Lemma 10.130.7. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $K/k$ be a field extension, and set $S_ K = K \otimes _ k S$. Let $\mathfrak q \subset S$ be a prime of $S$. Let $\mathfrak q_ K \subset S_ K$ be a prime of $S_ K$ lying over $\mathfrak q$. Then $S_{\mathfrak q}$ is Cohen-Macaulay if and only if $(S_ K)_{\mathfrak q_ K}$ is Cohen-Macaulay.

Proof. During the course of the proof we may (finitely many times) replace $S$ by $S_ g$ for any $g \in S$, $g \not\in \mathfrak q$. Hence using Lemma 10.115.5 we may assume that $\dim (S) = \dim _{\mathfrak q}(S/k) =: d$ and find a finite injective map $k[x_1, \ldots , x_ d] \to S$. Note that this also induces a finite injective map $K[x_1, \ldots , x_ d] \to S_ K$ by base change. By Lemma 10.116.6 we have $\dim _{\mathfrak q_ K}(S_ K/K) = d$. Set $\mathfrak p = k[x_1, \ldots , x_ d] \cap \mathfrak q$ and $\mathfrak p_ K = K[x_1, \ldots , x_ d] \cap \mathfrak q_ K$. Consider the following commutative diagram of Noetherian local rings

$\xymatrix{ S_{\mathfrak q} \ar[r] & (S_ K)_{\mathfrak q_ K} \\ k[x_1, \ldots , x_ d]_{\mathfrak p} \ar[r] \ar[u] & K[x_1, \ldots , x_ d]_{\mathfrak p_ K} \ar[u] }$

By Lemma 10.130.1 we have to show that the left vertical arrow is flat if and only if the right vertical arrow is flat. Because the bottom arrow is flat this equivalence holds by Lemma 10.100.1. $\square$

Lemma 10.130.8. Let $R$ be a ring. Let $R \to S$ be of finite type. Let $R \to R'$ be any ring map. Set $S' = R' \otimes _ R S$. Denote $f : \mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$ the map associated to the ring map $S \to S'$. Set $W$ equal to the set of primes $\mathfrak q$ such that the fibre ring $S_{\mathfrak q} \otimes _ R \kappa (\mathfrak p)$, $\mathfrak p = R \cap \mathfrak q$ is Cohen-Macaulay, and let $W'$ denote the analogue for $S'/R'$. Then $W' = f^{-1}(W)$.

Proof. Trivial from Lemma 10.130.7 and the definitions. $\square$

Lemma 10.130.9. Let $R$ be a ring. Let $R \to S$ be a ring map which is (a) flat, (b) of finite presentation, (c) has Cohen-Macaulay fibres. Then we can write $S = S_0 \times \ldots \times S_ n$ as a product of $R$-algebras $S_ d$ such that each $S_ d$ satisfies (a), (b), (c) and has all fibres equidimensional of dimension $d$.

Proof. For each integer $d$ denote $W_ d \subset \mathop{\mathrm{Spec}}(S)$ the set defined in Lemma 10.130.5. Clearly we have $\mathop{\mathrm{Spec}}(S) = \coprod W_ d$, and each $W_ d$ is open by the lemma we just quoted. Hence the result follows from Lemma 10.24.3. $\square$

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