The Stacks project

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. $\square$

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. $\square$

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. $\square$

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

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

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. $\square$

Proof. The set, call it $W$, is open by Lemma 10.130.4. 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. $\square$

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

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. $\square$

Proof. Trivial from Lemma 10.130.6 and the definitions. $\square$

Proof. For each integer $d$ denote $W_ d \subset \mathop{\mathrm{Spec}}(S)$ the set defined in Lemma 10.130.4. 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. $\square$