
Lemma 10.129.1. 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 10.124.1.

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 10.124.4 for example. Moreover, the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ 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 10.111.9 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 10.115.3 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 10.115.3 we conclude that $\dim (S_{\mathfrak q}) = \dim (k[y_1, \ldots , y_ d]_{\mathfrak p})$. Hence Lemma 10.127.1 applies and we see that $\mathfrak q$ is an element of the left hand side. $\square$

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