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. $\square$