Lemma 10.115.5. Let $k$ be a field. Let $S$ be a finite type $k$ algebra and denote $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak q$ be a prime of $S$, and let $x \in X$ be the corresponding point. There exists a $g \in S$, $g \not\in \mathfrak q$ such that $\dim (S_ g) = \dim _ x(X) =: d$ and such that there exists a finite injective map $k[y_1, \ldots , y_ d] \to S_ g$.

Proof. Note that by definition $\dim _ x(X)$ is the minimum of the dimensions of $S_ g$ for $g \in S$, $g \not\in \mathfrak q$, i.e., the minimum is attained. Thus the lemma follows from Lemma 10.115.4. $\square$

Comment #941 by JuanPablo on

Maybe there should be references to tags 00OK and 00OP, to see that the number $d$ of $y$ is the same as the dimension of $S_g$.

Comment #942 by JuanPablo on

Ah! nevermind, that is included in the last lemma.

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