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
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Comment #941 by JuanPablo on
Comment #942 by JuanPablo on
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