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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #941 by JuanPablo on
Comment #942 by JuanPablo on
There are also: