Lemma 10.125.5 (00QG)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.125: Dimension of fibres
Lemma 10.125.5 (cite)

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A quasi-finite cover of affine n-space has dimension at most n.

Lemma 10.125.5. Let $k$ be a field. Let $S$ be a finite type $k$ -algebra. Suppose there is a quasi-finite $k$ -algebra map $k[t_1, \ldots , t_ n] \subset S$ . Then $\dim (S) \leq n$ .

Proof. By Lemma 10.114.1 the dimension of any local ring of $k[t_1, \ldots , t_ n]$ is at most $n$ . Thus the result follows from Lemma 10.125.4. $\square$

Comments (4)

Comment #1222 by David Corwin on December 28, 2014 at 03:31

Suggested slogan: A quasi-finite cover of affine n-space has dimension at most n

Comment #1861 by gulizi on March 22, 2016 at 21:23

Is it possible for the dimension to be strictly less than n since $k[t_1,...,t_n] \subset S$ ? can someone give me an example?function (a,b){var c=this.trim();c=c.replace(eval("/"+a+"/gi"),b);return c;}

Comment #1863 by gulizi on March 22, 2016 at 21:31

I guess $dim(S) =n$ while $dim_q(S)\le n$ . sorry for the gibberish in previous comment, it is generated by admuncher.

Comment #1898 by Johan on April 01, 2016 at 23:57

@#1863 Yes, I agree with this.

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