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. $\square$

Comment #1222 by David Corwin on

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

Comment #1861 by gulizi on

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

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

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