Lemma 10.116.1 (00P0)—The Stacks project

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Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.116: Dimension of finite type algebras over fields, reprise
Lemma 10.116.1 (cite)

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Lemma 10.116.1. Let $k$ be a field. Let $S$ be a finite type $k$ algebra which is an integral domain. Let $K$ be the field of fractions of $S$. Let $r = \text{trdeg}(K/k)$ be the transcendence degree of $K$ over $k$. Then $\dim (S) = r$. Moreover, the local ring of $S$ at every maximal ideal has dimension $r$.

Proof. We may write $S = k[x_1, \ldots , x_ n]/\mathfrak p$. By Lemma 10.114.3 all local rings of $S$ at maximal ideals have the same dimension. Apply Lemma 10.115.4. We get a finite injective ring map

\[ k[y_1, \ldots , y_ d] \to S \]

with $d = \dim (S)$. Clearly, $k(y_1, \ldots , y_ d) \subset K$ is a finite extension and we win. $\square$

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