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Zariski's lemma says that if $k$ is a field and $K$ is a finitely generated $k$-algebra and also a field, then $K/k$ is a finite field extension.

In this wikipedia article there is a proof of the lemma using Noether's Normalization. The argument is essentially this: let $K=k[x_1,...,x_n]$ for some $x_i\in K$, then by Noether's Normalization, there are algebraically independent $y_1,...,y_d$ such that $K$ is a finitely generated $k[y_1,...,y_d]$-module. But since $\dim_{Krull}K=0$ and $\dim_{Krull}k[y_1,...,y_d]=d$, one concludes $d=0$.

I don't understand why the fact that $K$ is a finitely generated $k[y_1,...,y_d]$-module allows us to compare the Krull dimensions of $K$ and $k[y_1,...,y_d]$. I spent some good time trying to justify this formally, but I was totally stuck.

I've tried to look at the surjective ring homomorphism $\varphi:k[y_1,...,y_d]\to k[x_1,...,x_n]$ to show that a chain of primes in $k[y_1,...,y_d]$ would generate a chain of primes in $K$, but apparently this doesn't work.

How could I justify it?

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If you don't want to use the whole Cohen-Seidenberg Theorem, then the result you want follows also from the following elementary, straightforward to prove lemma:

Let $R'/R$ be an integral extension of domains. Then $R'$ is a field if and only if $R$ is.

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  • $\begingroup$ yes, that is a lot simpler, thank you $\endgroup$ – rmdmc89 Feb 3 '18 at 19:06
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Since $K$ is a finitely generated $k[y_1, \ldots, y_d]$-module it is in particular an integral extension of $k[y_1, \ldots, y_d]$. It is a well-known result (following from the Cohen-Seidenberg Theorems) that integral extensions preserve Krull dimension. For a proof, see this post or Proposition 9.2 and Corollary 9.3 (p. 227) of Eisenbud's Commutative Algebra.

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  • $\begingroup$ I have question about the definition of extension: in order to say that $K$ is an integral extension of $k[y_1,...,y_d]$, shouldn't we have an injective ring homomorphism $k[y_1,...,y_d]\hookrightarrow K$ in the first place? $\endgroup$ – rmdmc89 Feb 3 '18 at 20:35
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    $\begingroup$ $y_1, \ldots, y_d$ are (algebraically independent) elements of $K$, so we have $k \subseteq k[y_1, \ldots, y_d] \subseteq K$. The injective homomorphism is just the inclusion map. The fact that $y_1, \ldots, y_d$ are elements of $K$ doesn't seem to be stated clearly in the article for Zariski's Lemma, but it is in the Noether normalization article. $\endgroup$ – André 3000 Feb 3 '18 at 23:37

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