Using Noether's Normalization to prove Zariski's lemma

Question

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?

Answer 1

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.

Answer 2

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.