Zariski's Lemma

Theorem
Let $L/k$ be a field extension.

Let $L$ be finitely generated as an algebra over $k$.

Then $L/k$ is a finite field extension.

Proof
By Noether Normalization Lemma, we find a finite and injective morphism:


 * $\alpha: k \left[{x_1, \ldots, x_n}\right] \to L$

If we can prove that $n = 0$, the proof is complete.

Let $n > 0$.

Then:
 * $x_1 \in k \left[{x_1, \dotsc, x_n}\right]$

and:
 * $\alpha \left({x_1}\right) \ne 0$

We have that $\alpha \left({x_1}\right)^{-1}$ is integral over $k \left[{x_1, \dotsc, x_n}\right]$.

Thus there exists a $m \in \N$ and $a_0, \dotsc, a_{m-1} \in k \left[{x_1, \dotsc, x_n}\right]$ such that:


 * $\displaystyle \alpha \left({x_1}\right)^{-m} + \sum_{i \mathop = 0}^{m-1} \alpha \left({a_i}\right) \alpha \left({x_1}\right)^{-i} = 0$

If we multiply this by $\alpha \left({x_1}\right)^m$, we find that:

and thus, since $\alpha$ is injective, we find that:


 * $\displaystyle 1 = x_1 \left({- \sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} }\right)$

which means that $x_1$ is invertible.

This contradiction shows that $n = 0$.