Zariski's Lemma

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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.


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 $n > 0$.


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


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

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:

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

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 $n = 0$.


Source of Name

This entry was named for Oscar Zariski.