Zariski's Lemma

From ProofWiki
Jump to navigation Jump to search


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:

$\alpha: k \sqbrk {x_1, \dotsc, x_n} \to L$

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

Aiming for a contradiction, suppose $n > 0$.


$x_1 \in k \sqbrk {x_1, \dotsc, x_n}$


$\map \alpha {x_1} \ne 0$

We have that $\map \alpha {x_1}^{-1}$ is integral over $k \sqbrk {x_1, \dotsc, x_n}$.

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

$\ds \map \alpha {x_1}^{-m} + \sum_{i \mathop = 0}^{m - 1} \map \alpha {a_i} \map \alpha {x_1}^{-i} = 0$

Multiplying by $\map \alpha {x_1}^m$:

\(\ds 0\) \(=\) \(\ds 1 + \sum_{i \mathop = 0}^{m - 1} \map \alpha {a_i} \map \alpha {x_1}^{m - i}\)
\(\ds \) \(=\) \(\ds \map \alpha {1 + x_1 \paren {\sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} } }\)

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

$\ds 1 = x_1 \paren {-\sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} }$

which means that $x_1$ is invertible.

This contradiction shows that $n = 0$.


Source of Name

This entry was named for Oscar Zariski.