Zariski lemma

Proposition 1.

Let $R\subseteq S\subseteq T$ be commutative rings. If $R$ is noetherian, and T finitely generated as an $R$ -algebra and as an $S$ -module, then $S$ is finitely generated as an $R$ -algebra.

Lemma 1 (Zariski’s lemma).

Let $(L:K)$ be a field extension and $a_{1},\ldots,a_{n}\in L$ be such that $K(a_{1},\ldots,a_{n})=K[a_{1},\ldots,a_{n}]$ . Then the elements $a_{1},\ldots,a_{n}$ are algebraic over $K$ .

Proof.

The case $n=1$ is clear. Now suppose $n>1$ and not all $a_{i},1\leq i\leq n$ are algebraic over $K$ .
Wlog we may assume $a_{1},\ldots,a_{n}$ are algebraically independent and each element $a_{r+1},\ldots,a_{n}$ is algebraic over $D:=K(a_{1},\ldots,a_{r})$ . Hence $K[a_{1},\ldots,a_{n}]$ is a finite algebraic extension of $D$ and therefore is a finitely generated $D$ -module.
The above proposition applied to $K\subseteq D\subseteq K[a_{1},\ldots,a_{n}]$ shows that $D$ is finitely generated as a $K$ -algebra, i.e $D=K[d_{1},\ldots,d_{n}]$ .

Let $d_{i}=\frac{p_{i}(a_{1},\ldots,a_{n})}{q_{i}(a_{1},\ldots,a_{n})}$ , where $p_{i},q_{i}\in K[x_{1},\ldots,x_{n}]$ .
Now $a_{1},\ldots,a_{n}$ are algebraically independent so that $K[a_{1},\ldots,a_{n}]\cong K[x_{1},\ldots,x_{n}]$ , which is a UFD (http://planetmath.org/UFD).
Let $h$ be a prime divisor of $q_{1}\cdots q_{r}+1$ . Since $q$ is relatively prime to each of $q_{i}$ , the element ${q(a_{1},\ldots,a_{n})}^{-1}\in D$ cannot be in $K[d_{1},\ldots,d_{n}]$ . We obtain a contradiction. ∎

Title	Zariski lemma
Canonical name	ZariskiLemma
Date of creation	2013-03-22 17:18:11
Last modified on	2013-03-22 17:18:11
Owner	polarbear (3475)
Last modified by	polarbear (3475)
Numerical id	7
Author	polarbear (3475)
Entry type	Derivation
Classification	msc 12F05
Classification	msc 11J85