# 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 ZariskiLemma 2013-03-22 17:18:11 2013-03-22 17:18:11 polarbear (3475) polarbear (3475) 7 polarbear (3475) Derivation msc 12F05 msc 11J85