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.∎

单词	ZariskiLemma
释义	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.∎
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