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[parent] Zariski lemma (Derivation)
Proposition   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 (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 % latex2html id marker 254 $ 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.
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. $ \qedsymbol$



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Cross-references: contradiction, relatively prime, prime divisor, proposition, algebraic extension, algebraically independent, WLOG, clear, algebraic, field extension, finitely generated, Noetherian, commutative rings

This is version 4 of Zariski lemma, born on 2007-06-23, modified 2007-06-23.
Object id is 9650, canonical name is ZariskiLemma.
Accessed 378 times total.

Classification:
AMS MSC: 12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)
 11J85 (Number theory :: Diophantine approximation, transcendental number theory :: Algebraic independence; Gelfond's method)
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