Zariski lemma

Proposition 1.

Let RST be commutative rings. If R is noetherianPlanetmathPlanetmathPlanetmath, and T finitely generatedMathworldPlanetmathPlanetmath as an R-algebraPlanetmathPlanetmath 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 a1,,anL be such that K(a1,,an)=K[a1,,an]. Then the elements a1,,an are algebraic over K.


The case n=1 is clear. Now suppose n>1 and not all ai,1in are algebraic over K.
Wlog we may assume a1,,an are algebraically independentMathworldPlanetmath and each element ar+1,,an is algebraic over D:=K(a1,,ar). Hence K[a1,,an] is a finite algebraic extensionMathworldPlanetmath of D and therefore is a finitely generated D-module.
The above propositionPlanetmathPlanetmath applied to KDK[a1,,an] shows that D is finitely generated as a K-algebra, i.e D=K[d1,,dn].

Let di=pi(a1,,an)qi(a1,,an), where pi,qiK[x1,,xn].
Now a1,,an are algebraically independent so that K[a1,,an]K[x1,,xn], which is a UFD (
Let h be a prime divisorPlanetmathPlanetmathPlanetmath of q1qr+1. Since q is relatively prime to each of qi, the element q(a1,,an)-1D cannot be in K[d1,,dn]. We obtain a contradictionMathworldPlanetmathPlanetmath. ∎

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 DerivationPlanetmathPlanetmath
Classification msc 12F05
Classification msc 11J85