proof of the weak Nullstellensatz

Let K be an algebraically closed field, let n0, and let I be an ideal in the polynomial ringMathworldPlanetmath K[x1,,xn]. Suppose I is strictly smaller than K[x1,,xn]. Then I is contained in a maximal idealMathworldPlanetmath M of K[x1,,xn] (note that we don’t have to accept Zorn’s lemma to find such an M, since K[x1,,xn] is NoetherianPlanetmathPlanetmathPlanetmath by Hilbert’s basis theorem), and the quotient ringMathworldPlanetmath


is a field. We view K as a subfieldMathworldPlanetmath of L via the natural homomorphismMathworldPlanetmathPlanetmath KL, and we denote the images of x1,,xn in L by x¯1,,x¯n. Let {t1,,tm} be a transcendence basis of L over K; it is finite since L is finitely generatedMathworldPlanetmathPlanetmathPlanetmath as a K-algebraMathworldPlanetmathPlanetmath. Now L is an algebraic extensionMathworldPlanetmath of K(t1,,tm). By multiplying the minimal polynomial of x¯i over K(t1,,tm) by a suitable element of K[t1,,tm] for each i, we obtain non-zero polynomialsMathworldPlanetmathPlanetmath fiK[t1,,tm][X] with the property that fi(x¯i)=0 in L:

fi=ci,0+ci,1X++ci,diXdi  (1in)

for certain integers di>0 and polynomials ci,jK[t1,,tm] with ci,di0. Since K is algebraically closedMathworldPlanetmath (hence infinite), we can choose u1,,unK such that ci,di(u1,,um)0 for all i. We define a homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath


by taking ϕ to be the identityPlanetmathPlanetmathPlanetmathPlanetmath on K and sending tj to uj. Let N be the kernel of this homomorphism. Then ϕ can be extended to the localizationMathworldPlanetmath K[t1,,tm]N of K[t1,,tm]. Since ci,diN for all i, the x¯i are integral over this ring. Since K is algebraically closed, the extension theorem for ring homomorphisms implies that ϕ can be extended to a homomorphism


Because L is an extension fieldMathworldPlanetmath of K and ϕ is the identity on K, we see that ϕ is actually an isomorphismMathworldPlanetmathPlanetmath, that m=0, and that N is the zero idealMathworldPlanetmathPlanetmath of K. Now let a1=ϕ(x¯1),,an=ϕ(x¯n). Then for all polynomials f in the ideal I we started with, the fact that fM implies


We conclude that the zero setMathworldPlanetmath V(I) of I is not empty.

Title proof of the weak Nullstellensatz
Canonical name ProofOfTheWeakNullstellensatz
Date of creation 2013-03-22 15:27:43
Last modified on 2013-03-22 15:27:43
Owner pbruin (1001)
Last modified by pbruin (1001)
Numerical id 4
Author pbruin (1001)
Entry type Proof
Classification msc 13A10
Classification msc 13A15