proof of the weak Nullstellensatz

Let $K$ be an algebraically closed field, let $n\geq 0$, and let $I$ be an ideal in the polynomial ring $K[x_{1},\ldots,x_{n}]$. Suppose $I$ is strictly smaller than $K[x_{1},\ldots,x_{n}]$. Then $I$ is contained in a maximal ideal $M$ of $K[x_{1},\ldots,x_{n}]$ (note that we don’t have to accept Zorn’s lemma to find such an $M$, since $K[x_{1},\ldots,x_{n}]$ is Noetherian by Hilbert’s basis theorem), and the quotient ring

 $L=K[x_{1},\ldots,x_{n}]/M$

is a field. We view $K$ as a subfield of $L$ via the natural homomorphism $K\hookrightarrow L$, and we denote the images of $x_{1},\ldots,x_{n}$ in $L$ by $\bar{x}_{1},\ldots,\bar{x}_{n}$. Let $\{t_{1},\ldots,t_{m}\}$ be a transcendence basis of $L$ over $K$; it is finite since $L$ is finitely generated as a $K$-algebra. Now $L$ is an algebraic extension of $K(t_{1},\ldots,t_{m})$. By multiplying the minimal polynomial of $\bar{x}_{i}$ over $K(t_{1},\ldots,t_{m})$ by a suitable element of $K[t_{1},\ldots,t_{m}]$ for each $i$, we obtain non-zero polynomials $f_{i}\in K[t_{1},\ldots,t_{m}][X]$ with the property that $f_{i}(\bar{x}_{i})=0$ in $L$:

 $f_{i}=c_{i,0}+c_{i,1}X+\cdots+c_{i,d_{i}}X^{d_{i}}\qquad(1\leq i\leq n)$

for certain integers $d_{i}>0$ and polynomials $c_{i,j}\in K[t_{1},\ldots,t_{m}]$ with $c_{i,d_{i}}\neq 0$. Since $K$ is algebraically closed (hence infinite), we can choose $u_{1},\ldots,u_{n}\in K$ such that $c_{i,d_{i}}(u_{1},\ldots,u_{m})\neq 0$ for all $i$. We define a homomorphism

 $\phi\colon K[t_{1},\ldots,t_{m}]\longrightarrow K$

by taking $\phi$ to be the identity on $K$ and sending $t_{j}$ to $u_{j}$. Let $N$ be the kernel of this homomorphism. Then $\phi$ can be extended to the localization $K[t_{1},\ldots,t_{m}]_{N}$ of $K[t_{1},\ldots,t_{m}]$. Since $c_{i,d_{i}}\not\in N$ for all $i$, the $\bar{x}_{i}$ are integral over this ring. Since $K$ is algebraically closed, the extension theorem for ring homomorphisms implies that $\phi$ can be extended to a homomorphism

 $\phi\colon(K[t_{1},\ldots,t_{m}]_{N})[\bar{x}_{1},\ldots,\bar{x}_{n}]=L% \longrightarrow K.$

Because $L$ is an extension field of $K$ and $\phi$ is the identity on $K$, we see that $\phi$ is actually an isomorphism, that $m=0$, and that $N$ is the zero ideal of $K$. Now let $a_{1}=\phi(\bar{x}_{1}),\ldots,a_{n}=\phi(\bar{x}_{n})$. Then for all polynomials $f$ in the ideal $I$ we started with, the fact that $f\in M$ implies

 $f(a_{1},\ldots,a_{n})=\phi(f(x_{1},\ldots,x_{n})+M)=0.$

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

Title proof of the weak Nullstellensatz ProofOfTheWeakNullstellensatz 2013-03-22 15:27:43 2013-03-22 15:27:43 pbruin (1001) pbruin (1001) 4 pbruin (1001) Proof msc 13A10 msc 13A15