proof of the weak Nullstellensatz

Let $K$ be an algebraically closed field, let $n\ge 0$, and let $I$ be an ideal in the polynomial ring $K[x_1,\mathrm{\dots },x_n]$. Suppose $I$ is strictly smaller than $K[x_1,\mathrm{\dots },x_n]$. Then $I$ is contained in a maximal ideal $M$ of $K[x_1,\mathrm{\dots },x_n]$ (note that we don’t have to accept Zorn’s lemma to find such an $M$, since $K[x_1,\mathrm{\dots },x_n]$ is Noetherian by Hilbert’s basis theorem), and the quotient ring

$$L=K[x_1,\mathrm{\dots },x_n]/M$$

is a field. We view $K$ as a subfield of $L$ via the natural homomorphism $K\hookrightarrow L$, and we denote the images of $x_1,\mathrm{\dots },x_n$ in $L$ by ${\overline{x}}_1,\mathrm{\dots },{\overline{x}}_n$. Let $\{t_1,\mathrm{\dots },t_m\}$ 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 $K(t_1,\mathrm{\dots },t_m)$. By multiplying the minimal polynomial of ${\overline{x}}_i$ over $K(t_1,\mathrm{\dots },t_m)$ by a suitable element of $K[t_1,\mathrm{\dots },t_m]$ for each $i$, we obtain non-zero polynomials $f_i\in K[t_1,\mathrm{\dots },t_m][X]$ with the property that $f_i({\overline{x}}_i)=0$ in $L$:

$$f_i=c_{i,0}+c_{i,1}X+\mathrm{\cdots }+c_{i,d_i}{X}^{d_i}\mathit{\hspace{1em}\hspace{1em}}(1\le i\le n)$$

for certain integers $d_i>0$ and polynomials $c_{i,j}\in K[t_1,\mathrm{\dots },t_m]$ with $c_{i,d_i}\ne 0$. Since $K$ is algebraically closed (hence infinite), we can choose $u_1,\mathrm{\dots },u_n\in K$ such that $c_{i,d_i}(u_1,\mathrm{\dots },u_m)\ne 0$ for all $i$. We define a homomorphism

$$\varphi :K[t_1,\mathrm{\dots },t_m]⟶K$$

by taking $\varphi$ to be the identity on $K$ and sending $t_j$ to $u_j$. Let $N$ be the kernel of this homomorphism. Then $\varphi$ can be extended to the localization $K{[t_1,\mathrm{\dots },t_m]}_N$ of $K[t_1,\mathrm{\dots },t_m]$. Since $c_{i,d_i}\notin N$ for all $i$, the ${\overline{x}}_i$ are integral over this ring. Since $K$ is algebraically closed, the extension theorem for ring homomorphisms implies that $\varphi$ can be extended to a homomorphism

$$\varphi :(K{[t_1,\mathrm{\dots },t_m]}_N)[{\overline{x}}_1,\mathrm{\dots },{\overline{x}}_n]=L⟶K.$$

Because $L$ is an extension field of $K$ and $\varphi$ is the identity on $K$, we see that $\varphi$ is actually an isomorphism, that $m=0$, and that $N$ is the zero ideal of $K$. Now let $a_1=\varphi ({\overline{x}}_1),\mathrm{\dots },a_n=\varphi ({\overline{x}}_n)$. Then for all polynomials $f$ in the ideal $I$ we started with, the fact that $f\in M$ implies

$$f(a_1,\mathrm{\dots },a_n)=\varphi (f(x_1,\mathrm{\dots },x_n)+M)=0.$$

We conclude that the zero set $V(I)$ 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