proof of the weak Nullstellensatz

Let $K$ be an algebraically closed field, let $n\ge 0$, and let $I$ bean ideal in the polynomial ring $K[x_1,\mathrm{\dots },x_n]$. Suppose $I$ isstrictly smaller than $K[x_1,\mathrm{\dots },x_n]$. Then $I$ is contained in amaximal ideal $M$ of $K[x_1,\mathrm{\dots },x_n]$ (note that we don’t have toaccept Zorn’s lemma to find such an $M$, since $K[x_1,\mathrm{\dots },x_n]$ isNoetherian 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 naturalhomomorphism $K\hookrightarrow L$, 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 isfinite since $L$ is finitely generated as a $K$-algebra. Now $L$ isan algebraic extension of $K(t_1,\mathrm{\dots },t_m)$. By multiplying theminimal polynomial of ${\overline{x}}_i$ over $K(t_1,\mathrm{\dots },t_m)$ by asuitable element of $K[t_1,\mathrm{\dots },t_m]$ for each $i$, we obtainnon-zero polynomials $f_i\in K[t_1,\mathrm{\dots },t_m][X]$ with theproperty 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 algebraicallyclosed (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 beextended 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 algebraicallyclosed, 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 $m=0$, andthat $N$ 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 theideal $I$ 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.