proof of Hilbert’s Nullstellensatz

Let $K$ be an algebraically closed field, let $n\ge 0$, and let $I$ be an ideal of the polynomial ring $K[x_1,\mathrm{\dots },x_n]$. Let $f\in K[x_1,\mathrm{\dots },x_n]$ be a polynomial with the property that

$$f(a_1,\mathrm{\dots },a_n)=0\text{ for all }(a_1,\mathrm{\dots },a_n)\in V(I).$$

Suppose that $f^r\notin I$ for all $r>0$; in particular, $I$ is strictly smaller than $K[x_1,\mathrm{\dots },x_n]$ and $f\ne 0$. Consider the ring

$$R=K[x_1,\mathrm{\dots },x_n,1/f]\subset K(x_1,\mathrm{\dots },x_n).$$

The $R$-ideal $RI$ is strictly smaller than $R$, since

$$RI=\bigcup _{r=0}^{\mathrm{\infty }}{f}^{-r}I$$

does not contain the unit element. Let $y$ be an indeterminate over $K[x_1,\mathrm{\dots },x_n]$, and let $J$ be the inverse image of $RI$ under the homomorphism

$$\varphi :K[x_1,\mathrm{\dots },x_n,y]\to R$$

acting as the identity on $K[x_1,\mathrm{\dots },x_n]$ and sending $y$ to $1/f$. Then $J$ is strictly smaller than $K[x_1,\mathrm{\dots },x_n,y]$, so the weak Nullstellensatz gives us an element $(a_1,\mathrm{\dots },a_n,b)\in K^{n+1}$ such that $g(a_1,\mathrm{\dots },a_n,b)=0$ for all $g\in J$. In particular, we see that $g(a_1,\mathrm{\dots },a_n)=0$ for all $g\in I$. Our assumption on $f$ therefore implies $f(a_1,\mathrm{\dots },a_n)=0$. However, $J$ also contains the element $1-yf$ since $\varphi$ sends this element to zero. This leads to the following contradiction:

$$0=(1-yf)(a_1,\mathrm{\dots },a_n,b)=1-bf(a_1,\mathrm{\dots },a_n)=1.$$

The assumption that $f^r\notin I$ for all $r>0$ is therefore false, i.e. there is an $r>0$ with $f^r\in I$.

Title	proof of Hilbert’s Nullstellensatz
Canonical name	ProofOfHilbertsNullstellensatz
Date of creation	2013-03-22 15:27:46
Last modified on	2013-03-22 15:27:46
Owner	pbruin (1001)
Last modified by	pbruin (1001)
Numerical id	4
Author	pbruin (1001)
Entry type	Proof
Classification	msc 13A10