# proof of Hilbert’s Nullstellensatz

Let $K$ be an algebraically closed field, let $n\geq 0$, and let $I$ be an ideal of the polynomial ring  $K[x_{1},\ldots,x_{n}]$. Let $f\in K[x_{1},\ldots,x_{n}]$ be a polynomial  with the property that

 $f(a_{1},\ldots,a_{n})=0\hbox{ for all }(a_{1},\ldots,a_{n})\in V(I).$

Suppose that $f^{r}\not\in I$ for all $r>0$; in particular, $I$ is strictly smaller than $K[x_{1},\ldots,x_{n}]$ and $f\neq 0$. Consider the ring

 $R=K[x_{1},\ldots,x_{n},1/f]\subset K(x_{1},\ldots,x_{n}).$

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

 $RI=\bigcup_{r=0}^{\infty}f^{-r}I$

does not contain the unit element. Let $y$ be an indeterminate over $K[x_{1},\ldots,x_{n}]$, and let $J$ be the inverse image of $RI$ under the homomorphism         $\phi\colon K[x_{1},\ldots,x_{n},y]\to R$

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

 $0=(1-yf)(a_{1},\ldots,a_{n},b)=1-bf(a_{1},\ldots,a_{n})=1.$

The assumption that $f^{r}\not\in I$ 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 ProofOfHilbertsNullstellensatz 2013-03-22 15:27:46 2013-03-22 15:27:46 pbruin (1001) pbruin (1001) 4 pbruin (1001) Proof msc 13A10