## Slick proof of Hilbert’s Nullstellensatz

It is a well known fact that there are multiple proofs for the Nullstellensatz which do not use Noether’s normalization lemma. Serge Lang’s proof(in his book) and Zariski’s proof both fall under this category. In fact, Daniel Allcock of UT Austin posted a proof which essentially builds from the edifice of Zariski’s proof(see that here). He claims that it is the simplest proof for the Nullstellensatz and frankly this is quite true considering the proof uses nothing more than basic field theory, is only one page long and just requires some familiarity with transcendence bases, and the transcendence degree. Yet in the true spirit of simplicity, T. Tao has presented a proof which uses nothing more than the extended Euclidean algorithm and “high school algebra” albeit with a lengthy case-by-case analysis(see the proof here).

Most of these proofs(except Tao’s) go about proving the ‘Weak’ Nullstellensatz and obtain the ‘Strong’ Nullstelensatz through the famous Rabinowitsch trick.

But a few days, I found something truly magnificent, a proof by Enrique Arrondo in the American Mathematical Monthly which proves the Nullstellensatz using a weaker version of Noether normalization and techniques similar to that of Tao, Ritrabata Munshi. The proof is essentially a simplification of a proof by R. Munshi.

Here, I present a special case of the normalization lemma.

### Lemma

If $F$ is an infinite field and $f \in F[x_{1},x_{2},\cdots ,x_{n}]$ is a non-constant polynomial and $n \geq 2$ whose total degree is $d$, then there exists $a_{1},a_{2},\cdots ,a_{n-1} \in F$ such that the coefficient of $x_{n}^{d}$ in

$f(x_{1}+a_{1}x_{n},x_{2}+a_{2}x_{n},\cdots ,x_{n-1}+a_{n-1}x_{n},x_{n})$

is non-zero.

### Proof:

Let $f_{d}$ represent the homogenous component of $f$ of degree $d$.So, the coefficient of $x_{d}^{n}$ in $f(x_{1}+a_{1}x_{n},x_{2}+a_{2}x_{n},\cdots,x_{n-1}+a_{n-1}x_{n},x_{n})$ is $f_{d}(a_{1},\cdots,a_{n-1},1)$. As a polynomial in $F[x_{1},\cdots,x_{n-1}]$, there is some point $(a_{1},\cdots,a_{n-1} \in F^{n-1}$ where $f_{d}(a_{1},\cdots,a_{n-1},1)$ doesn’t vanish. Choose such a point to establish the $a_{i}$ and this guarantees a non-zero coefficient of $x_{n}^{d}$.