If $\mathbb{k}$ is an infinite field, $p \in \mathbb{k}[X_1, \dots, X_n]$ is zero iff $x \mapsto p(x)$ is the zero function.

Question

As the title states, I'm trying to prove the nontrivial implication, that is, if $p \in \mathbb{k}[X_1, \dots , X_n]$ is such that $p(x) = 0$ for all $x \in \mathbb{k}^n$ and $\mathbb{k}$ is infinite, then $p = 0$.

My idea is to do induction in the number of variables, but I'm not sure on how to proceed. For the base case, since $\mathbb{k}$ is infinite, $p$ having an infinite amount of roots would imply $x-r | p$ for infinitely many $r$. Thus, $p$ cannot be non-null, because it would have finite degree and so a bounded amount of roots.

Fixing a variable to be zero, the polynomial $p(0, \dots, X_n)$ on $n-1$ variables is zero as a a function, and so by an inductive reasoning it should be the zero polynomial. Thus, all coefficients of terms that do not contain $X_1$ must be zero. Doing this with each variable, the only possible terms to be nonzero are the ones that have all variables.

Can I finish this argument by an additional convenient evaluation on $p(X_1 \dots X_n)$? An immediate requirement is for the coefficients to 'remain separate' (e.g. $p(t, \dots ,t)$ is the zero function and so as a polynomial it is zero, but its coefficients are linear combinations of the original one, so I still can say that the original coefficients are themselves zero.).

score 4 · Accepted Answer · answered Sep 03 '18 at 20:06

4

Here are some old notes from a course I took recently, I think these apply:

answered Sep 03 '18 at 20:06

J. Moeller

3,044

Great, that's what I was trying to say but couldn't find the way of saying. Thanks! – qualcuno Sep 03 '18 at 20:10
2

It is an interesting statement, because the statement is false for finite fields. – J. Moeller Sep 03 '18 at 20:26

score 4 · Answer 2 · answered Sep 03 '18 at 20:46

Here is a stronger result without assuming infinitude of $|\mathbb{k}|$. You can prove this by induction on the number of variables $n$.

Lemma. Let $d_1,d_2,\ldots,d_n$ be nonnegative integers such that, for every $i=1,2,\ldots,n$, $d_i$ is greater than or equal to the degree of $X_i$ in $p\left(X_1,X_2,\ldots,X_n\right)$. Suppose that $$|\mathbb{k}|>\max\left\{d_1,d_2,\ldots,d_n\right\}\,.$$ Let $S_1,S_2,\ldots,S_n\subseteq \mathbb{k}$ be such that $\left|S_i\right|>d_i$ for each $i=1,2,\ldots,n$. If $$p\left(x_1,x_2,\ldots,x_n\right)=0\text{ for all }\left(x_1,x_2,\ldots,x_n\right)\in S_1\times S_2\times\ldots\times S_n\,,$$ then $p\equiv 0$.

This is a lemma that can be used to prove the Combinatorial Nullstellensatz, by the way. — Batominovski, Sep 03 '18 at 20:58

score 2 · Answer 3 · answered Sep 03 '18 at 20:19

2

This is an answer to a similar question

answered Sep 03 '18 at 20:19

Jens Schwaiger

4,418

If $\mathbb{k}$ is an infinite field, $p \in \mathbb{k}[X_1, \dots, X_n]$ is zero iff $x \mapsto p(x)$ is the zero function.

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