Zeros of polynomial over an infinite field

Solution 1:

Let $k$ be an infinite field, $K/k$ any field extension, and $f\in K[X_1,\ldots, X_n]$ a non-constant polynomial. Let $S$ be the set of all $m\in\{0,1,\ldots,n\}$ such that there exist $c_1,\ldots, c_n\in K$ such that $f(c_1,\ldots, c_n)\ne 0$ and $c_i\in k$ for $i\le m$. We already know that $0\in S$. Let $s=\max S$. So there exist $c_1,\ldots, c_n\in K$ such that $f(c_1,\ldots, c_n)\ne 0$ and $c_i\in k$ for $i\le s$.

Assume $s<n$ and let $g(X)=f(c_1,\ldots, c_s,X,c_{s+2},\ldots, c_n)\in K[X]$. By maximality of $s$, $g(x)=0$ for all $x\in k$. But then $g$ has infinitely many roots and must be the zero polynomial. On the other hand, $g(c_{s+1})\ne0$, contradiction. We conclude that $s=n$, i.e., there exist $c_1,\ldots, c_n\in k$ such that $f(c_1,\ldots, c_n)\ne 0$.

Solution 2:

The result follows from the following slightly more general result.

Theorem Let $K$ be an infinite field and $k$ an infinite subset, let $n\in\mathbb{N}$ and $f\in K[X_1,X_2,\ldots,X_n]$, the ring of polynomials in $n$ indeterminates over $k$. Then $f=0$ iff $f(c_1,c_2,\ldots,c_n)=0$ for all $c=(c_1,c_2,\ldots,c_n)\in k^n$.

The proof uses induction. For $n=1$ this follows from the fact that any polynomial in one variable $\not=0$ has at most $m$ roots where $m$ denotes the degree of the nonzero polynomial. Now let $f$ be as above. Then $$f=\sum_{i=0}^N f_i(X_1,X_2,\ldots,X_{n-1}) X_n^i\text{ with polynomials } f_i\in K[X_1,X_2,\ldots,X_{n-1}].$$ Now choose $c_1,c_2,\ldots,c_{n-1}\in k$. Then $f(c_1,c_2,\ldots,c_{n-1},X_n)$ is a polynomial in one variable which vanishes for all $c_n\in k$ implying that $f_i(c_1, c_2, \ldots, c_{n-1})=0$ for all $i$. This holds true for all $(c_1, c_2, \ldots, c_{n-1})\in k^{n-1}$. Thus by hypothesis all $f_i=0$, i.e. $f=0$. The reverse assertion holds true trivially.