Prove that for $f,g\in F[x]$, where $F$ is an infinite field, if $f(a)=g(a)$ for infinitely many elements $a\in F$, then $f=g$

Question

Prove that for $f,g\in F[x]$, where $F$ is an infinite field, if $f(a)=g(a)$ for infinitely many elements $a\in F$, then $f=g$.

I'm not entirely sure how to tackle the "infinitely many elements of a of field F" portion of this proof. How would you even account for all those a's?

Answer 1

Actually, the number of zeros of a non-zero polynomial is less then or equal to its degree. This is a general theorem which is true for any polynomial over any field. (Note that the number can indeed be less then the degree, for example $p(x)=x^2+1$ has no zeros at all in $\mathbb{R})$.

Now, assuming the above result, reaching a contradiction is easy: assume $f(x)\neq g(x)$ are two different polynomials. Then their difference $f(x)- g(x)$ is a non-zero polynomial.

Hence, its number of roots is bounded by its degree and in particular, is finite: $\Rightarrow$ contradiction.

Answer 2

The main fact is that if $a \in F$ is a zero of $f$, then we can write $f(x)=(x-a)q(x)$. This is a consequence of the division algorithm for polynomials because in general we have $f(x)=(x-a)q(x)+f(a)$.

This implies that if $f$ has $m$ zeros, then the degree of $f$ is at least $m$, unless $f=0$.

Apply this to $f-g$ and conclude that $f-g=0$ because it has infinitely many zeros.