Field theory: prove that $(\forall x \in \mathbb{F}_p : f(x) = g(x)) \iff f - g \in \mathbb{F}_p[X](X^p - X)$.

Question

Let $p$ be a prime and $f, g \in \mathbb{F}_p [X]$. Prove: $(\forall x \in \mathbb{F}_p : f(x) = g(x)) \iff f - g \in \mathbb{F}_p[X](X^p - X)$.

I have proved the implication ($\implies$). However, for the other implications I am having struggles. Can anyone help me?

Answer 1

Suppose that $f - g \in \mathbb{F}_p[X](X^p-X)$.Then $f(X) - g(X) = h(X)(X^p-X)$, for some $h \in \mathbb{F}_p[X]$. But the polynomial function $X^p-X$ is identically $0$, then $f(x) = g(x)$, for all $x \in \mathbb{F}_p$