Kronecker's theorem explanation.

Question

Can someone explain why $a-b \in F$ means that $a-b = 0$?

Answer 1

"$a-b$ must be a multiple of the polynomial $p(x)$" means $a-b = q(x) p(x)$ for some $q(x) \in F[x]$. If $q(x) \neq 0$ then deg $p \geq 1$ implies deg $(q p) \geq 1$. But if $a-b \in F$, then deg $pq = 0$, so $q(x)=0$ and hence $a-b=0.$

Answer 2

Note that $a$ and $b$ are elements of the field $F$, so $a-b\in F$ and thus $a-b$ is a constant polynomial in $F[x]$. Now, let's suppose that $a-b\neq 0$, because $p(x)\mid a-b$ there is some polynomial $q(x)\in F[x]$ such that $a-b=p(x)q(x)$ and hence $$\deg(a-b)=0=\deg(pq)=\deg(p)+\deg(q)\ge 1,$$ which is an absurd. Therefore $a-b=0$ (every non-zero polynomial divides the zero polynomial).