Let $p(x)$ be an irreducible polynomial of degree $n$ over $F$. Let $c$ denote a root of $p(x)$ is some extension field of $F$ as in the basic theorem on field extensions. If $s(c) = t(c)$ in $F (c)$, where $s(x)$ and $t(x)$ have degree $< n$, prove that $s(x) = t(x)$.
I am trying to consider their difference and show it is equal to $0$, as this is a popular technique in mathematics. However, I'm a little confused on where to begin. Thanks!
Let $f(x)=s(x)-t(x)\in F[x]$. Suppose $f\neq 0$. Then $\deg f(x)<n$. Also, $c$ is a root of $f(x)$. Let $a$ be the leading coefficient of $p(x)$. Then $p(x)=am_{c,F}(x)$. Also, $\deg m_{c,F}(x)=n$ and $m_{c,F}(x)|f(x)$. This is a contradiction. So $s(x)=t(x)$.