Let F be a field and p(t) an irreducible polynomial on F[t]. Let g(t) be an arbitrary polynomial in F[t]. Suppose that there exists an extension field E of F and an element 'a' in E which is a root of both p(t) and g(t). Prove that p(t) divides g(t) in F[t].
Would anyone know how to approach this problem/solve it?
Consider the ideal $I = \{f \in F[t]|f(a)=0\}$(Check that this is indeed an ideal.)
As $F[t]$ is a PID, $I$ is principal. So $I$ must be generated by some polynomial, say $I=(f(t))$. As $I$ is a proper ideal, $f(t)$ is not a unit. Now $p(t) \in I$, thus $f(t) | p(t)$. By irreducibility, we must have $f(t) = p(t)$ (up to a unit.) Thus $I = (p(t))$ But $g(t) \in I$, so $p(t)|g(t)$.