Let $F$ be a field and $f(x)$ an irreducible polynomial in $F[x]$ such that $\alpha$ is a root of it: $f(\alpha)=0$. Now, let $(f(x))\subset F[x]$ denote the ideal generated by $f(x)$. My question is: given an arbitrary polynomial $h(x)\in F[x]$, is it true that $h(x)\in (f(x)) \Leftrightarrow h(\alpha)=0$?
The $\Rightarrow$ is easy. But what about the $\Leftarrow$?
Thanks for reading!
Consider the evaluation homomorphism $\text{ev}_\alpha:F[x] \rightarrow F[\alpha]$ where $g(x) \mapsto g(\alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.
The kernel of this homomorphism is an ideal in $F[x]$, which will be the set of all polynomials with $\alpha$ as a root. Since $F$ is a field, then $F[x]$ is a principal ideal domain. That is, every ideal in $F[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $\ker(\text{ev}_\alpha)$ must be generated by a single element.
Can you see why that element must be $f$? And why $h(x) \in \langle f(x) \rangle$?