Prove that $p(x)$ is irreducible in $F[x]$

Question

Let $F$ be a field and let $K$ be an extension of $F$. Let $\alpha$ be algebraic over $F$. Let $p(x)$ be the polynomial of minimal degree having $\alpha$ as a root. Prove that $p(x)$ is irreducible in $F[x]$. This is a new concept for me about field extension and i am having difficulty with this and understanding the notation because i kept confusing myself with quotient. I was thinking is it safe to say that $$F[x]/(p(x)) \cong F$$ since $F$ is a field thus $F[x]/(p(x))$ is also a field and also a finite integral domain thus $p(x)$ is maximal, hence prime thus in an integral domain a prime element is always irreducible. Therefore $p(x)$ is irreducible.

Answer 1

Hints:

== $\;p(x)\;$ is irreducible $\;\implies p(x)=g(x)h(x)\implies g(x)\,\,or\,\,h(x)\;$ is constant

== $\;p(x)\;$ is the minimal polynomial of $\;\alpha\;$ over $\;\Bbb F\implies\;\forall\;h(x)\in\Bbb F[x]\;$ , then $\;h(\alpha)=0\iff p(x)\mid h(x)\;$ .

Further hint for the second hint: divide with residue (Euclidean Algorithm) $\;h(x)\;$ by $\;p(x)\;$ and apply the condition $\;h(\alpha)=0\;$ and minimality of $\;p(x)\;$.

Answer 2

Suppose $p(x)=q(x)r(x)$ with both $q$ and $r$ of positive degree, necessarily strictly smaller than the degree of $p$. Then $q(\alpha)r(\alpha)=0$, so either $q(\alpha)=0$ or $r(\alpha)=0$, contradicting the assumption that the degree was minimal.