Prove that $m(x)$ is is the minimum polynomial of $\alpha$ over $K$ ff $m(x)$ is irred in $K[x]$

Question

I was wondering if anyone could help me prove the last part of the following theorem,

Let $F$ be an extension of $K$ and $\alpha \in F$ be algebraic over $K$. Let $m(x)\in K[x]$, $m(x)$ monic and $m(\alpha)=0$. Prove that $m(x)$ is the minimum polynomial of $\alpha$ over $K$ $\iff$ $m(x)$ is irreducible in $K[x]$.

My proof follows as:

$(\rightarrow)$ Suppose that $m(x)$ is the min poly of $\alpha$ over $K$ then $\deg m(x)\ge 1$. Now suppose that $m(x)$ is reducible in $K[x]$, then $\exists f(x),g(x)\in K[x]$ with $1\le \deg(f(x),g(x))<\deg m(x)$ such that $m(x)=f(x)g(x)$ evaluating at $\alpha$ gives that $0=m(\alpha)=f(\alpha)g(\alpha)\implies f(\alpha)=0$ or $g(\alpha)=0$ this contradicts the definition of $m(x)$ thus $m(x)$ is irreducible in $K[x]$.

$(\leftarrow)$ Suppose $m(x)$ is irreducible in $K[x]$ then $\deg(m(x))\ge 1$ we want to show that $m(x)$ is the minimum polynomial of $\alpha$ over $k$. So let $z(\alpha)$ be the min poly of $\alpha$ over $K$, then This is where i'm stuck, could someone help me finish the proof?

Answer 1

You need to use that the minimal polynomial of $\alpha$ divides any other polynomial with $\alpha$ as a root.

Proof: let $p$ be the minimal polynomial, $q$ another with $\alpha$ as a root. By the division algorithm, there are polynomials $Q$, $R$ with $\deg{R}<\deg{p}$ so that $$ q(x) = Q(x)p(x)+R(x). $$ Evaluating this at $\alpha$, we find $$ 0 = q(\alpha) = Q(\alpha)p(\alpha)+R(\alpha) = R(\alpha). $$ But this contradicts that $p$ has minimal degree unless $R(x) = 0$.

If $m(x)$ is monic and irreducible, the only monic polynomial that divides $m(x)$ is $m(x)$ itself, so $m(x)$ is minimal.