The theorem is:
Let $K$ be an extension field of $F$, and let $\alpha \in K$ be algebraic. Suppose that $f(x) \in F[x]$. Then, the following are equivalent:
1) $f(x)$ is monic, irreducible, and $f(\alpha) = 0$.
2) $f(x)$ is monic, $f(\alpha) = 0$, and $f(x)$ divides any $g(x)$ with $g(\alpha) = 0$.
I need to prove that 2) $\implies$ 1), but I don't really see what I need to prove. The monic and $f(\alpha)$ = $0$ parts are included in both statements, so does that mean I just need to prove the division part $\implies$ irreducibility? Any help would this problem would be greatly appreciated. Thank you.
Yes, this is what you have to prove. To prove $(2)$ implies $(1)$, suppose $f(x) = g(x)h(x)$. Then $f(\alpha) = g(\alpha)h(\alpha) = 0$, so $g(\alpha) = 0$ or $h(\alpha) = 0$ (since $K$ is a field); WLOG, suppose $g(\alpha) = 0$. Then by hypothesis, $g(x) = f(x)k(x)$ for some $k(x) \in F[x]$. Then $f(x) = h(x)k(x)f(x)$, and since $f$ is monic (and therefore nonzero), $h(x)k(x) = 1$, so $h(x)$ is a unit, i.e. $f$ is irreducible.
To prove $(1)$ implies $(2)$, suppose $f(x)$ is irreducible, and consider the ideal $I = \{\langle g(x) \in F[x] \mid g(\alpha) = 0\}$; note that $I$ contains $f$. Since $F[x]$ is a PID, $I$ is principal, generated by some $h(x) \in F[x]$. Since $f \in I$, $h$ divides $f$, and so must be a unit or $f$ (up to multiplication by a unit); but no unit vanishes at $\alpha$, so we must have $h = cf$ for some $c \in F^{\times}$, and thus $f$ divides every $g \in F[x]$ such that $g(\alpha) = 0$.