Equivalence of Definitions of Minimal Polynomial

Theorem
Let $L / K$ be a field extension.

Let $\alpha \in L$ be algebraic over $K$.

1 equals 2
By Annihilating Polynomial of Minimal Degree is Irreducible, it follows that the two are equal.

1 equals 3
Let $f \in K[x]$ be the monic polynomial of smallest degree such that $f \left({\alpha}\right) = 0$.

Let $g \in K[x]$ be a polynomial.

If $f\mid g$, then $g = qf$ for some $q\in K[x]$.

Thus $g(\alpha) = f(\alpha) q(\alpha) = 0$.

Conversely, suppose $g(\alpha)=0$.

By the Division Theorem for Polynomial Forms over Field, there exists $q, r \in K \left[{X}\right]$ such that:
 * $g = q f + r$

and:
 * $r=0$ or $\deg r < \deg f$.

Evaluating this expression at $\alpha$ we find that:
 * $g \left({\alpha}\right) = q \left({\alpha}\right) f \left({\alpha}\right) + r \left({\alpha}\right) \implies r \left({\alpha}\right) = 0$

since $f \left({\alpha}\right) = g \left({\alpha}\right) = 0$.

But $f$ has minimal degree among the non-zero polynomials that are zero at $\alpha$.

Therefore $r = 0$.

Therefore:
 * $g = q f$

That is, $f$ divides $g$.