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 Minimal Polynomial is Irreducible, it follows that the two are equal.

1 equals 3
Let $f \in K \sqbrk x$ be the monic polynomial of smallest degree such that $\map f \alpha = 0$.

Let $g \in K \sqbrk x$ be a polynomial.

If $f \divides g$, then $g = q f$ for some $q \in K \sqbrk x$.

Thus $\map g \alpha = \map f \alpha \, \map q \alpha = 0$.

Conversely, suppose $\map g \alpha = 0$.

By the Division Theorem for Polynomial Forms over Field, there exists $q, r \in K \sqbrk x$ such that:
 * $g = q f + r$

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

Evaluating this expression at $\alpha$ we find that:
 * $\map g \alpha = \map q \alpha \, \map f \alpha + \map r \alpha \implies \map r \alpha = 0$

since $\map f \alpha = \map g \alpha = 0$.

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 $g$.