Minimal Polynomial is Unique

Theorem
Let $L/K$ be a field extension and $\alpha\in L$ be algebraic over $K$.

Then the minimal polynomial of $\alpha$ over $K$ is unique.

Proof
Let $f$ be a minimal polynomial of $\alpha$ over $K$.

By Minimal Polynomial is Irreducible, we have that $f$ is irreducible over $K$.

Let $g$ be another polynomial in $K[x]$ such that $g(\alpha)=0$.

By the definition of minimal polynomial, $\operatorname{deg}(f)\leq\operatorname{deg}(g)$.

By the Division Theorem for Polynomial Forms over a Field, there exists polynomials $q,r\in K[x]$ such that:


 * $g = qf + r$ and $\operatorname{deg}(r)<\operatorname{deg}(f)$.

Suppose $\operatorname{deg}(r)>0$.

Then evaluating both sides of the equation above at $\alpha$ we have $r(\alpha)=0$, which contradicts the minimality of the degree of $f$.

Thus, $r$ is constant and equal to $0$.

We have now shown that $f$ divides all polynomials in $K[x]$ which vanish at $\alpha$.

By the monic restriction, it then follows that $f$ is unique.