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 $g(\alpha)=0$.

It must be the case that $f$ divides $g$ in $K[x]$.

Otherwise, $f$ and $g$ are relatively prime in $K[x]$.

It follows that there exist polynomials $h_1,h_2\in K[x]$ such that
 * $fh_1 + gh_2 = 1$.

Application of the evaluation homomorphism at $\alpha$ leads to a contradiction.

Thus, if $f$ and $g$ were both minimal polynomials of $\alpha$ over $K$, we must have


 * $f = cg$ for some $c\in K$.

However, both $f$ and $g$ are monic by definition, which implies that $c=1$.