Minimal Polynomial is Unique

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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 $K \left[{x}\right]$ such that $g \left({\alpha}\right) = 0$.


By the definition of minimal polynomial, $\operatorname{deg} \left({f}\right) \le \operatorname{deg} \left({g}\right)$, where $\operatorname{deg}$ denotes degree.

By Division Theorem for Polynomial Forms over Field, there exist polynomials $q, r \in K \left[{x}\right]$ such that:

$g = q f + r$ and $\operatorname{deg} \left({r}\right) < \operatorname{deg} \left({f}\right)$.


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

Then evaluating both sides of the equation above at $\alpha$, we obtain $r \left({\alpha}\right) = 0$.

This 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 $K \left[{x}\right]$ which vanish at $\alpha$.

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

$\blacksquare$