Algebraic Element of Field Extension is Root of Unique Monic Polynomial of Minimal Degree

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Let $L / K$ be a field extension.

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

Then there exists a unique monic polynomial $f \in K \sqbrk x$ of smallest degree such that $\map f \alpha = 0$, called the minimal polynomial.



Follows from Well-Ordering Principle and because $K$ is a field.



Let $f$ and $g$ be two such monic polynomial.


$\map f \alpha - \map g \alpha = 0$

Aiming for a contradiction, suppose $f - g \ne 0$.

Let $a$ be the leading coefficient of $f-g$.

Then $\dfrac 1 {a} \paren {f - g}$ is another such polynomial of strictly smaller degree.

This is a contradiction.

Thus $f - g = 0$.