Annihilating Polynomial of Minimal Degree is Irreducible

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Let $L / K$ be a field extension.

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

Let $f \in K[x]$ be a nonzero polynomial of minimal degree such that $f(\alpha) = 0$.

Then $f$ is irreducible in $K[x]$.


Aiming for a contradiction, suppose that $f$ is not irreducible.

Then there exist non-constant polynomials $g, h \in K \left[{x}\right]$ such that $f = g h$.

We have:

$0 = f \left({\alpha}\right) = g \left({\alpha}\right) h \left({\alpha}\right)$

Since $L$ is a field, it is an integral domain.

Therefore, as $g \left({\alpha}\right), h \left({\alpha}\right) \in L$, either $g \left({\alpha}\right) = 0$ or $h \left({\alpha}\right) = 0$.

This contradicts the minimality of the degree of $f$.

Hence the result, by Proof by Contradiction.


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