# Annihilating Polynomial of Minimal Degree is Irreducible

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## Theorem

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 $\map f \alpha = 0$.

Then $f$ is irreducible in $K \sqbrk x$.

## Proof

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

Then there exist non-constant polynomials $g, h \in K \sqbrk x$ such that $f = g h$.

We have:

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

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

Therefore, as $\map g \alpha, \map h \alpha \in L$, either $\map g \alpha = 0$ or $\map h \alpha = 0$.

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

Hence the result, by Proof by Contradiction.

$\blacksquare$