Minimal Polynomial is Irreducible

Theorem

Let $L / K$ be a field extension.

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

Then the minimal polynomial in $\alpha$ over $K$ is unique and irreducible.

Proof

Let $\map f x$ be a minimal polynomial of $\alpha$ over $K$ of degree $n$.

From Minimal Polynomial is Unique we have that $\map f x$ is unique.

Since $K$ is a field, we may assume that the coefficient of $x^n$ is $1$.

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

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

By definition of $f$ as the minimal polynomial in $\alpha$:

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

From Field is Integral Domain, $L$ 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.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $71 \ \text{(i)}$