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 $f \left({x}\right)$ be a minimal polynomial of $\alpha$ over $K$ of degree $n$.

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

Let $g \left({x}\right)$ be another nonzero polynomial over $K$ with $g(\alpha)=0$ of degree $n$.

Then we may assume that the coefficient of $x^n$ in $g \left({x}\right)$ is also $1$.

But then $f \left({x}\right) - g \left({x}\right)$ is a polynomial in $\alpha$ of degree less than $n$.

The only possibility is that $f \left({x}\right)$ and $g \left({x}\right)$ coincide.

So $f \left({x}\right)$ is unique.

Suppose that $f$ is not irreducible.

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

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


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

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 $f$.