Minimal Polynomial is Irreducible

Theorem
Let $L / K$ be a field extension and let $\alpha \in L$ be algebraic over $K$. Then the minimal polynomial of $\alpha$ over $K$ is unique and irreducible.

Proof
Let $f(x)$ 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 $1$. If $g(x)$ is another non-zero polynomial of $\alpha$ over $K$ of degree $n$, then we may assume that the coefficient of $x^n$ is $1$ also in $g(x)$. But then $f(x)-g(x)$ is polynomial of $\alpha$ of degree $<n$. The only possibility is that $f(x)$ and $g(x)$ coincide. So $f(x)$ 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 of $\alpha$:


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

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 $f$.