Minimal Polynomial Exists/Proof 2

Proof
According to definition 2 of minimal polynomial, we ought to find $f \in K \sqbrk x$ such that:


 * $f \in K \sqbrk x$ is an irreducible, monic polynomial such that $\map f \alpha = 0$

Since $\alpha$ is algebraic over $K$, there is some $f \in K \sqbrk x$ such that $\map f \alpha = 0$.

By Polynomial Forms over Field form Unique Factorization Domain, $\map f x$ has a complete factorization:


 * $\map f x = a \cdot \map {g_1} x \cdots \map {g_n} x$

where $a \in K, a \ne 0$, and the $g_i$ are irreducible and monic.

Since $\map f \alpha = 0$, it follows that for some $g_i$:


 * $\map {g_i} \alpha = 0$

as required.