Minimal Polynomial Exists/Proof 1

Proof
According to definition 1 of minimal polynomial, we ought to find $f \in K \sqbrk x$ such that:


 * $f \in K \sqbrk x$ is a monic polynomial of smallest degree such that $\map f \alpha = 0$

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

Therefore we can define:


 * $n = \min \set { \deg f : f \in K \sqbrk x, \map f \alpha = 0 }$

where $\deg f$ is the degree of $f$.

It follows that there is some $f \in K \sqbrk x$ such that $\deg f = n$.

Then $f$ is of smallest degree such that $\map f \alpha = 0$.

Let the first coefficient of $f$ be $a \in K$.

Since $K$ is a field, it follows that $g := f / a$ is monic.

Therefore $g$ is the sought monic polynomial of smallest degree such that $\map g \alpha = 0$.