Minimal Polynomial is Unique/Proof 1

Proof
Let $f$ be a minimal polynomial of $\alpha$ over $K$ according to definition 1.

By Minimal Polynomial is Irreducible, we have that $f$ is irreducible over $K$.

Let $g$ be another polynomial in $K \sqbrk x$ such that $\map g \alpha = 0$.

By the definition of minimal polynomial, $\map \deg f \le \map \deg g$, where $\deg$ denotes degree.

By Division Theorem for Polynomial Forms over Field, there exist polynomials $q, r \in K \sqbrk x$ such that:


 * $g = q f + r$ and $\map \deg r < \map \deg f$.

Suppose $\map \deg r > 0$.

Then evaluating both sides of the equation above at $\alpha$, we obtain $\map r \alpha = 0$.

This contradicts the minimality of the degree of $f$.

Thus, $r$ is constant and equal to $0$.

We have now shown that $f$ divides all polynomials in $K \sqbrk x$ which vanish at $\alpha$.

By the monic restriction, it then follows that $f$ is unique.