Minimal Polynomial is Unique
Let $f$ be a minimal polynomial of $\alpha$ over $K$.
Let $g$ be another polynomial in $K \sqbrk x$ such that $\map g \alpha = 0$.
- $g = q f + r$ 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 $\alpha$.