Minimal Polynomial is Unique/Proof 2

Proof
Let $f, g \in K \sqbrk x$ be minimal polynomials for $\alpha$ according to definition 2.

That is, let $f$ and $g$ be irreducible monic polynomials in $K \sqbrk x$ with $\map f \alpha = \map g \alpha = 0$.

Suppose $f$ and $g$ are distinct.

Then $f$ and $g$ are coprime.

Thus there exist polynomials $a, b \in K \sqbrk x$ with $a f + b g = 1$.

Taking the evaluation homomorphism in $\alpha$, we obtain the contradiction that $0=1$.