Minimal Polynomial is Unique/Proof 2

Theorem

Let $L / K$ be a field extension.

Let $\alpha \in L$ be algebraic over $K$.

Then the minimal polynomial of $\alpha$ over $K$ is unique.

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 $0=1$.

$\blacksquare$