# If $\phi$ is an F-map from $K$ to $E$, both field extensions of $F$, then $\alpha \in K$ and $\phi (\alpha)$ have the same minimum polynomial

Definition of an F-map:

If $$K$$ and $$E$$ are field extensions of $$F$$, an F-map is a homomorphism, $$\phi: K \rightarrow E$$ such that $$F$$ is fixed.

I'm reading over a proof of why the number of distinct F-maps from $$K$$ to $$E$$ is less than or equal to $$[K:F]$$ (assuming finite) and the fact that $$\alpha$$ and $$\phi (\alpha)$$ have the same minimum polynomial is one of the facts stated, though I can't see why.

• Because, if you exapand, you'll notice that $f(\phi(\alpha))=\phi(f(\alpha))$ for all $f\in F[x]$. – Saucy O'Path Apr 14 at 19:57

Let $$p(X)=\sum a_kX^k\in F[X]$$. Then $$\phi(p(\alpha))=\sum \phi(a_k)\phi(\alpha)^k=\sum a_k\phi(\alpha)^k=p(\phi(\alpha))$$ Conclude that $$p(\alpha)=0\iff p(\phi(\alpha))=0.$$