# Evaluation map in field extensions. $F$-algebra homomorphism.

Let $K$ be a field extension of $F$, and let $a \in K$. Show that the evaluation map ev$_{a}: F[x] \longrightarrow K$ given by ev$_{a}(f(x)) = f(a)$ is ring and an $F$-vector space homomorphism.

I already showed that it is a ring homomorphism (I can write the proof), but I couldn't prove that is a $F$-vector space homomorphism. Someone can helps me? I like any hint.

Hint : Since you already know that $ev_a(f+g)=ev_a(f)+ev_a(g)$, you only need to show that $ev_a(\lambda f)=\lambda ev_a(f)$ for $\lambda$ in $F$ and $f\in F[x]$. But you already know that $ev_a(\lambda f)=ev_a(\lambda)ev_a(f)$, so what you actually need to show is that $ev_a(\lambda)=\lambda$ for any $\lambda$ in $F$.
View $\lambda\in F$ as a constant polynomial in $F[x]$.