Well, first I write some definitions:
Let $K|k$ be a field extension. Then $k(a)$ denotes $$ k(a)=\bigcap\{F:k\subseteq F \subseteq K \,\ a\in F \} $$ and is the smallest field of the extension $K|k$ such that $a$ belongs to it.
With this, and given $a$ in $K$, I set $$ \begin{array}{cccc} \phi:& k[x]&\to & K\\ & p(x)& \to & \phi(p(x))=p(a) \end{array} $$ This function is a ring homomorphism, such that $\phi(\alpha)=\alpha, \ \forall \alpha \in k$ and $\phi(x)=a$. What I don't get is the following:
$$\text{Im}(\phi)\subseteq k(a)$$
I'd appreciate any hint to prove this.
Hint
Show that $\mathrm{Im(\phi)}=\{p(a)\ |\ p\in k[x]\}$ is a field that contains $k$ and $a$ and show that every field contains $k$ and $a$ then contains $\mathrm{Im(\phi)}$.