Show that $\varphi_a: F[x] \to E$ given by $\varphi_a(f)=f(a)$ is a ring homomorphism with kernel $\langle p(x) \rangle$. 
Let $F$ be a field and let $p(x)$ be irreducible over $F$. If $E$ is a field which contains $F$ and there is an element $a$ in E such that $p(a)=0$, show that $\varphi_a: F[x] \to E$ given by $\varphi_a(f)=f(a)$ is a ring homomorphism with kernel $\langle p(x) \rangle$.

I've showed that $\varphi_a$ is a ring homomorphism, but I cannot show that $\ker \varphi_a=\langle p(x)\rangle$.
If $g \in \langle p(x)\rangle$, then $g(x)=p(x)h(x)$ for $h(x) \in F[x]$ and $\varphi_a(g)=p(a)h(a)=0 \implies g \in \ker \varphi_a$.
I don't know how to conclude the other inclusion. If $f \in \ker \varphi_a$, then $f(a)=0$ so $(x-a)$ is a factor of $f$. We can write $f(x)=(x-a)q(x)+r(x)$ for $q,r \in F[x]$.
What I need to show is that $f$ can be written as a product $p(x)a(x)$ for some $a \in F[x]$, but how can I get here?
 A: Note that since $F$ is a field, $F[X]$ is a Euclidean domain. Now suppose $q \in \ker \phi_a$. Then let $r$ be the gcd of $p, q$. Since $p$ is irrecucible and $r \mid a$, there are two possibilities up to multiplication by a unit. First, we could have $r = p$, in which case $q \in (p)$. Second, we could have $r = 1$. Then $1 = r \in \ker \phi_a$, since $r$ can be written as a linear combination of $p, q$. Then $1 = \phi_a(1) = 0$. This contradicts that $E$ is a field. This establishes $\ker \phi_a \subseteq (p)$. The other inclusion is trivial.
A: Hint
Wlog, we can assume $p$ is monic (since $F$ is a field).
Why not divide $m(x)$, where $m(x)$ is the minimal polynomial of $a$, into $p(x)$ instead?
We get $p(x)=q(x)m(x)+r(x)$, with $\rm{deg}(r(x))\lt\rm{deg}(m(x))$.  Then we get $r(a)=0$.  Hence $r\equiv 0$.  So, $p=qm$.
That's a contradiction.  We conclude that $p=m$.
Now you can finish your proof.
A: *

*$\ker \varphi_a$ is an ideal of $F[x]$


*$F[x]$ is a Principal Ideal Domain ( as $F$ is a field )


*$\ker \varphi_a=\langle m_a(x)\rangle$ $\quad$ $[m_a(x) $ minimal polynomial of $a$ in $F]$


*$m_a(x) \mid p(x) $ and $m_a(x), p(x) $ both are irreducible implies $p(x)=m_a(x) \cdot u$ where $u\in F$ is unit. In other words $p(x), m_a(x) $ are associate in $F[x]$


*$m_a(x) \mid p(x) $ implies $\langle p(x) \rangle\subset \langle m_a(x) \rangle$ and $p(x) \mid m_a(x) $ implies $\langle m_a(x) \rangle\subset \langle p(x)  \rangle$


*$\ker \varphi_a=\langle m_a(x) \rangle= \langle p(x)\rangle$
