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?

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.

• Why can you conclude that $r \equiv 0$ just from $r(a)=0$? Also with respect to what is this a contradiction? Aug 11, 2022 at 9:08
• Because $r$ has smaller degree than $m$. But $m$ was assumed to have minimal degree among polynomials with $p(a)=0$. The contradiction is that $p=qm$, but $p$ is irreducible. Aug 11, 2022 at 15:35

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 irreducible 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.

• $q\in p$? Maybe a little nitpicky but pretty clear you mean the ideal generated by $p$. Nice solution though. Aug 12, 2022 at 1:12
• @Cpc Indeed, nice catch on the typo. Aug 12, 2022 at 1:14
1. $$\ker \varphi_a$$ is an ideal of $$F[x]$$

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

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

4. $$m_a(x) \mid p(x)$$ 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 $$F[x]$$

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

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