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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?

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3 Answers 3

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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.

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  • $\begingroup$ Why can you conclude that $r \equiv 0$ just from $r(a)=0$? Also with respect to what is this a contradiction? $\endgroup$
    – Walker
    Aug 11, 2022 at 9:08
  • $\begingroup$ 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. $\endgroup$
    – me too
    Aug 11, 2022 at 15:35
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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.

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  • $\begingroup$ $q\in p$? Maybe a little nitpicky but pretty clear you mean the ideal generated by $p$. Nice solution though. $\endgroup$
    – me too
    Aug 12, 2022 at 1:12
  • $\begingroup$ @Cpc Indeed, nice catch on the typo. $\endgroup$ Aug 12, 2022 at 1:14
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  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. $\ker \varphi_a=\langle m_a(x) \rangle= \langle p(x)\rangle$

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