# $\langle x\rangle$ is a maximal ideal of $R[x]$, where $R$ is a field

Show that $\langle x\rangle$ is a maximal ideal of $R[x]$, where $R$ is a field.

This is one of my assignment questions. It is supposed to be hard but what I did below is so easy but I couldn't find anything wrong. Can anybody help me check it out?

Since $R$ is a field and $f(x) = x$ is irreducible in $R[x]$, so $R[x]/\langle x\rangle$ is a field, and $\langle x\rangle$ is a ideal of $R[x]$, so $\langle x\rangle$ is maximal.

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You are correct that if you can prove that $R[x]/\langle x\rangle$ is a field, then you can conclude that $\langle x\rangle$ is maximal.

However, it is not true in general that if $a$ is irreducible in a domain $D$, then $D/\langle a\rangle$ is a field. For example, $2$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$, but $\langle 2\rangle$ is not a maximal ideal of $\mathbb{Z}[\sqrt{-5}]$. So you need more than just "$f(x) = x$ is irreducible in $R[x]$" to conclude "$R[x]/\langle x\rangle$ is a field."

(For another, simpler example, consider the case of $\langle x \rangle$ in $R[x,y]$. You can still say that "since $R$ is a field, $x$ is irreducible", but it is no longer true that $R[x,y]/\langle x\rangle$ is a field; in fact, $\langle x\rangle$ is not maximal in this ring, since it is properly contained in the proper ideal $\langle x,y\rangle$.)

But perhaps you can show that the homomorphism $R[x]\to R$ given by "evaluation at $0$" is onto, and has kernel $\langle x\rangle$? That will show that $R[x]/\langle x\rangle \cong R$ is a field.

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True, since $\mathbb{Z}[\sqrt-5]$ is not a UFD –  Belgi Apr 9 '12 at 22:12
@Belgi: True; but you need more than just UFD. $x$ is irreducible in the UFD $\mathbb{R}[x,y]$, but $\langle x\rangle$ is not maximal. In general, is $a$ is irreducible in a domain, then $\langle a\rangle$ is maximal among principal ideals. But it's possible that it may be nonmaximal; PID suffices for the implication, which will happen here, but the proffered reasons are insufficient. –  Arturo Magidin Apr 9 '12 at 22:14
You take me back to the lectures in ring theory :) again you are right –  Belgi Apr 9 '12 at 22:16
@ArturoMagidin I did show that R[x]/<x> is isomorphic to R is a field, then I noticed that there is actually a theorem that says if F is a field and monic p(x) in F[x] is irriducible in F[x], then F[x]/<p(x)> is a field. –  Shannon Apr 9 '12 at 22:22
@Shannon: Then you need to invoke the theorem explicitly; like I said, what you write is not sufficient on its face, since there are situations where "$a$ is irreducible" does not imply $D/\langle a\rangle$ is a field. I would suggest just showing that $R[x]/\langle x\rangle$ is isomorphic to $R$ and be done. –  Arturo Magidin Apr 9 '12 at 22:24

If you know that $R[x]$ is a PID when $R$ is a field, then any ideal above $\langle x\rangle$ must be generated by some monic polynomial $f$ which divides $x$. Hence $f=x$ or $f=1$ and so the ideal is either $\langle x\rangle$ or $R[x]$, and $\langle x\rangle$ is maximal.

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Hint $\rm\ f\not\in\! (x)\ \Rightarrow\ (x,f) = (x,\: f\ mod\ x) = (x,f(0)) = (1)\:$ by $\rm\:f(0)\neq 0\ \Rightarrow\ f(0)$ unit

Or: $\rm\ mod\ x\!:\:\ x\equiv 0\ \Rightarrow\ f(x)\equiv f(0) =$ unit or $0,\:$ so $\rm\:R[x]\ mod\ x\:$ is a field.

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