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.


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.

  • $\begingroup$ True, since $\mathbb{Z}[\sqrt-5]$ is not a UFD $\endgroup$ – Belgi Apr 9 '12 at 22:12
  • 1
    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Apr 9 '12 at 22:14
  • $\begingroup$ You take me back to the lectures in ring theory :) again you are right $\endgroup$ – Belgi Apr 9 '12 at 22:16
  • $\begingroup$ @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. $\endgroup$ – Shannon Apr 9 '12 at 22:22
  • $\begingroup$ @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. $\endgroup$ – 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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.