MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(This is inspired by the answer to my earlier question.)

Does there exist

a field $F$ $\:$ and $\:$ two ideals $I$ and $J$ of $F[x]$ $\:$ and $\:$ a ring isomorphism $\: \phi : F[x]/I \to F[x]/J$

such that when $\: q_i : F[x] \to F[x]/I \:$ and $\: q_j : F[x] \to F[x]/J \:$ are the quotient maps,
there does not exist an endomorphism $\: \psi : F[x] \to F[x] \:$ such that $\;\; \phi \circ q_i \: = \: q_j \circ \psi \;\;$?

share|cite|improve this question
Do you demand that $\phi$ be $F$-linear? – S. Carnahan Oct 12 '12 at 6:47
@S. Carnahan: If so, $\psi$ would always exist even for $\phi$ without an inverse because $F[x]$ is free. – Noah Stein Oct 12 '12 at 7:37
Yes, I would also like to know if $\phi$ is an isomorphism of rings, abelian groups, sets, $F[x]$-modules, or something else. Right now, the question is rather underspecified. – S. Carnahan Oct 12 '12 at 7:56
up vote 1 down vote accepted

The answer is yes, if $\phi$ is an isomorphism of rings:

Take $F$ to be the reals, and let $I$ and $J$ both be the ideal of $F[x]$ generated by $x^2+1$. Let $q=q_i=q_j$ be the projection map. The quotient $E:=q(F)$ is isomorphic to the complex numbers. Let $\phi$ be any automorphism of $E$ taking $2^{1/4}$ to $2^{1/4}i$, where $2^{1/4}$ is a real fourth root of 2 and $i$ is $q(x)$.

If $\psi$ is any endomorphism of $F[x]$ then $$q(\psi(2^{1/4}))=q(\pm2^{1/4})=\pm2^{1/4},$$ whereas $$\phi(q(2^{1/4}))=\phi(2^{1/4})=2^{1/4}i.$$ Thus $q\circ\psi\ne\phi\circ q$.

The first equation follows from the fact that $\psi$ preserves $\mathbb{Q}$-conjugates, and also $\psi$ must map $F$ into $F$ because $\psi$ preserves units.

share|cite|improve this answer
So, the answer is yes, if Boolean Prime Ideal Theorem. $\:$ However, that's a simple enough use of it for me to suspect that an example can be proven in ZF. $\;\;$ – Ricky Demer Oct 12 '12 at 9:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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