(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 $\;\; \phi \circ q_i \: = \: q_j \circ \psi \;\;$?

  • 1
    $\begingroup$ Do you demand that $\phi$ be $F$-linear? $\endgroup$ – S. Carnahan Oct 12 '12 at 6:47
  • $\begingroup$ @S. Carnahan: If so, $\psi$ would always exist even for $\phi$ without an inverse because $F[x]$ is free. $\endgroup$ – Noah Stein Oct 12 '12 at 7:37
  • 1
    $\begingroup$ 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. $\endgroup$ – S. Carnahan Oct 12 '12 at 7:56

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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\;\;$ $\endgroup$ – user5810 Oct 12 '12 at 9:20

Your Answer

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