Are there ever exotic isomorphisms between quotients of F[x]? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:37:11Z http://mathoverflow.net/feeds/question/109436 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109436/are-there-ever-exotic-isomorphisms-between-quotients-of-fx Are there ever exotic isomorphisms between quotients of F[x]? Ricky Demer 2012-10-12T06:33:04Z 2012-10-12T09:14:06Z <p>(This is inspired by the answer to my <a href="http://mathoverflow.net/questions/65109/checking-if-fx-i-is-isomorphic-to-fx-j" rel="nofollow">earlier question</a>.) <br><br><br> Does there exist</p> <p>a field $F$ $\:$ and $\:$ two ideals $I$ and $J$ of $F[x]$ $\:$ and $\:$ a ring isomorphism $\: \phi : F[x]/I \to F[x]/J$</p> <p>such that when $\: q_i : F[x] \to F[x]/I \:$ and $\: q_j : F[x] \to F[x]/J \:$ are the quotient maps, <br> there does <em>not</em> exist an endomorphism $\: \psi : F[x] \to F[x] \:$ such that $\;\; \phi \circ q_i \: = \: q_j \circ \psi \;\;$?</p> http://mathoverflow.net/questions/109436/are-there-ever-exotic-isomorphisms-between-quotients-of-fx/109441#109441 Answer by SJR for Are there ever exotic isomorphisms between quotients of F[x]? SJR 2012-10-12T08:24:17Z 2012-10-12T08:24:17Z <p>The answer is yes, if $\phi$ is an isomorphism of rings:</p> <p>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 $q(x)$.</p> <p>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 $q\circ\psi\ne\phi\circ q$.</p> <p>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.</p>