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 related to my earlier question on isomorphism of general quotients of $\:F\hspace{.02 in}[x]\:$.

Let $F$ be a field, let $p$ and $q$ be (non-zero) monic irreducible polynomials, let $I$ and $J$ be the ideals generated by $p$ and $q$ respectively, and assume the fields $\:F\hspace{.02 in}[x]/I\:$ and $\:F\hspace{.02 in}[x]/J\:$ are isomorphic.
Let $\: i : F\to F\hspace{.02 in}[x] \:$ be the canonical inclusion. $\;\;\;$ Does it follow that there are elements $a$ and $b$ of $F$ such that, for $\: L : F\hspace{.02 in}[x] \to F\hspace{.02 in}[x] \:$ the homomorphism given by $\;\; L\circ i \: = \: i \;\;$ and $\;\; L\hspace{.02 in}(x) \: = \: (a\hspace{-0.05 in}\cdot \hspace{-0.05 in}x)+b \;\;$,
one has $\;\; L\hspace{.02 in}(\hspace{.03 in}p\hspace{-0.02 in}) = q \:\:\:$?

share|cite|improve this question
What Francesco's example shows is that you should be asking not whether $L(p)$ equals $q$, but whether the ideal $\langle L(p) \rangle$ equals the ideal $\langle q \rangle$. Obviously for $a=1/\sqrt{2}$ and $b=0$ in Francesco's example, $L$ does satisfy the condition on ideals. – Jason Starr Apr 10 '14 at 18:00
up vote 7 down vote accepted

In general, it seems to me that the answer is no.

In fact, take $F=\mathbb{R}, \,$ $I=(x^2+1)$ and $J=(x^2+2)$.

Then $\mathbb{R}[x]/I$ and $\mathbb{R}[x]/J$ are both isomorphic to $\mathbb{C}$.

On the other hand, for any $a, b \in \mathbb{R}$ we have $$(ax+b)^2+1 = a^2 x^2 + 2ab x +b^2+1.$$ If the right-hand side were equal to $x^2+2$ then we should have $b ^2 = \pm 1$, hence $a=0$ and this gives a contradiction.

share|cite|improve this answer
A sort of generic example in this style is $F=k((u,v))$ (field of rational functions in two variables over a field $k$) with $p(x)=x^2-u$, $q(x)=x^2-v$; in this case both $F[x]/(p)$ and $F[x]/(q)$ are isomorphic to $F$, but the embeddings cannot be matched, essentially by the same argument as in the answer – მამუკა ჯიბლაძე Apr 4 '14 at 19:17
(Sorry, the notation should be $k(u,v)$, as $k((u,v))$ denotes something else) – მამუკა ჯიბლაძე Apr 5 '14 at 4:39

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.