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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 $\;\; L\hspace{.02 in}(\hspace{.03 in}p\hspace{-0.02 in}) = q \:\:\:$?

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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 at 18:00

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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.

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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 at 19:17
    
(Sorry, the notation should be $k(u,v)$, as $k((u,v))$ denotes something else) –  მამუკა ჯიბლაძე Apr 5 at 4:39

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