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

notwhether $L(p)$ 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