This is related to [my earlier question](https://mathoverflow.net/q/65109/5810) 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 $\;\; L\hspace{.02 in}(\hspace{.03 in}p\hspace{-0.02 in}) = q \:\:\:$?
  
  




Can non-isomorphic field extensions be isomorphic fields?
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.


