Elementary problem about proving non-equivalence of representations I'm trying to solve the following exercise.
Exercise. Let $E=F(x)$ where $x$ is transcendental over $F$. Prove the group (I assume this means subgroup of $\mathrm{GL}_2(E)$) generated by $\begin{pmatrix}
1 & x \\
0 & 1 \end{pmatrix}$ and $ \begin{pmatrix}
1 & 1 \\
0 & 1 \end{pmatrix} $ is not equivalent to any subgroup of $\mathrm{GL}_2(K)$ where $K$ is a finite extension of $F$.
I haven't studied representations before so I'm a bit lost here. First of all, I take it "equivalent groups" means equivalent as linear groups.


*

*Since isomorphic representations must have linearly isomorphic underlying spaces, I guess I'm (not) looking for an intertwining $F$-linear isomorphism $f:F(x)\oplus F(x)\cong K\oplus K$. Is that correct?

*I guess I'm supposed to reach a contradiction here, so I should suppose there's an intertwining $F$-linear isomorphism $f$ as above i.e satisfying $g\cdot f(v)=f(g\cdot v)$. I don't understand how to get a contradiction with the transcendence of $x$ over $F$.

Context. The exercise appears in a set of notes written by the lecturer in a basic course "topics in linear groups". Paraphrasing, the relevant section begins like this: Given a field extension $E/F$ and a group $G$, an $F$-linear rep of $G$ extends to an $E$-linear rep of $E$. Now I quote the first line that confuses me:

If a representation $\psi:G\to \mathrm{GL}_n(E)$ is equivalent to a representation $\rho\to \mathrm{GL}_n(F)$ then...

The author motivates the section by noting that for equivalent such representations (over $E,F$ with $E/F$), the traces take values in $F$, and raises the question of the converse. That is, if the traces of an $E$-linear rep take values in $F$, is the $E$-linear rep equivalent to some $F$-linear rep?
"Equivalent representations" were only defined when they're over the same field (just isomorphic reps). Maybe the meaning of "equivalence" over difference fields should have been my question.

Added. As suggested by Eric Wofsey - a reasonable definition would be "having conjugate images inside the general linear group of the vector space with scalars extended to an algebraic closure of $E$".
 A: I'm not completely sure without more context, but I would guess that "equivalent groups" here means conjugate subgroups of $GL_2(\bar{E})$ where $\bar{E}$ is an algebraic closure of $E$.  So, writing $G$ for the subgroup described, the goal is to prove that there does not exist any finite extension $K$ of $F$ and $a\in GL_2(\bar{E})$ such that $aga^{-1}$ has entries in $K$ for all $g\in G$.
As a hint for how to do this, the matrices in $G$ become easier to deal with when you subtract the identity matrix from them.  A full solution is hidden below.

 Let $g=\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ and $h=\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix}$.  Then $g,h\in G$, $g,h\neq I$, and $x(g-I)=h-I$.  This equation is preserved by conjugation, so any subgroup equivalent to $G$ contains elements $g',h'\neq I$ such that $x(g'-I)=h'-I$  But then any field containing the entries of both $g'$ and $h'$ must contain $x$, since a nonzero entry of $h'-I$ divided by the corresponding entry of $g'-I$ is $x$.  In particular, it is impossible for both $g'$ and $h'$ to have entries in $K$ if $K$ is a finite extension of $F$.

