An absolutely irreducible SL2 representation has a quadratic trace-field model

Find a noncentral semisimple word and a word meeting both eigenspaces. Some group matrix has trace different from $2,-2$. Otherwise, if all matrices were scalar the representation would be reducible; choose a nonscalar $D=ϵI+N$ with $ϵ=\pm 1$, $N\mathrm{\ne }0$ and $N^2=0$ by [F1]. For every group matrix $C$ and integer $m\ge 0$, $\mathrm{tr}(D^{m}C)={ϵ}^m(\mathrm{tr}C+m{ϵ}^{-1}\mathrm{tr}(NC))\mathrm{.}$ All these traces belong to $\{2,-2\}$ by the supposition. In characteristic zero a nonconstant affine function on the nonnegative integers takes infinitely many values, so $\mathrm{tr}(NC)=0$ for every $C$. In a basis over $F$ with $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ this forces the lower-left entry of every $C$ to be zero. Its first coordinate line would be invariant, a contradiction. Thus choose a word $A$ with trace different from $\pm 2$, and adjoin one root $\xi$ of its characteristic polynomial. Its two distinct eigenvalues are $\xi ,{\xi }^{-1}$, so it is diagonalizable already over $F(\xi )$. In that eigenbasis some group word $B$ has both off-diagonal entries nonzero. Indeed the elements preserving the first and second eigenlines form two proper subgroups of $\mathrm{\Gamma }$ by absolute irreducibility. A group cannot be the union of two proper subgroups: for elements in their respective differences, their product belongs to neither. Choose $B$ outside that union. These are finitely many existential selections, not an infinite choice principle.

Normalize and recover every generator by trace equations. In the chosen $A$ eigenbasis write $B_{12}=u\mathrm{\ne }0$ and conjugate by $\mathrm{diag}(u^{-1},1)$ to make this entry one. The lower-left entry becomes the nonzero product $q=B_{12}{B}_{21}$. Put $a=\mathrm{tr}A$, $b=\mathrm{tr}B$, $c=\mathrm{tr}(AB)$. In the normalized basis $A=\left(\begin{array}{cc}\xi & 0\\ 0& {\xi }^{-1}\end{array}\right),B=\left(\begin{array}{cc}{d}_1& 1\\ q& d_2\end{array}\right),d_1=\frac{c-{\xi }^{-1}b}{\xi -{\xi }^{-1}},d_2=b-d_1,q=d_1{d}_2-1\mathrm{\ne }0.$ All these entries lie in $K^{\mathrm{\prime }}=K(\xi )$. For each generator $C$ put $e=\mathrm{tr}C$, $f=\mathrm{tr}(AC)$, $g=\mathrm{tr}(BC)$ and $h=\mathrm{tr}(ABC)$, which belong to $K$ by its definition. Its diagonal entries are $c_{11}=\frac{f-{\xi }^{-1}e}{\xi -{\xi }^{-1}},c_{22}=e-c_{11}\mathrm{.}$ Its off-diagonal entries are uniquely solved from $c_{21}+q{c}_{12}=g-d_1{c}_{11}-d_2{c}_{22},\xi c_{21}+{\xi }^{-1}q{c}_{12}=h-\xi d_1{c}_{11}-{\xi }^{-1}{d}_2{c}_{22}\mathrm{.}$ The determinant of this two-by-two linear system is $q({\xi }^{-1}-\xi )\mathrm{\ne }0$. Thus every normalized generator matrix has entries in $K^{\mathrm{\prime }}$. These are the actual conjugated original matrices in $F(\xi )$, not merely formal matrices with the correct traces: the normalization was performed first and the displayed linear equations uniquely recover their entries. Therefore their determinant-one equations and every group relation hold in $K^{\mathrm{\prime }}$, by its injection into $F(\xi )$. The conjugated representation is an actual model over $K^{\mathrm{\prime }}$, with no new relator-existence assumption.

The quadratic swap is explicit conjugation. If $K^{\mathrm{\prime }}=K$ there is nothing to prove. Otherwise its automorphism $\tau$ exchanges $\xi$ and ${\xi }^{-1}$ and fixes every word trace. The formulas of step 2.1 give $\tau (d_1)=d_2$, $\tau (q)=q$. Put $P=\left(\begin{array}{cc}0& 1\\ q& 0\end{array}\right),P^{-1}=\left(\begin{array}{cc}0& q^{-1}\\ 1& 0\end{array}\right)\mathrm{.}$ Direct multiplication gives $PA{P}^{-1}=\tau A$ and $PB{P}^{-1}=\tau B$. For each generator $C$, both $PC{P}^{-1}$ and $\tau C$ have traces $e,f,g,h$ when paired respectively with $I,\tau A,\tau B,(\tau A)(\tau B)$: conjugation preserves trace, and $\tau$ fixes the four word traces. The diagonal/off-diagonal recovery in step 2.1 applies to this swapped pair as well; its determinant is still nonzero. Uniqueness therefore gives $PC{P}^{-1}=\tau C$ for every generator, hence for all group words. This proves simultaneous conjugacy over $K^{\mathrm{\prime }}$ itself, without an appeal to pointwise complex-character classification.

Projective descent for compatible invariant constructions. Under the hypotheses on $z$, applying $\tau$ to its construction yields the point constructed from $\tau \rho$, while conjugation invariance and step 3.1 identify that point with $z$. Hence $\tau z=z$. Choose one nonzero homogeneous coordinate $z_j$ and normalize it to $1$. Equality of projective points implies that each ratio $z_i\mathrm{/}{z}_j$ is fixed by $\tau$. Every element of the quadratic field has the unique form $u+v\xi$ with $u,v\in K$; being fixed gives $v(\xi -{\xi }^{-1})=0$, so $v=0$. Thus every normalized coordinate lies in $K$, proving that $z$ is a $K$-point. In the trivial-extension case this is immediate. The argument uses only finite matrix and field operations; it neither asserts that an arbitrary merely pointwise invariant assignment is rational nor supplies any analytic or chain-complex prerequisite for constructing $z$.