TITLE: Do surface groups embed into PSL_2 over a real quadratic integer ring?
QUESTION [7 upvotes]: $\DeclareMathOperator\PSL{PSL}$ Let $ \mathbb{Z} $ be the ring of integers and $ \mathbb{R} $ the field of real numbers. Let $ \Sigma_g $ be a surface of genus $ g \geq 2 $. Let $ \pi_1(\Sigma_g) $ be the fundamental group of the surface. There are many way to embed $ \pi_1(\Sigma_g) $ into $\PSL_2(\mathbb{R}) $. There are, however, no ways to embed $ \pi_1(\Sigma_g) $ into $ \PSL_2(\mathbb{Z}) $. Given some $ g \geq 2 $, is there a good way (an algorithm) to find a real algebraic integer $ \alpha $  such that $ \pi_1(\Sigma_g) $ embeds in $ \PSL_2(R) $? Here $ R $ is the ring
$$
R:=\mathbb{Z}[\alpha]
$$
(Preferably $ \alpha $ is just a (real) quadratic extension. That is, $ \alpha $ is the root of some polynomial $ x^2+bx+c $ where $ b^2-4c \geq 0 $ and $ b,c \in \mathbb{Z} $.)
History of the question: The original question claimed that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ and asked for embeddings into $ \PSL_2(\mathbb{F}) $ where $ \mathbb{F} $ is a finite degree field extension of $ \mathbb{Q} $. The claim that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ is false. In fact there are many such embeddings.
The first edit of the question fixed this and asked instead for an embedding into $ \PSL_2(R) $ for $ R $ a finite rank extension of $ \mathbb{Z} $ by algebraic integers. The current version of the question is the second edit.

REPLY [4 votes]: Here is, at Moishe Kohan's request, an optimal answer to the original question (which asked for a representation over any number field; as i noted in the comments there exists plenty of surface group representations with rational coefficients, as follows from a theorem of Takeuchi).

To prove the existence of single a $\mathrm{PSL}_2(\mathbb Q)$-representation of a surface group is quite simpler than the full proof of Takeuchi's theorem (which is not super hard itself in the torsion-free case). Of course it suffices to prove it in genus 2. To do so observe first that it is possible to find two matrices $A_1, B_1 \in \mathrm{SL}_2(\mathbb Q)$ such that they generate a discrete subgroup of $\mathrm{SL}_2(\mathbb R)$ and
$$ A_1B_1A_1^{-1}B_1^{-1} = \left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right) $$
for some $a \in \mathbb Q \setminus \{0, \pm 1\}$ and the quotient $\langle A_1, B_1 \rangle \backslash \mathbb H^2$ is a one-holed torus (the set of matrices $A, B$ satisfying the last condition is an open set, so rational matrices are dense there, and we can always conjugate to diagonalise the commutator over $\mathbb Q$). Geometrically, in the half-plane model, this means that a fundamental domain for the action of $\langle A_1, B_1 \rangle$ on the convex hull of its limit set is a polygon with one side on the geodesic $c$ the geodesic from 0 to $\infty$ and the adjacent sides are orthogonal to $c$; the boundary component of the quotient torus (the convex core of $\langle A_1, B_1 \rangle \backslash \mathbb H^2$, let us call it $T$) is the image of $c$.
Now let $\sigma$ be the reflexion in $c$, and $A_2, B_2$ be the conjugates of $A_1, B_1$ by $\sigma$. Then $\langle A_2, B_2 \rangle \subset \mathrm{PSL}_2(\mathbb Q)$, and the quotient of $\mathbb H^2$ by $\Gamma = \langle A_1, B_1, A_2, B_2 \rangle$ is the double of $T$, that is a genus 2 surface. Hence $\Gamma$ is a genus 2 surface group inside $\mathrm{PSL}_2(\mathbb Q)$ which is discrete in $\mathrm{PSL}_2(\mathbb R)$.

Takeuchi's argument for the full theorem is similar to this, except using Calabi--Weil rigidity instead of the geometric argument, and being more precise about commutators. It is also possible to prove it using Fenchel--Nielsen coordinates.