$\DeclareMathOperator\PSL{PSL}$ 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.
