Let $\alpha\_0$ be the unique non-trivial character satisfying $\alpha\_0^2=1$ of the split torus $\mathrm{T} \subset \mathrm{SL}(2,q)$ and denote by $\mathrm{R}(\alpha\_0)$ the character of $\mathrm{SL}(2,q)$ obtained by first extending $\alpha\_0$ to the Borel subgroup $\mathrm{B} \subset \mathrm{SL}(2,q)$ and then by inducing to the whole $\mathrm{SL}(2,q)$. It turns out that $\mathrm{R}(\alpha\_0)$ is a sum of two irreducible characters $\mathrm{R}\_{\pm}(\alpha\_0)$. The values of these characters can be explicitly computed and lie in $\mathbb{Q}\_p(\sqrt{q})$ (we consider representations over $\overline{\mathbb{Q}}\_p$). What can be said about the minimal field of definition of the representations corresponding to these characters? 


I am mainly interested in the case $q=p^2$ for odd $p$. In that case, the characters $\mathrm{R}\_{\pm}(\alpha\_0)$ take values in $\mathbb{Q}\_p$ (in fact even in $\mathbb{Z}$). Can the corresponding representations be defined over $\mathbb{Q}\_p$? Or at least over some non-ramified extension of $\mathbb{Q}\_p$?



Fields of definition of parabolically induced representations of $\mathrm{SL}(2,q)$
The answer is yes, the representation is defined over $\mathbb{Q}\_p(\sqrt{q})$. We first claim that $R\_{+}(\alpha\_0)$ and $R\_{-}(\alpha\_0)$ are not isomorphic. This can easily be seen by using the Bruhat decomposition and concluding that $dim(End(R(\alpha\_0)))=2$. 
Second, let $\psi$ be the character of $R\_{+}(\alpha\_0)$, and let us write $G=SL(2,q)$ which is a finite group. Then we have the idempotent $$e=\frac{1}{|G|}\sum\_{g\in G}\psi(g^{-1})g$$ in the group ring of $G$. This idempotent is already defined over $\mathbb{Q}\_p(\sqrt{q})$. One can now easily show that $R\_{+}(\alpha\_0) = eR(\alpha\_0)$. Since $R(\alpha\_0)$ and $e$ are defined over $\mathbb{Q}\_p(\sqrt{q})$, the same is true for $R\_{+}(\alpha\_0)$.


