Non-split Cartan subgroup of $\text{GL}_{2}(\mathbb{F}_{p})$ can not have an element with distinct rational eigenvalues

Question

Let $p$ be a prime. My goal is to prove that any Non-split Cartan subgroup of $\text{GL}_{2}(\mathbb{F}_{p})$ can not have an element with distinct rational eigenvalues. In particular, that it can not have an element with eigenvalues $1$ and $-1$.

In that sense, I have tried to prove that a Non-split Cartan subgroup of $\text{GL}_{2}(\mathbb{F}_{p})$ is isomorphic to $\mathbb{F}_{p^{2}}^{\times}$ but I do not even know if this would help at all.

To add some context, I am using the following definition of Non-split Cartan subgroup of $\text{GL}_{2}(\mathbb{F}_{p})$:

A Non-split Cartan subgroup of $\text{GL}_{2}$ is a group conjugate to $$G=\left\{\begin{pmatrix} a & \delta b\\ b & a \end{pmatrix}\right\}\subset\text{GL}_{2}(\mathbb{F_{p}}),$$ for some $\delta$ such that $\left(\frac{\delta}{p}\right)=-1$.

I have also been looking for some bibliography treating the theory behind this kind of groups in a detailed way, but I have not found anything so far.

If someone could help me, I would appreciate it a lot!

Thanks in advance!

Answer 1

Here is a shorter, better proof. Let $g=\begin{pmatrix}a\\&b\end{pmatrix}\in G=\mathrm{GL}_2(k)$ be a matrix such that $a\ne b\in k^\times$ and let $T\ni g$ be a maximal torus. Then since $T$ is a maximal torus, we have $C_G(T)=T$, where $C_G(T):=\{g\in G:gt=tg,\forall t\in T\}$ is the centralizer of $T$. The inclusion $g\in T$ implies the reverse inclusion $C_G(g)\supseteq C_G(T)=T$. But we see that $C_G(g)=\{\begin{pmatrix}a\\&b\end{pmatrix}:a,b\in k^\times\}$, which is itself a maximal torus. Thus, by the maximality of $T$, we have $T=C_G(g)$ is the split maximal torus. Thus, we have shown that any torus containing an element with distinct rational eigenvalues is split, and conversely, any non-split torus does not contain such an element.

Answer 2

Claim 1: The maximal tori in $\mathrm{GL}_2(k)$ are the split torus $k^\times\times k^\times=\{\begin{pmatrix}a\\&b\end{pmatrix}:a,b\in k^\times\}$ and the anisotropic torus $\ell^\times\subset\mathrm{GL}_2(k)$, where $\ell/k$ is the unique quadratic extension and $\ell^\times$ acts on the $2$-dimensional $k$-vector space $\ell$ in the obvious way. (as algebraic groups, they are $\mathbb G_m\times\mathbb G_m$ and $R_{\ell/k}\mathbb G_m$, respectively.)

Indeed, for any maximal torus $T\subset\mathrm{GL}_2$ over $k$, the base change $T_{\overline k}\subset \mathrm{GL}_2$ over $\overline k$ is conjugate to the split torus $T_0:=\mathbb G_m\times\mathbb G_m$, carrying the Frobenius $F_{T_0}\colon\begin{pmatrix}x\\&y\end{pmatrix}\mapsto\begin{pmatrix}x^q\\&y^q\end{pmatrix}$. Thus, there exists a $g\in\mathrm{GL}_2(\overline k)$ such that $T=gT_0g^{-1}$. Now since $T$ is $k$-rational, in particular $F(T)=T$, and hence $F(g)T_0F(g)^{-1}=gT_0g^{-1}$, i.e., $g^{-1}F(g)\in N_{\mathrm{GL_2}}(T_0)=T_0\rtimes C_2$. Now $g^{-1}F_Tg\colon T_0\to T_0$ acts as conjugation by $g^{-1}F(g)$, composed with $F_{T_0}$.

Thus if $g^{-1}F(g)\in T_0$ then $g^{-1}F_Tg$ acts as $F_{T_0}$ on $T_0$, hence by Galois descent, we have $T\cong\mathbb G_m\times\mathbb G_m$ over $k$. On the other hand, if $g^{-1}F(g)\in wT_0$ where $w=\begin{pmatrix}&1\\1\end{pmatrix}$ is a lift of the nontrivial element in the Weyl group of $T_0$, then $g^{-1}F_Tg$ acts as $wF_{T_0}$ on $T_0$, i.e., by $\begin{pmatrix}x\\&y\end{pmatrix}\mapsto \begin{pmatrix}y^q\\&x^q\end{pmatrix}$. By Galois descent, this gives the restriction of scalars $R_{\ell/k}\mathbb G_m$ (with $k$-rational points $\ell^\times$).

Claim 2: $\ell^\times\subset\mathrm{GL}_2(k)$ contains no element with distinct rational eigenvalues. This follows from the following observation: the eigenvalues of $x\in \ell^\times$ viewed as an endomorphism $\ell\to\ell$ is $x$ and $x^q$. But if $x$ is rational, then $x=x^q$, a contradiction.