adjoint map and number field Let $\mathbb{k}$ be a totally real number field and let 
$$\text{SL}^{\pm}(n+1,\mathbb{R}) = \{ A \in \text{GL}(n+1,\mathbb{R}) | \det A = \pm 1 \}.$$
Let
$$\text{SL}^{\pm}(n+1,\mathbb{k}) = \{ A \in \text{GL}(n+1,\mathbb{k}) | \det A = \pm 1 \}.$$
Consider the adjoint representation as a map from $\text{SL}^{\pm}(n+1,\mathbb{R})$ to the automorphism group of the Lie algebra $\mathfrak{sl}(n+1,\mathbb{R})$. The kernel of this map is $\{ I, -I \}$. Let $\gamma \in \text{SL}^{\pm}(n+1,\mathbb{k})$. Then since $\text{Ad}$ is a $\mathbb{Q}$-morphism, clearly the image $\text{Ad}(\gamma)$ is an $n^{2}+2n$ by $n^{2}+2n$ matrix whose entries are in $\mathbb{k}$. 
Now conversely suppose that $Ad(\gamma)$ is an $n^{2}+2n$ by $n^{2}+2n$ matrix whose entries are in $\mathbb{k}$. Then does $\gamma$ have to be in $\text{SL}^{\pm}(n+1,\mathbb{k})$? 
If not, can anyone give an counter example?
 A: The answer is Yes. The only reason for the appearance of formally real fields and $\mathbb R$ in the question is to rule out roots of unity; in fact, we have the following (cleaner and more general) statement:
Theorem 1. Let $L$ be a field of characteristic $0$, and $K$ be a subfield of $L$. Let $N\in\mathbb N$. Let $U\in\mathrm{GL}_N\left(L\right)$ be such that every $V\in\mathfrak{sl}_N\left(K\right)$ satisfies $UVU^{-1}\in \mathrm{M}_N\left(K\right)$. Then, there exists a $\lambda\in L^{\times}$ such that $\lambda U \in\mathrm{GL}_N\left(K\right)$.
[You might need a moment to see why this answers your question. Here is a dictionary: My $N$ is your $n+1$. My $U$ is your $\gamma$. My $K$ is your $\mathbb k$. My $L$ is your $\mathbb R$. Your condition that $\mathrm{Ad}\left(\gamma\right)$ is my condition that every $V\in\mathfrak{sl}_N\left(K\right)$ satisfies $UVU^{-1}\in \mathrm{M}_N\left(K\right)$. I'm leaving to you to check that, in your case, the $\lambda$ must be $\pm 1$; this is where $\mathbb R$ and the determinant of $\gamma$ being $\pm 1$ come into play, and it is completely trivial.]
Proof of Theorem 1. First, let us show that
(1) every $V\in M_N\left(K\right)$ satisfies $UVU^{-1}\in \mathrm{M}_N\left(K\right)$.
Proof of (1). Since $K$ has characteristic $0$ (this is the only place where we are using this assumption!), we have $\mathrm{M}_N\left(K\right)=\mathfrak{sl}_N\left(K\right)+K\cdot I_N$, where $I_N$ is the $N\times N$ identity matrix. Since (1) is linear in $V$, we are therefore done with the proof of (1) once we have shown that (1) holds for all $V\in\mathfrak{sl}_N\left(K\right)$ and for all $V\in K\cdot I_N$. But (1) holds for all $V\in\mathfrak{sl}_N\left(K\right)$ by assumption, and for all $V\in K\cdot I_N$ by inspection. Thus, (1) is proven.
Now, consider the map $r : M_N\left(K\right) \to M_N\left(K\right)$ which maps every $V$ to $UVU^{-1}$. This $r$ is well-defined due to (1), and a $K$-algebra isomorphism as can be easily seen; hence, $r$ is a $K$-algebra automorphism of $M_N\left(K\right)$. But it is known (see, e. g., the line after Corollary 2.12 in Chapter IV of Milne's Class Field Theory, or see Corollary 1 in http://ysharifi.wordpress.com/2011/01/26/automorphisms-of-central-simple-algebras-2/ , or any other source on central simple algebras and Brauer groups) that all $K$-algebra automorphisms of $M_N\left(K\right)$ are inner. Hence, $r$ is inner, so there exists some $P\in\mathrm{GL}_N\left(K\right)$ such that every $V\in\mathrm{M}_N\left(K\right)$ satisfies $r\left(V\right) = PVP^{-1}$. So every $V\in\mathrm{M}_N\left(K\right)$ satisfies
$PVP^{-1} = r\left(V\right) = UVU^{-1}$.
This can be easily rewritten as $U^{-1}PV\left(U^{-1}P\right)^{-1}=V$. In other words, $U^{-1}PV=VU^{-1}P$. Since this holds for all $V\in\mathrm{M}_N\left(K\right)$, it must also hold for all $V\in\mathrm{M}_N\left(L\right)$ (because it is a linear equation in $V$, so it is enough to check it on an $L$-basis of $\mathrm{M}_N\left(L\right)$, but such a basis can be chosen to lie in $\mathrm{M}_N\left(K\right)$). In other words, the matrix $U^{-1}P$ lies in the center of $\mathrm{M}_N\left(L\right)$. But this center is known to be $L\cdot I_N$ (this is, again, a known fact from the theory of central simple algebras, but it also appears in most linear algebra books). Thus, $U^{-1}P\in L\cdot I_N$. In other words, there exists some $\lambda\in L$ such that $U^{-1}P = \lambda I_N$. This $\lambda$ is nonzero (else, $P$ would be zero, contradicting $P\in\mathrm{GL}_N\left(K\right)$), so this becomes $P=\lambda U$. Hence, $\lambda U=P\in\mathrm{GL}_N\left(K\right)$, proving Theorem 1.
