# Central simple algebras give rise to algebraic groups

Let $D$ be a central division algebra over a field $k$ of dimension $n^2$. I have heard that the functor

$$R \mapsto (D \otimes_k R)^{\ast}$$

going from commutative $k$-algebras to groups is representable by an affine $k$-scheme $G$, and moreover that the group scheme $G$ is a connected, reductive group over $k$. Where can I read more about this?

In fact, there is (almost) nothing special about $D$ being a central division algebra here. Let $F$ be a field, and let $A$ be a unital, associative $F$-algebra of finite dimension $n$. Then the functor
$$\mathrm{GL}_{1}(A) \colon \mathrm{Alg}_{F} \to \mathrm{Groups}, R \mapsto (A \otimes_{F} R)^{\times}$$
is an affine group scheme over $F$. The representing algebra is given by $B = S(A^{\ast})[1/N]$, where $A^{\ast}$ is the dual of $A$, $S(A^{\ast})$ is the symmetric algebra of $A^{\ast}$, and $N \colon A \to F$ is the norm map, considered as an element of $S^{n}(A^{\ast})$. The comultiplication is $c \colon B \to B \otimes_{F} B$ is induced by the map $A^{\ast} \to A^{\ast} \otimes_{F} A^{\ast}$ dual to the multiplication map $A \otimes_{F} A \to A$.
One specific case which you are almost certainly familiar with is the case $A = \mathrm{End}_{F}(V)$, where $V$ is a finite dimensional vector space over $F$. In this case, $\mathrm{GL}_{1}(A)$ is just $\mathrm{GL}(V)$, the general linear group scheme associated to $V$. If $V = F^{n}$, then $\mathrm{GL}_{1}(A) = \mathrm{GL}_{n}$, and the representing algebra can be more explicitly identified as the polynomial ring over $F$ in $n^{2}$ variables $X_{ij}, 1 \leqslant i, j \leqslant n$, with the determinant of the matrix $X = (X_{ij})$ inverted.
It depends on what details you're looking for, but a good reference for this is Chapters VI and VII of The Book of Involutions by Knus, Merkurjev, Rost, and Tignol. For instance, it is also proved, or at least mentioned, that $\mathrm{GL}_{1}(A)$ is connected and smooth. It is also proved that $H^{1}(F, \mathrm{GL}_{1}(A)) = 1$, which is the most general formulation of Hilbert's Theorem 90 that I am aware of. It does not mention whether $\mathrm{GL}_{1}(A)$ is reductive in general (this is my "almost" caveat above), but perhaps it can be deduced in your specific case from other statements available in the book.