Let $A$ be a central simple algebra over some field $k$, with degree $n$. There is a functor $F$ defined by the assignment, for a commutative ring $R$, $$ F(R)=\operatorname{Iso}_{R\text{-alg}}(A\otimes_k R,M_n(R)). $$
I'm writing $\operatorname{Iso}_{R\text{-alg}}(A\otimes_k R,M_n(R))$ to be the set of algebra isomorphisms $A\otimes_k R\to M_n(R)$.
Is there a nice way to see that this is in fact a scheme over $k$?
Proposition: Let $k$ be a commutative ring and let $A,B$ be two $k$-algebras whose underlying $k$-modules are finitely generated free. Then the functor $F : \mathsf{CAlg}_k \to \mathsf{Set}$ defined by $R \mapsto \mathrm{Isom}_{\mathsf{Alg}_R}(A \otimes_k R,B \otimes_k R)$ on objects and by $(R \to S) \mapsto (f \mapsto f \otimes_R S)$ on morphisms is a representable functor, i.e. an affine scheme over $k$.
Proof (sketch). Let us look at the functor $F_{A,B}$ defined by $F_{A,B}(R)= \mathrm{Hom}_{\mathsf{Alg}_R}(A \otimes_k R,B \otimes_k R)$. Assume that $F_{A,B}$ is representable for all $A,B$. Then also $F_{B,A}$ is representable. It follows that $F_{A,B} \times F_{B,A}$ is representable, too (namely, by the coproduct of the commutative $k$-algebras representing $F_{A,B}$ and $F_{B,A}$). Now, $F$ is the subfunctor $F_{A,B} \times F_{B,A}$ consisting of those pairs $(f,g)$ for which $f \circ g = \mathrm{id}$ and $g \circ f = \mathrm{id}$. Thus, $F$ is really the equalizer of two morphisms $F_{A,B} \times F_{B,A} \rightrightarrows F_{A,A} \times F_{B,B}$. It follows that $F$ is representable, too (namely, via some coequalizer).
Therefore, it suffices to prove that $F_{A,B}$ is representable. Choose a presentation $A=\langle (X_i)_{i \in I} : (P_j)_{j \in J} \rangle_{\mathsf{Alg}_k}$. Here, each $P_j$ is a non-commutative polynomial in the variables $X_i$ and with coefficients in $k$. We may assume that $I$ is finite. We compute: $$F_{A,B}(R) = \mathrm{Hom}_{\mathsf{Alg}_{\Large R}}(A \otimes_k R,B \otimes_k R) \cong \mathrm{Hom}_{\mathsf{Alg}_{\Large k}}(A,B \otimes_k R)$$ $$\cong \{c \in (B \otimes_k R)^I : \forall j \in J \, (P_j(c)=0)\}$$ Choose a finite $k$-basis $\{b_l : l \in L\}$ the underlying $k$-module of $B$. Then $(B \otimes_k R)^I \cong R^{I \times L}$ as $R$-modules. The proof will be finished when we show that each $P_j(c)=0$ is a polynomial equation in the entries of $c \in R^{I \times L}$.
To see this, we might as well replace $B$ over $k$ by $B \otimes_k R$ over $R$ and thus assume $k=R$. (This simplifies the notation a little bit.) So we have some $R$-algebra $B$ whose underlying $R$-module is freely generated by $\{b_l : l \in L\}$. Also, we have a polynomial $P$ in the non-commuting variables $(X_i)_{i \in I}$ with coefficients in $R$. Let $c \in B^I$. Write $c = (\sum_l r_{i,l} \cdot b_l)_{i \in I}$. We want to show that the condition $P(c)=0$ is a polynomial equation in coefficients $r_{i,l} \in R$.
I will show this via some simple example, because the general case should then be clear (if not, let me know!), only blown up with indices and ugly notation ...
Let's look at $I=L=\{1,2\}$, $c_1 = r b_1$ and $c_2 = s b_2$. Thus, our equation is $(r b_1) \cdot (s b_2)=0$, i.e. $(rs) b_1 \cdot b_2=0$. Write $b_1 \cdot b_2$ as a linear combination of $b_1,b_2$, for example $b_1 \cdot b_2 = u b_1 + v b_2$. Then our equation becomes $rs u = 0$ and $rs v = 0$. These are polynomial equations in $r,s$ with coefficients in $u,v \in R$. In the general case, the coefficients can be generated by the structure constants $c_{l,l',l''} \in R$, which are defined by $b_l \cdot b_{l'} = \sum_{l''} c_{l,l',l''} \cdot b_{l''}$. $\checkmark$