$A$...$R$-algebra $\Rightarrow M_n(A)\cong M_n(R)\otimes A$ I would like to prove the following statement: if $A$ is an $R$-algebra (for a commutative ring $R$ with $1$), then the $R$-algebras $M_n(A)$ and $M_n(R)\otimes A$ are isomorphic.
By a proposition (Grillet, Abstract Algebra, p. 529) , we have the situation 

    where $\varphi((r_{i,j})_{i,j=1}^n):=(r_{i,j}\cdot 1_A)_{i,j=1}^n$ and $\psi(a):= (\text{matrix with }a\text{ in } 1,1\text{-th entry and }0\text{ elsewhere})$ and $\iota,\kappa$ are canonical. By the proposition, we have a unique algebra homomorphism $\chi$, such that $\chi\circ\iota=\varphi$ (1), $\chi\circ\kappa=\psi$ (2) and $\forall (r_{i,j})\!\otimes\!a: \chi((r_{i,j})\!\otimes\!a)=\varphi((r_{i,j}))\cdot\psi(a)$ (3). Let us prove that $\chi$ is bijective (and therefore an isomorphism).
surjective: By (1) and (2) we know that $im(\varphi)\subseteq im(\chi)$ and $im(\psi)\subseteq im(\chi)$. Let $E_{i,j}$ denote the matrix with $1_R$ at $i,j$-th entry and $0$ elsewhere. Since $E_{i,j}\in im(\varphi)$, $aE_{1,1}\in im(\psi)$ and $E_{i,j}E_{k,l}=\delta_{j,k}E_{i,l}$, it follows that $aE_{i,j}\in im(\chi)$, hence all matrices with entries in $A$ are in $im(\chi)$.
How can I prove that $\chi$ is injective?
 A: You can argue more generally that $\hom_R(P,Q)\otimes_RA\cong\hom_A(P\otimes_RA,Q\otimes_RB)$ when $P$ is finitely presented over $R$ and that the isomorphism is compatible with composition of maps, and then take $P=Q=R^n$.
A: Instead of showing that $\chi$ is injective, you could attempt to find $\hat\chi$ such that $\hat\chi\circ\chi$  is the identity-maps, then $\chi(u)=\chi(v)$ will imply that $u = \hat\chi\circ \chi(u) = \hat\chi\circ\chi (v) = v$, in other words $\chi$ is injective.
For instance (actually not for instance since it should be the only possibility), consider
$$ \hat\chi \colon M_n(A) \to  M_n(R)\otimes A \colon (c_{ij}) \mapsto \sum_{ij} E_{ij} \otimes c_{ij}. $$
-- Edit: Note that you should take $\psi(a)$ equal to $a\cdot 1$ (which is the diagonal matrix with all $a$'s) for this to work. The problem is that if you do it the way you did, I'm not sure that $im(\psi)$ and $im(\varphi)$ will commute, which they should if you want to apply the universal property. If you take $\psi(a)=a\cdot 1$ they will, because $1$ is central.
