# $A$…$R$-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?

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ugh, help, how can I write diagrams? – Leon Apr 2 '11 at 1:31

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.

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Excellent, a very good idea. Just one tiny question: why is $\chi\circ\hat{\chi}=id$ already enough and not $\chi\circ\hat{\chi}=id=\hat{\chi}$\circ\chi? – Leon Apr 2 '11 at 15:43
@Leon Lampreet: Actually $\chi\circ\hat\chi$ is not enough, but $\hat\chi\circ\chi$ is. This is because if you know that $\hat\chi\circ \chi = id$ you can proof injectivity of $\chi$ like this: $\chi(u)=\chi(v)\implies\hat\chi(\chi(u))=\hat\chi(\chi(v))\implies u=v$. Now we have that $\chi$ is surjective and injective indeed, therefore $\chi$ is a bijective morphism, as requested. – Myself Apr 2 '11 at 15:50
thank you, it was quite educational for me :) – Leon Apr 2 '11 at 17:17

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 $P=Q=R^n$.

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(Where A = B?) Is this general case easier then? – Myself Apr 2 '11 at 2:07
@Myself: there are less things that you can do wrong :) – Mariano Suárez-Alvarez Apr 2 '11 at 2:15