Matrix Algebra Over the Opposite Ring I'm trying to show that $M_n(R^{opp}) \cong M_n(R)^{opp}$ as rings. My thought was that $\varphi : M_n(R)^{opp} \to M_n(R^{opp})$ defined by $\varphi (A) = A^{T}$ is the desired isomorphism. But I'm a little confused as to what there is to verify...
Clearly $\varphi^{-1} = \varphi$, so it is bijective, $\varphi (A+B) = (A+B)^{T} = A^{T} + B^{T}$, and $\varphi (A \star B) = \varphi (BA) = (BA)^{T} = A^{T} B^{T} = \varphi (A) \varphi (B)$, where $\star$ is the opposite product of $M_n(R)^{opp}$.
Is that it? Seems like a cheesy proof. I tried consulting some of the users in the chatroom. The only one to proffer a critique was Thorgott. He said that "I don't think $\varphi^{-1} = \varphi$, given that codomain and domain are different." After thinking about this for some time, I don't think this is a problem, because although $M_n(R^{opp})$ and $M_n(R)^{opp}$ denote different rings, the underlying sets are the same.
 A: The entire argument is hidden in the step $(BA)^T = A^T B^T$ and the notation obscures what's going on slightly; on the LHS the matrices are being multiplied in $M_n(R)$ and on the RHS the matrices are being multiplied in $M_n(R^{op})$ (and the statement is false if $R$ is noncommutative and both matrix multiplications are interpreted in the same matrix ring). Once you write out what's going on here in coordinates it should be clear what's going on and what you have to prove.
Maybe the cleanest way to write this proof is to introduce explicit notation for the "identity" map $(-)^{op} : M_n(R) \to M_n(R^{op})$. $\varphi(A)$ should really be defined as $(A^{op})^T = (A^T)^{op}$ (the "opposite transpose"). Now you're no longer sweeping anything under the rug: you need to prove that
$$((BA)^T)^{op} = (A^T)^{op} (B^T)^{op}$$
(now the op's have been properly inserted!) which will then give $\varphi(A \star B) = \varphi(A) \varphi(B)$.
As for Thorgott's objection, this is a notational issue again; write $\varphi_R$ for the transpose map $M_n(R)^{op} \to M_n(R^{op})$ and you'll see that the inverse of $\varphi_R$ is $(\varphi_{R^{op}})^{op}$, where the "op" here has yet another meaning: if $f : R \to S$ is a ring homomorphism then it has an opposite ring homomorphism $f^{op} : R^{op} \to S^{op}$ which on underlying sets is just $f$ again.
A: You might consider that $M_n(R)$ is (isomorphic to) the ring of endomorphisms of $R_R^n$ (right $R$-module).
Thus $M_n(R^{\mathrm{op}})$ is the ring of endomorphisms of $(R^{\mathrm{op}})^n_{R^{\mathrm{op}}}$, which is the same as the ring of endomorphisms of $_RR^{n}$ (left $R$-module), which is $M_n(R)^{\mathrm{op}}$.
A: It's usually best to avoid unnecessary detail, but in this case, since there are so many multiplication operations to confuse (in $R$, $R^{op}$, $M_n(R^{op})$, $M_n(R)$ and $M_n(R)^{op}$) it might be worth just writing the details.  (There's no ambiguity with the addition operation, of course.)
I'd like to use, respectively, juxtaposition, $\ast$, $\star$, $\circ$, $\bullet$ for these operations.
We'll abbreviate matrices by writing the $i,j$ entry between brackets.
Given two matrices $[a_{ij}]$, $[b_{ij}]$ we would have, by definition
$\phi([a_{ij}])=[a_{ji}]$
$[a_{ij}]\circ[b_{ij}]=[\sum_k a_{ik}b_{kj}]$
$[a_{ij}]\star[b_{ij}]=[\sum_k a_{ik}\ast b_{kj}]$
With that in mind
$$
\phi([a_{ij}]\bullet[b_{ij}])=\phi([b_{ij}]\circ[a_{ij}])=\phi([\sum_k b_{ik}a_{k,j}])\\ =[\sum_k b_{jk}a_{ki}]=[\sum_k a_{ki}\ast b_{jk}]=\phi([a_{ij}])\star\phi([b_{ij}])
$$
One computation, five operations!  Possibly not aesthetically pleasing, but at least straightforward bookkeeping.
