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If $R$ is a unital ring and $M_{2\times 2}(R)$ is a commutative ring, then $R$ is a trivial ring because if $$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & 0\\0 & 0\end{pmatrix}, $$ then $0=1.$

However, $M_{2\times 2}(R)$ can be commutative for possibly non-unital rings of any size. That is, for any cardinal number $\kappa$ (finite or not), there exists an abelian group of order $\kappa.$ Equipping this group with the zero multiplication gives a rng $R$ such that $M_{2\times 2}(R)$ has zero multiplication and so is commutative. The trivial ring is also in this class.

Are there any examples of rngs $R$ whose multiplication is non-zero and such that $M_{2\times 2}(R)$ is commutative?

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What does it mean to equip a group with zero multiplication to obtain a rng? – anon Jun 30 '12 at 12:51
@anon For an abelian group $(A,+,0)$, define $\cdot$ by $x\cdot y=0$. This multiplication is associative and distributes over $+$ (on both sides). Unless $A=\{0\}$ though, there is no identity element in the structure we obtain, so it is a rng. – user23211 Jun 30 '12 at 12:53
up vote 3 down vote accepted

No. Expanding the first cell of the product of arbitrary matrices, we would have to have $aa'+bc'=a'a+b'c$ for any $a,b,c,a',b',c'$, so it suffices to look at the particular case $a=a'=b'=c=0$:

$$\begin{pmatrix}0&x\\0&0\end{pmatrix}\begin{pmatrix}0&0\\y&0\end{pmatrix}=\begin{pmatrix}xy&0\\0&0\end{pmatrix}\\ \begin{pmatrix}0&0\\y&0\end{pmatrix}\begin{pmatrix}0&x\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&yx\end{pmatrix}$$ Therefore $xy=0$ for any $x,y\in R$, and the family of such rngs is exactly the abelian additive groups equipped with zero multiplication.

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Right, thanks a lot! – user23211 Jun 30 '12 at 13:29

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