Basic facts about matrix algebras

Let $R$ be a ring, and let $m,n$ be positive integers.

Proposition 1. The $R$ -algebras $M_m(M_n(R))$ and $M_{mn}(R)$ are isomorphic.

The obvious map between is clearly a bijection and can be seen to be an $R$ -algebra homomorphism by observing that multiplication works as expected in each.

Proposition 2. If $S$ is an $R$ -algebra, then there is a natural isomorphism $M_n(R)\otimes_R S\to M_n(S).$

Let $\varphi:M_n(R)\otimes S\to M_n(S)$ be defined by $m\otimes_R s\mapsto sm.$ This map is clearly an $s$ -algebra homomorphism.

Let $E^{a,b}$ be the $n\times n$ matrix with $(E^{a,b})_{i,j}=1$ if $(i,j)=(a,b)$ and $(E^{a,b})_{i,j}=0$ otherwise. Observe that the map $\varphi$ is onto as the $E^{a,b}$ matrices generate $M_n(S)$ as an $S$ -algebra.

Suppose that $m\otimes_R s\mapsto 0.$ Then either $m=0$ , $s=0$ , or $sm_{i,j}=0$ for all $i,j.$ Writing $m=\sum_{i,j}m_{i,j}E^{i,j}$ we see that $(m\otimes_R s)=\sum_{i,j} (m_{i,j}E^{i,j}\otimes_R s)=\sum_{i,j} (E^{i,j}\otimes_R sm_{i,j})=0.$ Thus $\varphi$ is injective and hence an isomorphism.

Proposition 3. If $I$ is an ideal of $R$ , let $M_n(I)$ denote the subset of $M_n(R)$ consisting of matrices with entries in $I.$ The identification $I\leftrightarrow M_n(I)$ is a bijection between the set of two-sided ideals of the ring $R$ and the set of two-sided ideals of $M_n(R).$

First observe that if $I$ is a two sided ideal of $R$ then $M_n(I)$ as defined above is clearly a two-sided ideal of $M_n(R).$ Now suppose that $J$ is a two sided ideal of $M_n(R)$ and define $J^*=\{r\in R : r=m_{i,j} \text{ for some } m\in J\}.$ It is easy to see that $J^*$ is a two sided ideal of $R$ since $J$ is a two sided ideal of $M_n(R)$ as follows. Suppose that $j\in J^*$ and $r\in R.$ By the definition of $J^*$ there exists a matrix $m(j)\in J$ with $j$ as an entry. We then easily define a matrix with one nonzero entry equal to $r$ such that $rj$ is an entry of $m(r)m(j).$ Hence $rj\in J^*.$ A similar argument shows $jr\in J^*.$

It now suffices to show that $M_n(I)^*=I$ and $M_n(J^*)=J$ for any two sided ideals $I\subseteq R$ and $J\subseteq M_n(R).$ It is immediate from definitions that $M_n(I)^*=I.$ To show the second equality, first observe that $J\subset M_n(J^*)$ by definition. For any $s\in J^*$ , by matrix multiplication we can construct a matrix $sE^{i,j}$ (with $E^{i,j}$ defined as above). Hence we can generate all matrices in $M_n(J^*).$ we can construct a matrix

Proposition 4. Let $\text{Id}$ denote the identity matrix in $M_n(R).$ The map $R\to M_n(R)$ defined by $r\mapsto r\text{Id}$ identifies $R$ with the centre of $M_n(R).$

It is easy to see that if a matrix has nozero entries off of the diagonal, then we may construct a matrix that does not commute with it. If a diagonal matrix has two differring entries along the diagonal, then this matrix does not commute with the matrix consisting of a single nonzero column of $1$ ’s in the position matching one of the distinct entries. Thus the only matrices with a chance of commuting are diagonal matrices of the form $r\text{Id}.$ Thus the map identifies the centre $R$ with the centre of $M_n(R).$

Proposition 5. If $A$ is a central simple algebra over a field $k$ , then $M_n(A)$ is also a central simple algebra over $k.$

By proposition 3 we see that $M_n(A)$ is simple iff $A$ is simple. By proposition 4 we see that we may identify $A$ with the subring $A\text{Id}$ of $M_n(A)$ and that $M_n(A)$ has center equal to the center of $A.$ Thus if $A$ is a CSA over $k$ then so is $M_n(A).$