Basic facts about matrix algebras

Posted on May 25, 2019

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

Proposition 1. Show that 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. Show that 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 (iii) we see that M_n(A) is simple iff A is simple. By (iv) 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).