Center of Matrix Algebra / Matrix Ring is Scalar Matrices

We will prove that the center Z(M_n(A))=Z(A)I_n, where A is an R-algebra (or ring with unity).

One direction is pretty clear. Let X\in Z(A)I_n, Y\in M_n(A). Then X=zI_n for some z\in Z(A). XY=zI_n Y=zY, YX=YzI_n=zY, so X is in the center Z(M_n(A)).

The other direction will require the use of E_{ij} matrices, which is a n by n matrix with (i,j) entry 1, and rest zero.

Let X=(x_{ij})\in Z(M_n(A)). We can write X=\sum x_{ij}E_{ij}. Our key step is compute E_{pq}X=\sum x_{qj}E_{pj}=XE_{pq}=\sum x_{ip}E_{iq}. Thus we may conclude that

(E_{pq}X)_{ij}=\begin{cases}x_{qj}&\text{if}\ i=p\\  0&\text{otherwise}  \end{cases}

(XE_{pq})_{ij}=\begin{cases}x_{ip}&\text{if}\ j=q\\  0&\text{otherwise}  \end{cases}

Plugging in some convenient values like i=p, j=q, we can conclude that x_{qq}=x_{pp} for all p,q, i.e. all diagonal entries are equal.

Plug in i=p, j\neq q gives us x_{qj}=0 for all j\neq q.

Thus X is a scalar matrix, i.e. X=\alpha I_n for some \alpha\in A.

Observing that (\beta I_n)X=\beta\alpha I_n=X(\beta I_n)=\alpha\beta I_n, we conclude that \beta\alpha=\alpha\beta for all \beta in A. So \alpha has to be in the center Z(A).


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