Centre of matrix ring over skew field

Question

Let $R$ be a semisimple ring. Show that $R$ is simple iff the centre of $R$ is a field.

Book's solution:

If $R$ is simple, it has the form $\mathfrak{M}_n(K)$ for a skew field $K$, and its centre is the centre of $K$, a field. Otherwise it is a direct product of more than one matrix ring and its centre is a direct product of more than one field.

Why would the centre of $\mathfrak{M}_n(K)$ be a field? I get that the centre of matrix ring isomorphic to centre of ring, and hence if $K$ is a commutative field then the centre of $\mathfrak{M}_n(K)$ is $K$ (and hence a field). But why should this be true if $K$ is a skew field?

Answer 1

Why would the centre of $\mathfrak{M}_n(K)$ be a field?

To commute with all of the unit matrices, an element of the center must be contained in the set $D=\{kI_n\mid k\in K\}$. This is already isomorphic to $K$.

Then in order for an element to commute with all the elements of $D$, the scalar multiplier must be in the center of $K$ (which is sometimes denoted $Z(K)$). So in fact, the center of $M_n(K)$ is $\{kI_n\mid k\in Z(K)\}$. This is easily seen to be isomorphic to $Z(K)$ which is, of course, a field.

Answer 2

Take $L_{ij}=e^ie_j$ and suppose that $M$ is in the center of $\mathfrak{M}(K)$. Then $ML_{ij}e_i=0$ and $L_{ij}Me_i=M_{il}e_l=M_{ij}$ if $i\neq j$, so that $M_{ij}=0$. Thus $M$ is a diagonal matrix. Now $L_{ij}+L_{ji}$ shows that all the diagonal elements coincide, so that the matrix is a multiple of the identity. Now clearly the identity matrix commutes with everything, and so do $Z(K)$, multiples of it, so that the center contains $Z(K)I$, and checking against other multiples of the identity shows that only these are in the center. Thus $Z(\mathfrak{M}(K))=Z(K)I$. Now, we conclude since the center of a Skew-field is a field.

Answer 3

That is because for any ring $A$, the centre of $M_n(A)$ is made up of scalar matrices with coefficients in the centre of $A$. Cf. Bourbaki, Algebra, Ch. 8, Modules and semi-simple rings, § 5 Commutation, no 3, cor. 2 of theorem 2.