Jacobson Radical, Matrix Ring

Jacobson Radical of the Matrix Ring

If J is jac(R), then the n×n matrices over J form the jacobson radical of the n×n matrices over R. Note that R need not be commutative.

Let y be a matrix that is everywhere 0, except for a single entry that lies in J. We will show (below) that y lies in the jacobson radical, using condition 2. This holds for each such y, and since jacobson radical is an ideal, the matrices over J all lie in the jacobson radical.

Consider 1-xy for any matrix x. This gives a column with entries in J, subtracted from the identity matrix. The point of intersection is a unit in R. For convenience, premultiply by an invertible matrix that scales that row by the inverse of the unit. Now all the diagonal entries are 1. Negate the off diagonal entries to build a left inverse, thus 1-xy is left invertible, and y lies in the jacobson radical.

Conversely, let y lie in the jacobson radical of the matrix ring, which happens to be an ideal. Multiply by matrices on either side to extract any nonzero entry of y, and move it to the upper left. These entries can be added and scaled, giving a two sided ideal I in R. Thus the jacobson radical of the matrix ring is the n×n matrices over some ideal I.

Let y be the identity matrix scaled by anything in I. Let x be the identity matrix scaled by anything in R. Now 1-xy is a matrix that is constant down the main diagonal and zero elsewhere. Let the diagonal entries equal u. Let w be the left inverse of 1-xy, hence w, scaled by u on the right, equals the identity matrix. Thus w is also diagonal, and its entries are the left inverse of u. This makes u left invertible, and since x was arbitrary, everything in I lies in J. That completes the proof. The jacobson radical of the matrix ring is the matrices over J.

A ring R is jacobson semisimple iff the matrices over R are jacobson semisimple.

General Linear Quotient

Let R be a ring with jacobson radical J, and let I be a two sided ideal of R. This defines a ring homomorphism R/I, which extends to a ring homomorphism h on the n×n matrices over R, wherein all entries are reduced mod I.

Consider GLn(R), the invertible matrices over R. These remain invertible when h is applied. Thus GLn(R) maps into GLn(R/I). (In this context, h becomes a group homomorphism.) When I lies in J, the map is surjective.

Let the matrix x represent an element of GLn(R/I), which means x is invertible in the quotient ring. Let y represent the inverse, and write h(xy) = 1. Thus xy equals the identity matrix plus a matrix drawn from J. This is a unit, hence x is invertible, and lives in GLn(R).