Semisimple and DCC

A ring R is left semisimple iff it is jacobson semisimple and it exhibits dcc on its principal left ideals.

Assume R is left semisimple. Let J be the jacobson radical, which acts as a left R module, and let U be the summand. (Every submodule has a summand; that's the definition of semisimple.) In other words, J*U = R.

Write 1 uniquely as e+f, where e ∈ J and f ∈ U. If x is an element of J, then x*1 = x, hence xe = x and xf = 0. Similarly, if y is in U, ye = 0 and yf = y. Thus e and f implement projections from R onto the summands J and U.

In fact J is a principal submodule, generated by e, and U is generated by f. This happens whenever R is a finite direct sum of modules; each module, each left ideal, is principal.

With e in J, 1-e is a unit. Thus f is a unit, and generates all of R. With U = R, J = 0, and R is jacobson semisimple.

Combine this with the fact that a left semisimple ring is left artinian, and the principal left ideals are dcc. That completes one direction of the proof.

For the converse, note that every nonzero left ideal contains a minimal principal left ideal. This is in fact a minimal left ideal, for a smaller left ideal would contain another principal left ideal.

Let U1 be a minimal left ideal in R. If you can't find one, then every nonzero element spans 1, and is invertible, and R is a division ring, which is semisimple. So let's move on and assume U1 exists.

Since U1 is principal and minimal it is simple. Thus any left ideal in R is going to include all or none of U1. Since J = 0, let M1 be a maximal left ideal that misses U1. Note that M1 and U1 are disjoint. (In the context of groups and modules, disjoint subgroups or submodules still share the identity element.)

If U1 and M1 do not span all of R, then together they make a larger proper left ideal, which is impossible. Thus R = U1*M1.

Inside M1, choose a minimal U2. This misses some maximal left ideal M2, so that U2*M2 = R.

Let V2 = M1∩M2. Since V2 is part of M2, it is disjoint from U2. Take any x in M1 and write it as e+f, for e ∈ M2 and f ∈ U2. Now e is in M2 and M1, hence in V2. Therefore M1 = U2*V2.

Let U3 be a minimal principal left ideal in V2, and let M3 be maximal, with U3*M3 = R. Let V3 = V2∩M3. Verify that V2 = U3*V3. Then find U4 in V3, and so on, for as long as possible.

At each step, R is the direct sum U1*U2*U3*…Un*Vn. When R is the direct sum of left ideals, each left ideal is principal. It is generated by the image of 1 in that particular R module. Therefore M1 V2 V3 V4 … forms a descending chain of principal left ideals, and by assumption, such a chain is finite. It must end in 0. In other words, the last Un completes R, as an R module.

Now R is the direct sum of minimal principal left ideals. Each of these is an R module with no smaller modules inside; thus a simple left R module. Therefore, R is the finite direct sum of simple left R modules, and R is left semisimple.

Symmetry

Let R be jacobson semisimple. Call upon a theorem that equates left an right semisimple. Now R is left artinian iff it is left semisimple iff it is right semisimple iff it is right artinian.