Primitive Rings, The Density Theorem

Topology on the Ring of Endomorphisms

Let V be a bimodule, with R acting on the left, and S acting on the right. Let E be the ring of all S module endomorphisms of V, acting on the left. If x is an element of R, x plays the role of one of the endomorphisms in E. See bimodules for more details. Some of the elements of R kill V, and correspond to the 0 endomorphism of E. Call these elements K. This is the kernel of a ring homomorphism φ from R into E. Thus φ(x) = x*V.

Give E a topology, with a base as follows. Establish a finite set of elements of V and call this T₁. Establish another finite set of elements of V and call this T₂. The set of endomorphisms that map T₁ to T₂ is a base open set. If two base open sets intersect in a function f, that maps T₁ to T₂ and U₁ to U₂ simultaneously, then all the functions that map T₁ to T₂, and U₁ to U₂, form a base open set that is contained in the two original base open sets. The base criterion is satisfied, and we have a valid topology.

Continuous Operations

If we consider addition in E, an endomorphism mapping T to T₃ may be the sum of two endomorphisms mapping T to T₁ and T₂ respectively. In other words, T₁ + T₂ = T₃, as vectors drawn from V. These endomorphisms imply base open sets in E cross E that map, via addition, into the base open set from T to T₃. Therefore addition is continuous.

Negate the endomorphisms in a base open set to show addditive inverse is continuous. Combine this with the above, and subtraction is continuous.

Multiplication in the ring E is really function composition. Let a base open set in the product map T to W. A point in the preimage is two endomorphisms, mapping T to U and U to W. Put everything in a base open set, and the preimage is open cross open. Thus multiplication is continuous, and E is a topological ring.

Apply the same topology to the unit subspace of E, and map each endomorphism to its inverse. Consider all the functions that take T to U, and select a particular function in the preimage, that takes U to T. This establishes a base open set that is also in the preimage. Thus multiplicative inverse, and division by a unit, is continuous.

All this continuity is not necessary for the density theorem, but it's pretty.

Act Densly

R acts densly on V if the image of φ(R) is dense in E. In other words, φ(R) intersects each base open set. For every finite correspondence from some T to some U, that can be realized by some S endomorphism f ∈ E, there is some element x in R such that xV maps T to U. That is, xV agrees with f(v) on T.

Summand

Continuing the above, let S be the ring of R endomorphisms of V, written on the right. Thus S is hom(V,V), a dual of V with respect to itself. Then E, the S endomorphisms of V, written on the left, becomes hom(V,V) when V is viewed as a right S module. As before, R maps into E.

Let W be a summand of V as an R module. There is then a projection p(V) that preserves W and kills the rest of V. Note that p is an element of S.

Let f be an element of E, whence (f*W)*p = f*(W*p) = f*W. Thus f maps W into W. View V as an E module, and W becomes an E submodule of V.

If V is a semisimple R module, then V is the direct sum of simple R modules, and V, as an E module, is the direct sum of E submodules. Each summand is a simple R module, and since E includes the action of R, each summand is a simple E module. Therefore V is a semisimple E module, with the same simple decomposition.

The Density Theorem

Let V be a semisimple R module, and let S be the ring of R endomorphisms of V, written on the right. Let E be the ring of S endomorphisms of V, written on the left, so that φ(R) maps into E. As shown above, V is a semisimple E module.

Let f be a member of E that maps a set of n elements T to another set of elements U. Then build a new R module V′, the direct sum of n copies of V. This is a semisimple R module. Let S′ be the ring of endomorphisms of V′. Such an endomorphism is given by its actions on components, projected to components, or an n×n matrix over S. Similarly, E′ consists of n×n matrices over E. This works because V′ is a direct sum of E modules, just as it is a direct sum of R modules.

Let f′ be the diagonal map that applies f to each component of V′. We need to show f′ is an S′ endomorphism. Apply a matrix from S′ to the right of a column vector in V′, then apply f′ on the left. Consider the i^th component, which comes from the i^th row of the matrix on the right. The column vector consists of elements of V, then various R endomorphisms from S are applied, and the results are added together. Then f is applied. Taking things in the other order, f is applied to each entry in the column vector, then this is dotted with the i^th row of the matrix. These are the aforementioned R endomorphisms from S. Since f can be applied to V before or after these endomorphisms, and since f respects addition, the result is the same. Therefore f′ is an S′ endomorphism of V′, and a member of E′.

Let C be the cyclic left R module generated by the elements of T, one element per component of V′. The finite list of elements T has been spread out into a column vector, a member of V′. Since V′ is semisimple, C is a summand. That makes C a summand as an E module. Therefore, f′*C lies in C. Apply f′ to the generator of C, i.e. the elements in T, and get the corresponding elements of U. Remember that this lies in C. Thus there is some x in R that accomplishes the same thing. Therefore, R acts densly on V.

Further assume V is a finitely generated S module. Now f, an arbitrary S module homomorphism in E, is completely defined by its image on n generators. If xV agrees with f(V) on these generators, then they are the same function. Therefore, φ(R) maps onto E.

Without Semisimple

Let's see what goes wrong when V is not semisimple. Let V = Q, the rationals, which is a left Z module. The simple Z modules are Z_p, and the semisimple modules are direct sums of same. Since Q has characteristic 0, it is not semisimple.

The Z endomorphisms of Q are isomorphic to Q, determined by the image of 1. That builds S on the right.

Since Q has become a right Q vector space, the endomorphisms of Q on the left follow 1, and are isomorphic to Q. We have E = Q, and R = Z, and since nothing in R kills V, R embeds in E. However, R does not act densely. Let f divide Q by 2. There is no x in R that can accomplish this, for any set of points in V.