Given n independent vectors, you can map these anywhere in V, then squash the rest of the basis down to 0. Thus there is at least one linear transformation that accomplishes any of these maps. (In practice there are lots of them.) Again, a set of transformations is dense if it includes at least one such transformation for each set of n independent vectors mapping to another set of n vectors. Let's show this definition is consistent with the one given in the previous section.
Let V be a bimodule, with R on the left and K on the right, where K is a division ring. This makes V a right K vector space. Let E be the endomorphisms of V, written on the left. These are every possible linear transformation from V into itself. Remember that φ(R) maps into E - as R acts on V from the left.
Let the image φ(R) be the aforementioned set of linear transformations. According to the previous definition, R is dense in E if it moves every set of n elements of V to any other set of n elements, provided the same map can be realized by something in E. This implies the set is dense as described above.
Conversely, assume φ(R) is a dense set. Let T be a set of n elements of V, and map them over to U. Let the first m elements of T be linearly independent. The other n-m elements go along for the ride. Since the set is dense, it includes a function f that maps the first m elements of T over to the first m elements of U. Since f is a K endomorphism, the remaining elements must reach their destinations as well. Thus f satisfies our mapping criteria. Since f comes from R, R is dense in E. The definitions coincide.
Since V is faithful, nothing in R kills V. Thus R embeds in E.
If V is finite dimensional, R maps onto E. In this case R = E. And what is E? E is the set of all possible K linear transformations of V, which are represented by the n×n matrices over K.
You may recognize this as a simmple ring. A column vector of length n acts as a simple faithful left R module. The diagonal matrices, isomorphic to K, become the R endomorphisms, written on the right. This turns V into a right K vector space, and E becomes the n×n matrices over K, just like R. Note that R is left and right artinian.
What happens when V is infinite dimensional? Let b1 b2 b3 … be a countable set of independent vectors from V. Given n, let Rn be the subring of R that maps b1 through bn into the subspace spanned by b1 through bn. Let An be the subset of R that drives b1 through bn to 0. Once bi is killed it is killed, hence Am is a left ideal in R. Also, An is a left ideal in Rn. Now b1 through bn span a subspace of V, that I will call S, and S is a well defined Rn/An module. In fact, S is faithful; we already pushed all the annihilators into An. Therefore Rn/An is a subring of the K endomorphisms of S, which are in turn the n×n matrices over K. Since R acts Densely on V, this map is onto. In other words, we can find a transformation for every matrix, (basis elements beyond bn can go wherever they like), and something in R makes this happen. Therefore, Rn/An is the complete ring of matrices over K. For every n, some subring of R has the n×n matrices over K as its homomorphic image.
Return to the fact that R acts densely on V. Something in R kills b1, and maps b2 to b2. This is in A1, but not A2. Similarly, something in R kills b1 and b2, but not b3. This is in A2, but not A3. Continue this chain forever, whence R is not left artinian. Therefore the left artinian primitive rings have been characterized; they are the same as the left artinian simple rings.
Suppose R exhibits dcc on its principal left ideals, even though it is not left artinian. Remember that primitive implies semiprimitive, which is the same as jacobson semisimple. Combine this with dcc and R becomes a semisimple ring. These are artinian, so we're back where we started. Hence the top three entries in our dcc classification diagram become equivalent.
In order to show the converse, let V be a right K vector space, let E be the K endomorphisms of V (linear transformations), and let R be a subring of E. Since R embeds, V is a faithful R module. Now R is transitive iff every vector in V generates V iff V is a simple R module. Thus R transitive implies R is left primitive, implies R is dense.
There is one catch however. Perhaps there are more R module endomorphisms than just scaling V by K. As it turns out, 2 transitive prevents any more endomorphisms from creeping in, whence K is still the ring of all possible R endomorphisms on the right, and E is still the ring of K endomorphisms on the left. Let l be a nonzero right R endomorphism, acting nontrivially on a nonzero vector x in V. Suppose l moves x to some other vector, i.e. xl is not a K multiple of x. Being 2 transitive, there is some f in R such that f*x = 0, while f*(x*l) = x. Thus 0*l = x, which is impossible. Since l scales x, write x*l = x*a for some a in K. We want to show a is constant across all vectors in V. Let w be another vector in V, and use the fact that R is 1 transitive to write w = f*x, whence w*l = f*x*l = f*x*a = w*a. The R endomorphisms are indeed K.