Jacobson Radical, Von Neumann Ring

Von Neumann Ring

A von neumann ring (biography), also called von neumann regular or von neumann semisimple, has some x, depending on y, satisfying yxy = y. Here x is like a pseudo inverse of y.

If y has left or right inverse x, then yxy = y, and x will suffice. If y is a unit then multiply by y inverse on the left and right, and x has to ve the inverse of y.

Principal Left Ideal

Let H be a principal left ideal generated by y. Note that xy also generates H, and is an idempotent. Thus every principal left ideal is generated by an idempotent.

Conversely, let y generate H, and let the idempotent e generate H. For some x and z, xy = e, and ze = y. Write yxy = ye = zee = ze = y. Therefore, R is von neumann iff every principal left ideal is generated by an idempotent. By symmetry, the same holds for principal right ideals.

Finitely Generated

Let a left ideal H in a von neumann ring R have two generators e and f. Assume, without loss of generality, that e and f are idempotent.

Set c = f-f*e. Note that c and e span f (and hence H), and c*e = 0.

Let b be the idempotent of R*c, so that b = xc, and b*e = xc*e = 0. Consider e+b-eb. Premultiply by e and b, and get e and b respectively. Thus e+b-eb generates H, and is contained in H, and H is principal.

Repeat this n times and every finitely generated left ideal is principal. By symmetry, the same holds for finitely generated right ideals.

Von Neumann implies Jacobson Semisimple

Let J be jac(R). Let y belong to J, and review condition 4. Write y = yxy, hence y*(1-xy) = 0, and y = 0.

Semisimple = Von Neumann with ACC

Review the structure of a semisimple ring. Every left ideal is generated by an idempotent, and R is von neumann.

Conversely, assume R is von neumann and left noetherian. Every left ideal is generated by an idempotent, making it a summand. This makes R a semisimple ring.

With R von neumann, any of the four chain conditions, left acc, right acc, left dcc, right dcc, makes R semisimple, whence all four chain conditions apply.

Combining Von Neumann Rings

The rationals are a field, hence von neumann, but the integers are not. The subring of a von neumann ring need not be von neumann. However, an ideal H in a von neumann ring R, treated as a subring without 1, is von neumann. A left principal ideal in H, generated by y, implies a left principal ideal Ry, which leads to an idempotent e, which exists in H. Left principal ideals are generated by idempotents, and H is von neumann.

If S is the quotient of a von neumann ring R, pull y ∈ S back to y′ in R, find x′ in R, and map this forward to x in S, making S von neumann.

If R is the direct product of von neumann rings, select an xi for each yi, building an x satisfying yxy = y. Thus R is von neumann. The direct sum is also von neumann, though an infinite direct sum does not contain 1.

Endomorphisms of M

If M is a semisimple module, let R be the ring of endomorphisms of M. Given f, we need g such that f = fgf. Let K be the kernel of f, with complement U. Thus f is 1-1 on U, with image V. Let g(V) map back onto U, the reverse of f. Let g be anything on the complement of V. Verify fgf = f, on K, on U, and on M. Thus the ring of endomorphisms is von neumann, where the selection of g is not unique.

Let M be a module over a division ring K. In other words, M is a K vector space. If U is a subspace, assign it a basis, then extend the basis to all of M. The "rest" of the basis defines V, with U+V = M. Therefore M is semisimple.

Let M = Kn, and the endomorphisms of M, or the n×n matrices in K, forms a von neumann ring. Take finite direct products of such rings to build a semisimple ring. This is a round about proof that semisimple implies voneuman.

Take an infinite direct product of matrices over K, to find nonartinian voneuman rings, which are not semisimple.

c(S)

Recall the ring of continuous functions from a compact hausdorff space into the reals, denoted c(S). Each maximal ideal vanishes at a given point in the space. The intersection of these ideals contains functions that vanish everywhere, namely 0. Therefore c(S) is jacobson semisimple - though it is rarely von neumann.

Let the space be the interval [0,1] and suppose c(S) is von neumann. Let f(x) = x. Fix a nonzero x, and we're talking about the reals, whence g(x) has to be the inverse of f(x). Therefore g(x) = 1/x on (0,1], g is unbounded, and g cannot be continuous. The ring is not von neumann after all.