Radicals, Brown Macoy Radical

Brown Macoy Radical

If brown(R) does not contain jac(R), then take x in the former and y in the latter, such that x and y span 1. Since both jac(R) and brown(R) are ideals, we may as well say x+y = 1. Thus x = 1-y, and 1-y is a unit, hence x is not in a maximal ideal after all. This is a contradiction, therefore R properly contains brown(R) contains jac(R) contains onenil(R) contains upnil(R) contains lownil(R) contains 0.

In the above chain, every inclusion, except possibly kothe's, can be made strict by selecting an appropriate ring.

• By assumption, our rings contain 1. An application of zorn's lemma builds a maximal ideal, or a maximal left ideal, missing 1. Therefore all the radicals are proper ideals of R.

• Let's build a ring where brown(R) properly contains jac(R). Let R be the ring of endomorphisms on a countably infinite dimensional K vector space. Every K vector space is a semisimple K module, hence R is von neumann, and R is jacobson semisimple.

  Now R contains a nontrivial ideal H consisting of the endomorphisms with finite dimensional images. If f is a map with infinite dimensional image, We need to show f spans every endomorphism g in H. Let g map the basis b of our vector space into a finite vector space C. Given f, start with infinitely many independent vectors in the range, and pull back to infinitely many independent vectors in the domain. Premultiply by an endomorphism, in fact a monomorphism, that carries b onto the aforementioned independent vectors in the domain of f. then postmultiply by an endomorphism that moves the images to the appropriate vectors in C. The composition reproduces g, hence the ideal generated by f includes g, and H.

  A finite dimensional endomorphism can be squashed down to one basis vector mapping to one nonzero image z, then expanded so that any subset of b maps to z. Premultiply by scaling factors, and b can be mapped, arbitrarily, into the one dimensional subspace spanned by z. Postmultiply by an isomorphism that moves z to any other nonzero vector. Linear combinations of these build every transformation with a finite dimensional image.

  Every nonzero endomorphism f, with infinite or finite image, spanns all of H. Therefore H lies in every nonzero ideal, and in every maximal ideal. Brown(R) is nonzero while jac(R) = 0.

• A dvr has one maximal ideal, which becomes the jacobson radical, yet there are no nil ideals.

• Onenil(R) > upnil(R) is an open question.

• See the previous theorem for a ring with a nontrivial upper nil radical, and a 0 lower nil radical.

• Z_x/x² has a nonzero lower nil radical generated by x.

If R is commutative, R > brown(R) = jac(R) ≥ onenil(R) = upnil(R) = lownil(R) ≥ 0.

If R is noetherian, the same containment rules apply, although brown(R) ≥ jac(R) might drop to equality; I really don't know. The earlier example, consisting of the endomorphisms on a K vector space, shows brown(R) > jac(R), but it is not left noetherian. Each subspace defines a left ideal, the endomorphisms that map into that subspace, and a tower of increasing subspaces builds an ascending chain of left ideals in R.