Radicals, m System, n System

m-system

Let S be a nonempty subset of a ring R. If, for every x and y in S, there is w in R, such that xwy is in S, then S is an m-system. The "m" is suppose to remind one of multiplicatively closed sets. In fact a multiplicatively closed set is an m-system - set w = 1.

Let P be an ideal in R. Recall that P is prime iff, for every x and y, xwy in P for all w in R implies either x or y lies in P. Therefore P is prime iff the complement S = R-P is an m-system. This is analogous to the commutative case: P is prime iff S is multiplicatively closed.

If an m-system S misses an ideal H, use zorn's lemma to find a maximal ideal P missing S, and containing H. Suppose ideals A and B do not lie in P, but AB does. Thus P+A intersects S in some element p₁+x. Similarly P+B meets S in p₂+y. For some w, (p₁+x)*w*(p₂+y) lies in S. Expand this product, and each term lies in P, hence the product lies in P, which is a contradiction. Therefore P is prime. This is analogous to the commutative case: a maximal ideal missing a multiplicatively closed set is prime.

n-system

The "n" in n-system stands for nilpotent. For any x in an n-system, there is w such that xwx lies in S. The complement of an ideal C is an n-system iff C is semiprime.

If an ideal H misses an n-system S, drive H up to a maximal ideal missing S, giving a semiprime ideal. The proof is as above.

m-system inside an n-system

Let x₁ be an element in a nonempty n-system. There exists a w₁ with x₁w₁x₁ in S. Let x₂ = x₁w₁x₁. Let x₃ = x₂w₂x₂, and so on, building an infinite sequence. We want to show x₁ x₂ x₃ etc is an m-system.

For i ≤ j, note that x_j is in the left ideal generated by x_i, and in the right ideal generated by x_i. If x_j is on the right, write x_iy = x_j. Set w = yw_j, and x_iwx_j = x_j+1. If x_j is on the left, find y such that x_j = yx_i, and set w = w_jy. Once again, x_jwx_i = x_j+1. That establishes the m-system.