Jacobson Radical, Nakiama's Lemma

Nakiama's Lemma

Let H be a left ideal inside the jacobson radical of R, and let M be a finitely generated R module. If H*M = M then M = 0. This is Nakiama's lemma.

Select the fewest generators sufficient to span M.

Start with 0 and build an ascending chain of proper submodules of M. If such a chain is infinite, let U be the union of these submodules. Now U spans all of M only if it includes all the generators, and if that happens, these generators must appear somewhere in the chain. We have satisfied the conditions of zorn's lemma, and can now select T, a maximal proper submodule of M.

The quotient M/T is a simple R module. Since H maps M onto M, H maps M/T onto M/T.

Replace M/T with R/G, the left cosets of a maximal ideal G in R. We can do this because M/T is simple. Now H lies in J, which lies in G. The action of H drives everything into G. Thus H*(M/T) = 0. This means M/T = 0, whence T is all of M, which is a contradiction. Therefore M = 0.

When H is Nilpotent

Here is a variation on the theme. If H is a nilpotent ideal, i.e. some power of H is 0, then M = HM = H²M = H³M = … = 0.

Applying Nakiama to Quotients

If Q is a quotient of M it is also finitely generated. If H*Q = Q then Q = 0, and the kernel is all of M.

When Nakiama Doesn't Work

Let R be the localization of Z about p. In other words, R is the fractions without p in the denominator. This is a local ring whose maximal ideal J (also the jacobson radical) is generated by p.

Let Q be the rationals, with R acting on Q via multiplication. Since everything in Q is divisible by p, JQ = Q. Yet Q is obviously nonzero. Since Nakiama fails, J is not nilpotent, and Q is not a finitely generated R module.