Local Rings, Indecomposable, ACC, and DCC

Indecomposable, ACC, and DCC

Let M be acc, dcc, and indecomposable. Let E be the endomorphism ring of M. We will prove E is local, hence M is strongly indecomposable.

Let f be an endomorphism in E. By fitting's theorem, there is some n such that fⁿ splits M into kernel and image. Of course M is indecomposable, so one or the other is trivial.

If the image is trivial then f is nilpotent. If the kernel is trivial then f is injective, and that makes f an automorphism. An automorphism is a unit in E. Therefore the nonunits of E are all nilpotent. This makes E a local ring, and M is strongly indecomposable.

We have seen that Z is not strongly indecomposable, hence dcc is a necessary condition for this theorem.

Jacobson Radical is Nilpotent

Let J be the jacobson radical of E. We showed above that J is a nil ideal, but it is also a nilpotent ideal.

Let l be the length of M, i.e. the length of its decomposition series.

Let f be an endomorphism in J. If, as a set, the image of f⁷ equals the image of f⁸, then applying f again and again produces the same set. This contradicts the fact that f is nilpotent. Images must decrease, and we can only take l steps down. Therefore f^l = 0.

Consider J^l. Take l functions from J and apply them, one after the other. Suppose f₂, applied to f₁, does not reduce the size of f₁(M). Apply f₂ again and again, and we still have f₁(M). Since M contains f₁(M), f₂ is not nilpotent on M. This is a contradiction, hence each new function takes a step down. After l steps, we have to be at 0. Each product of l functions, drawn from J, becomes 0, and all finite sums thereof are 0. Therefore J^l = 0, and J is nilpotent.

The earlier example of Z_p^∞ is a module that is not acc. It happens to be strongly indecomposable, but its endomorphism ring is the p-adic numbers, and the jacobson radical is generated by p. This is not a nilpotent ideal, or even a nil ideal. The theorems on this page do indeed require acc and dcc.

Local Artinian Rings

If R is left artinian, R is local iff R has no nontrivial idempotents.

One direction is easy. We already showed, in the introduction, that a local ring has no idempotents. So assume R has no idempotents (other than 0 and 1).

Recall that left artinian implies left noetherian. Thus R is a left R module with acc and dcc.

Now E, the endomorphism ring, is isomorphic to R, according to the image of 1. Since E has no nontrivial idempotents, R is indecomposable. By the theorem at the top of this page, R is strongly indecomposable, and local.

Apply the second theorem, and the jacobson radical is nilpotent, with exponent bounded by the length of the composition series of R.