Chains of Modules, Endomorphisms

Endomorphisms

Let M₁ = f(M), let M₂ = f(M₁, let M₃ = f(M₂, and so on. Suppose M₁ ⊂M. In other words, f is not onto.

At the j^th step, M_j is contained in M_j-1. When we apply f, all of M_j-1 winds up in M_j, so when f is restricted to M_j, that image is certainly contained in M_j. Therefore M_j+1 is contained in M_j. We have a chain of descending modules.

Since M is artinian, the chain stops. Let j be the first index where M_j+1 = M_j. Remember that f is 1-1, and can be reversed. The set M_j+1 has M_j as its preimage, and both sets are the same. Reverse f again to show M_j-1 = M_j. This is impossible, unless j = 0 and M = M₁. Therefore f maps M onto M, and f is a module automorphism.

Next let M be noetherian and let f be an M endomorphism that is onto. Suppose f is not injective. Find x and y such that f(x) = f(y). Subtract, and f maps x-y to 0. Of course f maps 0 to 0 as well.

Let M₀ = 0 and run backwards, running f in reverse. Thus M₁ is the preimage of 0 under f. Now M₁ is a submodule of M that properly contains M₀.

Suppose M_j properly contains M_j-1. Let M_j+1 be the preimage of M_j under f. Since M_j includes M_j-1, taking the preimage includes the preimage of M_j-1, which is M_j. Each submodule contains the previous.

If M_j+1 = M_j then f(M_j) should also equal M_j, yet M_j-1 is smaller. Therefore we know M_j+1 is larger. We have built an infinite ascending chain of submodules. Since M is noetherian, this is impossible. Therefore f is injective, and an automorphism on M.