关于完全可约模含不可约子模证明中有限生成假设的疑问

Hey Joe, great question—this is one of those subtle, foundational points that can feel confusing at first, but it all boils down to leveraging simpler proofs for finite cases and reducing the general problem to that scenario. Let’s break it down step by step:

Why We Can Safely Assume V is Finitely Generated #

First, recall that a completely reducible (semisimple) module has a key property: every submodule is a direct summand. Now, take any non-zero element v ∈ V (since V is non-zero, such an element exists). Consider the cyclic submodule <v> generated by v—this is finitely generated (obviously, by just one element!).

Since V is completely reducible, <v> must be a direct summand of V, meaning V = <v> ⊕ W for some submodule W. Crucially, direct summands of completely reducible modules are themselves completely reducible. So now we’ve got a finitely generated, non-zero, completely reducible submodule <v> inside V.

If we can prove that every finitely generated non-zero completely reducible module contains an irreducible submodule, then <v> has such a submodule—and that submodule is automatically a submodule of V too. So we don’t need to tackle the entire possibly infinite V; we just need to handle the finite generated case, then pull the result back to V.

The Finite Generated Case: Simpler Proof Without Choice #

For finitely generated modules, we can use induction on the number of generators, which avoids needing advanced set theory like Zorn’s Lemma:

• Base case: If V is cyclic (1 generator), either it’s already irreducible (done!) or it splits into a direct sum of two non-zero submodules (by complete reducibility). Each of those submodules has fewer generators, so we apply induction. Eventually, we’ll hit an irreducible submodule.
• Inductive step: Assume all finitely generated completely reducible modules with k generators have irreducible submodules. For a module with k+1 generators, split it into a direct summand (a cyclic submodule) plus another submodule, then apply the inductive hypothesis to either part.

What’s "Different" About Infinite Generated Modules? #

Infinite generated completely reducible modules do still contain irreducible submodules—after all, by definition, they’re direct sums of irreducible submodules, so you can just pick any of those summands! The "problem" isn’t that the conclusion fails; it’s that proving it directly might require Zorn’s Lemma (a form of the Axiom of Choice) to find a minimal non-zero submodule (which is irreducible).

Many textbooks prefer to avoid invoking choice early on, so they use the finite generation reduction to keep the proof elementary. Without the reduction, you’d have to argue using Zorn’s Lemma: consider the set of all non-zero submodules of V, ordered by reverse inclusion (so smaller submodules are "higher" in the order). Any chain of submodules here has a non-zero intersection (or you can use the complete reducibility property to find a lower bound), so Zorn’s Lemma gives a minimal element—an irreducible submodule.

But since we can reduce the infinite case to the finite one using the direct summand property, there’s no need to bring in choice here. That’s why the textbook starts with the finite generation assumption—it’s a clever shortcut to a simpler proof.

内容的提问来源于stack exchange，提问作者Joe