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This theorem was in Keith Conrad note on Noetherian module and I read the proof. Three conditions in the theorem are equivalent, but I want to find the way to prove $(1) \Leftrightarrow (3)$ directly.

Eric Wofsey
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Minh Nguyễn Hoàng
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1 Answers1

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$(3) \Rightarrow (1)$:

Suppose (3), and for a contradiction suppose that we have a submodule $S$ of $M$ which is not finitely generated. Let $\{s_1,s_2, ...\}$ be some infinite countable subset of $S$, with $<s_1, ..., s_k> \subsetneq <s_1, ..., s_{k+1}>$ (can always find as $S$ not finitely generated). Then consider the collection $\{<s_1, ..., s_k> \mid k \in \mathbb{N}\}$ of submodules of $M$. By (3), this must contain a maximal element, which is a contradiction to the construction of the set $\{s_1,s_2, ...\}$.

$(1) \Rightarrow (3)$:

Suppose all submodules of $M$ are finitely generated. Suppose that (3) is false, so we can find a collection $S$ of submodules of $M$ with no maximal element. Pick $M_1 \in S$. $M_1$ not maximal, so there is some $M_2$ with $M_1 \subsetneq M_2$. Continue to construct the chain $M_1 \subsetneq M_2 \subsetneq M_3 ...$ . Let N be the union of these $M_i$. This is a submodule of $M$, and not finitely generated as if $N = < n_1, ... , n_r>$, then as all $n_i \in N_k$ for some $k$, we must have $N_k = N_{k+1} = ...$. So $N$ not being finitely generated contradicts 1.

(Notes: The union of submodules is not typically a submodule, but is here as our submodules form a chain by inclusion. Also our proof $(1) \Rightarrow (3)$ basically had to go via (2), so I'd be interested to see a more direct way!)

James
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