# Question about direct sum of Noetherian modules is Noetherian

Here is a corollary from Atiyah-Macdonald:

Question 1: The corollary states that finite direct sums of Noetherian modules are Noetherian. But they prove that countably infinite sums are Noetherian, right? (so they prove something stronger)

Question 2: I have come up with the following proof of the statement in the corollary, can you tell me if it's correct? Thank you:

Assume $M_i$ are Noetherian and let $(\bigoplus_i L_i)_n$ be an increasing sequence of submodules in $\bigoplus_i M_i$. Then in particular, $L_{in}$ is an increasing sequence in $M_i$ and hence stabilises, that is, for $n$ greater some $N_i$, $L_{in} = L_{in+1} = \dots$. Now set $N = \max_i N_i$. Then $(\bigoplus_i L_i)_n$ stabilises for $n> N$ and is equal to $\bigoplus_i L_i$, where $L_i = L_{iN_i}$.

This proves that finite direct sums of Noetherian modules are Noetherian so it's a bit weaker. But if it's correct it proves the corollary.

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Your proof is not complete, because there may be submodules of $\oplus M_i$ that don't decompose as $\oplus L_i$ where the $L_i$ are submodules of the $M_i$, think of the diagonal in the $2$-dimensional integer grid $\lbrace (n,n)|n\in\mathbb Z\rbrace\subset \mathbb Z\oplus\mathbb Z$. – Olivier Bégassat Jul 21 '12 at 14:10
@ClarkKent Will take a look. – Dylan Moreland Jul 21 '12 at 16:23

Countably infinite direct sums won't be Noetherian, unless almost all of the $M_n$ are trivial: look at the ascending chain $M_1 \subset M_1 \oplus M_2 \subset M_1 \oplus M_2 \oplus M_3 \subset \cdots$.

As for the alternative proof, you have to be a bit more careful. Even in the case of $n = 2$ it is not true that a submodule of $M_1 \oplus M_2$ must have the form $L_1 \oplus L_2$ for $L_i \subset M_i$. For example, the submodule $L$ of $M = \mathbb Z^2$ generated by $(2, 2)$ cannot be written in this way.

However, if you are working over a principal ring then you do have the elementary divisors theorem. In the above example, $\{(1, 0), (1, 1)\}$ is a basis for $M$ and if I'm allowed to instead write $M = \mathbb Z(1, 0) \oplus \mathbb Z(1, 1)$ then I can say something nice about $L$. I don't see a way of doing this in general.

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I suspect one could also write your counter example like this: $M \subset M \oplus M \subset M \oplus M \oplus M \subset \dots$? – Rudy the Reindeer Jul 21 '12 at 17:51
@Clark Well, I wanted to be general. If I take all the $M_n$ to be the same non-zero module then yes, that's a counterexample! – Dylan Moreland Jul 21 '12 at 20:57

You are getting mixed up: The proof is saying that given any $n$ number of Noetherian modules, the direct sum of these finitely many $n$ number of modules is Noetherian.

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As pointed out by others, the submodules of $M=M_1\oplus M_2$ are not necessarily direct sums of submodules of $M_1$ and $M_2$. Nevertheless, you always have the exact sequence

$$0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0$$

Then $M$ is Noetherian if (and only if) $M_1$ and $M_2$ are Noetherian. One direction is trivial (if M is Noetherian then $M_1$ and $M_2$ are Noetherian). I prove the other direction here:

Assume $M_1$ and $M_2$ are Noetherian. Given nested submodules $N_1\subset N_2\subset \cdots$ in $M$, we can see their images stabilize in $M_1$ and $M_2$. More precisely, the chain $$(N_1\cap M_1)\subset (N_2\cap M_1)\subset\cdots$$ terminates in $M_1$, say at length $j_1$ and so does $$(N_1+M_1)/M_1\subset (N_2+M_1)/M_1\subset\cdots$$ in $M_2$, say at length $j_2$. Set $j=\max\{j_1,j_2\}$ to get $$(N_{j}\cap M_1)=(N_{j+1}\cap M_1)=\cdots$$ in $M_1$ and $$(N_{j}+M_1)/M_1=(N_{j+1}+M_1)/M_1=\cdots$$ in $M_2$. But $N_j$'s are nested modules. Hence the above equalities can occur if and only if $N_j=N_{j+1}=\cdots$. To check this claim, pick $n\in N_{j+1}$. Then $m:=n'-n\in M_1$ for some $n'\in N_{j}$. But $m\in N_{j+1}$ as well. Hence $m\in N_{j+1}\cap M_1$, that is $m\in N_{j}\cap M_1$ by the first equality above. So $m\in N_j$, that is $n\in N_j$, giving us $N_{j+1}\subset N_j$.

So the chain $N_1\subset N_2\subset ...$ terminates in $M$.

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The above proof is based on Steve Kleiman's notes (chapter 16, chain conditions), which you can find online on his webpage. – H.R Nov 12 '12 at 18:55