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. 

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.
