Artinian, but not Noetherian module: a wonderful example. I found this wonderful example here. I did not understand some details, could you help me understand? The questions will be asked within the example.
Let $p\in\mathbb{N}$ be a prime number. Consider the set $$G:=\bigg\{\frac{a}{p^n}\in\mathbb{Q}\;\bigg|\; a\in\mathbb{Z},\;n\ge 0\bigg\}.$$ For all $n\ge 0$ consider
$$G_n:=\bigg\{\frac{a}{p^n}\in\mathbb{Q}\;\bigg|\; a\in\mathbb{Z}\bigg\}.$$
Now,
$$\mathbb{Z}\subset G_0\subset G_1\subset\cdots$$
I must want to prove that $G/\mathbb{Z}$ is Artinian, but it is not Noetherian.
Let $\pi\colon G\to G/\mathbb{Z}$ be the canonical projection. Then $G′_n$=$\pi(G_n)$ is a submodule of $G/\mathbb{Z}$
and
$$0=G′_0\subset G′_1\subset G′_2\subset G′_3\subset G′_4\subset\cdots.$$
The inclusions are proper, because for any $n>0$
we have
$$G′_{n+1}/G′_n\cong (G_{n+1}/\mathbb{Z})/(G_n/\mathbb{Z})\cong G_{n+1}/G_n\ne 0,$$
due to Third Isomorphism Theorem for modules. This shows, that $G/\mathbb{Z}$ is not Noetherian.
In order to show that $G/\mathbb{Z}$
is Artinian, we will show, that each proper submodule of $G/\mathbb{Z}$ is of the form $G′_n$.
Let $N\subseteq G/\mathbb{Z}$ be a proper submodule. Assume that for some $a\in\mathbb{Z}$ and $n\ge0$
we have $$\frac{a}{p^n}+\mathbb{Z}\in N.$$
We may assume that $\gcd(a,p^n)=1$. Therefore there are $\alpha,\beta\in\mathbb{Z}$
such that $$1=\alpha a+\beta p^n.$$
Now, since $N$ is a $\mathbb{Z}$-module we have
$$\frac{\alpha a}{p^n}+\mathbb{Z}\in N$$
and since $0+\mathbb{Z}=\beta+\mathbb{Z}=\frac{\beta p^n}{p^n}+\mathbb{Z}\in N$
we have that
$$\frac{1}{p^n}+\mathbb{Z}=\frac{\alpha a+\beta p^n}{p^n}+\mathbb{Z}\in N.$$
Now, let $m>0$ be the smallest number, such that $\frac{1}{p^m}+\mathbb{Z}\notin N$.

What we showed is that $$N=G′_{m−1}=\pi(G_{m−1}),$$
because for every $0\le n\le m−1$ (and only for such $n$) we have $\frac{1}{p^n}+\mathbb{Z}\in N$  >and thus $N$ is a image of a submodule of $G$, which is generated by $\frac{1}{p^n}$ and this is >precisely $G_{m−1}$
Question 1. I cannot understand this passage. Why is this true?

Now, let
$$N_1\supseteq N_2\supseteq N_3\supseteq\cdots $$
be a chain of submodules in $G/\mathbb{Z}$
Then there are natural numbers $n_1,n_2,\dots$ such that $N_i=G′_{n_i}$. Note that $G′_k\supseteq G′_s$ if and only if $k\ge s$
In particular we obtain a sequence of natural numbers
$$n_1\ge n_2\ge n_3\ge\cdots $$

This chain has to stabilize.
Question 2. Why does it stabilize?

 A: Question 1
It's difficult to know what you are missing since you are not specific.
I would guide you on this route of ideas:

*

*notice that if $N$ contains something of the form $\frac{x}{p^n}$, with $x$ coprime to $p$, then in fact it also contains $\frac{1}{p^n}$. (Hint: first write $ax+bp^n=1$ for some integers $a,b$.)


*Furthermore continuing the last point, $N$ would then also contain $\frac{1}{p^k}$ for the $k\leq n$.


*If the collection of $n$ such that $\frac{1}{p^n}\in N$ is unbounded, then in fact $N=G/\mathbb Z$.


*When the collection of $n$ such that $\frac{1}{p^n}\in N$ is bounded with greatest element $m$, say, then $N=\langle \frac{1}{p^m}+\mathbb Z\rangle$


*Conclude the subgroups of $G/\mathbb Z$ are linearly ordered, having generators $\frac{1}{p^k}+\mathbb Z$ for $k\in \mathbb N$.
Question 2

Why does [a descending chain of natural numbers] stabilize?

That's just the well-ordering principle for natural numbers right?
This whole example is based on the fact that the submodules are order isomorphic to $\mathbb N$, which obviously has infinite ascending chains but no infinitely decreasing chains.
