MHB Finitely Generated Modules and Ascending Chains

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we find Proposition 2.4 on finitely generated modules and chain conditions. I need help with some aspects of the proof.

Proposition 2.4 reads as follows:https://www.physicsforums.com/attachments/3177
View attachment 3178

I need help (at first anyway - the second part of the proof is even more challenging) with showing that the following implication holds:"A module $$M$$ is finitely generated $$ \ \Longrightarrow \ $$ the union of every ascending chain of proper submodules is proper … "
Cohn begins the proof of this implication by assuming that M is finitely generated.

He then assumes that we have a ascending chain of submodules $$C_1 \subset C_2 \subset C_3 \subset … \ … \ …$$ whose union is $$M$$ (i.e. the union is not a proper submodule)

He then deduces that not all the $$C_i$$ are proper submodules … … ...I am having problems in seeing how this proves the above implication … can someone explain the logic involved?It seems to me (but I am uncertain!) that Cohn's strategy is something like the following …

1. Assume M is finitely generated …2. … then seek to prove that:

if we have an ascending chain of proper submodules
then its union is a proper submodule

3. Prove the contrapositive of 2, so that we have that 2 follows from assumption 1Is this a correct description/interpretation of Cohn's proof?

Hope someone can clarify the above … …[Surely there is a more straightforward proof!]

Peter

 

[Euge]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we find Proposition 2.4 on finitely generated modules and chain conditions. I need help with some aspects of the proof.

Proposition 2.4 reads as follows:https://www.physicsforums.com/attachments/3177
View attachment 3178

I need help (at first anyway - the second part of the proof is even more challenging) with showing that the following implication holds:"A module $$M$$ is finitely generated $$ \ \Longrightarrow \ $$ the union of every ascending chain of proper submodules is proper … "
Cohn begins the proof of this implication by assuming that M is finitely generated.

He then assumes that we have a ascending chain of submodules $$C_1 \subset C_2 \subset C_3 \subset … \ … \ …$$ whose union is $$M$$ (i.e. the union is not a proper submodule)

He then deduces that not all the $$C_i$$ are proper submodules … … ...I am having problems in seeing how this proves the above implication … can someone explain the logic involved?It seems to me (but I am uncertain!) that Cohn's strategy is something like the following …

1. Assume M is finitely generated …2. … then seek to prove that:

if we have an ascending chain of proper submodules
then its union is a proper submodule

3. Prove the contrapositive of 2, so that we have that 2 follows from assumption 1Is this a correct description/interpretation of Cohn's proof?

Hope someone can clarify the above … …[Surely there is a more straightforward proof!]

Peter

Hi Peter,

I can see how you're having difficulty following Cohn's proof. Hopefully I can clear things up. In the theorem, $M$ is an $R$-module.

The proof has two components. The forward direction has a contradiction argument and the reverse direction has a contrapositive argument. To prove the forward direction, suppose $M$ is a finitely generated $R$-module such that for some ascending chain $C_1 \subset C_2 \subset \cdots$ of proper submodules of $M$, the union $\cup C_i = M$. Let $M$ be generated by $u_1,\ldots, u_n$. Since the $u_j$ belong to $\cup C_i$, each $u_j$ belongs to at least one $C_i$. So, for each $j \in \{1,2,\ldots, n\}$, there corresponds an index $r_j$ such that $u_j \in C_{r_j}$. If $r$ is the largest of the indices $r_1,\ldots, r_n$, then the ascending chain implies $u_j \in C_r$ for all $j$. Consequently, $\sum u_i R = C_r$. Since $M$ is generated by the $u_i$, $M = \sum u_i R$ and thus $M = C_r$. We have found an element of the $C$-chain that is not proper. This is a contradiction. Now for the reverse direction, we show that if $M$ is not finitely generated, then there is some ascending chain of proper submodules of $M$ whose union is not proper. Here we inductively construct a sequence $C_n$ of finitely generated submodules of $M$ such that $C_0 = 0$ and $C_{n+1} = C_n + a_{n+1}R$ where $a_{n+1} \in M \setminus C_n$. Take $a_1 \neq 0$ and define $C_1 = a_1 R$. Having chosen $a_1,\ldots, a_n$, choose $a_{n+1} \in M \setminus C_n$ and let $C_{n+1} = C_n + a_{n+1} R$. By construction, $C_n$ is a strict, ascending chain of submodules of $M$. Since $M$ is not finitely generated, each $C_n$ is proper. However, $\cup C_n = M$.

 

[Math Amateur]

Hi Peter,

I can see how you're having difficulty following Cohn's proof. Hopefully I can clear things up. In the theorem, $M$ is an $R$-module.

The proof has two components. The forward direction has a contradiction argument and the reverse direction has a contrapositive argument. To prove the forward direction, suppose $M$ is a finitely generated $R$-module such that for some ascending chain $C_1 \subset C_2 \subset \cdots$ of proper submodules of $M$, the union $\cup C_i = M$. Let $M$ be generated by $u_1,\ldots, u_n$. Since the $u_j$ belong to $\cup C_i$, each $u_j$ belongs to at least one $C_i$. So, for each $j \in \{1,2,\ldots, n\}$, there corresponds an index $r_j$ such that $u_j \in C_{r_j}$. If $r$ is the largest of the indices $r_1,\ldots, r_n$, then the ascending chain implies $u_j \in C_r$ for all $j$. Consequently, $\sum u_i R = C_r$. Since $M$ is generated by the $u_i$, $M = \sum u_i R$ and thus $M = C_r$. We have found an element of the $C$-chain that is not proper. This is a contradiction. Now for the reverse direction, we show that if $M$ is not finitely generated, then there is some ascending chain of proper submodules of $M$ whose union is not proper. Here we inductively construct a sequence $C_n$ of finitely generated submodules of $M$ such that $C_0 = 0$ and $C_{n+1} = C_n + a_{n+1}R$ where $a_{n+1} \in M \setminus C_n$. Take $a_1 \neq 0$ and define $C_1 = a_1 R$. Having chosen $a_1,\ldots, a_n$, choose $a_{n+1} \in M \setminus C_n$ and let $C_n = C_{n-1} + a_n R$. By construction, $C_n$ is a strict, ascending chain of submodules of $M$. Since $M$ is not finitely generated, each $C_n$ is proper. However, $\cup C_n \neq M$.

Thanks for the help Euge … at first glance your proof is much clearer … just working through the details now …Thanks again,

Peter***EDIT*** Indeed your proof is very clear ... I wish Cohn would proceed by such clear arguments ...

Thanks again for your help ...

 

[Deveno]

Just to add something, here:

If $C_0$ is a submodule of $M$ with $C_0 = \langle S\rangle$ and $a_1 \not\in C_0$, then $C_1 = \langle a_1,S\rangle$ is a submodule of $M$ that $C_0$ is properly contained in.

If doing this process terminates in a finite number of steps with $C_k = M$, clearly $M$ is generated by $\{a_1,\dots,a_k\} \cup S$. In particular, if $S$ is finite (or empty), then $M$ is finitely-generated.

So if $M$ is not finitely-generated, the above process cannot possibly terminate.

Conceivably, one might imagine that $M$ is such a module that an infinite union of infinitely many ascending submodules is all of $M$, but that none of these submodules by itself is all of $M$. For example, $M$ might be the direct sum of countably infinitely many copies of $\Bbb Z$.

Here we could take:

$\displaystyle C_k = \bigoplus_{i = 1}^k \Bbb Z$

and while $\displaystyle M = \bigcup_{k = 1}^{\infty} C_k$ (since every element of $M$ has all 0's after some coordinate, that is, lies in some $C_k$) it is clear that no FINITE sum of a finite number of elements of $M$ (no matter which ones we choose) can account for EVERY element of $M$ (we can look at the maximum position of a non-zero coordinate in the set of generators, and all the elements of $M$ with non-zero coordinates past this position are not in the generated submodule).

In this particular example, we see that $M$ cannot be finitely-generated (this is very much like saying polynomials can have no maximum degree).

Cohn's theorem says: "this example is not too special", that non-finitely-generated modules behave the same way, and that finitely-generated modules "can't do this".

Something for you to ponder:

$\Bbb Q$ is an abelian group ($\Bbb Z$-module) under normal addition with the scalar multiplication:

$k \cdot \dfrac{a}{b} = \dfrac{ka}{b}$ for $k,a \in \Bbb Z$, and $b \in \Bbb Z^{\ast}$.

Is this module finitely-generated?

 

[Math Amateur]

Just to add something, here:

If $C_0$ is a submodule of $M$ with $C_0 = \langle S\rangle$ and $a_1 \not\in C_0$, then $C_1 = \langle a_1,S\rangle$ is a submodule of $M$ that $C_0$ is properly contained in.

If doing this process terminates in a finite number of steps with $C_k = M$, clearly $M$ is generated by $\{a_1,\dots,a_k\} \cup S$. In particular, if $S$ is finite (or empty), then $M$ is finitely-generated.

So if $M$ is not finitely-generated, the above process cannot possibly terminate.

Conceivably, one might imagine that $M$ is such a module that an infinite union of infinitely many ascending submodules is all of $M$, but that none of these submodules by itself is all of $M$. For example, $M$ might be the direct sum of countably infinitely many copies of $\Bbb Z$.

Here we could take:

$\displaystyle C_k = \bigoplus_{i = 1}^k \Bbb Z$

and while $\displaystyle M = \bigcup_{k = 1}^{\infty} C_k$ (since every element of $M$ has all 0's after some coordinate, that is, lies in some $C_k$) it is clear that no FINITE sum of a finite number of elements of $M$ (no matter which ones we choose) can account for EVERY element of $M$ (we can look at the maximum position of a non-zero coordinate in the set of generators, and all the elements of $M$ with non-zero coordinates past this position are not in the generated submodule).

In this particular example, we see that $M$ cannot be finitely-generated (this is very much like saying polynomials can have no maximum degree).

Cohn's theorem says: "this example is not too special", that non-finitely-generated modules behave the same way, and that finitely-generated modules "can't do this".

Something for you to ponder:

$\Bbb Q$ is an abelian group ($\Bbb Z$-module) under normal addition with the scalar multiplication:

$k \cdot \dfrac{a}{b} = \dfrac{ka}{b}$ for $k,a \in \Bbb Z$, and $b \in \Bbb Z^{\ast}$.

Is this module finitely-generated?

Thanks Deveno … your thoughts in this post were extremely important in enhancing my understanding of finitely generated modules!

Peter

 

[Math Amateur]

Hi Peter,

I can see how you're having difficulty following Cohn's proof. Hopefully I can clear things up. In the theorem, $M$ is an $R$-module.

The proof has two components. The forward direction has a contradiction argument and the reverse direction has a contrapositive argument. To prove the forward direction, suppose $M$ is a finitely generated $R$-module such that for some ascending chain $C_1 \subset C_2 \subset \cdots$ of proper submodules of $M$, the union $\cup C_i = M$. Let $M$ be generated by $u_1,\ldots, u_n$. Since the $u_j$ belong to $\cup C_i$, each $u_j$ belongs to at least one $C_i$. So, for each $j \in \{1,2,\ldots, n\}$, there corresponds an index $r_j$ such that $u_j \in C_{r_j}$. If $r$ is the largest of the indices $r_1,\ldots, r_n$, then the ascending chain implies $u_j \in C_r$ for all $j$. Consequently, $\sum u_i R = C_r$. Since $M$ is generated by the $u_i$, $M = \sum u_i R$ and thus $M = C_r$. We have found an element of the $C$-chain that is not proper. This is a contradiction. Now for the reverse direction, we show that if $M$ is not finitely generated, then there is some ascending chain of proper submodules of $M$ whose union is not proper. Here we inductively construct a sequence $C_n$ of finitely generated submodules of $M$ such that $C_0 = 0$ and $C_{n+1} = C_n + a_{n+1}R$ where $a_{n+1} \in M \setminus C_n$. Take $a_1 \neq 0$ and define $C_1 = a_1 R$. Having chosen $a_1,\ldots, a_n$, choose $a_{n+1} \in M \setminus C_n$ and let $C_{n+1} = C_n + a_{n+1} R$. By construction, $C_n$ is a strict, ascending chain of submodules of $M$. Since $M$ is not finitely generated, each $C_n$ is proper. However, $\cup C_n \neq M$.

Sorry Euge, I was too quick in claiming that I understood your proof … I have been reviewing your proof (and working on a problem that Deveno set me) and now have a concern ...

For the reverse direction in your proof we have to show the following:

The union of every ascending chain of proper submodules of a module M is a proper submodule $$\Longrightarrow$$ M is finitely generated

Now we choose to demonstrate the contrapositive which is as follows:

M is NOT finitely generated $$\Longrightarrow$$ There exists an
ascending chain of proper submodules of M whose union is NOT proper

Now in your proof you construct an ascending chain $$C_n$$ of proper submodules of M and then state the following:

"However, $\cup C_n \neq M$."

BUT … then the union of the $$C_n$$ IS PROPER …

Didn't we want to show that there existed such a chain whose union was NOT proper i.e. $\cup C_n = M$?

Can you help?

Peter

 

[Euge]

Sorry Euge, I was too quick in claiming that I understood your proof … I have been reviewing your proof (and working on a problem that Deveno set me) and now have a concern ...

For the reverse direction in your proof we have to show the following:

The union of every ascending chain of proper submodules of a module M is a proper submodule $$\Longrightarrow$$ M is finitely generated

Now we choose to demonstrate the contrapositive which is as follows:

M is NOT finitely generated $$\Longrightarrow$$ There exists an
ascending chain of proper submodules of M whose union is NOT proper

Now in your proof you construct an ascending chain $$C_n$$ of proper submodules of M and then state the following:

"However, $\cup C_n \neq M$."

BUT … then the union of the $$C_n$$ IS PROPER …

Didn't we want to show that there existed such a chain whose union was NOT proper i.e. $\cup C_n = M$?

Can you help?

Peter

That was a typo. Sorry about that! I've made the correction.

 

[Math Amateur]

Hi Peter,

I can see how you're having difficulty following Cohn's proof. Hopefully I can clear things up. In the theorem, $M$ is an $R$-module.

The proof has two components. The forward direction has a contradiction argument and the reverse direction has a contrapositive argument. To prove the forward direction, suppose $M$ is a finitely generated $R$-module such that for some ascending chain $C_1 \subset C_2 \subset \cdots$ of proper submodules of $M$, the union $\cup C_i = M$. Let $M$ be generated by $u_1,\ldots, u_n$. Since the $u_j$ belong to $\cup C_i$, each $u_j$ belongs to at least one $C_i$. So, for each $j \in \{1,2,\ldots, n\}$, there corresponds an index $r_j$ such that $u_j \in C_{r_j}$. If $r$ is the largest of the indices $r_1,\ldots, r_n$, then the ascending chain implies $u_j \in C_r$ for all $j$. Consequently, $\sum u_i R = C_r$. Since $M$ is generated by the $u_i$, $M = \sum u_i R$ and thus $M = C_r$. We have found an element of the $C$-chain that is not proper. This is a contradiction. Now for the reverse direction, we show that if $M$ is not finitely generated, then there is some ascending chain of proper submodules of $M$ whose union is not proper. Here we inductively construct a sequence $C_n$ of finitely generated submodules of $M$ such that $C_0 = 0$ and $C_{n+1} = C_n + a_{n+1}R$ where $a_{n+1} \in M \setminus C_n$. Take $a_1 \neq 0$ and define $C_1 = a_1 R$. Having chosen $a_1,\ldots, a_n$, choose $a_{n+1} \in M \setminus C_n$ and let $C_{n+1} = C_n + a_{n+1} R$. By construction, $C_n$ is a strict, ascending chain of submodules of $M$. Since $M$ is not finitely generated, each $C_n$ is proper. However, $\cup C_n = M$.

Thanks for the correction Euge, but can you explain how it is that we get $\cup C_n = M$?

Peter

 

[Math Amateur]

Just to add something, here:

If $C_0$ is a submodule of $M$ with $C_0 = \langle S\rangle$ and $a_1 \not\in C_0$, then $C_1 = \langle a_1,S\rangle$ is a submodule of $M$ that $C_0$ is properly contained in.

If doing this process terminates in a finite number of steps with $C_k = M$, clearly $M$ is generated by $\{a_1,\dots,a_k\} \cup S$. In particular, if $S$ is finite (or empty), then $M$ is finitely-generated.

So if $M$ is not finitely-generated, the above process cannot possibly terminate.

Conceivably, one might imagine that $M$ is such a module that an infinite union of infinitely many ascending submodules is all of $M$, but that none of these submodules by itself is all of $M$. For example, $M$ might be the direct sum of countably infinitely many copies of $\Bbb Z$.

Here we could take:

$\displaystyle C_k = \bigoplus_{i = 1}^k \Bbb Z$

and while $\displaystyle M = \bigcup_{k = 1}^{\infty} C_k$ (since every element of $M$ has all 0's after some coordinate, that is, lies in some $C_k$) it is clear that no FINITE sum of a finite number of elements of $M$ (no matter which ones we choose) can account for EVERY element of $M$ (we can look at the maximum position of a non-zero coordinate in the set of generators, and all the elements of $M$ with non-zero coordinates past this position are not in the generated submodule).

In this particular example, we see that $M$ cannot be finitely-generated (this is very much like saying polynomials can have no maximum degree).

Cohn's theorem says: "this example is not too special", that non-finitely-generated modules behave the same way, and that finitely-generated modules "can't do this".

Something for you to ponder:

$\Bbb Q$ is an abelian group ($\Bbb Z$-module) under normal addition with the scalar multiplication:

$k \cdot \dfrac{a}{b} = \dfrac{ka}{b}$ for $k,a \in \Bbb Z$, and $b \in \Bbb Z^{\ast}$.

Is this module finitely-generated?

Thanks again for the help Deveno ...

You write:

" … … Something for you to ponder:

$\Bbb Q$ is an abelian group ($\Bbb Z$-module) under normal addition with the scalar multiplication:

$k \cdot \dfrac{a}{b} = \dfrac{ka}{b}$ for $k,a \in \Bbb Z$, and $b \in \Bbb Z^{\ast}$.

Is this module finitely-generated?"Well, let us follow the process Euge (and Cohn) outlined and construct a strictly ascending sequence of finitely generated (and hence proper) submodules of $$\mathbb{Q}$$ such that:

$$C_0 = 0$$ and $$C_{n+1} = C_n + a_{n+1} \mathbb{Z} $$

where $$a_{n+1} = s_{n+1}/t_{n+1}$$ and $$s_{n+1}, t_{n+1} \in \mathbb{Z} so a_{n+1} \in \mathbb{Q}$$ and, further where $$a_{n+1} \in \mathbb{Q}\ C_n$$Then we have:

$$C_0 = 0$$$$C_1 = C_0 + a_1 \mathbb{Z}$$ where $$a_1 =s_1/t_1$$ and $$a_1 \in \mathbb{Q}$$

so that $$C_1$$ is generated by $$a_1$$
$$C_2 = C_1 + a_2 \mathbb{Z}$$ where $$a_2 =s_2/t_2$$ and $$a_2 \in \mathbb{Q}$$

so that $$C_2 $$ is generated by $$a_1, a_2$$and so on …Then we have generated a strictly increasing (and infinite) chain of proper submodules …Now given the chain is strictly increasing and infinite we have $$\sum_n C_N = \mathbb{Q}$$ …BUT note I am VERY uncertain of this last step asserting that $$\sum_n C_N = \mathbb{Q}$$ … can you help with the explanation?

Could you please critique my analysis above and/or confirm it is OK?

Peter

 

[Deveno]

Thanks again for the help Deveno ...

You write:

" … … Something for you to ponder:

$\Bbb Q$ is an abelian group ($\Bbb Z$-module) under normal addition with the scalar multiplication:

$k \cdot \dfrac{a}{b} = \dfrac{ka}{b}$ for $k,a \in \Bbb Z$, and $b \in \Bbb Z^{\ast}$.

Is this module finitely-generated?"Well, let us follow the process Euge (and Cohn) outlined and construct a strictly ascending sequence of finitely generated (and hence proper) submodules of $$\mathbb{Q}$$ such that:

$$C_0 = 0$$ and $$C_{n+1} = C_n + a_{n+1} \mathbb{Z} $$

where $$a_{n+1} = s_{n+1}/t_{n+1}$$ and $$s_{n+1}, t_{n+1} \in \mathbb{Z} so a_{n+1} \in \mathbb{Q}$$ and, further where $$a_{n+1} \in \mathbb{Q}\ C_n$$Then we have:

$$C_0 = 0$$$$C_1 = C_0 + a_1 \mathbb{Z}$$ where $$a_1 =s_1/t_1$$ and $$a_1 \in \mathbb{Q}$$

so that $$C_1$$ is generated by $$a_1$$
$$C_2 = C_1 + a_2 \mathbb{Z}$$ where $$a_2 =s_2/t_2$$ and $$a_2 \in \mathbb{Q}$$

so that $$C_2 $$ is generated by $$a_1, a_2$$and so on …Then we have generated a strictly increasing (and infinite) chain of proper submodules …Now given the chain is strictly increasing and infinite we have $$\sum_n C_N = \mathbb{Q}$$ …BUT note I am VERY uncertain of this last step asserting that $$\sum_n C_N = \mathbb{Q}$$ … can you help with the explanation?

Could you please critique my analysis above and/or confirm it is OK?

Peter

Your argument is incomplete: how are you choosing the $a_j$, specifically?

In order for the inclusion of $C_j \subset C_{j+1}$ to be STRICT, we had better ensure $a_{j+1} \not\in C_j$.

Now, you would do well to consider $C_1 = a_1\Bbb Z$. What does this submodule look like?

It's fair to assume that $a_1 = \dfrac{s_1}{t_1}$ where $\text{gcd}(s_1,t_1) = 1$.

How could we ensure $a_2 = \dfrac{s_2}{t_2} \neq k\dfrac{s_1}{t_1}$ for ANY $k \in \Bbb Z$?

I'll give you a hint: $\Bbb Z$ is a UFD, with unique factorization into ____?

To be a little more obvious, suppose we take $C_1 = \left(\frac{1}{2}\right)$.

Is $\frac{1}{3} \in C_1$?

(You might want to look up "fractional ideals").

 

[Math Amateur]

Your argument is incomplete: how are you choosing the $a_j$, specifically?

In order for the inclusion of $C_j \subset C_{j+1}$ to be STRICT, we had better ensure $a_{j+1} \not\in C_j$.

Now, you would do well to consider $C_1 = a_1\Bbb Z$. What does this submodule look like?

It's fair to assume that $a_1 = \dfrac{s_1}{t_1}$ where $\text{gcd}(s_1,t_1) = 1$.

How could we ensure $a_2 = \dfrac{s_2}{t_2} \neq k\dfrac{s_1}{t_1}$ for ANY $k \in \Bbb Z$?

I'll give you a hint: $\Bbb Z$ is a UFD, with unique factorization into ____?

To be a little more obvious, suppose we take $C_1 = \left(\frac{1}{2}\right)$.

Is $\frac{1}{3} \in C_1$?

(You might want to look up "fractional ideals").

Hi Deveno … thanks for the help … just a quick thought before I go and look up fractional ideals …

Would a solution be as follows?:

$$C_0 = 0$$

$$C_1 = <1/2>$$

$$C_2 = <1/2, 1/3>$$

$$C_3 = <1/2, 1/3, 1/5>$$

… … and so on ...

So the sequence $$a_n $$ is $$1/n$$ where n is the sequence of ascending primes … and so the sequence C_n is an infinite strictly ascending chain of submodules

***EDITS***

BUT …

1. … … possibly something wrong with this as I haven't used fact the $$\mathbb{Z}$$ is a UFD!

2. … further, I have not followed your advice to consider $$C_1 + a_1 \mathbb{Z}, C_2 + a_2 \mathbb{Z}, C_3 + a_3 \mathbb{Z}$$, … .. and so on … but maybe we could amend above 'solution' to read as follows:

MATH]C_0 = 0$$

$$C_1 = <1/2> \mathbb{Z}$$

$$C_2 = <1/2, 1/3> \mathbb{Z}$$

$$C_3 = <1/2, 1/3, 1/5> \mathbb{Z} $$

… … and so on …BUT … sadly, apart from your advice, I do not know exactly why I am suggesting this … unless it has to do with assuring that $$\sum_n C_n = M$$ …Can you clarify?Peter

 

[Deveno]

Well, let's see if:

$\langle \frac{1}{n}\rangle = \frac{1}{n}\Bbb Z$.

Since $1 \in \Bbb Z$, clearly the RHS contains the LHS, since $\langle \frac{1}{n}\rangle$ is the smallest submodule of $\Bbb Z$ containing $\frac{1}{n}$.

On the other hand, if $k \in \Bbb Z$, then:

$\dfrac{k}{n} = \dfrac{1}{n} + \cdots + \dfrac{1}{n}$ ($k$ summands), which is in $\langle \frac{1}{n}\rangle$ by closure (of addition).

Now, suppose that $\text{gcd}(a,b) = 1$ for $a,b \in \Bbb Z$ (so that there is $r,s \in \Bbb Z$ with $ra + sb = 1$). I claim that:

$\langle \frac{1}{a},\frac{1}{b}\rangle = \langle \frac{1}{ab}\rangle$.

It should be easy to see that the LHS contains the RHS, since:

$\dfrac{1}{ab} = \dfrac{ra + sb}{ab} = \dfrac{s}{a} + \dfrac{r}{b}$

On the other hand:

$\langle \frac{1}{a},\frac{1}{b}\rangle = \frac{1}{a}\Bbb Z + \frac{1}{b}\Bbb Z$, so a typical element is of the form:

$\dfrac{k}{a} + \dfrac{m}{b} = \dfrac{ma + kb}{ab} \in \frac{1}{ab}\Bbb Z = \langle \frac{1}{ab}\rangle$.

So now let's prove that:

$\displaystyle \Bbb Q = \sum_{p \text{ prime}} \left\langle \frac{1}{p} \right\rangle$.

Let $\dfrac{a}{b} \in \Bbb Q$ with $\text{gcd}(a,b) = 1$.

Factoring $b$ into primes, we have:

$\dfrac{a}{b} = \dfrac{a}{p_1p_2\cdots p_k}$.

Induction on $k$ show we can write this as:

$a\left(\dfrac{c_1}{p_1} + \cdots + \dfrac{c_k}{p_k}\right)$ for some integers $c_1,\dots,c_k$ (see the "two-factor case" above).

For example, let $a = 7,b = 30$. We have:

1 = 8(2) - 1(15), so $\frac{1}{30} = \frac{16}{30} - \frac{15}{30} = \frac{8}{15} - \frac{1}{2}$.

Similarly 1 = 2(5) - 3(3), so $\frac{1}{15} = \frac{10}{15} - \frac{9}{15} = \frac{2}{3} - \frac{3}{5}$.

So $\frac{7}{30} = 7(\frac{16}{3} - \frac{24}{5} - \frac{1}{2})$, that is:

$c_1 = -1,c_2 = 16,c_3 = -24$.

Now if we denote the $k$-th prime by $p_k$, and set:

$C_0 = \{0\}$

$C_1 = \langle \frac{1}{2}\rangle$

$C_{k+1} = C_k + \langle \frac{1}{p_{k+1}}\rangle =$$\displaystyle \sum_{j=1}^{k+1} \left\langle \frac{1}{p_j}\right\rangle$

to conclude this is an INFINITE proper chain, we need to know that there are infinitely many primes.