# I Finitely Generated Modules and Their Submodules ...

1. Oct 1, 2016

### Math Amateur

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with the proof of Lemma 1.2.21 ...

Lemma 1.2.21 and its proof reads as follows:

Question 1

In the above text by Berrick and Keating, we read the following:

"... ... Since $M$ is finitely generated, there is a minimal subset $\{ x_0, \ ... \ ... \ , x_s \}$ of $M$ such that

$x_0 R + \ ... \ ... \ , x_s R + L = M. \ ... \ ... \ ...$ "

My problem is as follows:

I cannot see exactly why there exists a minimal subset $\{ x_0, \ ... \ ... \ , x_s \}$ of $M$ such that

$x_0 R + \ ... \ ... \ , x_s R + L = M$. ... ... ...

Can someone please demonstrate, rigorously and formally, that there exists a minimal subset $\{ x_0, \ ... \ ... \ , x_s \}$ of $M$ such that

$x_0 R + \ ... \ ... \ , x_s R + L = M$?

Question 2

In the above text by Berrick and Keating, we read the following:

"... ... Let $S$ be the set of submodules $X$ of M that contain $x_1 R + \ ... \ ... \ , x_s R + L$ but do not contain $x_0$. It is obvious that $S$ is inductive ... ...

Can someone please explain exactly why $S$ is inductive ... ... ?

Hope someone can help ...

Peter

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B&K's definition of "inductive" is contained in section 1.2.18 ... ... . which reads as follows:

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2. Oct 1, 2016

### Staff: Mentor

To your first question.

$M$ is finitely generated, so it can be written $x_0R+\dots +x_mR=M$. Furthermore $L \subseteq M$, so $M=x_0R +\dots +x_mR +L$. (I simply added it, not bothering about minimality or so.) This means $M$ can be written in such a way. Now chose among all possible ways to do this a shortest $\{x_0, \dots , x_s\}$, that is with the smallest $s$ needed. It doesn't have to be unique, only with the fewest elements such that $M=x_0R +\dots +x_sR +L$.
$m$ would do, but it might not be the smallest number. However, we only have numbers $1$ to $m$. Somewhere has to be the smallest which we call $s$.

To the second part.

We have submodules that contain $\{x_1 , \dots , x_s\}$ and $L$. $S$ is the set of these submodules. $S$ is partially ordered by inclusion. Either $X \subseteq Y$ or $X \nsubseteq Y$ for any two $X,Y \in S$.
Therefore we can build chains $X_0 \subseteq X_1 \subseteq \dots \subseteq X_n$ in $S$. The condition to belong to $S$ is to contain $x_1, \dots , x_s$ and $L$ and not $x_0$. This is true for all $X_i$ and it is also true for $\cup_i X_i$. Therefore $\cup_i X_i \in S$, which is the condition to be inductive. Because we only deal with indices $1, \dots ,s$ we can take $\Lambda =\{1, \dots ,s\}$ as ordered index set.

3. Oct 2, 2016

### Math Amateur

Thanks for the reply, fresh_42 ... ... but i do not follow you when you write:

"... ... $M$ is finitely generated, so it can be written $x_0R+\dots +x_mR=M$. Furthermore $L \subseteq M$, so $M=x_0R +\dots +x_mR +L$. ... ...

Why is this true ... ?

Specifically, how, exactly, do we get from $L \subseteq M$ to the conclusion that $M=x_0R +\dots +x_mR +L$ ... how do we justify this ...

Peter

4. Oct 2, 2016

### Staff: Mentor

What does it mean that $M$ is finitely generated?

5. Oct 2, 2016

### Math Amateur

$M$ is finitely generated if we have $M = x_0 R + x_1 R + \ ... \ ... \ + x_m R$ for some finite set $X = \{ x_0, x_1, \ ... \ ... \, x_m \}$

Peter

6. Oct 2, 2016

### Staff: Mentor

Yes, so $M = x_0 R + x_1 R + \ ... \ ... \ + x_m R$ and $L$ is a submodule of $M$.
So $L \subseteq M = x_0 R + x_1 R + \ ... \ ... \ + x_m R$ and $M=M+L= x_0 R + x_1 R + \ ... \ ... \ + x_m R+L$.

The crucial point here is $M=M+L$, which is true because
a) $M \subseteq M+L$
b) $L \subseteq M \, \wedge \, M \subseteq M \, \Rightarrow \, M+L \subseteq M$ since $M$ is closed under addition. You can't escape $M$ by addition.

Edit: A simple example is $R=\mathbb{Z}\, , \,M=6\mathbb{Z}\, , \,L=36\mathbb{Z}$. What is a maximal submodule of $M=6\mathbb{Z}$ that isn't $M$ itself and contains $L=36\mathbb{Z}$? And is there only one maximal submodule?

Last edited: Oct 2, 2016
7. Oct 2, 2016

### Math Amateur

Thanks again, fresh_42 ... ...

You write: "" ... ... The crucial point here is $M=M+L$ ... ...

Indeed that is crucial ... ... would not have guessed that ...

Now, catching on to your thoughts ... but just a quick question ...

You write:

"... ... $L \subseteq M \, \wedge \, M \subseteq M$ ... ...

what do you mean by $M \, \wedge \, M$ ... ?

Peter

8. Oct 2, 2016

### Staff: Mentor

No, the $\wedge$ isn't prior to $\subseteq$.
It is a logical symbol and stands for $\text{ and }$.

So it reads $(L \subseteq M) \, \text{ and } \, (M \subseteq M) \, \Longrightarrow \, (L+M) \subseteq M$.

(For the record: $\vee$ stands for the logical $\text{ or }$. The similarity to $\cap$ and $\cup$ isn't accidental.)

9. Oct 2, 2016

### Math Amateur

Thanks for your help fresh_42 ... I doubt I would have understood this issue without your help ...

Most grateful ...

Peter

10. Oct 2, 2016

### Staff: Mentor

You're welcome, Peter!
But have a thought or two on my example:

11. Oct 2, 2016

### Math Amateur

Hi fresh_42 ...

I have to confess that I have no idea how to, formally and rigorously determine the maximal submodule of $6 \mathbb{Z}$ ... but intuitively it seems that the maximal submodule would be $N = 12 \mathbb{Z}$ ... is that correct ...?

How would you go about formally and rigorously determining the maximal submodule of $6 \mathbb{Z}$ ... ?

Peter

12. Oct 2, 2016

### Staff: Mentor

It is about the maximal submodules, that contain $L$.
Since each element of $M=6\mathbb{Z}$ is a multiple of six, submodules can also only contain multiples of six. Starting with six, we get $M$. So other multiples, like $N=12\mathbb{Z}$ are proper submodules. And $N$ contains $L=36\mathbb{Z}$, that is also correct, since multiples of $36$ are also multiples of $12$. But there is another proper submodule, $N'$, which also contains $L$. And it is maximal, too. It should show you, that maximal submodules of $M$ which contain $L$ don't need to be unique.

Maximality (of $N$) in general is shown by the pattern $N \subseteq P \subseteq M \, \Longrightarrow \, P=N \text{ or } P=M$ that is nothing fits between a maximal element and the entire module. In the case of integers it is quite simple because one must only deal with multiples and divisors and all can be generated by only one element.

13. Oct 2, 2016

### Math Amateur

Thanks for the helpful example fresh_42 ...

I suspect that the other maximal submodule of $M$ containing $L$ is $18 \mathbb{Z}$ ... is that correct?

Peter

14. Oct 2, 2016

### Staff: Mentor

You get for each prime number $p$ a maximal submodule $6p\mathbb{Z} \subseteq 6\mathbb{Z}$. Of course only $p\in\{2,3\}$ contain $L=36\mathbb{Z}$ because all other primes don't contribute to the second $6$ which is needed for $36$.