# I Finitely Generated Modules ... another question

1. Oct 2, 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:

In the above text (in the proof) by Berrick and Keating we read the following:

"... ... Any submodule properly containing $X'$ must contain $x_0$, also, so $X'$ is maximal among all the proper submodules of $M$. ... ...

My first question is ... why must any submodule containing $X'$ also contain $x_0$ ... ?

My second question is ... why does $X$' containing $x_0$ allow us to conclude that
$X'$ is maximal among all the proper submodules of $M$. ... ... ?

Hope someone can help ...

Peter

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2. Oct 2, 2016

### andrewkirk

Let $N$ be a submodule of $M$ that properly contains $X'$.
Then we have $x_1R+....+x_sR+L\subseteq X'\subsetneq N$
So if $x_0\notin N$, we have $N\in S$, which contradicts our assumption that $X'$ is maximal in $S$.

Hence we must have $x_0\in N$.

But that means that $N\supseteq x_0R+(x_1R+....+x_sR+L)=M$. So $M$ is maximal because any submodule $N$ of $M$ that properly contains it must be equal to $M$.

3. Oct 2, 2016

### Staff: Mentor

Let us proceed by the pattern I told you.

Firstly $X'$ is chosen as a maximal element of the set
$$S := \{ N \,\vert \, N\subsetneq M \text{ submodule and } x_1R+\dots +x_sR+L \subseteq N \text{ and } x_0 \notin N \}$$
Secondly, we know that $M$ is finitely generated by $\{x_0,x_1, \dots , x_s\}$.

Now the pattern goes:

Assume we have a submodule $P$, which lies between $X'$ and $M$, that is $X' \subseteq P \subseteq M$.
We have to show that either $P=X'$ or $P=M$.

1.) If $X' \subsetneq P$ is a proper inclusion, then $x_0$ has to be in $P$ because all other generators are already in $X'$ and $x_0$ is the only one which is left and can be added to create a proper inclusion. But then $P=M$. *)

2.) If $X' \subseteq P$ and $P \subsetneq M$ is a proper submodule of $M$, then $P$ belongs to $S$. But $X'$ is a maximal element of $S$. So the only way out is $P=X'$.

Thus we have shown that there cannot be a submodule $P$ between $X'$ and $M$, which makes $X'$ a maximal submodule.

*) Edit: In contrast to the example with the integers, you can't escape by taking elements like $x_0^2$ or similar instead of $x_0$, because $x_0$ belongs to $M$ and there is no multiplication within $M$ and $x_0$ is not in $R$.

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

### Math Amateur

Thanks ... very clear and very helpful ...

Pity the text couldn't be as clear ...

Peter

5. Oct 3, 2016

### Staff: Mentor

I'm afraid I made a little mistake in my cases. Given $X' \subseteq P \subseteq M$ then one has to distinguish
1.) $x_0 \in P$ and
2.) $x_0 \notin P$.

$X' \subsetneq P$ being a proper submodule does not imply $x_0 \in P$. E.g. $2x_0$ could be a generator of $P$ which is not $x_0$ ; and $2$ might not be a unit of $R$, so it cannot be divided.

However, the cases above work. And one of them has to be true: $x_0$ is either in $P$ or it is not.
So if 1.) $x_0 \in P$ then $M = X' + x_0R \subseteq P$ and $P=M$,
and if 2.) $x_0 \notin P$ then $P \in S$ and $X' \subseteq P$ implies $P=X'$ by the maximality of $X'$ in $S$.

Thus we have $X' \subseteq P \subseteq M\, \Longrightarrow \, P=X' \text{ or } P=M$, i.e. $X' \subseteq M$ is maximal.

6. Oct 3, 2016

### Math Amateur

Well ... have to confess ... I did not notice that subtlety ...

... ... hmmm ... still reflecting on it ...

... well ... learning quite a lot from you ...

Thanks so much for your help and insights ...

Peter