Noetherian Modules, Submodules and Factor Modules .... Problem/Exercise

[Math Amateur]

Problem/Exercise

\(\displaystyle M\) is an R-module.
\(\displaystyle N\) is a submodule of M.
\(\displaystyle N\) and \(\displaystyle M/N\) are Noetherian

Show that \(\displaystyle M\) is Noetherian ...

====================================

Progress so far ...Let \(\displaystyle K\) be a submodule of \(\displaystyle M\) ... must show \(\displaystyle K\) is fingen ...

Consider the mapping \(\displaystyle \pi \ : \ M \to M/N\) where \(\displaystyle \pi (x) = \overline{x}\)

We then have \(\displaystyle \pi (K) = K/N\) ... ... ... suspect \(\displaystyle K/N\) is fingen ... but how to prove that ... and indeed ... how would that help ... ?
Not sure of direction here ... but following hint from steenis ...
Consider \(\displaystyle x \in K\) ... then \(\displaystyle \overline{x} \in M/N\) ...

But ... \(\displaystyle M/N\) is fingen ... so \(\displaystyle \overline{x} = \sum \overline{ a_i } r_i\) where \(\displaystyle \overline{ a_i } \in M/N\) so that \(\displaystyle a_i \in M\) ...

Thus \(\displaystyle \overline{x} - \sum \overline{ a_i } r_i = \overline{ 0 } \)

Hence \(\displaystyle x - \sum a_i r_i \in N\) ... ... ... ... ... (1) ... ... (is this step correct?)[Now ... a bit lost in overall direction ... but will now attempt to show that that \(\displaystyle x - \sum a_i r_i \in K \cap N\) ... ] But it also follows that \(\displaystyle x \in K\) ...

... implies \(\displaystyle \overline{x} \in K/N\) ... and so \(\displaystyle \sum \overline{a_i} r_i \in K/N\)

\(\displaystyle \Longrightarrow \sum a_i r_i \in K\) ... ... ... but how do I justify this ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K\) since \(\displaystyle K\) is submodule ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K \cap N\) ... ... see (1) above ...
Is the above correct ...?

I have been guided by hints from steenis but am lacking a clear overall strategy for the proof ...I suspect that \(\displaystyle K/N\) is fingen because \(\displaystyle K/N \subseteq M/N\) and we have that \(\displaystyle M/N\) is fingen .. and I suspect \(\displaystyle K \cap N\) is fingen because \(\displaystyle N\) is fingen ... BUT proofs ... ?

... also I suspect if we show \(\displaystyle K/N\) and \(\displaystyle K \cap N\) to be fingen then we can show that \(\displaystyle K\) is fingen ... but not sure how ,,,
Can someone clarify the above, correct any errors/weaknesses and indicate the way forward ...
Hope someone can help ...

Peter
===============================================================================***EDIT***


A further thought ...


We have shown that \(\displaystyle x - \sum a_i r_i \in K \cap N\) ...

... and ...

because \(\displaystyle K \cap N \subseteq N\) and \(\displaystyle N\) is fingen we have that \(\displaystyle K \cap N\) is fingen ... ( it that sufficient for a proof ...? )

... hence ...

\(\displaystyle x - \sum a_i r_i = \sum b_i r'_i\) ...

so ... \(\displaystyle x = \sum a_i r_i + \sum b_i r'_i\) ...

that is \(\displaystyle K\) is finitely generated by \(\displaystyle a_1, a_2, \ ... \ ... \ , a_r, b_1, b_2, \ ... \ ... \ , b_s\) ... ...

so ... \(\displaystyle M\) is Noetherian ...
But do we need a more rigorous proof of the fact that \(\displaystyle K \cap N\) is fingen ...

Peter

 

[steenis]

You are ready, well done.

Recapitulation:
$M$ is a right $R$-module, $N \leq M$, $N$ and $M/N$ are noetherian. To prove: $M$ is noetherian,
We use that if every submodule of $M$ is fingen, then $M$ is noetherian.
Let $K \leq M$ be a submodule of $M$.
$N \cap K \leq N$, $N$ is noetherian, so $N \cap K$ is fingen, say $N \cap K = \langle b_1, \cdots, b_s \rangle$ (because every submodule of a noetherian module is fingen)
$M/N$ is noetherian, so it is fingen, say $M/N = \langle \bar{a_1}, \dots, \bar{a_r} \rangle$, $a_i \in M$

Take $x \in K$, then $\bar{x} = \Sigma \bar{a_i} r_i \in M/N$
It follows that $\bar{x} - \Sigma \bar{a_i} r_i = \bar{0}$ and $x - \Sigma a_i r_i \in N$
$x \in K$ thus $x - \Sigma a_i r_i \in N \cap K$

Therefore $x - \Sigma a_i r_i = \Sigma b_j s_j$, so $x = \Sigma a_i r_i + \Sigma b_j s_j$,
concluding that $K = \langle a_1, \dots, a_r, b_1, \cdots, b_s \rangle$ is fingen and $M$ is noetherian.

Dp you post this, in your own words, on PF to amaze them ?

 

[Math Amateur]

You are ready, well done.

Recapitulation:
$M$ is a right $R$-module, $N \leq M$, $N$ and $M/N$ are noetherian. To prove: $M$ is noetherian,
We use that if every submodule of $M$ is fingen, then $M$ is noetherian.
Let $K \leq M$ be a submodule of $M$.
$N \cap K \leq N$, $N$ is noetherian, so $N \cap K$ is fingen, say $N \cap K = \langle b_1, \cdots, b_s \rangle$ (because every submodule of a noetherian module is fingen)
$M/N$ is noetherian, so it is fingen, say $M/N = \langle \bar{a_1}, \dots, \bar{a_r} \rangle$, $a_i \in M$

Take $x \in K$, then $\bar{x} = \Sigma \bar{a_i} r_i \in M/N$
It follows that $\bar{x} - \Sigma \bar{a_i} r_i = \bar{0}$ and $x - \Sigma a_i r_i \in N$
$x \in K$ thus $x - \Sigma a_i r_i \in N \cap K$

Therefore $x - \Sigma a_i r_i = \Sigma b_j s_j$, so $x = \Sigma a_i r_i + \Sigma b_j s_j$,
concluding that $K = \langle a_1, \dots, a_r, b_1, \cdots, b_s \rangle$ is fingen and $M$ is noetherian.

Dp you post this, in your own words, on PF to amaze them ?

Thanks steenis ... really neat proof ..

Just a clarification ...

I am still a bit puzzled as to how/why exactly $x - \Sigma a_i r_i \in N \cap K$

We have \(\displaystyle x \in K\) ... but don't we need to show somehow that \(\displaystyle \Sigma a_i r_i \in K\) ...
Can you help ...

Peter

 

[steenis]

I am very embarrassed but I do not know. I took it for granted, but thinking about it I can not find it.

Maybe you have a point.

Do not ask mathwonk yet, I need som time to think about it, maybe we misread the post of mathwonk

 

[steenis]

For now I think that we must not say that $N \cap K$ is fingen, but $N$ is fingen, because $N$ is noetherian. So let $N = \langle b_1, \cdots, b_s \rangle$.
$x - \Sigma a_j r_j \in N$ then $x - \Sigma a_i r_i = \Sigma b_j s_j$
and
$x = \Sigma a_i r_i + \Sigma b_j s_j$
etc.

Something like this ?

 

[Math Amateur]

I am very embarrassed but I do not know. I took it for granted, but thinking about it I can not find it.

Maybe you have a point.

Do not ask mathwonk yet, I need som time to think about it, maybe we misread the post of mathwonk

Is my reasoning below incorrect? (From Post #1...)
But it also follows that \(\displaystyle x \in K\) ...

... implies \(\displaystyle \overline{x} \in K/N\) ... and so \(\displaystyle \sum \overline{a_i} r_i \in K/N\)

\(\displaystyle \Longrightarrow \sum a_i r_i \in K\) ... ... ... but how do I justify this ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K\) since \(\displaystyle K\) is submodule ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K \cap N\) ... ... see (1) above ...
Peter

 

[steenis]

I want some time to think.

I think mathwonk meant $N$ instead of $N \cap K$, read his post again.

You could ask mathwonk why he mentioned $N \cap K$, what it is doing in his proof.

Sorry, we continue tomorrow, tasmanian and dutch time.

 

[Math Amateur]

For now I think that we must not say that $N \cap K$ is fingen, but $N$ is fingen, because $N$ is noetherian. So let $N = \langle b_1, \cdots, b_s \rangle$.
$x - \Sigma a_j r_j \in N$ then $x - \Sigma a_i r_i = \Sigma b_j s_j$
and
$x = \Sigma a_i r_i + \Sigma b_j s_j$
etc.

Something like this ?

I know you need time to think ... but just a quick point ...

Surely $N \cap K$ is fingen ... it is after all a submodule of a Noetherian module ...

Peter

 

[steenis]

yes.

 

[Math Amateur]

yes.

Sorry if I made that point too stridently or with too heavy an emphasis ... I certainly didn't mean to offend ...

Perhaps I worded it carelessly ...

Peter

 

[steenis]

No, I am sorry my answer was so short, I was in a hurry. I try to be more polite in future.

 

[Math Amateur]

No, I am sorry my answer was so short, I was in a hurry. I try to be more polite in future.

No problem at all ... I was just worried and concerned that you felt that I had been offensive ...

I'm relieved ...

Peter

 

[steenis]

Is my reasoning below incorrect? (From Post #1...)

But it also follows that \(\displaystyle x \in K\) ...

... implies \(\displaystyle \overline{x} \in K/N\) ... and so \(\displaystyle \sum \overline{a_i} r_i \in K/N\)

\(\displaystyle \Longrightarrow \sum a_i r_i \in K\) ... ... ... but how do I justify this ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K\) since \(\displaystyle K\) is submodule ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K \cap N\) ... ... see (1) above ...

Peter

I do not know yet, I am doubting. I will wait for the answer of mathwonk.

 

[Math Amateur]

I do not know yet, I am doubting. I will wait for the answer of mathwonk.

OK ... understand...

Nothing from mathwonk yet ...

Peter

 

[steenis]

Is my reasoning below incorrect? (From Post #1...)

But it also follows that \(\displaystyle x \in K\) ...

... implies \(\displaystyle \overline{x} \in K/N\) ... and so \(\displaystyle \sum \overline{a_i} r_i \in K/N\)

\(\displaystyle \Longrightarrow \sum a_i r_i \in K\) ... ... ... but how do I justify this ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K\) since \(\displaystyle K\) is submodule ...

\(\displaystyle \Longrightarrow x - \sum a_i r_i \in K \cap N\) ... ... see (1) above ...

Peter

Dear Peter, I have made a BIG mistake: $K/N$ only exists when $N \subset K$, which is not given in this case. So if $x \in K$, then we cannot say $\bar{x} \in K/N$, however $\bar{x} \in M/N$. I am very sorry, but I am learning from this too.

 

[Math Amateur]

Dear Peter, I have made a BIG mistake: $K/N$ only exists when $N \subset K$, which is not given in this case. So if $x \in K$, then we cannot say $\bar{x} \in K/N$, however $\bar{x} \in M/N$. I am very sorry, but I am learning from this too.

Oh! Of course you are right ...

I will have to acknowledge this on my post to mathwonk ...

Peter

 

[steenis]

You are too friendly ...

 

[Math Amateur]

You are too friendly ...

Have acknowledged the error ... and in doing so mentioned that you pointed out the error ...

Problem/issue ... now we do not have a solution ... :( ...

Peter

 

[steenis]

You are still worried about $K \cap N$, I think we do not need it. I think it was a mistake of mathwonk and I hope he will answer to our questions on PF soon.

Meanwhile I will post a corrected proof in the next post.

 

[Math Amateur]

You are still worried about $K \cap N$, I think we do not need it. I think it was a mistake of mathwonk and I hope he will answer to our questions on PF soon.

Meanwhile I will post a corrected proof in the next post.

Great ...

Peter

 

[steenis]

Corrected proof.

$M$ is a right $R$-module, $N \leq M$.
$N$ and $M/N$ are noetherian, to prove $M$ is noetherian.
We use that if every submodule of $M$ is fingen, then $M$ is noetherian.
So let $K \leq M$, we have to prove that $K$ is fingen.

$M/N$ is noetherian, so it is fingen: $M/N = \langle \bar{a_1}, \dots, \bar{a_r} \rangle$, with $a_i \in M$.
$N$ is noetherian, so it is fingen: $N = \langle b_1, \cdots, b_s \rangle$.

Take $x \in K$, then $\bar{x} \in M/N$, so $\bar{x} = \Sigma \bar{a_i} r_i \in M/N$.
Then $x + N = \Sigma a_i r_i + N$ and $x - \Sigma a_i r_i \in N$.
Therefore $x - \Sigma a_i r_i = \Sigma b_j s_j$, so $x = \Sigma a_i r_i + \Sigma b_j s_j$.
Concluding that $K = \langle a_1, \dots, a_r, b_1, \cdots, b_s \rangle$, thus $K$ is fingen, and $M$ is noetherian.

 

[Math Amateur]

Corrected proof.

$M$ is a right $R$-module, $N \leq M$.
$N$ and $M/N$ are noetherian, to prove $M$ is noetherian.
We use that if every submodule of $M$ is fingen, then $M$ is noetherian.
So let $K \leq M$, we have to prove that $K$ is fingen.

$M/N$ is noetherian, so it is fingen: $M/N = \langle \bar{a_1}, \dots, \bar{a_r} \rangle$, with $a_i \in M$.
$N$ is noetherian, so it is fingen: $N = \langle b_1, \cdots, b_s \rangle$.

Take $x \in K$, then $\bar{x} \in M/N$, so $\bar{x} = \Sigma \bar{a_i} r_i \in M/N$.
Then $x + N = \Sigma a_i r_i + N$ and $x - \Sigma a_i r_i \in N$.
Therefore $x - \Sigma a_i r_i = \Sigma b_j s_j$, so $x = \Sigma a_i r_i + \Sigma b_j s_j$.
Concluding that $K = \langle a_1, \dots, a_r, b_1, \cdots, b_s \rangle$, thus $K$ is fingen, and $M$ is noetherian.

Yes, that is it ... I had just realized that because N was fingen that \(\displaystyle x - \sum a_i r_i \in N\) gives us the answer ...

But then ... I was thinking that because you had suggested a focus on \(\displaystyle N\) not \(\displaystyle K \cap N\) ...

Great work!

Peter