Noetherian Modules: Direct Sums & Bland Proposition 4.2.7

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some further help to fully understand the proof of part of Proposition 4.2.7 ... ...

Proposition 4.2.7 reads as follows:https://www.physicsforums.com/attachments/8209In the above proof, when Bland is dealing with the converse, we read the following:

" ... ... Then the short exact sequence

$$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n \longrightarrow 0$$ ... ... "Now I understand the induction argument that Bland goes on to talk about ... but we need to establish that the sequence of modules given is indeed a short exact sequence ... so ... some questions follow ... ... When an author gives a short exact sequence such as

$$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n \longrightarrow 0$$

is he/she implying that the R-linear mappings (R-module homomorphisms ...) involved are obvious ... ?If that is the case then what are the obvious R-module homomorphisms in the case of

$$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \stackrel{ f }{ \longrightarrow } \bigoplus_{ i = 1 }^{ n } M_i \stackrel{ g }{ \longrightarrow } M_n \longrightarrow 0$$

and how does $$\text{ I am } f = \text{ Ker } g$$ ...Peter

 

[steenis]

Yes, you can assume that the R-maps are obvious, in this case;

$f = j_{n-1}$ is the inclusion $\bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i$

and $g = p_n$ is the projection $\bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n$

Of course $p_n \circ j_{n-1} = 0$, so jou have to prove that $\text{ker } p_n \subset \text{im } j_{n-1}$

Recall Example 6 p.47, before you go further.

 

[Math Amateur]

Yes, you can assume that the R-maps are obvious, in this case;

$f = j_{n-1}$ is the inclusion $\bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i$

and $g = p_n$ is the projection $\bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n$

Of course $p_n \circ j_{n-1} = 0$, so jou have to prove that $\text{ker } p_n \subset \text{im } j_{n-1}$

Recall Example 6 p.47, before you go further.

Thanks Steenis ...

Appreciate your help and guidance...

Peter

 

[Math Amateur]

Yes, you can assume that the R-maps are obvious, in this case;

$f = j_{n-1}$ is the inclusion $\bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i$

and $g = p_n$ is the projection $\bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n$

Of course $p_n \circ j_{n-1} = 0$, so jou have to prove that $\text{ker } p_n \subset \text{im } j_{n-1}$

Recall Example 6 p.47, before you go further.

Consider ...

$$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \stackrel{ f }{ \longrightarrow } \bigoplus_{ i = 1 }^{ n } M_i \stackrel{ g }{ \longrightarrow } M_n \longrightarrow 0$$

We have $$f \equiv j_{ n - 1 } \ : \ \bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i$$

where $$j_{ n - 1 } ( x_1, x_2, \ ... \ ... \ , x_{ n-1 } ) = ( x_1, x_2, \ ... \ ... \ , x_{ n-1 } , 0 )$$

... note that $$j_{ n - 1 }$$ is a canonical injection ( see Bland page 39)
$$g \equiv p_n \ : \ \bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n $$

where $$p_n ( x_1, x_2, \ ... \ ... \ , x_n ) = x_n $$

... note that $$p_n$$ is a canonical projection ( see Bland page 39) Now ... $$\text{ I am } f = \text{ I am } j_{ n - 1 } = M_1 \times M_2 \times \ ... \ ... \ M_{ n-1 } \times 0$$

and

$$\text{ Ker } g = \text{ Ker } p_n = = M_1 \times M_2 \times \ ... \ ... \ M_{ n-1 } \times 0$$... so ... $$\text{ I am } f = \text{ Ker } g$$Hence $$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n \longrightarrow 0$$

is a short exact sequence ...
Is that correct?

Peter

 

[steenis]

Yes, that is correct, because $j_{n-1}$ is injective en $p_n$ is surjective.

But you are not ready yet.

 

[Math Amateur]

Yes, that is correct, because $j_{n-1}$ is injective en $p_n$ is surjective.

But you are not ready yet.

I assume that when you say: "But you are not ready yet." ... that you mean I need to show that the sequence ...$$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \stackrel{ j_{ n - 1} }{ \longrightarrow } \bigoplus_{ i = 1 }^{ n } M_i \stackrel{ p_n }{ \longrightarrow } M_n \stackrel{ s }{ \longrightarrow } 0$$... is not only exact at $$\bigoplus_{ i = 1 }^{ n } M_i$$ (which I have shown ... ) ... but .. is also exact at $$\bigoplus_{ i = 1 }^{ n-1 } M_i$$ and at $$M_n$$ ... ... Is that correct ...?If that is correct then note that ...

$$\text{ Ker } j_{ n-1 } = (0, 0, \ ... \ ... \ , 0)$$ ($$n -1$$ elements) $$= \text{ I am } 0$$ ...

... and ...

$$\text{ I am } p_n = M_n = \text{ Ker } s$$ ...So ... $$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \stackrel{ j_{ n - 1} }{ \longrightarrow } \bigoplus_{ i = 1 }^{ n } M_i \stackrel{ p_n }{ \longrightarrow } M_n \stackrel{ s }{ \longrightarrow } 0$$... is an exact sequence ...Is that correct?

Peter

 

[steenis]

Yes it is correct. No it is not what I meant.

You still have to prove the proposition: if $M_i$ is noetherian then $\bigoplus_{ i = 1 }^{ n } M_i$ is noetherian.

 

[Math Amateur]

Yes it is correct. No it is not what I meant.

You still have to prove the proposition: if $M_i$ is noetherian then $\bigoplus_{ i = 1 }^{ n } M_i$ is noetherian.

Hi Steenis ... thanks for your help ...

I was happy with Bland's proof ... but anyway ...To prove ... if $M_i$ is Noetherian then $\bigoplus_{ i = 1 }^{ n } M_i$ is Noetherian.Assume each $$M_i$$ is Noetherian ...Now enter induction process ...

We have that $\bigoplus_{ i = 1 }^{ 1 } M_i = M_1 $ is Noetherian ...

Assume that $$\bigoplus_{ i = 1 }^{ m } M_i$$ is Noetherian for each integer $$n$$ such that $$1 \lt m \lt n$$ ...

Then in the short exact sequence ...

$$0 \longrightarrow \bigoplus_{ i = 1 }^{ n-1 } M_i \longrightarrow \bigoplus_{ i = 1 }^{ n } M_i \longrightarrow M_n \longrightarrow 0$$ we have

$$\bigoplus_{ i = 1 }^{ n-1 } M_i$$ is Noetherian because of the assumption in the induction process ...

... and ...

$$M_n$$ is Noetherian by assumption ...

so that $$\bigoplus_{ i = 1 }^{ n } M_i$$ is Noetherian by Corollary 4.2.4 ...Thus the induction process gives us our desired result ...Is that correct?

Peter

 

[steenis]

Yes that is correct. I am sorry Peter, I thought it was an exercise. I didnot sleep good this night.

 

[Math Amateur]

Yes that is correct. I am sorry Peter, I thought it was an exercise. I didnot sleep good this night.

That is OK, Steenis ...

Good exercise anyway ... :) ...

Sorry that you didn't sleep well ...

Thanks again for your help ... it is much appreciated...

Peter