Exact Sequences and Noetherian Rings - Proposition 3.1.2 - Berrick and Keating

[Math Amateur]

I am reading the book "An Introduction to Rings and Modules with K-theory in View" by A.J. Berrick and M.E. Keating ... ...

I am currently focused on Chapter 3; Noetherian Rings and Polynomial Rings.

I need help with the proof of Proposition 3.1.2.

The statement and proof of Proposition 3.1.2 reads as follows (pages 109-110):https://www.physicsforums.com/attachments/4839
https://www.physicsforums.com/attachments/4840In the above text (at the start of the proof), Berrick and Keating write:" ... ... Suppose that \(\displaystyle M\) is Noetherian. A submodule of \(\displaystyle M'\) is isomorphic to a submodule of \(\displaystyle M\), and so is finitely generated. ... ... "I have two questions ... ...

Question 1

How do we demonstrate, formally and rigorously, that there exists a submodule of \(\displaystyle M'\) that is isomorphic to a submodule of \(\displaystyle M\) ... ... ?Question 2

How, exactly (that is, formally and rigorously), do we know that the submodule of \(\displaystyle M'\) (which is isomorphic to a submodule of \(\displaystyle M\)) is finitely generated ... (I know it sounds plausible ... but ... what is the formal demonstration of this fact) ... ..Hope someone can help ...

Peter

 

[Deveno]

By exactness, $\text{ker }\alpha = \{0\}$, so $\alpha$ is injective. That is: $\alpha(M') \cong M'$.

So if $N'$ is a submodule of $M'$, then $\alpha(N')$ is a submodule of $\alpha(M') \subseteq M$ isomorphic to $N'$.

It sounds as if your text author is using the *definition*:

A right $R$-module is Noetherian if any submodule is finitely-generated.

Since $M$ is Noetherian, and $\alpha(N')$ is a submodule of $M$, it follows that $\alpha(N')$ is finitely-generated, and so if $\{m_1,\dots,m_n\}$ is a finite generating set for $\alpha(N')$, then $\{\alpha^{-1}(m_1),\dots,\alpha^{-1}(m_n)\}$ is likewise a finite generating set for $N'$.

 

[Math Amateur]

By exactness, $\text{ker }\alpha = \{0\}$, so $\alpha$ is injective. That is: $\alpha(M') \cong M'$.

So if $N'$ is a submodule of $M'$, then $\alpha(N')$ is a submodule of $\alpha(M') \subseteq M$ isomorphic to $N'$.

It sounds as if your text author is using the *definition*:

A right $R$-module is Noetherian if any submodule is finitely-generated.

Since $M$ is Noetherian, and $\alpha(N')$ is a submodule of $M$, it follows that $\alpha(N')$ is finitely-generated, and so if $\{m_1,\dots,m_n\}$ is a finite generating set for $\alpha(N')$, then $\{\alpha^{-1}(m_1),\dots,\alpha^{-1}(m_n)\}$ is likewise a finite generating set for $N'$.

Thanks Deveno ... will be working through this shortly ...

Appreciate your help ...

I obviously need to revise exact sequences ...

Peter

 

[Math Amateur]

By exactness, $\text{ker }\alpha = \{0\}$, so $\alpha$ is injective. That is: $\alpha(M') \cong M'$.

So if $N'$ is a submodule of $M'$, then $\alpha(N')$ is a submodule of $\alpha(M') \subseteq M$ isomorphic to $N'$.

It sounds as if your text author is using the *definition*:

A right $R$-module is Noetherian if any submodule is finitely-generated.

Since $M$ is Noetherian, and $\alpha(N')$ is a submodule of $M$, it follows that $\alpha(N')$ is finitely-generated, and so if $\{m_1,\dots,m_n\}$ is a finite generating set for $\alpha(N')$, then $\{\alpha^{-1}(m_1),\dots,\alpha^{-1}(m_n)\}$ is likewise a finite generating set for $N'$.

Hi Deveno,

Just reflecting on your post ... making sure I fully understand what you have said ...

You write:

"... ... By exactness, $\text{ker }\alpha = \{0\}$, so $\alpha$ is injective. That is: $\alpha(M') \cong M'$. ... ... "

Could you explain exactly how/why it is the case that $\alpha$ being injective leads to $\alpha(M') \cong M'$ ... ...

Sorry to be a bit slow, but do need some clarification/explanation ... ...

Peter

 

[Math Amateur]

Hi Deveno,

Just reflecting on your post ... making sure I fully understand what you have said ...

You write:

"... ... By exactness, $\text{ker }\alpha = \{0\}$, so $\alpha$ is injective. That is: $\alpha(M') \cong M'$. ... ... "

Could you explain exactly how/why it is the case that $\alpha$ being injective leads to $\alpha(M') \cong M'$ ... ...

Sorry to be a bit slow, but do need some clarification/explanation ... ...

Peter

***EDIT***

Hmm ... reflecting some more ...

To try to answer my own question ...

... ... to show that $\alpha(M') \cong M'$ we need to demonstrate an injective and surjective mapping between \(\displaystyle \alpha(M')\) and \(\displaystyle M'\) ... ...

BUT ... \(\displaystyle \alpha\) itself is such a mapping ... since it is injective by definition of the exactness of the sequence concerned ... and, further, is of course, surjective on its image \(\displaystyle \alpha(M')\) ...

Is that correct?

Peter

 

[Deveno]

Yup.

You see, the "exactness" of a short exact sequence:

$0 \to A \to B \to C \to 0$

tells us a great deal.

First of all, $R$-modules have a "zero object" (the trivial module), for which there is only ONE possible $R$-linear map:

$0 \to A$ (namely, sending the only element of the trivial module to the additive identity of $A$).

There is also only one possible $R$-linear map $C \to 0$ (namely, the one that kills everything).

Exactness at the portion $0 \to A \to B$, means the map from $A$ to $B$ is injective, since the image of $A$ is the kernel of the map $0 \to A$, which is the $0$ of $A$.

Exactness of the portion $B \to C \to 0$ means the map $B \to C$ is surjective, since the kernel of the map $C \to 0$ is all of $C$, and this is (by exactness) the image of $B$.

Finally, exactness at the portion $A \to B \to C$ tells us that $C \cong B/\text{im }A$, by the fundamental isomorphism theorem.

In essence then, a short exact sequence embodies (up to isomorphism) the "canonical" short exact sequence:

$0 \to N \to M \to M/N \to 0$ (where $N \subseteq M$).

Typically, this let's us "transfer" information about $M$ to any $R$-module isomorphic to a submodule of $M$, and in certain cases, allows us to "recover" (or "reconstitute") information about a homomorphic image of $M$ by looking at pre-images under the homomorphism.

 

[Math Amateur]

Yup.

You see, the "exactness" of a short exact sequence:

$0 \to A \to B \to C \to 0$

tells us a great deal.

First of all, $R$-modules have a "zero object" (the trivial module), for which there is only ONE possible $R$-linear map:

$0 \to A$ (namely, sending the only element of the trivial module to the additive identity of $A$).

There is also only one possible $R$-linear map $C \to 0$ (namely, the one that kills everything).

Exactness at the portion $0 \to A \to B$, means the map from $A$ to $B$ is injective, since the image of $A$ is the kernel of the map $0 \to A$, which is the $0$ of $A$.

Exactness of the portion $B \to C \to 0$ means the map $B \to C$ is surjective, since the kernel of the map $C \to 0$ is all of $C$, and this is (by exactness) the image of $B$.

Finally, exactness at the portion $A \to B \to C$ tells us that $C \cong B/\text{im }A$, by the fundamental isomorphism theorem.

In essence then, a short exact sequence embodies (up to isomorphism) the "canonical" short exact sequence:

$0 \to N \to M \to M/N \to 0$ (where $N \subseteq M$).

Typically, this let's us "transfer" information about $M$ to any $R$-module isomorphic to a submodule of $M$, and in certain cases, allows us to "recover" (or "reconstitute") information about a homomorphic image of $M$ by looking at pre-images under the homomorphism.

Thanks for the brief tutorial, Deveno ...

Most helpful indeed ...

Thanks again,

Peter

 

[Math Amateur]

By exactness, $\text{ker }\alpha = \{0\}$, so $\alpha$ is injective. That is: $\alpha(M') \cong M'$.

So if $N'$ is a submodule of $M'$, then $\alpha(N')$ is a submodule of $\alpha(M') \subseteq M$ isomorphic to $N'$.

It sounds as if your text author is using the *definition*:

A right $R$-module is Noetherian if any submodule is finitely-generated.

Since $M$ is Noetherian, and $\alpha(N')$ is a submodule of $M$, it follows that $\alpha(N')$ is finitely-generated, and so if $\{m_1,\dots,m_n\}$ is a finite generating set for $\alpha(N')$, then $\{\alpha^{-1}(m_1),\dots,\alpha^{-1}(m_n)\}$ is likewise a finite generating set for $N'$.

Hi Deveno,

Just a further point of clarification and assurance ...

You write:

" ... ... if $N'$ is a submodule of $M'$, then $\alpha(N')$ is a submodule of \(\displaystyle \alpha(M')\) ... ... "I have worked through the details and convinced myself that this is true ...

BUT ... you go on to claim that \(\displaystyle \alpha(N')\) is a submodule of \(\displaystyle M\) ...

My question is ... where/how have we shown that \(\displaystyle \alpha(N')\) is a submodule of \(\displaystyle M\) ... ?... ... ... ...

... reflecting ... presumably ... a submodule of a submodule of \(\displaystyle M\) ... is a submodule of \(\displaystyle M\) ... ... (it seems so from the definition of a submodule) ...

Is that correct?

Peter

 

[Deveno]

Yes.