MHB 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 $$M$$ is Noetherian. A submodule of $$M'$$ is isomorphic to a submodule of $$M$$, and so is finitely generated. ... ... "I have two questions ... ...

Question 1

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

How, exactly (that is, formally and rigorously), do we know that the submodule of $$M'$$ (which is isomorphic to a submodule of $$M$$) 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 $$\alpha(M')$$ and $$M'$$ ... ...

BUT ... $$\alpha$$ 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 $$\alpha(M')$$ ...

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 $$\alpha(M')$$ ... ... "I have worked through the details and convinced myself that this is true ...

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

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

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

Is that correct?

Peter

 

[Deveno]

Yes.