Exact Sequences - Isomorphisms Resulting from Exact Sequences

[Math Amateur]

I am reading Dummit and Foote Section 10.5 Exact Sequences - Projective, Injective and Flat Modules.

I need help with some of the conclusions to Example 2, D&F Section 10.5, pages 379-380 - see attached. However, note that the question is essentially about isomorphisms. However, I would like someone to confirm the correctness of the exact sequences context.

To establish the context of the example as I interpret it ...

On pages 378-379 D&F, as I see it, establish the following:

If we have

\(\displaystyle A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C \)

where \(\displaystyle \psi \) is an injective homomorphism

and \(\displaystyle \phi \) is a surjective homomorphism and \(\displaystyle \text{ ker } \phi = \psi (A) \)

then

(1) \(\displaystyle A \cong \psi (A) \) because the homomorphism \(\displaystyle \psi \ : \ A \to \psi (A) \) is both injective and surjective

and

(2) by the First Isomorphism Theorem \(\displaystyle C \cong B/ \text{ker } \phi \text{ or } C \cong B/ \psi (A) \)

so that C is isomorphic to the quotient of B by an isomorphic copy of A

or, in other words B is an extension of C by A.

Can someone please confirm that my interpretation of the situation regarding extensions of C by A as outlined by D&F is correct. I would definitely feel more confident if someone was to confirm the above as OK!

Now Example 2 is a special case of Example 1 - see bottom of page 379 and top of page 380 on the attachment.

In Example 2 D&F give two short exact sequences (see attachment).

The first exact sequence, I believe, establishes the following isomorphism:

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / \text{ ker } \phi \)

that is,

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \) ... ... ... (3)

The second exact sequence, I believe, establishes the following isomorphism

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / \text{ ker } \phi \)

that is

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / n ( \mathbb{Z} ) \) ... ... ... (4)Now it seems (intuitively) that given (3) and (4) above we should have

\(\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \cong \mathbb{Z} / n ( \mathbb{Z} ) \) ... ... ... (5)

since they are both isomorphic to \(\displaystyle \mathbb{Z} / n \mathbb{Z} \)

BUT

D&F say that this is explicitly not the case ... indeed, they write:

"Note that the modules in the middle of the previous two exact sequences are not isomorphic even though the respective "A" and "C" terms are isomorphic. Then there are at least two "essentially different" or "inequivalent" ways of extending \(\displaystyle \mathbb{Z} / n \mathbb{Z} \) by \(\displaystyle \mathbb{Z} \).

Can someone please clarify this situation for me.

Peter

 

[Math Amateur]

I am reading Dummit and Foote Section 10.5 Exact Sequences - Projective, Injective and Flat Modules.

I need help with some of the conclusions to Example 2, D&F Section 10.5, pages 379-380 - see attached. However, note that the question is essentially about isomorphisms. However, I would like someone to confirm the correctness of the exact sequences context.

To establish the context of the example as I interpret it ...

On pages 378-379 D&F, as I see it, establish the following:

If we have

\(\displaystyle A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C \)

where \(\displaystyle \psi \) is an injective homomorphism

and \(\displaystyle \phi \) is a surjective homomorphism and \(\displaystyle \text{ ker } \phi = \psi (A) \)

then

(1) \(\displaystyle A \cong \psi (A) \) because the homomorphism \(\displaystyle \psi \ : \ A \to \psi (A) \) is both injective and surjective

and

(2) by the First Isomorphism Theorem \(\displaystyle C \cong B/ \text{ker } \phi \text{ or } C \cong B/ \psi (A) \)

so that C is isomorphic to the quotient of B by an isomorphic copy of A

or, in other words B is an extension of C by A.

Can someone please confirm that my interpretation of the situation regarding extensions of C by A as outlined by D&F is correct. I would definitely feel more confident if someone was to confirm the above as OK!

Now Example 2 is a special case of Example 1 - see bottom of page 379 and top of page 380 on the attachment.

In Example 2 D&F give two short exact sequences (see attachment).

The first exact sequence, I believe, establishes the following isomorphism:

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / \text{ ker } \phi \)

that is,

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \) ... ... ... (3)

The second exact sequence, I believe, establishes the following isomorphism

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / \text{ ker } \phi \)

that is

\(\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / n ( \mathbb{Z} ) \) ... ... ... (4)Now it seems (intuitively) that given (3) and (4) above we should have

\(\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \cong \mathbb{Z} / n ( \mathbb{Z} ) \) ... ... ... (5)

since they are both isomorphic to \(\displaystyle \mathbb{Z} / n \mathbb{Z} \)

BUT

D&F say that this is explicitly not the case ... indeed, they write:

"Note that the modules in the middle of the previous two exact sequences are not isomorphic even though the respective "A" and "C" terms are isomorphic. Then there are at least two "essentially different" or "inequivalent" ways of extending \(\displaystyle \mathbb{Z} / n \mathbb{Z} \) by \(\displaystyle \mathbb{Z} \).

Can someone please clarify this situation for me.

Peter

I am making a quick reply to clarify my post ... I still need an issue clarified ... but I also believe I did not read D&F clearly enough ...

I still think that my contention in (5) in my previous post, namely:

\(\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \cong \mathbb{Z} / n ( \mathbb{Z} ) \) ... ... ... (5)

is correct. Do members agree?

However, D&F are correct in saying that the respective "A" and "C" terms in the two exact sequences are isomorphic - indeed they are the same (see attachment - bottom of page 379 and top of page 380) - respectively, they are \(\displaystyle \mathbb{Z} \) and \(\displaystyle \mathbb{Z} / n \mathbb{Z} \) in both cases. BUT also as D&F say, the modules in the middle of the two exact sequences (the "B" terms), namely \(\displaystyle \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) \) and \(\displaystyle \mathbb{Z} \) are different - not isomorphic. Yes, I can see this ... ... and that probably means that D&F's statement that "thus there are (at least) two "essentially different" or "inequivalent" ways of extending \(\displaystyle \mathbb{Z} / n \mathbb{Z} \) by \(\displaystyle \mathbb{Z} \)" is correct ...

BUT ... ... it still seems to me that the two (different?) extensions of \(\displaystyle \mathbb{Z} / n \mathbb{Z} \) by \(\displaystyle \mathbb{Z} \) ... ... that is the actual extensions (as distinct from the ways of extending) namely \(\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \) and \(\displaystyle \mathbb{Z} / n ( \mathbb{Z} ) \) are isomorphic ... ...

Is this correct?

Peter

 

[Deveno]

I think you are mis-understanding what is being said:

the extensions are, in the two cases:

$\Bbb Z \oplus \Bbb Z_n$ and $\Bbb Z$, and these two abelian groups are definitely NOT isomorphic.

In this scenario, "extension" actually means "reverse quotient" or "pre-image of a quotient".

It IS true that:

$(\Bbb Z \oplus \Bbb Z_n)/(\Bbb Z \oplus \{[0]\}) \cong \Bbb Z_n$

this is what D&F mean by "the $C$'s are isomorphic".

What is usually more useful is a "split extension":

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

where the composition of the arrows between $B$ and $C$ (starting at $C$) yields the identity on $C$.

 

[ThePerfectHacker]

On pages 378-379 D&F, as I see it, establish the following:

If we have

\(\displaystyle A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C \)

Quite simply an exact sequence is when the kernel agrees with the image. So look at the middle term, $B$. You have a map going into $B$ by $\psi$ and a map going out of $B$ by $\varphi$. To be an exact sequence you need that $\ker \varphi = \text{im} \psi$.

 

[blue_raver22]

Dear Peter,

Thank you for reaching out for clarification on the example in Dummit and Foote's book on exact sequences. Your interpretation of the context and the isomorphisms in the example seems to be correct. However, the statement that there are "at least two essentially different ways of extending \mathbb{Z}/n\mathbb{Z} by \mathbb{Z}" is referring to the fact that there are two different exact sequences that can be constructed using the same modules, but with different homomorphisms.

In the first exact sequence, the homomorphisms are defined as follows: \psi: \mathbb{Z} \to \mathbb{Z} \oplus (\mathbb{Z}/n\mathbb{Z}), \psi(x) = (x, 0) and \phi: \mathbb{Z} \oplus (\mathbb{Z}/n\mathbb{Z}) \to \mathbb{Z}/n\mathbb{Z}, \phi(x,y) = y. This results in the isomorphism (3) that you mentioned.

In the second exact sequence, the homomorphisms are defined as follows: \psi: \mathbb{Z} \to \mathbb{Z}, \psi(x) = nx and \phi: \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}, \phi(x) = \overline{x}. This results in the isomorphism (4) that you mentioned.

So while both sequences involve the same modules, the different homomorphisms used result in different exact sequences and therefore different isomorphisms. This is why the modules in the middle of the sequences (A and C) are not isomorphic, even though the terms A and C are isomorphic.

I hope this clarifies the situation for you. If you have any further questions, please don't hesitate to ask.

Best,