MHB How Does Condition (2) Imply Condition (3) in Bland's Proposition 3.2.7?

[Math Amateur]

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

I am trying to understand Section 3.2 on exact sequences in Mod_R and need help with the proof of Proposition 3.2.7.

Proposition 3.2.7 and its proof read as follows:View attachment 3612I am having trouble in understanding the proof that condition (2) implies condition (3).

Bland's argument of $$ (2) \Longrightarrow (3) $$ begins, of course, with the assumption that $$\text{ Ker } g$$ is a direct summand of $$M$$; and then Bland let's $$N$$ be a submodule of $$M$$ such that

$$M = \text{ Ker } g \ \oplus \ N $$

Given this and given that the sequence being considered is exact, we have

$$ M/ \text{ Ker } g \ \cong \ g(M) $$ by the First Isomorphism Theorem for R-modules.

Thus ... ...

$$M/ \text{ Ker } g \ \cong \ M_2$$

since $$g$$ is an epimorphism ... ...

But as you can see in the above text, Bland states that

$$ M_2 \ \cong \ M/ \text{ Ker } g \ \cong \ N $$ BUT ... ...

How does Bland deduce $$ M/ \text{ Ker } g \ \cong \ N $$?

That is why is $$ M/ \text{ Ker } g \ \cong \ N $$?

Peter

 

[Fallen Angel]

Hi Peter,

That's because the sum is direct, in general , given a modules direct sum $A=B\oplus C$, $A/B\cong C$

 

[steenis]

Hi, I am new to this forum. I am also trying to read this propostion of Bland. There is something mysterious because Rotman stated a counterexample of this proposition. See: Rotman - An Introduction to Homological Algebra, 2nd edition, 2009, example 2.29 p.54.

 

[Math Amateur]

Hi, I am new to this forum. I am also trying to read this propostion of Bland. There is something mysterious because Rotman stated a counterexample of this proposition. See: Rotman - An Introduction to Homological Algebra, 2nd edition, 2009, example 2.29 p.54.

Hi Steenis ... I am currently working on algebraic geometry... but will try to get time to look back at this issue shortly ...

Bland's book on Rings and Their Modules is a great and challenging book! His notation and his rigour are excellent as is the clarity of his arguments ... I really enjoy challenging myself with his book ...

You may get an answer on this issue from some of the knowledgeable algebraists on this forum ...

It might be an idea to scan and post the relevant pages from Bland and from Rotman ...

By the way, welcome to MHB ...

PeterPS Revising exact sequences ...*** EDIT ***

I am now posting Bland Proposition 3.2.7 (page 85), together with a note on terminology ... and Rotman (An Introduction to Homological Algebra) Example 2.29 ... ... so MHB members can see and understand Steenis' issue as outlined in his post above ... ... as follows:Bland: Proposition 3.2.7 (page 85)

View attachment 5808
Bland: Note on Terminology

View attachment 5809Rotman: Example 2.29

View attachment 5810

 

[steenis]

Peter, thank you for your answer.
I do not know how "to scan and post the relevant pages" from books.
Peter, can you show me how to learn to edit the mathematical notation in this forum?I want to add this to my former post:
See also prop.2.28 p.52 of Rotman (same book). This is the propostition 3.2.7 p.85 of Bland in the (=>)-direction. Just after the proof of prop.2.28 on p.53, Rotman says that the converse of prop.2.28 is not true, which is the (<=)-direction of the proposition of Bland.
To my knowledge, the (<=)-direction is not in other books.
I think that one correct formulation of the (<=)-direction of the proposition of Bland is given, for instance, by Grillet, proposition 1.5 on page 395 (Grillet - Abstract Algebra, 2nd edition 2007).

One more remark: can someone also look critically to the (3) => (1) proof of the same proposition 3.2.7 p.85 of Bland? I think it is not correct, but I cannot put my finger on it.

I agree that the book of Bland is challenging, but I am not yet convinced that it is a good book. An errata-sheet would be most helpful.

 

[Math Amateur]

Peter, thank you for your answer.
I do not know how "to scan and post the relevant pages" from books.
Peter, can you show me how to learn to edit the mathematical notation in this forum?I want to add this to my former post:
See also prop.2.28 p.52 of Rotman (same book). This is the propostition 3.2.7 p.85 of Bland in the (=>)-direction. Just after the proof of prop.2.28 on p.53, Rotman says that the converse of prop.2.28 is not true, which is the (<=)-direction of the proposition of Bland.
To my knowledge, the (<=)-direction is not in other books.
I think that one correct formulation of the (<=)-direction of the proposition of Bland is given, for instance, by Grillet, proposition 1.5 on page 395 (Grillet - Abstract Algebra, 2nd edition 2007).

One more remark: can someone also look critically to the (3) => (1) proof of the same proposition 3.2.7 p.85 of Bland? I think it is not correct, but I cannot put my finger on it.

I agree that the book of Bland is challenging, but I am not yet convinced that it is a good book. An errata-sheet would be most helpful.

Hi Steenis,

It is late here in Tasmania so I must be brief ... will answer further in the morning when I will, among other things, check out Grillet's text ...

I just want to mention Latex ... if you want to put mathematical notation into your posts ... and I'm sure you will then you need to encode using Latex ... it is not to difficult ... see MHB on Latex and find the Latex Guide under Tips and Tutorials ... will be in touch again when it is morning in Tasmania, Australia ...

Best Wishes,

Peter

 

[Deveno]

My two cents worth:

There is a qualitative difference between an exact short sequence:

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

where, in any case, we have $B \cong C/A$, and a *split* short exact sequence:

$0 \to A \to A \oplus B \stackrel{\leftarrow}{\to} B \to 0$

where we *still* have, of course $B \cong (A\oplus B)/A$

The underlying "counter-example" of Rotman is based on the exact sequence:

$0 \to \Bbb Z_2 \stackrel{f}{\to} \Bbb Z_4 \stackrel{g}{\to} \Bbb Z_2 \to 0$

Where $f$ is the map: $f([a]_2) = [2a]_4$, and $g$ is the map: $g([x]_4) = [x]_2$.

It is clear that this is exact, since:

$\text{ker }g = \{[0]_4, [2]_4\} = \text{im }f$, and $f$ is injective, whilst $g$ is surjective.

It is instructive to see how this sequence violates Brand's hypotheses:

It does not adhere to the statement hypothesis of the theorem, since it is not split (there is no mapping $h: \Bbb Z_2 \to \Bbb Z_4$ such that $g \circ h = \text{id}_{\Bbb Z_2}$-this can be verified by considering what happens to $[1]_2$).

It violates (1) and (3), because $\Bbb Z_2$ is not a summand of $\Bbb Z_4$ (otherwise we would have: $\Bbb Z_4 \cong \Bbb Z_2 \oplus \Bbb Z_2$, arguing by order alone, which is false).

It is clear that violation of (1) or (3) is thus equivalent (in this case) to violating (2).

Rotman then extends this to a direct sum exact sequence counter-example by considering a (countably) infinite-fold direct product of $A \oplus B$, using the fact that (up to isomorphism) direct sum is a commutative and associative operation.

In summary, just because we have an exact sequence:

$0 \to M \to M\oplus N \to N \to 0$

there is no reason to suppose the mapping $M \to M \oplus N$ is the canonical inclusion, nor the map $M \oplus N \to N$ is the canonical projection (which is essentially what happens when the sequence splits).

In "abelian categories" (such as AbGrp, $\ _R$Mod, or Vect$_k$) "split" means "direct sum" (and we have a symmetry between "left-split" and "right-split" short exact sequences). I note in passing this is *not* the case for the category Grp, where "left-split" short exact sequences lead to direct products, and "right-split" short exact sequences lead to semi-direct products.

 

[Math Amateur]

Peter, thank you for your answer.
I do not know how "to scan and post the relevant pages" from books.
Peter, can you show me how to learn to edit the mathematical notation in this forum?I want to add this to my former post:
See also prop.2.28 p.52 of Rotman (same book). This is the propostition 3.2.7 p.85 of Bland in the (=>)-direction. Just after the proof of prop.2.28 on p.53, Rotman says that the converse of prop.2.28 is not true, which is the (<=)-direction of the proposition of Bland.
To my knowledge, the (<=)-direction is not in other books.
I think that one correct formulation of the (<=)-direction of the proposition of Bland is given, for instance, by Grillet, proposition 1.5 on page 395 (Grillet - Abstract Algebra, 2nd edition 2007).

One more remark: can someone also look critically to the (3) => (1) proof of the same proposition 3.2.7 p.85 of Bland? I think it is not correct, but I cannot put my finger on it.

I agree that the book of Bland is challenging, but I am not yet convinced that it is a good book. An errata-sheet would be most helpful.

So that MHB readers can see what Steenis is referring to, I am posting Grillet, proposition 1.5 on page 395 (Grillet - Abstract Algebra, 2nd edition 2007) ... ... as follows:https://www.physicsforums.com/attachments/5811
View attachment 5812I hope that helps MHB members read and understand Steenis issue/problem ... ...

Peter

 

[Math Amateur]

My two cents worth:

There is a qualitative difference between an exact short sequence:

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

where, in any case, we have $B \cong C/A$, and a *split* short exact sequence:

$0 \to A \to A \oplus B \stackrel{\leftarrow}{\to} B \to 0$

where we *still* have, of course $B \cong (A\oplus B)/A$

The underlying "counter-example" of Rotman is based on the exact sequence:

$0 \to \Bbb Z_2 \stackrel{f}{\to} \Bbb Z_4 \stackrel{g}{\to} \Bbb Z_2 \to 0$

Where $f$ is the map: $f([a]_2) = [2a]_4$, and $g$ is the map: $g([x]_4) = [x]_2$.

It is clear that this is exact, since:

$\text{ker }g = \{[0]_4, [2]_4\} = \text{im }f$, and $f$ is injective, whilst $g$ is surjective.

It is instructive to see how this sequence violates Brand's hypotheses:

It does not adhere to the statement hypothesis of the theorem, since it is not split (there is no mapping $h: \Bbb Z_2 \to \Bbb Z_4$ such that $g \circ h = \text{id}_{\Bbb Z_2}$-this can be verified by considering what happens to $[1]_2$).

It violates (1) and (3), because $\Bbb Z_2$ is not a summand of $\Bbb Z_4$ (otherwise we would have: $\Bbb Z_4 \cong \Bbb Z_2 \oplus \Bbb Z_2$, arguing by order alone, which is false).

It is clear that violation of (1) or (3) is thus equivalent (in this case) to violating (2).

Rotman then extends this to a direct sum exact sequence counter-example by considering a (countably) infinite-fold direct product of $A \oplus B$, using the fact that (up to isomorphism) direct sum is a commutative and associative operation.

In summary, just because we have an exact sequence:

$0 \to M \to M\oplus N \to N \to 0$

there is no reason to suppose the mapping $M \to M \oplus N$ is the canonical inclusion, nor the map $M \oplus N \to N$ is the canonical projection (which is essentially what happens when the sequence splits).

In "abelian categories" (such as AbGrp, $\ _R$Mod, or Vect$_k$) "split" means "direct sum" (and we have a symmetry between "left-split" and "right-split" short exact sequences). I note in passing this is *not* the case for the category Grp, where "left-split" short exact sequences lead to direct products, and "right-split" short exact sequences lead to semi-direct products.

Thanks for the help, Deveno ... ... I roughly (actually, very roughly) followed the main idea of your post ...

However, to fully understand your post I need to do some revision ...[NOTE: Unlike Deveno, my knowledge of algebra is not at my fingertips nor completely front of mind ... ... :) ... ...]Thanks again, Deveno!

Peter

 

[Math Amateur]

Hi Steenis,

It is late here in Tasmania so I must be brief ... will answer further in the morning when I will, among other things, check out Grillet's text ...

I just want to mention Latex ... if you want to put mathematical notation into your posts ... and I'm sure you will then you need to encode using Latex ... it is not to difficult ... see MHB on Latex and find the Latex Guide under Tips and Tutorials ... will be in touch again when it is morning in Tasmania, Australia ...

Best Wishes,

Peter

Hi Steenis,

Further to my remarks on Latex there is a good tutorial or guide on the Math Help Forum here ... (see the attached file by Mathman)

LaTex Tutorial - Math Help ForumHope that helps,

Peter

 

[Math Amateur]

My two cents worth:

There is a qualitative difference between an exact short sequence:

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

where, in any case, we have $B \cong C/A$, and a *split* short exact sequence:

$0 \to A \to A \oplus B \stackrel{\leftarrow}{\to} B \to 0$

where we *still* have, of course $B \cong (A\oplus B)/A$

The underlying "counter-example" of Rotman is based on the exact sequence:

$0 \to \Bbb Z_2 \stackrel{f}{\to} \Bbb Z_4 \stackrel{g}{\to} \Bbb Z_2 \to 0$

Where $f$ is the map: $f([a]_2) = [2a]_4$, and $g$ is the map: $g([x]_4) = [x]_2$.

It is clear that this is exact, since:

$\text{ker }g = \{[0]_4, [2]_4\} = \text{im }f$, and $f$ is injective, whilst $g$ is surjective.

It is instructive to see how this sequence violates Brand's hypotheses:

It does not adhere to the statement hypothesis of the theorem, since it is not split (there is no mapping $h: \Bbb Z_2 \to \Bbb Z_4$ such that $g \circ h = \text{id}_{\Bbb Z_2}$-this can be verified by considering what happens to $[1]_2$).

It violates (1) and (3), because $\Bbb Z_2$ is not a summand of $\Bbb Z_4$ (otherwise we would have: $\Bbb Z_4 \cong \Bbb Z_2 \oplus \Bbb Z_2$, arguing by order alone, which is false).

It is clear that violation of (1) or (3) is thus equivalent (in this case) to violating (2).

Rotman then extends this to a direct sum exact sequence counter-example by considering a (countably) infinite-fold direct product of $A \oplus B$, using the fact that (up to isomorphism) direct sum is a commutative and associative operation.

In summary, just because we have an exact sequence:

$0 \to M \to M\oplus N \to N \to 0$

there is no reason to suppose the mapping $M \to M \oplus N$ is the canonical inclusion, nor the map $M \oplus N \to N$ is the canonical projection (which is essentially what happens when the sequence splits).

In "abelian categories" (such as AbGrp, $\ _R$Mod, or Vect$_k$) "split" means "direct sum" (and we have a symmetry between "left-split" and "right-split" short exact sequences). I note in passing this is *not* the case for the category Grp, where "left-split" short exact sequences lead to direct products, and "right-split" short exact sequences lead to semi-direct products.

Hi Deveno ... ... just a basic question ...You write:

" ... ... an exact short sequence:

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

where, in any case, we have $B \cong C/A$ ... ... "How, exactly, do we know that $B \cong C/A$ ... ... ?

Peter

 

[Math Amateur]

Hi Deveno ... ... just a basic question ...You write:

" ... ... an exact short sequence:

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

where, in any case, we have $B \cong C/A$ ... ... "How, exactly, do we know that $B \cong C/A$ ... ... ?

Peter

Hi Deveno ... I have been reflecting on my question above ...

I think that the answer is as follows:Given a short exact sequence

$$ 0 \to A \stackrel{ \psi}{\to} C \stackrel{ \phi }{\to} B \to 0 $$

we have that $$\phi \ : \ C \to B$$ is a surjective homomorphism

... ... and so the First Isomorphism applies and we have:$$B \cong C/ \text{ ker } \phi $$

But $$ \text{ ker } \phi = \text{ I am } \psi $$

... so $$B \cong C/ \text{ I am } \psi$$

But since $$\psi$$ is injective we have that $$\text{ I am } \psi \cong A$$ ...

... so we have $$B \cong C/A$$Could you please critique my analysis ... is the above basically correct?

Peter

 

[Deveno]

Hi Deveno ... ... just a basic question ...You write:

" ... ... an exact short sequence:

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

where, in any case, we have $B \cong C/A$ ... ... "How, exactly, do we know that $B \cong C/A$ ... ... ?

Peter

Exactness of $0 \to A \to C$ means that the kernel of the morphism $A \to C$ is the image of the only possible morphism $0 \to A$, the 0-map, thus we may regard $A$ as a sub-module of $C$ (since $A \to C$ is an injection), and the morphism $A \to C$ as the natural inclusion map (technically, the *image* of $A$ is a submodule of $C$, but we are going to identify isomorphic modules).

In the same manner exactness of $C \to B \to 0$ means that $B$ is the kernel of the map $B \to 0$ (another 0-map), so we see the map $C \to B$ must be surjective. By the fundamental homomorphism theorem, we have:

$C/(\text{im } A) \cong B$, which I wrote as $C/A \cong B$ as a common abuse of notation.

As a concrete example, consider the $\Bbb R$-modules: $A = \Bbb R,\ C = \Bbb R^2,\ B = \Bbb R$, with the mappings:

$f: \Bbb R \to \Bbb R^2$ given by $f(x) = (x,0)$, and

$g: \Bbb R^2 \to \Bbb R$ given by $g(x,y) = y$.

Then $0 \to A \stackrel{f}{\to} C \stackrel{g}{\to} C \to 0$ is short exact, since:

$\text{ker }g = \{(x,y) \in \Bbb R^2: y = 0\} = \text{im } f$.

Technically, we should write:

$\Bbb R^2/(\Bbb R \times \{0\}) \cong \Bbb R$

but it should be plain to see that writing $\Bbb R^2/\Bbb R \cong \Bbb R$ leads to no confusion.

It turns out that this example is "split", since $h: \Bbb R \to \Bbb R^2$ given by $h(y) = (0,y)$ has the required property that:

$g \circ h = 1_{\Bbb R}$

and indeed, $\Bbb R^2 \cong \Bbb R \oplus \Bbb R$.

 

[steenis]

Thank you Peter for your Latex help, I will use it.

Thank you Deveno for your clear explanation. I am not good in understanding examples and counter-examples. It has always been my mathematical weakness.

Tonight I came to the conclusion that the items (1) and (3) in prop.3.2.7. of Bland are not equivalent. From (1) it follows that the ses splits. That part of the proof of Bland is correct. According to Rotman, you cannot deduce from (3) that the ses splits. The question remains: what is wrong in the proof of (3) => (1) of Bland ?

 

[Math Amateur]

Peter, thank you for your answer.
I do not know how "to scan and post the relevant pages" from books.
Peter, can you show me how to learn to edit the mathematical notation in this forum?I want to add this to my former post:
See also prop.2.28 p.52 of Rotman (same book). This is the propostition 3.2.7 p.85 of Bland in the (=>)-direction. Just after the proof of prop.2.28 on p.53, Rotman says that the converse of prop.2.28 is not true, which is the (<=)-direction of the proposition of Bland.
To my knowledge, the (<=)-direction is not in other books.
I think that one correct formulation of the (<=)-direction of the proposition of Bland is given, for instance, by Grillet, proposition 1.5 on page 395 (Grillet - Abstract Algebra, 2nd edition 2007).

One more remark: can someone also look critically to the (3) => (1) proof of the same proposition 3.2.7 p.85 of Bland? I think it is not correct, but I cannot put my finger on it.

I agree that the book of Bland is challenging, but I am not yet convinced that it is a good book. An errata-sheet would be most helpful.

Hi Steenis,

I did eventually get to working through Bland Proposition 3.2.7 ... ... and I could not find anything wrong ... ... but I was alarmed at Rotman's remark after Proposition 2.28 where he says:

" ... ... We shall see in Example 2.29, that the converse of Proposition 2.28 is not true ... ... "

This is concerning as it appears to contradict the 'only if' part of Bland's Proposition 3.2.7 Condition (3) ... ...

What is needed is someone more erudite than I am to look at this issue ...

Note that I also have trouble in seeing exactly how Example 2.29 of Rotman actually contradicts Bland's Proposition 3.2.7 Condition (3) ... ...

So maybe someone more knowledgeable can help ...

I am providing copies of all relevant text below here in one place, for the convenience of MHB members and guests ...

First Bland's Proposition 3.2.7 ... as follows:View attachment 5819
Bland Proposition 3.2.7 above mentions Proposition 2.1.11 which reads as follows:
View attachment 5820Bland Proposition 3.2.7 above also mentions Proposition 3.2.3 which reads as follows:View attachment 5821Rotman Proposition 2.28 reads as follows:View attachment 5822
View attachment 5823Rotman Proposition 2.28 mentions Rotman Exercise 2.8, page 65 ... which reads as follows:
View attachment 5824 Rotman Proposition 2.28 also mentions Rotman Example 2.29, page 54... which reads as follows:View attachment 5825I hope that someone knowledgeable can resolve the problems and issues raised by Steenis ...

Peter

 

[steenis]

Thank you Peter, I also hope that someone more knowledgeable in this area can help me and the readers of Bland.

Is this true?
An ses $S: 0\longrightarrow A\longrightarrow _f A\oplus C\longrightarrow _g C\longrightarrow 0$ of (left) R-modules is split if and only if $f= i_1 = \mbox{the natural injection } a\longmapsto (a,0) \mbox{ and } g= \pi_2 = \mbox{the natural projection } (a,c)\longmapsto c$

If so, then is the (<=) part of prop.3.2.7 of Bland only true if $f= i_1 \mbox{ and } g= \pi_2 $, which is not always the case.

Compare this with Grillet, proposition 1.5 on page 395 (Grillet - Abstract Algebra, 2nd edition 2007).
This proposition of Grillet is already included above.

 

[steenis]

Note that I also have trouble in seeing exactly how Example 2.29 of Rotman actually contradicts Bland's Proposition 3.2.7 Condition (3) ... ...

Peter

@Peter
Bland claims, that given $M\cong M_1\oplus M_2$, the ses $0\longrightarrow M_1\longrightarrow _f M\longrightarrow _g M_2\longrightarrow 0$ splits. Using prop 2.1.11 of Bland, this is the same as saying that $0\longrightarrow N_1\longrightarrow _f N_1\oplus N_2 \longrightarrow _g N_2\longrightarrow 0$ is an split ses, where $N_1, N_2 \leq M$ and $N_1 \cong M_1$ and $N_2 \cong M_2$.
In the last 7 lines of his example 2.29, Rotman constructs an ses $0\longrightarrow A\longrightarrow A\oplus M\longrightarrow M\longrightarrow 0$ (where $A=Z_2$) that is not split, hence the contradiction. (The construction of Rotman is too complicated for me.)

 

[Math Amateur]

@Peter
Bland claims, that given $M\cong M_1\oplus M_2$, the ses $0\longrightarrow M_1\longrightarrow _f M\longrightarrow _g M_2\longrightarrow 0$ splits. Using prop 2.1.11 of Bland, this is the same as saying that $0\longrightarrow N_1\longrightarrow _f N_1\oplus N_2 \longrightarrow _g N_2\longrightarrow 0$ is an split ses, where $N_1, N_2 \leq M$ and $N_1 \cong M_1$ and $N_2 \cong M_2$.
In the last 7 lines of his example 2.29, Rotman constructs an ses $0\longrightarrow A\longrightarrow A\oplus M\longrightarrow M\longrightarrow 0$ (where $A=Z_2$) that is not split, hence the contradiction. (The construction of Rotman is too complicated for me.)

Steenis,

After doing some reading I can now agree with your concerns ... including those in your latest posts... but unfortunately I cannot see a resolution to the apparent contradictions... hopefully someone more knowledgeable will help ...

Peter

 

[Math Amateur]

Steenis,

After doing some reading I can now agree with your concerns ... including those in your latest posts... but unfortunately I cannot see a resolution to the apparent contradictions... hopefully someone more knowledgeable will help ...

Peter

Just a short note to say that I have now placed a plea for help with Steenis' issues with Bland Proposition 3.2.7/Rotman Example 2.29 on the Physics Forums under the sub-forum Linear and Abstract Algebra ...Here is a link to the post ... ...https://www.physicsforums.com/threads/split-exact-sequences-bland-proposition-3-2-7.881174/Hope to get some help from there as well as MHB ...

Peter

 

[Euge]

Hi Peter and steenis,

As Deveno discussed, the sequence $0 \to A \xrightarrow{i} B \xrightarrow{p} A \to 0$ is does not satisfy the hypotheses of Bland's Proposition 3.2.7, so it is not a counterexample. In fact, the two short exact sequences Rotman gives are not of the form $0 \to S \to S\oplus T\to T\to 0$ in the first place. On the other hand, Bland was not careful with his statement that $M\cong M_1 \oplus M_2$ -- if this isomorphism is only with respect to $R$-modules and not with respect to exact sequences, then he's missing compatibility conditions that this isomorphism must have with $f$ and $g$. The bottom line here is that an exact sequence $0 \to A \to M \to B \to 0$ of $R$-modules is split if and only if it is isomorphic to the exact sequence $0 \to A \xrightarrow{\iota} A\oplus B\xrightarrow{\pi} B \to 0$, where $\iota$ is the inclusion map and $\pi$ is the projection map onto the second factor.

In case you're not sure what I mean by isomorphic short exact sequences, let's start with two short exact sequences $0 \to A_1 \xrightarrow{\alpha_1} A \xrightarrow{\alpha_2} A_2 \to 0$ and $0 \to B_1 \xrightarrow{\beta_1} B\xrightarrow{\beta_2} B_2 \to 0$, which I'll denote by $[\mathbf{A}]$ and $[\mathbf{B}]$, respectively. A morphism from $[\mathbf{A}]$ to $[\mathbf{B}]$ is a triple $[\mathbf{f}] = (\gamma_1,\gamma,\gamma_2)$ of $R$-homomorphisms such that the following diagram commutes.

$$\require{AMScd}
\begin{CD}
0 @>>> A_1 @>{\alpha_1}>> A @>{\alpha_2}>> A_2 @>>> 0\\
@. @V{\gamma_1}VV @V{\gamma}VV @V{\gamma_2}VV @.\\
0 @>>> B_1 @>{\beta_1}>>B @>{\beta_2}>> B_2 @>>> 0
\end{CD}$$

If, in addition, the map $\gamma$ is an isomorphism, then we say that $[\mathbf{f}]$ is an isomorphism between $[\mathbf{A}]$ and $[\mathbf{B}]$ and that $[\mathbf{A}]$ is isomorphic to $[\mathbf{B}]$.

Sometimes, when the context is understood, a single module $M$ represents a short exact sequence. So the statement $M\cong M_1 \oplus M_2$ could mean precisely that the short exact sequence $0 \to M_1 \xrightarrow{f} M \xrightarrow{g} M_2 \to 0$ is isomorphic to the canonical sequence $0 \to M_1 \xrightarrow{i} M_1 \oplus M_2 \xrightarrow{\pi} M_2 \to 0$, i.e., the diagram

$$\begin{CD}
0 @>>> M_1 @>f>> M @>g>> M_2 @>>> 0\\
@. @| @V{\cong}VV @| @.\\
0 @>>> M_1 @>{\iota}>>M_1 \oplus M_2 @>{\pi}>> M_2 @>>> 0
\end{CD}$$

commutes.

 

[Math Amateur]

Hi Peter and steenis,

As Deveno discussed, the sequence $0 \to A \xrightarrow{i} B \xrightarrow{p} A \to 0$ is does not satisfy the hypotheses of Bland's Proposition 3.2.7, so it is not a counterexample. In fact, the two short exact sequences Rotman gives are not of the form $0 \to S \to S\oplus T\to T\to 0$ in the first place. On the other hand, Bland was not careful with his statement that $M\cong M_1 \oplus M_2$ -- if this isomorphism is only with respect to $R$-modules not with respect to exact sequences, then he's missing compatibility conditions that this isomorphism must have with $f$ and $g$. The bottom line here is that an exact sequence $0 \to A \to M \to B \to 0$ of $R$-modules is split if and only if it is isomorphic to the exact sequence $0 \to A \xrightarrow{\iota} A\oplus B\xrightarrow{\pi} B \to 0$, where $\iota$ is the inclusion map and $\pi$ is the projection map onto the second factor.

In case you're not sure what I mean by isomorphic short exact sequences, let's start with two short exact sequences $0 \to A_1 \xrightarrow{\alpha_1} A \xrightarrow{\alpha_2} A_2 \to 0$ and $0 \to B_1 \xrightarrow{\beta_1} B\xrightarrow{\beta_2} B_2 \to 0$, which I'll denote by $[\mathbf{A}]$ and $[\mathbf{B}]$, respectively. A morphism from $[\mathbf{A}]$ to $[\mathbf{B}]$ is a triple $[\mathbf{f}] = (\gamma_1,\gamma,\gamma_2)$ of $R$-homomorphisms such that the following diagram commutes.

$$\require{AMScd}
\begin{CD}
0 @>>> A_1 @>{\alpha_1}>> A @>{\alpha_2}>> A_2 @>>> 0\\
@. @V{\gamma_1}VV @V{\gamma}VV @V{\gamma_2}VV @.\\
0 @>>> B_1 @>{\beta_1}>>B @>{\beta_2}>> B_2 @>>> 0
\end{CD}$$

If, in addition, the map $\gamma$ is an isomorphism, then we say that $[\mathbf{f}]$ is an isomorphism between $[\mathbf{A}]$ and $[\mathbf{B}]$ and that $[\mathbf{A}]$ is isomorphic to $[\mathbf{B}]$.

Sometimes, when the context is understood, a single module $M$ represents a short exact sequence. So the statement $M\cong M_1 \oplus M_2$ could mean precisely that the short exact sequence $0 \to M_1 \xrightarrow{f} M \xrightarrow{g} M_2 \to 0$ is isomorphic to the canonical sequence $0 \to M_1 \xrightarrow{i} M_1 \oplus M_2 \xrightarrow{\pi} M_2 \to 0$, i.e., the diagram

$$\begin{CD}
0 @>>> M_1 @>f>> M @>g>> M_2 @>>> 0\\
@. @| @V{\cong}VV @| @.\\
0 @>>> M_1 @>{\iota}>>M_1 \oplus M_2 @>{\pi}>> M_2 @>>> 0
\end{CD}$$

commutes.

Thanks so much Euge ... Your post is EXTREMELY helpful ...

Thanks for taking the time to clarify this puzzling matter ...

The key point in resolving the puzzle, apart from Deveno's post, is your statement ... " ... On the other hand, Bland was not careful with his statement that $M\cong M_1 \oplus M_2$ -- if this isomorphism is only with respect to $R$-modules not with respect to exact sequences, then he's missing compatibility conditions that this isomorphism must have with $f$ and $g$. ... ... "... ... indeed, as you say, Bland was not careful in his statement of the Proposition!Thanks again,

Peter

 

[steenis]

Thank you Euge for this clear and comprehensive explanation.
(In fact, this coincides with my remark in post #5 when I referred to Grillet, proposition 1.5 on page 395 (Grillet - Abstract Algebra, 2nd edition 2007). But yours is better.)

Clever, how you make the diagrams.

Hi Peter and steenis,

In fact, the two short exact sequences Rotman gives are not of the form $0 \to S \to S\oplus T\to T\to 0$ in the first place.

I do not agree with this. In his example 2.29 Rotman constructs an ses $0 \to A \to A\oplus M\to M\to 0$ that is not split, $A = Z_2$. Read the last 7 lines of his example. The construction is too complicated for me, but I think the ses is not split because the maps $A \to A\oplus M$ and $A\oplus M\to M$ are not $i$ and $\pi$, respectively.

I think, the only thing we need to resolve this problem, is a clear counterexample of the implication (3) => the ses splits.

 

[Euge]

I do not agree with this. In his example 2.29 Rotman constructs an ses $0 \to A \to A\oplus M\to M\to 0$ that is not split, $A = Z_2$. Read the last 7 lines of his example. The construction is too complicated for me, but I think the ses is not split because the maps $A \to A\oplus M$ and $A\oplus M\to M$ are not $i$ and $\pi$, respectively.

The sequence was $0 \to A \to B\oplus M \to A\oplus M \to 0$, not $0 \to A \to A \oplus M \to M \to 0$. This point is minor, however. What Rotman really means is there are non-split exact sequences $0 \to S \to X \to T \to 0$ such that $X \cong S\oplus T$ as modules. In the example, although $B \oplus M \neq A \oplus (A \oplus M)$, it is true that $B\oplus M \approx A \oplus (A \oplus M)$, because $A \oplus (A\oplus M) \cong A \oplus M \cong M \cong B \oplus M$.

A splitting sequence $0 \to S \to S\oplus T \to T \to 0$ is not unique. For example, let $R = \Bbb Z$ (so we're considering abelian groups), and consider the sequence $0 \to \Bbb Z \xrightarrow{f} \Bbb Z \oplus \Bbb Z \xrightarrow{g} \Bbb Z \to 0$, where $f = \operatorname{id}\times -\operatorname{id}$ and $g = \pi_1 + \pi_2$, that is, $f(m) = (m,-m)$ and $g(m,n) = m + n$. Then $\operatorname{im}(f) = \{(m,-m):m\in \Bbb Z\} = \operatorname{ker}(g)$, so the sequence is short exact. Since $\{(m,-m):m\in \Bbb Z\} = \langle (1,-1)\rangle \cong \Bbb Z$, $\operatorname{im}(f)$ is a direct summand of $\Bbb Z \oplus \Bbb Z$; hence this sequence is split exact. As you can see, $f$ is not the inclusion map and $g$ is not the projection map.

I think, the only thing we need to resolve this problem, is a clear counterexample of the implication (3) => the ses splits.

I don't have the text to look through the chapter and context, but from the proof, it looks to me that he did assume compatibility of the isomorphism $M\cong M_1 \oplus M_2$ with the maps $f$ and $g$. So his proof of $(3)\implies (1)$ looks correct.

 

[steenis]

The sequence was $0 \to A \to B\oplus M \to A\oplus M \to 0$, not $0 \to A \to A \oplus M \to M \to 0$.

...I don't have the text to look through the chapter and context, but from the proof, it looks to me that he did assume compatibility of the isomorphism $M\cong M_1 \oplus M_2$ with the maps $f$ and $g$. So his proof of $(3)\implies (1)$ looks correct.

Rotman constructs an ses $0 \to A \to B\oplus M \to A\oplus M \to 0$. After that he states that $A\oplus M\cong M\cong B\oplus M$, i.e., $0 \to A \to A \oplus M \to M \to 0$. The text of the example is in post #15.

Part of the texts of Bland are also listed above, thanks to Peter, you can check the proof and the context. I really do not know what Bland assumes. To me it does not follow from the context.

Of course, assuming the compatibility, then (3) => the ses splits is true. I already mentioned that in my post #5, and you stated that in your former post #20. But what does Bland assume?

The proof of (1)=>(3) may be looking correct, but I want to know if it is correct. In contrary, the proof confuses me.

 

[Math Amateur]

Rotman constructs an ses $0 \to A \to B\oplus M \to A\oplus M \to 0$. After that he states that $A\oplus M\cong M\cong B\oplus M$, i.e., $0 \to A \to A \oplus M \to M \to 0$. The text of the example is in post #15.

Part of the texts of Bland are also listed above, thanks to Peter, you can check the proof and the context. I really do not know what Bland assumes. To me it does not follow from the context.

Of course, assuming that, then (3) = the ses splits is true. I already mentioned that in my post #5, and you stated that in your former post #20. But what does Bland assume?

The proof of (1)=>(3) may be looking correct, but I want to know if it is correct. In contrary, the proof confuses me.

Steenis, Euge

I have received posts on the Physics Forums (sub Forum Linear and Abstract Algebra) ... they seem to largely agree with Euge & Deveno ...

See the posts here ...

https://www.physicsforums.com/threads/split-exact-sequences-bland-proposition-3-2-7.881174/
Euge ... Bland's text on Exact Sequences prior to Proposition 3.2.7 reads as follows ... ...https://www.physicsforums.com/attachments/5829
https://www.physicsforums.com/attachments/5830
https://www.physicsforums.com/attachments/5831Hope that helps ...

Peter

 

[steenis]

Hi Peter,

It is too much work for me to also respond to the answers on the Physics Forum. But if you want, you can answer that they have to read example 2.29. Why does everybody only read the first 5 or 6 lines, while the important lines are the last 7 lines? That is where the counterexample comes from.

 

[steenis]

I found another counterexample of the implication (3)=>the ses splits. I did not invent it myself but go to

modules - Example of a non-splitting exact sequence $0 → M → M\oplus N → N → 0$ - Mathematics Stack Exchange

for the details.

An ses $S1: 0\to M_1 \to _f M_1\oplus M_2\to _g M_2\to 0$ is constructed that is not split and that is not equivalent / compatible (see post #20 of Euge or Grillet in post #8) to $S0: 0\to M_1 \to _i M_1\oplus M_2\to _\pi M_2\to 0$. It shows that compatibility to S0 is necessary to prove (3)=>(1). Therefore (3)=>(1) cannot be true.

One thing will make me happy: a direct counterexample of (3)=>(1).(Is this example the same as the example of Rotman ? It is better to understand, though ...)

 

[Euge]

Rotman constructs an ses $0 \to A \to B\oplus M \to A\oplus M \to 0$. After that he states that $A\oplus M\cong M\cong B\oplus M$, i.e., $0 \to A \to A \oplus M \to M \to 0$. The text of the example is in post #15.

I made this point, but when you write "i.e., $0 \to A \to A \oplus M \to M \to 0$," you mean $0 \to A \to A \oplus M \to M \to 0$ up to $R$-isomorphism. I wasn't arguing that his example in the last 7 lines was a bad example. My comments on that were based on technicality, not content.

Part of the texts of Bland are also listed above, thanks to Peter, you can check the proof and the context. I really do not know what Bland assumes. To me it does not follow from the context.

Certainly, my comments are based on the posts Peter has provided. However, the posts are not always enough for context. One time, Peter asked a question about something mathematically suspicious (I believe it was about direct sums) and provided the posts so we could addressed his question, but the answer to that question was based on context given in the introduction of the text, which he did not post.

Of course, assuming the compatibility, then (3) => the ses splits is true. I already mentioned that in my post #5, and you stated that in your former post #20. But what does Bland assume?

The proof of (1)=>(3) may be looking correct, but I want to know if it is correct. In contrary, the proof confuses me.

I think you misunderstood my comment. It's not about whether we assume compatibility, but whether Bland assumes compatibility. After reading the additional information that Peter posted, I've concluded that Bland 1) only meant $M \approx M_1 \oplus M_2$ with respect to $R$-modules and not with respect to short exact sequences, and 2) made an erroneous assumption, namely "If $w\in M_1$ is such that $f(w) = x$..."

The proof of $(1) \implies (3)$ is correct.

Why does everybody only read the first 5 or 6 lines, while the important lines are the last 7 lines? That is where the counterexample comes from.

I don't know what makes you say we're only reading the first 5 or 6 lines, when Deveno and I have addressed that very counterexample you mentioned.

Seeing that you have linked another counterexample, is it the case that you thought I didn't think that the statement

If $M \cong M_1 \oplus M_2$ as $R$-modules, then $\operatorname{Im}(f)$ is a direct summand of $M$

is false? If so, let me make it clear again that I find this statement false. After all, the short exact sequence $0 \to \Bbb Z \xrightarrow{2\iota_1} \Bbb Z \oplus \Bbb Z_2^\infty \xrightarrow{g} \Bbb Z_2^\infty \to 0$ with $2\iota_1(n) = (2n,0,0,0,\ldots)$ and $g(n,n_1,n_2,\ldots) = ([n]_2,n_1,n_2,\ldots)$, is a counterexample.

 

[steenis]

I think you misunderstood my comment. It's not about whether we assume compatibility, but whether Bland assumes compatibility. After reading the additional information that Peter posted, I've concluded that Bland 1) only meant $M \approx M_1 \oplus M_2$ with respect to $R$-modules and not with respect to short exact sequences, and 2) made an erroneous assumption, namely "If $w\in M_1$ is such that $f(w) = x$..."

The proof of $(1) \implies (3)$ is correct.

I agree, I Always agreed that (1)=>(3) is correct.
And I have said in former posts that there seems something wrong in his proof of (3)=>(1), but I cannot put my finger on it.

From your post #23

I don't have the text to look through the chapter and context, but from the proof, it looks to me that he did assume compatibility of the isomorphism $M\cong M_1 \oplus M_2$ with the maps $f$ and $g$. So his proof of $(3)\implies (1)$ looks correct.

I do not agree.

I don't know what makes you say we're only reading the first 5 or 6 lines, when Deveno and I have addressed that very counterexample you mentioned.

This was about a post Peter made in Physics Forum, not about you or Deveno, sorry about that.

Seeing that you have linked another counterexample, is it the case that you thought I didn't think that the statement

If $M \cong M_1 \oplus M_2$ as $R$-modules, then $\operatorname{Im}(f)$ is a direct summand of $M$

is false? If so, let me make it clear again that I find this statement false. After all, the short exact sequence $0 \to \Bbb Z \xrightarrow{2\iota_1} \Bbb Z \oplus \Bbb Z_2^\infty \xrightarrow{g} \Bbb Z_2^\infty \to 0$ with $2\iota_1(n) = (2n,0,0,0,\ldots)$ and $g(n,n_1,n_2,\ldots) = ([n]_2,n_1,n_2,\ldots)$, is a counterexample.

And now I am really confused, what do you mean: is (3)=>(1) false or not false ?

By the way, I made a mistake in my post #27. I did not prove that S1 and S0 are equivalent. Sorry about that. Who can prove this?

 

[Euge]

I Always agreed that (1)=>(3) is correct.

You wrote, "The proof of (1)=>(3) may be looking correct, but I want to know if it is correct. In contrary, the proof confuses me." Therefore, I added the point that $(1) \implies (3)$ is correct for your reassurance.

And I have said in former posts that there seems something wrong in his proof of (3)=>(1), but I cannot put my finger on it.

I already addressed this concern. I don't understand why you're repeating your disagreement with post #23, since I did not reiterate that. The comments from that post were not definitive. It came before the additional information Peter provided, so please ignore it.

And now I am really confused, what do you mean: is (3)=>(1) false or not false ?

If you're combining my recent comments with the comments of post #23, then that's probably why you're confused. Please just address the recent comments. I said that after receiving the addition info from Peter, I find $(3)\implies (1)$ to be false. I gave points as to why the proof was incorrect, and I even gave a counterexample, which turned out to be the same example as the one in the link you provided. :D

 

[steenis]

The proof of (1)=>(3) may be looking correct, but I want to know if it is correct. In contrary, the proof confuses me.

Ok, sorry, here I was wrong which might be clear, because I was quoting you.

So we agree that (3)=>(1) is wrong. What must we do to finish this thread? I think a proof of the non-equivalence of S1 and S0 in post #27. And, if possible, a direct and easy counterexample of (3)=>(1).

 

[Euge]

I already gave a direct counterexample to $(3)\implies (1)$, which is actually the same as the the one in the MSE link you provided.

 

[steenis]

Where is your counterexample, I cannot find it?

 

[Euge]

It's at the end of post #28, but it's the same as the k.stm's example in your link.

 

[steenis]

That specific example is a direct counterexample of "(3) => the ses splits" and therefore an indirect counterexample of (3) => (1). A direct counterexample of (3) => (1) is example in which (3) is true and (1) is not true, that is how I have learned it in my university years. Of course that all in the context of Bland's proposition and if possible very easy. It is not necessary, not mandatory, because the job is already done. It is just meant to enlighten that (3)=>(1) is faulty. I do not know if it is possible or it is too difficult.

I want to stop this discussion now, the only thing that is open is the non-equivalence of S1 and S0 in post #27.

 

[Euge]

That does not make sense, since any counterexample to $(3)\implies (1)$ is a a counterexample to the implication that $(3)$ implies the ses splits; for a split exact sequence necessarily satisfies $(1), (2)$, and $(3)$. Seeing that you want to stop the discussion, I'll leave it to other users address your concerns.

 

[Math Amateur]

That does not make sense, since any counterexample to $(3)\implies (1)$ is a a counterexample to the implication that $(3)$ implies the ses splits; for a split exact sequence necessarily satisfies $(1), (2)$, and $(3)$. Seeing that you want to stop the discussion, I'll leave it to other users address your concerns.

Thanks to Euge and Steenis for clarifying matters ...

There are some further helpful and informative posts on the Physics Forums in the sub-forum Linear and Abstract Algebra ... here

https://www.physicsforums.com/threa...-bland-proposition-3-2-7.881174/#post-5538854

Hope that helps ...

Peter