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If $0 \to L \to M \to N \to 0$ is a short exact sequence of $A$-modules then the following statements are equivalent:

  1. There is an isomorphism $M \cong L \oplus N$ under which $L \to M$ is given by $m \mapsto (m, 0)$ and $M \to N$ by $(m, n) \mapsto n$.
  2. There exists a section of $M \to N$.

I understand that the condition "there is an isomorphism $M \cong L \oplus N$" is much weaker than the condition 1. above, so

I wonder if there are some simple $M, L$ and $N$ such that $M \cong L \oplus N$, but there is no section of $M \to N$.

user26857
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user374573
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  • The section should be $N\to M$. – Arnaud D. Oct 22 '16 at 20:34
  • @ArnaudD. section of $M \to N$ means a map $N \to M$ such that... – user374573 Oct 22 '16 at 20:35
  • @user374573 That's weird...but I guess it is possible. Whose definition of "section" is that? – DonAntonio Oct 22 '16 at 20:36
  • @user374573 I hadn't noticed that "of", that makes more sense. – Arnaud D. Oct 22 '16 at 20:37
  • @DonAntonio I know it from M. Reid's "Commutative Algebra" – user374573 Oct 22 '16 at 20:38
  • @user374573 In 2.10, page 45 of Reid's "Undergraduate Commutative Algebra" I read " a section $;N\to M;$" , not the other way around. Now, the section is of the map $;\beta:M\to N;$ , which really makes sense. – DonAntonio Oct 22 '16 at 20:41
  • @DonAntonio looks like a legitimate form of verbiage. I see no difference in "the section of the map $\beta: M \to N$" and "the section of $M \to N$" – user374573 Oct 22 '16 at 20:46
  • @user374573 Perhaps, but the fact is it caused a little confusion among, at least, two of us, mathematicians with advanced degrees. Of course, it is only a matter of agreement, yet Reid uses it the way I wrote you above. – DonAntonio Oct 22 '16 at 20:55
  • @DonAntonio thank you. I will take this into account in the future. – user374573 Oct 22 '16 at 20:59
  • @user374573 In your question, do you mean whether there are $;M,L,N;$ s.t. $;M\cong L\oplus N;$ and also in the corresponding place in an exact sequence ? Or already without the exact sequence requirement? – DonAntonio Oct 22 '16 at 21:19
  • @DonAntonio also in the corresponding place in the exact sequence. I just want an example which shows, that a weaker version of statement 1. does not imply 2. – user374573 Oct 22 '16 at 21:37
  • @user374573 If I understand you correctly, that is impossible as conditions (1)-(2) are equivalent...and also equivalent to "there's a section $;M\to L;$" , which is even less used. – DonAntonio Oct 22 '16 at 21:38
  • @DonAntonio what is impossible? Yes 1. and 2. above are equivalent, but I want to make statement 1. a bit weaker and show (by an example) that it does not imply 2. By "a bit weaker" I mean taking "$M \cong L\oplus N$" instead of "$M \cong L\oplus N$ under which $L \to M$ is given by $m \mapsto (m, 0)$ and $M \to N$ by $(m, n) \mapsto n$". – user374573 Oct 22 '16 at 21:46
  • @user374573 Oh, I see. But I think that to make some sense all this it should be somehow related to the maps that make the short sequence short shouldn't it? – DonAntonio Oct 22 '16 at 21:51
  • @DonAntonio Yes, $ {0} \to L \to M \to N \to {0} $ must be a s.e.s. – user374573 Oct 22 '16 at 21:52

2 Answers2

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That's a great question!

Usually counterexamples are already easy to find for $\mathbb{Z}$-modules (also known as abelian groups), but sadly, in this case, if you look only at finitely generated abelian groups, then any short exact sequence $$0 \to A' \to A \to A'' \to 0$$ with $A \cong A'\oplus A''$ is actually split. (It is possible to see it by revising the classification of finitely generated abelian groups and calculations of the Yoneda $\operatorname{Ext}^1_\mathbb{Z} (A'',A')$.)

So we need to look for counterexamples among abelian groups that are not finitely generated. Maybe the easiest is the following. Take an infinite direct sum of copies of $\mathbb{Z}/n\mathbb{Z}$ and consider the map

\begin{align*} p\colon \mathbb{Z} \oplus \bigoplus_{i \ge 0} \mathbb{Z}/n\mathbb{Z} & \twoheadrightarrow \bigoplus_{i \ge 0} \mathbb{Z}/n\mathbb{Z},\\ (x,y_0,y_1,y_2,\ldots) & \mapsto (x \mod{n}, y_0, y_1, y_2, \ldots). \end{align*}

This is not the projection, but this is a surjective homomorphism, and we have a legitimate short exact sequence $$0 \to \mathbb{Z} \xrightarrow{x \mapsto (n x, ~ 0,0,0,\ldots)} \mathbb{Z} \oplus \bigoplus_{i \ge 0} \mathbb{Z}/n\mathbb{Z} \xrightarrow{p} \bigoplus_{i \ge 0} \mathbb{Z}/n\mathbb{Z} \to 0$$

But you can see that $p$ doesn't have a section.

  • Thank you, that is a great answer :) – user374573 Oct 22 '16 at 22:18
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    Someone already marked this question as a duplicate of another one, having... an identical answer :'( I copied mine from my homological algebra notes, and I think it is hard to come up with something easier. –  Oct 22 '16 at 22:23
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How about taking

$$ 0 \to \Bbb Z/2 \Bbb Z \to \Bbb Z/2 \Bbb Z $$

and

$$ \Bbb Z \xrightarrow{\cdot 2} \Bbb Z \to \Bbb Z / 2 \Bbb Z. $$

Both are short exact. But if you take the countable direct sum of both then the middle term is the direct sum of the both other terms, but there is no section.

user60589
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  • Nonse of these two short sequences seem to comply with the OP's intention. Read last comments below the question. – DonAntonio Oct 22 '16 at 21:56
  • @DonAntonio I want to combine both s.e.s. via direct product to a s.e.s. with $L=\bigoplus_{\Bbb N} \Bbb Z $, $N=\bigoplus_{\Bbb N} \Bbb Z/ 2 \Bbb Z $ and $M = \bigoplus_{\Bbb N} (\Bbb Z\oplus \Bbb Z/ 2 \Bbb Z)$. Since the second s.e.s. does not split the resulting sequence does not split. Where is my error? – user60589 Oct 22 '16 at 22:02
  • Both s.e.s in the OP begin and end with zero , and thus the middle maps are injective/surjective resp. (from left to right) . – DonAntonio Oct 22 '16 at 22:05
  • @DonAntonio Those too? – user60589 Oct 22 '16 at 22:06
  • @DonAntonio Decorate the formulas above with "$0\to$" and "$\to 0$" :-) –  Oct 22 '16 at 22:08