I just construct an exact sequence $0\to M\to M\oplus N\to N\to0$ of $\mathbb{Z}$-modules that does not split, where $M=\mathbb{Z}$, $N=(\mathbb{Z}/2\mathbb{Z})^\\mathbb{N}$, and the map from $M$ to $M\oplus N$ maps $n$ to $(2n,0)$. Do you know other examples? You are free to consider other categories. In particular, are there any example that is in some sense "finite"? I guess such sequence must split for finitely generated modules over PID. Is this true?

(The above questions were mostly answered in the question that Martin points out. In particular, the answer to the last question is affirmative, even for commutative Noetherian rings instead of PID.)

More generally, is it true (e.g. for f.g. modules over PID) that any two exact sequence with the same terms

$0\to M\to P\to N\to0$

$0\to M\to P\to N\to0$

are equivalent, in the sense that there exist vertical isomorphisms which make the diagram commute? In other words, are there two extensions of $N$ by $M$ which are nonequivalent but both yield $P$? If not, can you give any counterexample? Thanks.

(It was noted in Ayoub's paper that this is not true in the category of finite groups, where he exhibited three finite groups that both fit in a split exact sequence and a non-split one.)

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    $\begingroup$ Dublicate of mathoverflow.net/questions/80002/… $\endgroup$ Dec 9, 2011 at 8:01
  • $\begingroup$ Thank you for pointing out. My questions were mostly answered there, so I edit my question and add a generalization of the original question. $\endgroup$
    – Junyan Xu
    Dec 9, 2011 at 11:40
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    $\begingroup$ @Junayan: 1) You should not edit a question in a way that changes the essence. This makes it hard to understand potential answers and comments that were made to the original form. The point of editing is to correct typos or small errors. If comments or answers lead you to another question, you should ask that in a new question. 2) Having said that I don't see how this is a new question. If $P=M\oplus N$ and the sequence is not split, then it is not equivalent to the split sequence with the same terms. Is there anything else you want from these sequences? $\endgroup$ Dec 9, 2011 at 16:45
  • $\begingroup$ @Sándor: 1) I think it is a natural generalization because I have considered adding it before I got any comments. And perhaps I don't want to get the problem closed after the comment (maybe it's not a good idea). 2) It was proved that for f.g. modules over a commutative Noetherian ring, the former sequence must split, hence equivalent to the canonical one. So I wonder if this result can be extended to two sequences with same terms, not necessarily $P=M\oplus N$. I'm originally concerned with modules rather than (not necessarily abelian) groups. $\endgroup$
    – Junyan Xu
    Dec 10, 2011 at 3:35
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    $\begingroup$ @Junyan: 1) I understand your reasoning, but I still think that for archiving purposes it is better to post a new question when you have a new question. 2) You are right. It was not clear to me what exactly you're asking and that you want to restrict to finitely generated modules. The explanation you wrote in this comment would have been a perfect way to ask that question. It is important if you explain your thoughts that led you to ask the question so people understand better what you have in mind. Cheers! $\endgroup$ Dec 10, 2011 at 4:18

1 Answer 1


1) Concerning the terminology: Given a commutative diagramm $$\begin{array}{cccccccccccccc} 0 &\to & M & \to & P_1 & \to & N & \to & 0 \ \newline & & f\downarrow & & g\downarrow & & \downarrow h & & & \ \newline 0 &\to & M & \to & P_2 & \to & N & \to & 0 \ \newline \end{array}\hspace{20pt}(\ast)$$

the extensions are called "equivalent", if $f$ and $h$ are the identities (cf Hilton-Stammbach, A Course in Homological Algebra, III.1). If $f, g$ can be any automorphisms, I don't know if there is a standard name for the resulting equivalence relation. In a group theory paper $g$ was called a "strong isomorphism", indicating that in general an isomorphism doesn't fit into $(\ast)$. In a related question, I called the extensions "weakly equivalent", indicating that it's a weaker relation than eqivalence of extensions.

2) Concerning the question, when two extensions $0 \to M \to P \to N \to 0$ are weakly equivalent:

a) If $R$ is any ring and $N$ is projective, than the extension splits. Thus the extensions are equivalent.

b) Let $R$ be a PID and suppose that $P$ is torsion free and finitely generated. Then the extensions are weakly equivalent.

For, let $P$ have rank $n$ and denote the images of $M$ in $P$ by $U$ resp. $U'$. Then there are $r_1, ...,r_n \in R$ (uniquely determined by $N$) and there is a basis $\lbrace x_1,...,x_n \rbrace$ of $P$ such that the non-zero $r_1x_1,...,r_nx_n$ form a basis of $U$. Analogously, there is a basis $\lbrace x'_1,...,x'_n \rbrace$ of $P$ such that the non-zero $r_1x'_1,...,r_nx'_n$ form a basis of $U'$. Now $P \to P, x_i \mapsto x'_i$ defines an isomorphism that induces a commutative diagramm $(\ast)$ with vertical ismorphism.

c) If $P$ in b) has torsion, then the extensions are not weakly equivalent in general.

An example for $R = \mathbb{Z}$ is as follows: The subgroups $$U_1 := \langle \hspace{1pt} (2,2,0), (0,0,1) \hspace{1pt} \rangle, \quad\quad U_2 := \langle \hspace{1pt} (2,1,1), (4,0,0) \hspace{1pt} \rangle $$ of $A := \mathbb{Z}/8 \oplus \mathbb{Z}/4 \oplus \mathbb{Z}/2$ induce extensions $$ 0 \to U_i \to A \to A/U_i \to 0$$ with $U_i \cong \mathbb{Z}/4 \oplus \mathbb{Z}/2 \cong A/U_i.$ They fit into $(\ast)$ iff there is an automorphism $f$ of $A$ with $f(U_1) = U_2$. But by checking the possibilities for $f(1,0,0), f(0,1,0)$ one easily sees that such $f$ doesn't exist.

  • $\begingroup$ Thanks a lot! I think your example is equivalent to $U_1:=\mathbb{Z}/4\oplus0\oplus\mathbb{Z}/2, U_2:=\mathbb{Z}/2\oplus\mathbb{Z}/4\oplus0$, which is so simple that I should have discovered it if I have thought seriously. May I ask where does your example come from so that it takes such a complicated form? On the other hand, I indeed overlooked the distinction between two types of equivalence. If we conform to the "strict" equivalence, a counterexample is already present for $0\to\mathbb{Z}/2\to\mathbb{Z}/2\oplus\mathbb{Z}/4\to\mathbb{Z}/2\oplus\mathbb{Z}/2\to0$. $\endgroup$
    – Junyan Xu
    Dec 10, 2011 at 13:31
  • $\begingroup$ 1) I just used GAP to find the counterexample. 2) In the "strict" case there's an even simpler example: $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/3 \to 0$ with epi's $1 \mapsto \pm 1$ (compare to 2b). 3) I didn't check, but I'm pretty sure there is a bijection between the classes of weakly equivalent extensions and the orbits of the action of $Aut_R(M) \times Aut_R(N)$ on $Ext_R^1(N,M)$. $\endgroup$
    – Ralph
    Dec 10, 2011 at 19:37

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