Nonequivalent extensions with the same terms

Question

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.)

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.