Let $S,T$ be a partition of the primes into two nonempty sets (or if you prefer, the multiplicative sets generated by these); my . Localize at these sets and form the sequence $0\to\mathbb{Z}\to S^{-1}\mathbb{Z}\oplus T^{-1}\mathbb{Z}\to\mathbb{Q}\to 0$, where the first map is $n\mapsto (n,-n)$ and the second is $(a,b)\mapsto a+b$. (My comment on Ralph's answer was the case $T = \{p\}$. {p\}$.) Then the same partial fractions argument as in Ralph's answer shows that$0\to\mathbb{Z}\to S^{-1}\mathbb{Z}\oplus T^{-1}\mathbb{Z}\to\mathbb{Q}\to 0$this is an exact sequence which does not split. With Now let$U,V$be another such partition of the primes. Suppose there is an isomorphism$f: S^{-1}\mathbb{Z}\oplus T^{-1}\mathbb{Z}\to U^{-1}\mathbb{Z}\oplus V^{-1}\mathbb{Z}$making the corresponding exact sequences equivalent. Assume WLOG that$S$contains at least two elements$p,r\in S$and$p\in U$. For any$k\geq 1$, equivalence of the exact sequences gives$f(1/p^k,0) = (a_k,b_k)$where$a_k+b_k = 1/p^k$. Since$p\in U$and$U^{-1}\mathbb{Z}\cap V^{-1}\mathbb{Z} = \mathbb{Z}$, we get$(a_k,b_k) = (1/p^k + m_k,-m_k)$for some$m_k\in\mathbb{Z}$. The map$f$is a bit more work homomorphism, so$f(1,0) = (I'm happy to add 1 + p^km_k, -p^km_k)$. The value$k$was arbitrary, so the second component of$f(1,0)$is divisible by$p^k$for all$k\geq 1$and must be zero. Therefore$m_k = 0$and$f(1/p^k,0) = (1/p^k,0)$for all$k\geq 0$. The same argument if there shows that$f(1/r,0)$is interest)either$(1/r,0)$or$(0,1/r)$depending on whether$r\in U$or$r\in V$. The second case would make$f(1,0) = (0,1)$, contradicting the above, one can show that distinct partitions so$r\in U$. In this way we obtain$S\subseteq U$. The same arguments applied to the isomorphism$f^{-1}$yield inequivalent$U\subseteq S$, so$S=U$. Thus the exact sequences . This gives are equivalent if and only if$\{S,T\} = \{U,V\}$. There are uncountably many partitions of the primes into two nonempty sets, so there are uncountably many inequivalent non-split exact sequences as Mark Grant's comment on the original post suggested there should be$0\to\mathbb{Z}\to A\to\mathbb{Q}\to 0$. 1 Building on Ralph's answer a bit: Let$S,T$be a partition of the primes into two nonempty sets (or if you prefer, the multiplicative sets generated by these); my comment on Ralph's answer was the case$T = \{p\}$. Then the same partial fractions argument as in Ralph's answer shows that$0\to\mathbb{Z}\to S^{-1}\mathbb{Z}\oplus T^{-1}\mathbb{Z}\to\mathbb{Q}\to 0\$ is an exact sequence which does not split.