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Aug 24, 2012 at 21:56 comment added Ralph @Jason: Yes. Every torsion-free, divisible abelian group of card. $2^{\aleph_0}$ is isomorphic to $\mathbb{R}$. It's a nice excercise to show that $(\prod_p \mathbb{Z}_p)/\mathbb{Z}$ is torsion-free and divisible (hint: use Chinese Remainder Theorem to show divisibility).
Aug 24, 2012 at 20:55 comment added Ralph @Qfwfq: I'm not aware if order structures on the arguments extend in some way to Ext-groups. But in fact, the isomorphism (of abelian groups / rational vector spaces) nicely connects the most basic groups in math: $\mathbb{Z}, \mathbb{Q}, \mathbb{R}$.
Aug 24, 2012 at 19:30 comment added user1437 So this means that, $\mathbb{R}\cong (\prod_p \mathbb{Z}_p)/\mathbb{Z}$? A $\mathrm{lim}^1$ exact sequence in Weibel's book shows that $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})$ is isomorphic to the latter.
Aug 24, 2012 at 16:06 comment added Qfwfq @Ralph: Wow, then that construction can be seen as a pretty exotic way to construct $\mathbb{R}$ from the rationals! Can one recover the order structure of $\mathbb{R}$ too or only its abstract abelian group structure?
Aug 24, 2012 at 13:40 comment added Ralph Your result can be used to show $Ext(\mathbb{Q},\mathbb{Z}) \cong \mathbb{R}$ as stated in the paper referenced in Mark Grant's comment above: By taking a free resolution $0 \to K \to F \to \mathbb{Q} \to 0$ of countable ranks, the epimorphism $Hom(K,\mathbb{Z}) \to Ext(\mathbb{Q},\mathbb{Z})$ shows that $Ext(\mathbb{Q},\mathbb{Z})$ has cardinality at most $2^{\aleph_0}$. Hence by your result the cardinality is eactly $2^{\aleph_0}$. Since multiplication in $\mathbb{Q}$ makes $Ext(\mathbb{Q},\mathbb{Z})$ a $\mathbb{Q}$-vector space (of dimension $2^{\aleph_0}$), the assertion follows.
Aug 24, 2012 at 12:30 history edited Noah Stein CC BY-SA 3.0
Added argument for why these sequences are all inequivalent.
Aug 24, 2012 at 3:21 history answered Noah Stein CC BY-SA 3.0