# Non-split extension of the rationals by the integers

Can someone describe explicitly an abelian group $A$ such that the extension $$0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$$ doesn't split ?

Background: The Stein-Serre theorem (Hilton, Stammbach: A course in homol. algebra, Theorem 6.1) states that if $A$ is abelian of countable rank (=maximal number of linear independent elements), then $Ext(A,\mathbb{Z})=0$ implies $A$ free. When applied to $A= \mathbb{Q}$, I conclude $Ext(\mathbb{Q},\mathbb{Z})\neq 0$. Moreover, by interpreting $A \in Ext(\mathbb{Q},\mathbb{Z})$ as extension $0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$ of abelian groups, there must be an $A$ such that the extension doesn't split. The problem is that the proof of the theorem isn't constructive and doesn't show how to construct such an $A$.

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There should in fact be uncountably many of them! journals.cambridge.org/… What do you get if you quotient the $p$-adic integers by the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Z}_p$? –  Mark Grant Aug 22 '12 at 21:46
Mark -- something uncountable. –  algori Aug 22 '12 at 21:54
See also p. 5 of math.jhu.edu/~jmb/note/torext.pdf –  Damian Rössler Aug 22 '12 at 22:29
Consider the canonical exact sequence Z --> \prod_p Z_p --> M, where the product takes place over all primes p. Then M is a Q-vector space (exercise). Picking any non-zero element m in M gives an extension A of Q by Z via pullback; explicitly, A is the set of all elements in \prod_p Z_p that map to the Q-subspace generated by m in M. This extension is non-zero since Hom(Q,\prod_p Z_p) = \prod_p Hom(Q,Z_p) = 0. –  anon Aug 22 '12 at 22:34
@algori: Uncountable, but contains a copy of $\mathbb{Q}$. –  Kevin Ventullo Aug 23 '12 at 1:11

Nice question. For a prime $p$ let $\mathbb{Z}_{(p)} = \lbrace \frac{a}{b}\in \mathbb{Q}\mid p \nmid b\rbrace$ and $\mathbb{Z}[p^{-1}] = \lbrace \frac{a}{p^n}\in \mathbb{Q}\mid n \ge 0 \rbrace$. Then

$$A := \lbrace (x,y) \in \mathbb{Q} \times \mathbb{Z}[p^{-1}] \mid x-y \in \mathbb{Z}_{(p)} \rbrace$$ has the desired non-split extension. Informal $A$ consists of all pairs $(x,y) \in \mathbb{Q} \times \mathbb{Q}$ where $y$ is the $p$-part in the partial fraction decomposition of $x$ (up to an integer summand).

Proof: Let $\rho: A \to \mathbb{Q}$ be projection onto the first factor and $i: \mathbb{Z} \hookrightarrow A$ inclusion into the second factor. By partial fraction decomposition of the rationals, $\rho$ is surjective and $$\ker(\rho)=0 \times \big(\mathbb{Z}[p^{-1}] \cap \mathbb{Z}_{(p)}\big) = 0 \times \mathbb{Z} = \operatorname{im}(i).$$

Next, let $j: \mathbb{Q} \to A$ be a splitting hom. of $\rho$. Composing $j$ with the projection onto the second factor yields a hom. $f: \mathbb{Q} \to \mathbb{Z}[p^{-1}] \le \mathbb{Q}$. Since each endomorphism of $\mathbb{Q}$ is multiplication with some $q \in \mathbb{Q}$ we have $f(x)=qx \in \mathbb{Z}[p^{-1}]$ for all $x \in \mathbb{Q}$ which is only possible for $q=0$. Hence $j(x)=(x,0) \in A$ for all $x \in \mathbb{Q}$. Setting $x=1/p$ we obtain $1/p \in \mathbb{Z}_{(p)}$ which is the desired contradiction. QED

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Lovely! I think this is what I was going for with my inane comment above. –  Mark Grant Aug 23 '12 at 19:35
Very nice example! It's not much of a change, but the exact sequence $0\to\mathbb{Z}\to\mathbb{Z}_{(p)}\oplus\mathbb{Z}[p^{-1}]\to\mathbb{Q}\to 0$ is an example for the same reason and to my eyes it is more "symmetric". The second map is $n\mapsto(n,-n)$ and the third is $(s,t)\mapsto s+t$. –  Noah Stein Aug 23 '12 at 20:00
That's good! I searched long for the example, but not long enough to find this presentation :) Noah, thanks for showing me. –  Ralph Aug 23 '12 at 20:17
@Mark Opitz: By the paper linked by Damian, the connecting hom. $\delta: Hom(\mathbb{Q},\mathbb{Q}/\mathbb{Z}) \to Ext(\mathbb{Q},\mathbb{Z})$ is surjective. Let $\kappa: \mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ be the canonical hom. If $f \in Hom(\mathbb{Q},\mathbb{Q}/\mathbb{Z})$, then the extension corresponding to $\delta(f)$ is the pull-back of $(\kappa,f)$, i.e. $A=\lbrace (x,y) \in \mathbb{Q} \times \mathbb{Q} \mid f(x)=\kappa(y) \rbrace$. Hence it suffices to find a simple $f$. Let $\mathbb{Q}/\mathbb{Z}=\oplus_r V_r$ be the decomposition in r-torsion subgroups. –  Ralph Aug 23 '12 at 23:24
... Then I used $f: \mathbb{Q} \xrightarrow{\kappa} \mathbb{Q}/\mathbb{Z} \twoheadrightarrow V_p \hookrightarrow \mathbb{Q}/\mathbb{Z}$. Now $(x,y) \in A$ iff $\bar{x}_p = \bar{y}_p$ and $\bar{y}_r = 0$ if $r \neq p$ where $\kappa(x)=(\bar{x}_r)_r$. But this is equivalent to the description above. –  Ralph Aug 23 '12 at 23:27

Building on Ralph's answer a bit we can get uncountably many inequivalent examples as Mark Grant's comment on the original post suggested there should be.

Let $S,T$ be a partition of the primes into two nonempty sets (or if you prefer, the multiplicative sets generated by these). 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 $T = \{p\}$.) Then the same partial fractions argument as in Ralph's answer shows that this is an exact sequence which does not split.

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 homomorphism, so $f(1,0) = (1 + p^km_k, -p^km_k)$. 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 shows that $f(1/r,0)$ is 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, so $r\in U$. In this way we obtain $S\subseteq U$. The same arguments applied to the isomorphism $f^{-1}$ yield $U\subseteq S$, so $S=U$.

Thus the exact sequences 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 $0\to\mathbb{Z}\to A\to\mathbb{Q}\to 0$.

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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. –  Ralph Aug 24 '12 at 13:40
@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? –  Qfwfq Aug 24 '12 at 16:06
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. –  Jason Polak Aug 24 '12 at 19:30
@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: $\mathbb{Z}, \mathbb{Q}, \mathbb{R}$. –  Ralph Aug 24 '12 at 20:55
@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). –  Ralph Aug 24 '12 at 21:56

Let $A$ be a subgroup of $\mathbb{Q}^2$, generated by $e:=(0,1)$ and $f_n:=(1/n!,\alpha_n)$, where $\alpha_n = \frac{1}{n} (\alpha_{n-1} + \nu_n)$, $\alpha_1 = 1$, and $\nu_n \in \mathbb{Z}$, to be specified later.

First of all, note that for each $n$ the vectors $e$ and $f_n$ generate a subgroup $A_n$ that contains $A_{n-1}$, and that $A = \bigcup_n A_n$, so it can be easily checked that $A \cap (0 \times \mathbb{Q})$ is the subgroup generated by $e$, isomorphic to $\mathbb{Z}$. Furthermore, it is the kernel of the map that calculates the first coordinate, and the image is exactly $\mathbb{Q}$.

Now my aim is to choose $\nu_n$ in such a way that no element of $A$ is divisible enough. Clearly this can be done in many ways. Fix an $n$ for a moment, and notice that for $x \in A_n$ to be divisible by a large prime $p>n$ its second coordinate must equal $\frac{p!}{n!} \alpha_p$ modulo $p$. So by choosing $\nu_p$ we may ensure that a fixed $x$ is not divisible by $p$. What remains is to enumerate them carefully and choose $\nu_p$ in such a way that for every $n$ and every $x \in A_n$ there exists at least one $p$ such that $x$ is not divisible by $p$. Thus we need an injective map $(n,x) \mapsto p$ subject to $p > n$. It is easy but messy to write down... Just to elaborate the whole process: we choose $\nu_n$ in their usual order, and each time we run into a distinguished prime $p$ that is responsible for some $(n,x), n < p$, we should act accordingly using our knowledge of the previous $\alpha$'s.

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