I've been struggling a bit with the following problem, it must be not hard but I don't see it.

If $G$ is a torsion-free abelian group, then $\mathrm{Ext}_\mathbb{Z}^1(G, \mathbb Z) $ is divisible. "Conversely", if $G$ is divisible, then $\mathrm{Ext}_\mathbb{Z}^1(G, \mathbb Z)$ is torsion-free.

How should I proceed here? I've proved before that if $G$ is torsion-free and $H$ is divisible, then $\mathrm{Ext}_\mathbb{Z}^1(G, H) $ is divisible. But the key point in the problem above is that $\mathbb Z$ is not divisible, so this can't be applied.

Do you have any idea on how to tackle this problem? Regards


$\DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\Hom}{Hom}$ First, assume that $G$ is a torsion-free abelian group. I will make use of the following facts:

Lemma 1: A $\mathbb Z$-module is injective if and only if it is divisible.
Lemma 2: Any quotient of a divisible $\mathbb Z$-module is divisible.
The proofs are straight-forward. For the direction "divisible $\Rightarrow$ injective" you need to use Zorn’s Lemma, though.

Notice that $\mathbb Q$ is divisible. By Lemmas 1 and 2 we obtain an injective resolution $0\to \mathbb Z\to \mathbb Q \xrightarrow{d} \mathbb Q/\mathbb Z \to 0$ of $\mathbb Z$. By definition, we have $$ \Ext^1_{\mathbb Z} (G,\mathbb Z) = \Hom_{\mathbb Z}(G, \mathbb Q/\mathbb Z)/d_*(\Hom_{\mathbb Z}(G,\mathbb Q)). $$ By Lemma 2 and 1 again, it suffices to show that $\Hom_{\mathbb Z}(G, \mathbb Q/\mathbb Z)$ is divisible. This is, where we use that $G$ is torsion-free: Given any $n\in\mathbb N$, we have an exact sequence $0\to G\xrightarrow{\cdot n}G$. As $\mathbb Q/\mathbb Z$ is injective, the sequence $$ \Hom_{\mathbb Z}(G,\mathbb Q/\mathbb Z) \xrightarrow{\cdot n} \Hom_{\mathbb Z}(G,\mathbb Q/\mathbb Z)\longrightarrow 0 $$ is exact. As $n$ was arbitrary, this shows that $\Hom_{\mathbb Z}(G, \mathbb Q/\mathbb Z)$ is divisible. Hence, $\Ext^1_{\mathbb Z}(G, \mathbb Z)$ is divisible.

Now, assume that $G$ is divisible. Let $n\in\mathbb N$ be arbitrary. We want to prove that $\Ext^1_{\mathbb Z}(G,\mathbb Z) \xrightarrow{\cdot n} \Ext^1_{\mathbb Z}(G,\mathbb Z)$ is injective. Let $H_n:= \{g\in G \mid ng=0\}$. Then we have $\Hom_{\mathbb Z}(H_n,\mathbb Z) = 0$: Given $f\colon H_n\to \mathbb Z$, we have $0 = f(nh) = n\cdot f(h)$ and hence $f(h) = 0$, for all $h\in H_n$. From the short exact sequence $0\to H_n\to G \xrightarrow{\cdot n}G \to 0$ we deduce from the long exact sequence in cohomology an exact sequence $$ 0 = \Hom_{\mathbb Z}(H_n,\mathbb Z) \to \Ext^1_{\mathbb Z}(G,\mathbb Z) \xrightarrow{\cdot n} \Ext^1_{\mathbb Z}(G,\mathbb Z). $$ Therefore, $\Ext^1_{\mathbb Z}(G,\mathbb Z)$ contains no $n$-torsion for all $n\in\mathbb N$, and is thus torsion-free.

Notice that this argument shows that $\Ext^1_{\mathbb Z}(nG,\mathbb Z) \xrightarrow{\cdot n} \Ext^1_{\mathbb Z}(G,\mathbb Z)$ is always injective.

  • $\begingroup$ Sorry for the late response, I like your solution, it is good to know now how to solve that problem $\endgroup$ – EternalBlood Jan 15 '18 at 23:14

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