By wikipedia, suppose $A, B$ are left $R$-modules, one way to calculate $Ext_{R}^{1}(A, B)$ is to regard it as equivalent class of module extension of $A$ by $B$, in the sense that the diagram

$$\require{AMScd}\begin{CD} 0 @>>> B @>>> E @>>> A @>>> 0 \\ @. @| @VfVV @| \\ 0 @>>> B @>>> E' @>>> A @>>> 0 \end{CD}$$

is commutative. Further, the identity element in $Ext_{R}^{1}(A, B)$ is regarded as split extension $E=A\oplus B.$

Now, let's consider $Ext_{\mathbb Z}^{1}(S^1, \mathbb Z)$. To be specific, let $S^1=\{e^{2\pi i\theta}|0\le \theta\lt1\}$, which forms a group under complex multiplication $e^{2\pi ia}\cdot e^{2\pi ib}=e^{2\pi i(a+b)}$. Now, consider the projection map $$\phi: \mathbb R\to S^1$$ $$x\to e^{2\pi i[x]}.$$

with $\ker \phi=\mathbb Z$. And this gives a short exact sequence for abelian groups, and also for $\mathbb Z$-modules:

$$0\to \mathbb Z\to \mathbb R\xrightarrow{\phi}S^1\to 0.$$

Since taken any rational number $a$, which has finite order in $S^1$, so if $\sigma: S^1\to \mathbb R$, then $\sigma(a)=0$, this implies that there is no homomorphism $\sigma: S^1\to \mathbb R$, such that $\phi\circ \sigma=id_{S^1}.$Therefore the above extension is not split. Therefore we have shown that $Ext_{\mathbb Z}^{1}(S^1, \mathbb Z)$ is non-trivial. But, how can I compute the whole $Ext_{\mathbb Z}^{1}(S^1, \mathbb Z)$?

  • 1
    $\begingroup$ It's not true that any homomorphism from $S^1$ to $\mathbb{R}$ must be trivial. As an abstract abelian group, $\mathbb{R}$ is just an uncountable-dimensional $\mathbb{Q}$-vector space, and hence $\mathbb{R}/\mathbb{Z}$ is an uncountable-dimensional $\mathbb{Q}$-vector space plus $\mathbb{Q}/\mathbb{Z}$ (the subgroup of roots of unity). Killing this extra factor leaves you with lots of interesting homomorphisms $S^1 \to \mathbb{R}$ (almost none of which are continuous). $\endgroup$ Aug 25, 2015 at 5:01
  • $\begingroup$ @QiaochuYuan Thanks, you have made an interesting point. I see it, take $a$ is an irrational number, so $\sigma(a)$ can be non-zero element in $\mathbb R.$ $\endgroup$
    – AG learner
    Aug 25, 2015 at 5:26

1 Answer 1


$\text{Ext}^1(-, -)$ sends direct sums in the first variable to direct products, and as mentioned in the comments, as an abstract abelian group

$$S^1 \cong \left( \bigoplus_X \mathbb{Q} \right) \oplus \mathbb{Q}/\mathbb{Z}$$

where $X$ is uncountable. So it suffices to compute $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ and $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$. The former group turns out to be an uncountable-dimensional rational vector space, while the latter is a bit more complicated: see these notes and this paper for details.


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