# $\mathbb{Z}$-extensions of $\mathbb{Z}/n\mathbb{Z}$ [closed]

The $$\mathbb{Z}$$-extensions of $$\mathbb{Z}/n\mathbb{Z}$$ $$0\longrightarrow \mathbb{Z}\longrightarrow E\longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow0$$ are classifiied by $$\operatorname{Ext}_{\mathbb{Z}}^1(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z})$$. But one can compute $$\operatorname{Ext}_{\mathbb{Z}}^1(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z})=\mathbb{Z}/n\mathbb{Z}$$. Hence there are $$n$$ such extensions. But is there any way to explicitly write down those extensions?

• There is an exercise in Lang's Algebra that tells you exactly how to do this (can't look it up now, but should be easy to find, it's in the part dealing with homological algebra).
– R.P.
Mar 21, 2021 at 11:59

Yes. May be a general remark: the group $$\mathrm{Ext}^1$$ classifies the short exact sequences, which is more structure than remembering $$E$$ or even $$E$$ with the $$2$$-term filtration. In any case, $$0\rightarrow \mathbb Z \xrightarrow{\cdot n} \mathbb Z \rightarrow \mathbb Z/n\mathbb Z \rightarrow 0$$ is a generator of $$\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n\mathbb Z,\mathbb Z)$$, let's call it $$\alpha$$. Then $$k\cdot\alpha\in\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n\mathbb Z,\mathbb Z)$$ is obtained as pull-back of $$\alpha$$ via the multiplication by $$k$$: $$\mathbb Z/n\mathbb Z \xrightarrow{\cdot k} \mathbb Z/n\mathbb Z$$. The fiber product $$\mathbb Z\times_{\mathbb Z/n\mathbb Z,\cdot k} \mathbb Z/n\mathbb Z$$ is computed as $$\mathbb Z$$ plus the kernel of $$\ker(\cdot k)=\frac{n}{(n,k)}\mathbb Z/n\mathbb Z$$ and the extension we get looks as follows: $$0\rightarrow \mathbb Z \xrightarrow{} \mathbb Z\oplus\frac{n}{(n,k)}\mathbb Z/n\mathbb Z \rightarrow \mathbb Z/n\mathbb Z \rightarrow 0$$ (where the map $$\mathbb Z\oplus\frac{n}{(n,k)}\mathbb Z/n\mathbb Z \rightarrow \mathbb Z/n\mathbb Z$$ is identified with the projection $$\mathbb Z\times_{\mathbb Z/n\mathbb Z,\cdot k} \mathbb Z/n\mathbb Z\twoheadrightarrow \mathbb Z/n\mathbb Z$$). For example if $$k\in (\mathbb Z/n\mathbb Z)^\times$$ we get the usual sequence $$0\rightarrow \mathbb Z \xrightarrow{} \mathbb Z \rightarrow \mathbb Z/n\mathbb Z \rightarrow 0$$ but where the second map sends $$1$$ to $$k$$ instead of $$1$$.