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In this post, the OP stated $\mathbf{Ext}^1_\mathbb Z(\mathbb Q/\mathbb Z,\mathbb Z/n\mathbb Z) = \mathbb Z/n\mathbb Z$, which is something I can't really comprehend right now. (One of the answers in the post also uses this trick without justification.) I tried calculating it by taking the injective resolution of $\mathbb Z/n\mathbb Z$, and it ends up in a sequence

$$ 0\to \mathbf{Hom}(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)\xrightarrow{\cdot n} \mathbf{Hom}(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)\to 0 $$

However, it seems that I can't really understand the term $\mathbf{Hom}(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)$, and I can't get to the answer $\mathbb Z/n\mathbb Z$ here.

Eric Wofsey
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SummerAtlas
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  • Didnt you just post this earlier? – morrowmh Feb 16 '22 at 00:41
  • @MichaelMorrow Yes; I thought I understand it after a while so I deleted that post; unfortunately, that seems not to be the case – SummerAtlas Feb 16 '22 at 00:42
  • Oh I see. Perhaps include in this post the progress you've made since the earlier one (if any). I personally don't know the answer but someone on here should. – morrowmh Feb 16 '22 at 00:45

1 Answers1

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Probably the easiest way to compute this is by using the short exact sequence $$0\to \mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to 0.$$ This gives rise to a long exact sequence in Ext that goes $$\mathbf{Hom}(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})\to \mathbf{Hom}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\to \mathbf{Ext}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\to \mathbf{Ext}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z}).$$ Now for a key observation: multiplication by $n$ is an isomorphism $\mathbb{Q}\to\mathbb{Q}$, and multiplication by $n$ is the zero map $\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$. It follows that multiplication by $n$ is both an isomorphism and the zero map $\mathbf{Hom}(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})\to \mathbf{Hom}(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})$ and similarly for $\mathbf{Ext}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})$ (since it is induced by an isomorphism on the first coordinate but it is also induced by the zero map on the second coordinate). So, $\mathbf{Hom}(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})$ and $\mathbf{Ext}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})$ are both trivial. Thus our long exact sequence above gives an isomorphism between $\mathbf{Ext}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})$ and $\mathbf{Hom}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\cong \mathbb{Z}/n\mathbb{Z}$.

Eric Wofsey
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