Calculating $\mathbf{Ext}_\mathbb Z(\mathbb Q/\mathbb Z,\mathbb Z/n\mathbb Z)$ 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.
 A: 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 $\mathbf{Hom}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\cong \mathbb{Z}/n\mathbb{Z}$.
