Finding $\operatorname{Ext}^{1}(\Bbb Q,\Bbb Z)$ 
I am trying to compute $\operatorname{Ext}^{1}(\Bbb Q,\Bbb Z)$ explicitely.

Using $\Bbb Q/\Bbb Z$ I constructed a natural injective resolution of $\Bbb Z$, and I know that $\Bbb Q/\Bbb Z$ is injective. Please help after that. 

We have the following exact sequence: $$\operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Q)\to \operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Q/\Bbb Z)\to\operatorname{Ext}^1_{\Bbb Z}(\Bbb Q,\Bbb Z)\to\operatorname{Ext}^1_{\Bbb Z}(\Bbb Q,\Bbb Q).$$ Since $\operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Q)\cong\Bbb Q$ and $\operatorname{Ext}^1_{\Bbb Z}(\Bbb Q,\Bbb Q)=0$ we get $$\Bbb Q\to \operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Q/\Bbb Z)\to\operatorname{Ext}^1_{\Bbb Z}(\Bbb Q,\Bbb Z)\to 0,$$ so $\operatorname{Ext}^1_{\Bbb Z}(\Bbb Q,\Bbb Z)\cong M/\Bbb Q$, where $M=\operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Q/\Bbb Z)$. In order to determine $M$ recall that $\Bbb Q/\Bbb Z=\bigoplus_{p\text {prime}}\Bbb Z_{p^{\infty}}$. Now prove that $\operatorname{Hom}(\Bbb Q,\Bbb Z_{p^{\infty}})\cong\Bbb Q_p$. In the end we get $M=\prod_{p\text {prime}}\Bbb Q_p$.
