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I'm trying to find $\mathrm{Ext}_\mathbb{Z}^n(\mathbb{Z},\mathbb{Z})$, which involves computing the homologies of \begin{equation} 0\leftarrow \mathrm{Hom}_\mathbb{Z}(\mathbb{Q},\mathbb{Z})\leftarrow \mathrm{Hom}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z},\mathbb{Z})\leftarrow 0. \end{equation}

I got the complex above by taking the injective resolution $0\rightarrow \mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0$ and applying $\mathrm{Hom}_\mathbb{Z}(\_,\mathbb{Z})$.

I think $\mathrm{Hom}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z},\mathbb{Z})=0$ because every element in $\mathbb{Q}/\mathbb{Z}$ has finite order, while no nonzero element in $\mathbb{Z}$ does.

I also think $\mathrm{Hom}_\mathbb{Z}(\mathbb{Q},\mathbb{Z})=0$ because if we set $f(1)=a$ then for all $m\in \mathbb{Z}$ we have $mf(1/m)=a$, implying $a$ is divisible by every integer, forcing $a=0$ and hence $f=0$.

But, I thought that $\mathrm{Ext}^0(\mathbb{Z},\mathbb{Z})=\mathrm{Hom}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$. Any help identifying my mistake would be appreciated.

Jendrik Stelzner
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ChrisWong
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    You can either compute $\operatorname{Ext}^n_{\mathbb{Z}}(A,B)$ using a projective resolution of the first argument $A$ or an injective resolution of the second $B$. In your attempt, you used an injective resolution of the first argument, and this doesn't work. – Roland Jan 13 '23 at 21:48
  • Your sequence only proves that $Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z})\to Ext^1_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z},\mathbb{Z}) $ is injective (and in fact it is a bijection). So it can be use to compute this last Ext group, not to compute the Hom group, that it is trivially computed (as $\mathbb{Z}$ is free, hence projective). – Nulhomologous Jan 13 '23 at 22:42
  • Since $\hom_{\mathbb Z}(\mathbb Z,M) \cong M$ is an exact functor, you do not have to compute anything... – Pedro Jan 19 '23 at 11:43

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