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Let $G$ be a group and $M$ be a $G$-module. Then group cohomology $H^q(G,M)$ is defined as the right derived functor $\operatorname{Ext}^q_{\mathbb Z G}(\mathbb Z,M)$. Here $\mathbb Z$ is the trivial $G$-module.

Now if $M=\mathbb Z$ is the trivial $G$-module there is an injective resolution

$0\to \mathbb Z\to I_0\to I_1\to \cdots$

of $\mathbb Z$ as an abelian group. Now equip $I_q$ with the trivial $G$-action for every $q$. Then the homology of

$0\to I_0\to I_1\to \cdots$

is $H^q(G,\mathbb Z)=\begin{cases} \mathbb Z & q=0\\ 0 & \mbox{otherwise}. \end{cases}$

Now in Weibel's book homological algebra the coefficient $\mathbb Z$ appears in many examples. It does not produce trivial cohomology. What did I do wrong?

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    $\begingroup$ In general $I_q$ will not be injective as a $\mathbf Z[G]$-module. $\endgroup$ Jul 1, 2019 at 15:27
  • $\begingroup$ Thank you for the quick answer. $\endgroup$
    – EH2019
    Jul 1, 2019 at 16:00
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    $\begingroup$ The word "trivial" is misleading. Trivial should refer to the trivial module, which is $\{0\}$. But for various reasons it is usual to call "trivial module" the 1-dimensional vector space endowed with a trivial action. This object is actually not that trivial from the angle of (co)homology. $\endgroup$
    – YCor
    Jul 1, 2019 at 17:55

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