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?