Cohomology of the trivial action of $\mathbb{Z}_p$ on $\mathbb{Z}$ I'm wondering if the next exercise in 'An introduction to homological algebra' by Weibel is correct:
Let $G$ be the profinite group $\widehat{\mathbb{Z}}_p$. Show that 
$$H^i(G;\mathbb{Z}) = \begin{cases}
\widehat{\mathbb{Z}}_p & \text{ if $i$ even}\newline
0 & \text{ if $i$ odd}
\end{cases}$$
Is it possible that $\widehat{\mathbb{Z}}_p$ should be replaced by $\mathbb{Z}(p^\infty)$ ? I've found on the website of Weibel that it should be for $H^0(G;\mathbb{Z})=\mathbb{Z}$, $H^2(G;\mathbb{Z})=\mathbb{Z}(p^\infty)$ and for all other $i$ $H^i(G;\mathbb{Z})=0$, but shouldn't that be for odd $i$ and the rest still $\mathbb{Z}(p^\infty)$ ? Since the cohomology groups $H^i(\mathbb{Z}/m\mathbb{Z};\mathbb{Z})$ are $\mathbb{Z}/m\mathbb{Z}$ for $i$ even ?
 A: If you're using continuous cochain cohomology, then things work out as follows. For $H^0$, you get the invariants, all of $\mathbb{Z}$ in this case. For $H^1$, you get the group of continuous homomorpisms from $\mathbb{Z}_p$ to $\mathbb{Z}$. The image of such a homomorphism is necessarily finite (by continuity, compactness of $\mathbb{Z}_p$, and discreteness of $\mathbb{Z}$), but $\mathbb{Z}$ has no non-trivial finite subgroups, so you get zero. The cohomology sequence associated to $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}\rightarrow 0$ shows that $H^2(\mathbb{Z}_p,\mathbb{Z})\cong H^1(\mathbb{Z}_p,\mathbb{Q}/\mathbb{Z})$, which is the group of continuous homomorphisms from $\mathbb{Z}_p$ to $\mathbb{Q}/\mathbb{Z}$, i.e., the Pontryagin dual of $\mathbb{Z}_p$. This is isomorphic to $\mathbb{Q}_p/\mathbb{Z}_p$, or in your notation, $\mathbb{Z}(p^\infty)$. For $H^q(\mathbb{Z}_p,\mathbb{Z})$, $q>2$, all the groups are zero because the strict cohomological dimension of $\mathbb{Z}_p$ is $2$.
So, if Weibel intends the standard continuous cohomology of profinite groups, then what he says is not right. Otherwise I'm afraid I'm not sure. 
EDIT: It seems the correction on Weibel's website is correct, again, assuming continuous cochains.
