 Solution

Write $\mathfrak{p}=\pi \mathcal{O}$. Certainly multiplication by $\pi$ sends every $\mathcal{O}G$-homomorphism $W\to W^{\prime }$ to one with image in $\pi W^{\prime }$. Conversely, if $f:W\to \pi W^{\prime }$ is such a homomorphism, define $h(w)={\pi }^{-1}f(w)\in W^{\prime }$. This is well defined because $W^{\prime }$ is torsion-free. Cancellation of $\pi$ shows that $h$ is $\mathcal{O}$-linear and commutes with the $G$-action. Hence $f=\pi h$, andThis step in reduction of Hom from a projective group-algebra lattice does not itself require projectivity.