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