{c}

Put $I=\operatorname{Hom}_{\mathcal OG}(W,W')$. Since $W$ is a <projective module>, its <Hom functor> is exact. Applied to $0\to\pi W'\to W'\to W'/\pi W'\to0$, it gives
$$
0\longrightarrow\operatorname{Hom}_{\mathcal OG}(W,\pi W')\longrightarrow I\longrightarrow\operatorname{Hom}_{\mathcal OG}(W,W'/\pi W')\longrightarrow0.
$$
The kernel is $\pi I$ by (ii). Every map to $W'/\pi W'$ kills $\pi W$ and factors uniquely through $W/\pi W$; this gives the second isomorphism. Therefore
$$
\boxed{I/\pi I\cong\operatorname{Hom}_{\mathcal OG}(W,W'/\pi W')\cong\operatorname{Hom}_{kG}(W/\pi W,W'/\pi W')}.
$$
By (i), $I$ is finite free with rank $\dim_K\operatorname{Hom}_{KG}(M,M')$. Its reduction has dimension equal to that rank. Thus <reduction of Hom from a projective group-algebra lattice> yields
$$
\boxed{\dim_K\operatorname{Hom}_{KG}(M,M')=\dim_k\operatorname{Hom}_{kG}(k\otimes_{\mathcal O}W,k\otimes_{\mathcal O}W')}.
$$
The original PDF has the group-algebra subscripts used here; the supplied TeX drops or corrupts several of them.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 138 / 4 / iii / 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.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 138 / 4 / ii / Solution

Let $W,W'$ be choices of <integral form of a group representation> over a complete <p-modular system>. Then $I=\operatorname{Hom}_{\mathcal OG}(W,W')$ is finite free, and $K\otimes_{\mathcal O}I=\operatorname{Hom}_{KG}(K\otimes W,K\otimes W')$: clearing denominators proves the spanning assertion. Also $\operatorname{Hom}_{\mathcal OG}(W,\pi W')=\pi I$. If $W$ is a <projective module>, applying the <Hom functor> to $0\to\pi W'\to W'\to W'/\pi W'\to0$ yields
$$
I/\pi I\cong\operatorname{Hom}_{kG}(W/\pi W,W'/\pi W').
$$
Consequently these ordinary and modular Hom spaces have the same <dimension of a vector space>. Projectivity is essential: for $G=C_2$ over $\mathbb Z_2$, the trivial and sign lattices have zero Hom between them, whereas their reductions in characteristic $2$ coincide and have a one-dimensional Hom space.

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