torsion representation Let $\mathbb{Z}_p$ be te ring of p-adic integers and let $T$,$T'$ be two free $\mathbb{Z}_p$-module with a continuous action of $G_{\mathbb{Q}_p}$ (the absolute Galois group of $\mathbb{Q}_p$).
It is not hard to see that if $T$ and $T'$ are of rank one, then there is a non zero $G_{\mathbb{Q}_p}$-equivariant map $T \to T' \otimes \mathbb{Q}_p / \mathbb{Z}_p$ if and only if $T \simeq T'$ mod $p$ (just write down explicitely the characters).
Is this true in a more general case, namely when $T'$ and $T'$ mod $p$ are irreducible and $T$ is any $\mathbb{Z}_p$-representation ?
 A: Consider the exact sequence of trivial $G$-modules:
$$0 \rightarrow \mathbf{Z}/p \rightarrow
\mathbf{Q}_p/\mathbf{Z}_p \rightarrow \mathbf{Q}_p/\mathbf{Z}_p \rightarrow 0,$$
(where the second map is multiplication by $p$) and tensor with $T'$. Since $T'$ is free over $\mathbf{Z}_p$, it is flat, and so the sequence remains exact.
Thus $T'/p$ injects into $T' \otimes \mathbf{Q}_p/\mathbf{Z}_p$. In particular,
any non-trivial map $T/p \rightarrow T'/p$ extends to a non-trivial map
$$T \rightarrow T/p \rightarrow T'/p \rightarrow T' \otimes \mathbf{Q}_p/\mathbf{Z}_p.$$
In particular, for the "if" part, you only need the hypothesis that $T'$ is flat and there is a non-zero $G$-equivariant map from $T/p$ to $T'/p$ (for any $G$).
Conversely, any map $T \rightarrow T' \otimes \mathbf{Q}_p/\mathbf{Z}_p$ factors through $T/p^n$ for some $n$, and hence as
$$T \rightarrow T/p^n \rightarrow (T' \otimes \mathbf{Q}_p/\mathbf{Z}_p)[p^n]
= T'/p^n.$$
If $T'/p$ is irreducible, then the composition factors of the RHS are all of the form $T'/p$, and hence there is a surjective map $T/p \rightarrow T'/p$, which must be an isomorphism if $T/p$ is also irreducible.
