I have a discrete valuation ring $A$, an infinite collection $(M_i)_{i\in I}$ of non-zero $A$-modules and an injective co-generator $\Theta$ for the category $\text{Mod}_A$ of $A$-modules. Also set $M:=\bigoplus_{i\in I}M_i$, $D:=\text{Hom}(-,\Theta)$, the contravariant hom functor, and $$L:=\{\alpha\in DM:\alpha=0\text{ on } M_i\text{ for all but finitely many }i\}.$$
Since $\Theta$ is a co-generator, for each $i\in I$ we can choose non-zero $\alpha_i\in DM_i$. Now define $\alpha:M\to\Theta$ by $\alpha(m):=\sum_{i\in I}\alpha_i(\pi_i(m))$, so that $\alpha\in DM\setminus L$. Here is what I am stuck with:
I need to show that there exists $\beta\in D^2M$ with $\beta(\alpha)\neq 0$ and $\beta(L)=0$.
I guess I first need to show that $A\alpha\cap L=0$. Let $f\in A\alpha\cap L$. Then $f=a\alpha$ for some $a\in A$ and so for all but finitely many $i$ we have both $a\alpha(m_i)=0$ and $\alpha(m_i)\neq 0$. If $a=0$, then $f=0$ and clearly $a\notin A^\times$. So $a=p^ku$ for some irreducible $p\in A$, $u\in A^\times$ and $k\geq 1$. Then $p^k\alpha(m_i)=0$. But I'm not sure where to go from here.
This has been explained to me offline as follows:
Since $L\leq DM$, we have that $DM/L$ is an $A$-module. Since $\alpha\in DM\setminus L$, we have that $\alpha+L\neq L$, so that $DM/L$ is non-zero. Then since $\Theta$ is a co-generator, there exists a morphism $\gamma:A(\alpha+L)\to\Theta$ with $\gamma(\alpha+L)\neq 0$. Since $\Theta$ is injective, we can extend $\gamma$ to some $\widetilde{\gamma}:DM/L\to\Theta$ with $\widetilde{\gamma}(a+L)\neq 0$.
Now define $\beta:=\widetilde{\gamma}\circ q:DM\to\Theta$, where $q:DM\to DM/L$ is the quotient map. Then $\beta\in D^2M$, $\beta(\alpha)=\gamma(\alpha+L)\neq 0$ and $\beta(L)=0$, as required.