Uncountable injective submodule of quotient module of product of free modules

Question

Let $I$ be uncountable set. How to prove that $\left(\prod\limits_{i\in I}\mathbb{Z}x_i\right)/\sum\limits_{i\in I }\mathbb{Z}x_i$ always contain uncountable injective module ? Can we construct it explicitly ?

Answer 1

If I'm not mistaken this is already true for $I={\mathbb N}$.

For any abelian group $A$, the subgroup $D(A)$ of divisible elements (those $a\in A$ for which for any $n\in {\mathbb N}$ there exists some $b\in A$ such that $nb=a$) is the unique maximal injective subgroup of $A$. In your case of $A := {\mathbb Z}^{\mathbb N}/{\mathbb Z}^{({\mathbb N})}$ these are precisely the classes $[(a_n)_n]$ of those sequences $(a_n)_n$ for which for any $k\in{\mathbb N}$ the $a_n$ are eventually divisible by $k$. For example, $(n!)_n$ is such a sequence.

Now, given any countable sequence $(a^{(1)}_n)_n,(a^{(2)}_n)_n,\ldots$ of such sequences, you can use a diagonalization argument to construct another sequence in $D(A)$ which cannot be expressed as an integral combination of the $(a^{(i)}_n)_n$, not even up to ${\mathbb Z}^{({\mathbb N})}$: Roughly, for any prime power $q$, look at some $n_q\gg 0$ such that $a^{(i)}_k$ are divisible by $q$ for $i<q$ and $k>n_q$, and choose the $a_n$ to be non-divisible by $q$ for all $n\leq 2n_q$, say.

Do you want to work out the details yourself?