What is the injective envelope of a product of abelian groups? We know that any $R$-module has an injective envelope. Matlis has shown that any injective module over a Noetherian $R$ decomposes as a direct sum of indecomposable injectives, and that these are the injective modules $E(R/P),$ for $P$ a prime ideal of $R.$ In the specific case of the abelian group $\prod \mathbb{Z}(p),$ the product taken over the positive primes, do we know what the injective envelope of this group is as a direct sum of indecomposables?
The question boils down to what the torsion-free rank is and what the $p$-ranks are for the various primes $p.$ I suspect for the specific group in question, these ranks are all infinite. Are they countable or uncountable? This probably better captures the question I am asking.

I think the following should work, but take it with a grain of salt. Let $\mathbb Z(p^\infty)$ denote the $p$-Prüfer group, that is, the injective envelope of $\mathbb Z(p)$. The module $I=\prod_p \mathbb Z(p^\infty)$ is an injective module containing your module $M=\prod_p \mathbb Z(p)$ as a submodule.
Now consider the module
$$
E = M + \bigoplus_p \mathbb Z(p^\infty) \subset I.
$$
Then $E$ is divisible, hence injective, and $M$ is essential in $E$. So $E$ is an injective envelope of $M$.
To find the direct sum decomposition of $E$, we need to consider
$$E_{0} := E/\bigoplus_p \mathbb Z(p^\infty) \cong M/(M \cap \bigoplus_p \mathbb Z(p^\infty)) \cong M/\bigoplus_p \mathbb Z(p),$$
as $E = E_0 \oplus \bigoplus_p \mathbb Z(p^\infty)$ due to the injectivity of $\bigoplus_p \mathbb Z(p^\infty)$.
Now observe that $E_0$ is torsion-free. It is also injective (since factor modules of divisible modules are divisible).
Thus $E_0 \cong \mathbb Q^{(I)}$ for some set $I$. It only remains to find $|I|$. Since $M$ has the cardinality of the continuum, while $M \cap \bigoplus_p \mathbb Z(p^\infty)$ is countable, also $E_0$ has the cardinality of the continuum. Therefore the same is true for $I$, as $|\mathbb Q^{(I)}|=\max\{ \omega, I \}$.
So it looks like $E$ has uncountable (continuum) torsion-free rank, but $p$-rank $1$ for each prime $p$.
