I am studying from Jacobson's Basic Algebra II (2nd edition), and I am stuck on Exercise 6 of $\S$3.5 (page 122):
Exercise 3.5.6. Suppose $\{ M_\alpha \}$ is a set of irreducible modules. Is $\prod M_\alpha$ completely reducible?
If $\{ M_\alpha \}$ is a finite collection of irreducible modules, then $\prod M_\alpha = \bigoplus M_\alpha$, so it is completely reducible.
On the other hand, if $\{ M_\alpha \}$ is an infinite collection of irreducible modules, then I guess that $\prod M_\alpha$ is not completely reducible, since $\bigoplus M_\alpha \subsetneq \prod M_\alpha$. However, I am not able to give a clear argument for why $\prod M_\alpha$ is not completely reducible.
Jacobson has earlier proved the following characterisation of completely reducible modules (on page 121):
Theorem 3.10. The following conditions in a module $M$ are equivalent:
$M = \sum M_\alpha$, where the $M_\alpha$ are irreducible.
$M$ is completely reducible.
The lattice $L(M)$ of submodules of $M$ is complemented, that is, for every submodule $N$ of $M$ there exists a submodule $N'$ of $M$ such that $M = N \oplus N'$.
Perhaps one can show that for $N = \bigoplus M_\alpha$ there does not exist any submodule $N'$ of $\prod M_\alpha$ such that $N \oplus N' = M$, so that $(3)$ fails to hold? But, I'm not sure how to go about this.
Any help is appreciated.
Over the ring $\mathbb{Z}$, consider the product $\prod_p\mathbb{Z}/(p)$ where $p$ ranges over all the primes. Then each $\mathbb{Z}/(p)$ is irreducible, but the product is not completely reducible since a completely reducible $\mathbb{Z}$-module is torsion but the element of the product which is $1$ on all coordinates is not torsion.
Note, though, that an infinite product of irreducible modules may be completely reducible. For instance, over a field, every module is completely reducible.