$\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is a divisible abelian group

Question

I'm trying to prove that $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is a divisible $\mathbb{Z}$-module (p is prime, and the direct sum and direct product are taken over the set of all primes). It is an exercise from Rotman, An Introduction to Homological Algebra. Here's what I've done so far:

$\bigoplus \mathbb{Z_p}$ is the torsion submodule of $\prod \mathbb{Z_p}$, so the quotient is torsion-free. Since $\mathbb{Z}$ is a PID, then the quotient is flat.

How can flatness help me prove divisibility? Well, since $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is flat then $Hom_\mathbb{Z} \left( \frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}, \frac{\mathbb{Q}}{\mathbb{Z}}\right)$, the character module, is injective.

And I don't know how to continue. I don't think this is the way to go, but it's what I've tried. Another thing I've observed is that again, since $\mathbb{Z}$ is a PID, then $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is divisible iff it is injective, but once again, I don't know what to do with this.

Answer 1

What about the direct approach?

Suppose you have $(a_i)\in \prod \mathbb{Z}_p$ and an integer n, then modulo $\bigoplus \mathbb{Z_p}$ you may assume that $a_p=0$ for all the primes dividing $n$. $n$ is invertible in all the rest of $\mathbb{Z}_p$ so you can find $\frac{a_p}{n}$ in them.

So define $b_p = 0 $ for $p\mid n$ and $b_p = \frac{a_p}{n}$ for $p\not\mid n$ and then $n (b_p)+\bigoplus \mathbb{Z_p} = (a_n) + \bigoplus \mathbb{Z_p}$