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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.

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Just to be sure, do you mean $\mathbb{Z}_p$ as the p-adic integers or as $\mathbb{Z}/p$ the cyclic group with $p$ elements? –  Asaf Karagila Nov 12 '10 at 13:16
    
The second one: $\mathbb{Z}_p= \frac{\mathbb{Z}}{p\mathbb{Z}}$ –  Bruno Stonek Nov 12 '10 at 13:22
    
This question continues here math.stackexchange.com/questions/10007/… –  Bruno Stonek Nov 12 '10 at 15:03

1 Answer 1

up vote 10 down vote accepted

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 $n (b_p)+\bigoplus \mathbb{Z_p} = (a_n) + \bigoplus \mathbb{Z_p}$

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Perfect. Thank you. As usual, I was trying an obscure and overly-theoretical path instead of thinking what should be thought (in this case, what the quotient actually means). –  Bruno Stonek Nov 12 '10 at 13:45

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