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Consider the $\mathbb{Z}$-module $M=\bigoplus{\mathbb{Z}/p\mathbb{Z}}$, where the direct sum is taken over the set of all prime numbers. How do I show that the localizations $M_\mathfrak{p}$ are finitely generated $\mathbb{Z}_\mathfrak{p}$-modules for any prime ideal $\mathfrak{p}$ of $\mathbb{Z}$?

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Hint: If $q\neq p$ are primes, then $q$is a $p$-adic unit. What can you say about a module that is annihilated by a unit? –  Jyrki Lahtonen Sep 11 '11 at 19:50
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Recall the following facts:

(1) Direct limits, direct sums and tensor products commute up to isomorphism

(2) The localization of a $R$-module $M$ at a prime $\mathfrak{p}$ is isomorphic to the module obtained by extending $M$ to scalars in $R_{\mathfrak{p}}$

(3) $(\mathbb{Z}/p\mathbb{Z}) \otimes_{\mathbb{Z}}\mathbb{Z}_q$ is isomoprhic to $\mathbb{Z}/p\mathbb{Z}$ if $p=q$ and $0$ otherwise.

From these three facts, it follows that

$$M_{\mathfrak{p}} = M \otimes_{\mathbb{Z}} \mathbb{Z}_{\mathfrak{p}} = (\displaystyle\lim_{\rightarrow} \text{ } \displaystyle \bigoplus_{p<n} \text{ } \mathbb{Z}/p\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Z}_{\mathfrak{p}} \cong \displaystyle\lim_{\rightarrow} \text{ } \displaystyle \bigoplus_{p<n} \text{ } (\mathbb{Z}/p\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}_{\mathfrak{p}}) \cong \mathbb{Z}/\mathfrak{p}$$

and your claim follows.

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What about if we replace $M=\bigoplus{\mathbb{Z}/p\mathbb{Z}}$ with $N=\prod{\mathbb{Z}/p\mathbb{Z}}$? Are the localizations $N_\mathfrak{p}$ finitely generated $\mathbb{Z}_\mathfrak{p}$-modules? –  Dave Sep 11 '11 at 20:36
    
Are you working on an old Northwestern prelim? –  jspecter Sep 11 '11 at 20:38
    
$N_{\mathfrak{p}}$ is not finitely generated. Partition the primes of $\mathbb{Z}$ into countably many infinite sets $I_n$ and consider the $\mathbb{Z}$-module homomorphism $\phi_n:\mathbb{Z} \rightarrow \prod_{p\in I_n} \mathbb{Z}/p\mathbb{Z}$ which maps $\mathbb{Z}$ to the diagonal. As $|I_n|$ is infinite, $\phi_n$ is an embedding. It follows the homorphism $\oplus_{n\in\omega} \phi_n : \mathbb{Z}^\omega \rightarrow N$ is an embedding. –  jspecter Sep 11 '11 at 20:55
    
Thus the dimension of $N \otimes_{\mathbb{Z}} \mathbb{Q}_p \cong (N \otimes_{\mathbb{Z}} \mathbb{Z}_p) \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ as a $\mathbb{Q}_p$-vector space is at least countably infinite. And therefore $N \otimes_{\mathbb{Z}} \mathbb{Z}_p$ cannot be finitely generated as a $\mathbb{Z}_p$ module. –  jspecter Sep 11 '11 at 21:01
    
@jspecter Haha, I thought this as well. I did all of those problems last year and Robert's questions have been very familiar. –  Dylan Moreland Sep 12 '11 at 2:33
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