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Can anyone suggest a good thorough reference for $p$-localization and $p$-completion of the integers? I'm an algebraic topologist who's found himself washed up without any intuition.

In particular, here are the two questions I'm trying to answer right now:

Question 1: what is $\mathbb{Q}/\mathbb{Z}_p$ tensored with itself?

I mean $\mathbb{Z}_p$ to be the $p$-local integers. The tensor is over $\mathbb{Z}_p$ or $\mathbb{Z}$ - the answer should be the same.

I know $\mathbb{Q}/\mathbb{Z}_p$ is an injective $\mathbb{Z}_p$-module, and $\mathbb{Z}_p$ is a nice local ring, so maybe something can be said about a flat resolution of $\mathbb{Q}/\mathbb{Z}_p$?

Question 2: What is known about the cokernel of the map $\mathbb{Z}_p \rightarrow \mathbb{\hat{Z}}_p$?

I mean the map that $p$-completes the $p$-local integers. I think the cokernel is a rational vector space, but of finite or infinite dimension? Does it have any nice description?

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Note that number theorists normally notate $\mathbb{Z}_{(p)}$ to be the "$p$-local integers", and leave $\mathbb{Z}_p$ to mean the $p$-adic numbers, that is the completion of $\mathbb{Z}_{(p)}$. –  Daniel Loughran Apr 25 '11 at 13:08
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Remember that $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}$. (At one time I found this incredibly counterintuitive, and at slightly earlier times I thought the answer was a vector space of countably infinite dimension. But I have gotten used to it.) I think that should help you with Question 1. As for Question 2: well, the cokernel is big: continuum cardinality, anyway. To check whether it's a $\mathbb{Q}$-vector space, it's enough to check whether it is a uniquely divisible abelian group. –  Pete L. Clark Apr 25 '11 at 14:08
    
What are $p$-local integers? Those with powers of $p$ in the denominators, or those with denominators coprime to $p$? –  darij grinberg Apr 25 '11 at 14:45
    
@darij I think Luke means the localization of $\mathbb{Z}$ at the prime ideal $(p)$, so no $p$ in the denominator. –  Keenan Kidwell Apr 25 '11 at 15:21

1 Answer 1

up vote 7 down vote accepted

I'm going to switch to a more standard notation, where the localization is $\mathbb{Z}_{(p)}$ and the completion is $\mathbb{Z}_p$.

For your first question, the general rule that concerns us is that the tensor product of a $p$-divisible group with a $p$-torsion group is zero. The reason is that if $a$ is a $p^n$th power and $p^n b = 0$, then $a \otimes b = p^{-n}a \otimes p^n b = 0$. Since $\mathbb{Q}/\mathbb{Z}_{(p)}$ is both $p$-divisible and $p$-torsion, its tensor square is zero.

The cokernel of completion is a $\mathbb{Q}$-vector space, because it admits a $\mathbb{Z}_{(p)}$-action by multiplication, and it is uniquely $p$-divisible (Proof: lift any element to $\mathbb{Z}_p$, subtract the constant term of its $p$-adic expansion, divide by $p$, and check that different lifts yield the same answer). Its dimension over $\mathbb{Q}$ is the cardinality of the continuum, since that is its cardinality as a set.

I think most number theory texts have some discussion of $p$-adic numbers, and Gouvea even wrote a whole book about them.

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