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Let $C_*$ be a chain complex of abelian groups.

Is it true that $H_i(C_*\otimes \mathbb{Z}/p)=0$ for all $i$ if and only if $H_i(C_*\otimes \mathbb{Z}_p)=0$ for all $i$, where $\mathbb{Z}_p$ is localization of $\mathbb{Z}$ away from $p$?

And I want to know that the precise definition of localization of $\mathbb{Z}$ away from $p$. Is it equal to $\mathbb{Z}[\frac{1}{p}]$ or $(\mathbb{Z}-p\mathbb{Z})^{-1}\mathbb{Z}$?

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No. Consider for instance the chain complex $0 \to \mathbb{Z}_p \stackrel{p}{\to} \mathbb{Z}_p$. This is acyclic but its reduction mod $p$ isn't. The problem is that $\mathbb{Z}/p$ is not a flat module over the localization $\mathbb{Z}_p$.

As for your second question, it's the latter one. You are localizing away from $p$, so you invert the elements not in $(p)$.

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Dear Akhil, I'm the one who asked the question. Is it also false in the case that

$C_*$ is chain complex of free abelian groups? (In here, each $C_i$ is not necessarily finitely generated.)

Here, I think that if the additional condition that each $C_i$ is free abelian group is given, then we can use the Universal Coefficient theorem to get the desired result. Is it right?

More explicitly, $H_i(C_*\otimes \mathbb{Z}_p)=H_i(C_*)\otimes \mathbb{Z}_p$, because the localization is an exact functor. Hence, $H_i(C_*\otimes \mathbb{Z}_p)=0$ for every $i$ if and only if $H_i(C_*)$ is torsion group and every element has order relative prime to $p$ for every $i$.

And $H_i(C_*\otimes \mathbb{Z}/p)=H_i(C_*)\otimes \mathbb{Z}/p\oplus \operatorname{Tor}(H_{i-1}(C_*),\mathbb{Z}/p)=H_i(C_*)\otimes {}_{p}H_{i-1}(C_*)$, where ${}_{p}M=\lbrace x\in M | px=0\rbrace$.

Hence, $H_i(C_*\otimes \mathbb{Z}/p)=0$ for every $i$ if and only if $H_i(C_*)$ is torsion group and order of every element is relatively prime to $p$ for every $i$.

I think that your nice example shows that the condition that $C_*$ is free is needed, indeed. I hope that this is right.

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Notice this is not an answer to the question... – Mariano Suárez-Alvarez Feb 8 '11 at 4:24
In my web, I don't have a button "Add comment" in Akhil's answer and My question. Now I have only one add comment button in here after you make a comment.. T_T – user6775 Feb 8 '11 at 4:32
This is the downside of having a minimum rep requirement for commenting... – Aaron Mazel-Gee Feb 8 '11 at 18:29
or not registering... – Sean Tilson Feb 9 '11 at 1:34

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