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Is it true that $\mathbb{Z}[q^{-1}]$, for $q$ a prime, can be constructed as the colimit of the diagram $\mathbb{Z}\to\mathbb{Z}\to\ldots$ where the arrows are multiplication by $q$? If so, what is the main idea of showing this?

Thanks!

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    This should be clearer if you relabel the copies of $\mathbb{Z}$ as $\mathbb{Z} \to q^{-1}\mathbb{Z} \to q^{-2}\mathbb{Z} \to \ldots$, where now the arrows are inclusions.2012-01-20
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    Obviously, this colimit only works as an additive group, not as a ring, since the homomorphisms are not ring homomorphisms.2012-01-20
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    @ThomasAndrews Aha. Right, which is fine, since I'm trying to understand the generalization of this idea in stable homotopy. It definitely didn't look like we were taking multiplicative identity to multiplicative identity.2012-01-20
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    This is really only affecting the structure of these things as $\mathbb{Z}$-modules.2012-01-20

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By the comment of @Chris Eagle, and the fact that as $\mathbb{Z}$-modules $q^{-1}\mathbb{Z}\cong\mathbb{Z}$, the colimit may be easily written down as $\coprod_{i\geq 0}q^{-i}\mathbb{Z}$ which is easily seen to be isomorphic to the desired $\mathbb{Z}$-module.

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    The notation is misleading since $\coprod$ is usually reserved for colimits. $\cup$ would be a better choice to denote a nested union.2013-01-11