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!
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!
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.