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May 10, 2012 at 8:20 comment added Marc Palm Isn't the fact that your construction is the same is called strong approximation. The ring $\mathbb{Q} \otimes_{\mathbb{Z}} \widehat{\mathbb{Z}}$ seems not so explicit to my taste, but nice picture.
May 10, 2012 at 0:08 comment added paul garrett In the spirit of the situation, rather than literally saying that the opens are $q\times \hat{\mathbb Z}$, perhaps saying to give $\mathbb Q$ the discrete topology would be nicer, and then "tensor product" as linearizing bilinear maps, etc. Similarly, stylistically, if we've talked about topologies on tensor products for that reason, then we can avoid talking about choosing bases for $K$, etc. Just quibbles... :)
May 9, 2012 at 21:26 comment added Neil Strickland @Konrad: I edited the answer to include some comments about topologies.
May 9, 2012 at 21:25 history edited Neil Strickland CC BY-SA 3.0
Added comments about topologies
May 9, 2012 at 21:06 comment added Konrad Voelkel Thanks. To make that description work for me, there must be a natural topology on the tensor product of topological rings. What is it?
May 9, 2012 at 18:02 history answered Neil Strickland CC BY-SA 3.0