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The adic spectrum $Spa(\mathbb{Z},\mathbb{Z})$ looks as follows:

First, there is a point for every prime ideal $\mathfrak{p}\subset\mathbb{Z}$, corresponding to the valuation given by the composition of the quotient map $\mathbb{Z}\rightarrow \mathbb{Z}/\mathfrak{p}$ with the trivial valuation. Then there is an additional point for every $p$-adic valuation.

The points corresponding to maximal ideals are closed, a point corresponding to the $p$-adic valuation has the point corresponding to $(p)$ in its closure and the point corresponding to the zero ideal is a generic point.

It is known that this space is homeomorphic to the spectrum of a ring by a general result by M. Hoechster, found in "Prime Ideal Structure in Commutative Rings". However I am asking just out of curiosity and would like to avoid to get into the technical details of the construction and wondered if anybody has figured out this special case.

jorst
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  • I don't think it's fair to call it the ring with $\operatorname{Spa}(\mathbb Z, \mathbb Z) = \operatorname{Spec} R$; rather it is a ring with this property. Of course, you should keep in mind that Hochster's construction is not particularly meaningful for the arithmetic that we expect our space to capture. – Remy Nov 18 '15 at 13:51
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    As an example of how non-injective the $\operatorname{Spec}$ functor is: note that all the spaces $\operatorname{Spec} \mathcal O_K$ for number fields $K$ are homeomorphic, and they are also homeomorphic to $\operatorname{Spec} k(t)$ for any countable (including finite) field $k$. – Remy Nov 18 '15 at 14:07
  • thanks, that's a very good point. I edited the title accordingly. – jorst Nov 18 '15 at 15:30

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