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It is said, as far as I can tell that an arbitrary spectral space, i.e. a space that is $T_0$, sober and quasi-compact whose collection of quasi-compact open sets forms a basis and is closed under finite intersections, must be $Spec(R)$ for some ring $R$. Is there a canonical (or any) way of reconstructing $R$ from its collection of prime ideals and the Zariski topology thereupon? What is the relationship of the functor $Spec:Rng\to SpectralSpaces$ and the functor (if it exists) going the other way $SpectralSpaces\to Rng$?


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TThe set of prime ideals of any field is the one-point space. – user91132 Feb 16 '12 at 21:43
Good point. This gives some intuitive evidence for the fact that Hochester's construction can give a ring that is an algebra over any field (since every space is in some sense a module over the one-point space). – Jon Beardsley Feb 16 '12 at 22:15
up vote 4 down vote accepted

Check out Prime ideal structure in commutative rings by Melvin Hochester where the theorem you mentioned is proved and functoriality discussed.

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Thank you very much! – Jon Beardsley Feb 16 '12 at 19:58
One question. Hochster's construction is believable, but rather unpleasantly intricate. Is there any simpler way to do this? Is there a nice way of describing the necessary ring? I don't see any in Hochster's paper.. – Jon Beardsley Feb 18 '12 at 19:43

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