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 $SpectralSpaces\to Rng$?


  • 2
    $\begingroup$ TThe set of prime ideals of any field is the one-point space. $\endgroup$ – user91132 Feb 16 '12 at 21:43
  • $\begingroup$ 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). $\endgroup$ – Jonathan Beardsley Feb 16 '12 at 22:15

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

| cite | improve this answer | |
  • 1
    $\begingroup$ 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.. $\endgroup$ – Jonathan Beardsley Feb 18 '12 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.