Re: non-Hausdoff topology

[Steve Vickers <s.j.vickers@cs.bham.ac.uk>]

Dear Vaughan,

The Zariski topology on the prime spectrum is normally not T1.

In fact Hochster showed that any spectral space (the ones that
correspond to ordered Stone spaces in the Priestly duality) is
homeomorphic to the prime spectrum of some ring, with its Zariski
topology. In particular, this holds for Scott domains and some other
classes of spaces commonly arising in denotational semantics - though it
would be eccentric to treat them as spectra of rings.

Referring to the Wikipedia page on Zariski topology, the "classical"
definition gives a T1 topology,  but the "modern" definition adds extra
"generic" points corresponding to non-maximal prime ideals and they
create a non-discrete specialization order, T0 but not T1.

Regards,

Steve.

Vaughan Pratt wrote:
>
> On 7/7/2010 10:28 AM, Michael Barr wrote:
>> Not just in CS, but also central to algebraic geometry: the Zariski
>> topology is almost never hausdorff. But when topology is taught to
>> undergraduates, it is usually for the purposes of analysis and I don't
>> know if we could this point across.
>
> My understanding of Paul's complaint was with the passage not so much
> from T2 to T0 but from T1 to T0, needed for the Scott topology.  The
> Zariski topology is always T1.
>
> Vaughan Pratt
>



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