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This semester I am attending a reading seminar on non-archimedean analytic geometry (a subject I know nothing about), roughly following the notes of Conrad.

Reading Conrad's notes (and e.g. those of Bosch) it struck me that the prime spectrum of affinoid algebras never seems to appear, only the maximal spectrum. Can somebody explain the reason for this?

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    $\begingroup$ Because MaxSpec has more geometric appeal. Residue fields at closed pts are finite extns, so unique abs. value: $|f(x)|$ makes sense for max'l $x$. Not obvious how to include generic pts, is it? That's the answer. At other primes there are many ways to complete the residue field. For Berkovich spaces, Spec shows up. As prime spectrum consists of "all" field-valued pts of a ring, Berkovich spec consists of "all" non-arch field valued pts of Banach alg. Just like the utility of classical varieties when studying schemes, MaxSpec theory is not a waste of time (and technically easier to digest). $\endgroup$
    – BCnrd
    Commented Apr 19, 2010 at 18:11
  • $\begingroup$ Dear Brian, thanks for your answer, which I would accept if I could. The point about non-uniqueness of extension for non-maximal ideals is clear. I guess what I didn't grasp yet is why it isn't still useful to consider the non-maximal prime ideals, even in spite of this defect; Emerton's answer below gives an indication of this. $\endgroup$
    – user5117
    Commented Apr 20, 2010 at 7:02

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I am surprised that Brian got to this one first without making what I thought was another obvious comment: affinoids are Jacobson rings! A function which is zero at all points of an affinoid rigid space corresponds to an element of your affinoid algebra which is in all maximal ideals and hence (by Jacobson-ness) is nilpotent. For a general ring this certainly isn't true: the intersection of all prime ideals is the nilpotent elements, but the intersection of all maximal ideals might be bigger (think of a 1-dimensional local ring, for example).

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  • $\begingroup$ Dear Kevin, I think this was implicit in Brian's remark that MaxSpec is not a waste of time, following on from his allusion to classical varieties. Good to spell it out, though! $\endgroup$
    – Emerton
    Commented Apr 19, 2010 at 19:26
  • $\begingroup$ Kevin, there's only so much room in these comment boxes. :) Anyway, Emerton is right, and I'd hoped the OP had at least read section 1.1 of my notes, which mention what you say, to make up for it. But for wider MO purposes, it's good that you put up that post. If I'd ever bothered to register, I'd give you +1. $\endgroup$
    – BCnrd
    Commented Apr 19, 2010 at 19:29
  • $\begingroup$ Fair point Emerton. I just thought it was a bit strange that Brian seemed to be justifying it by saying "it works for varieties, so why shouldn't it work for rigid spaces?", whereas somehow he didn't say the reason it worked for varieties! I think his point that one can make sense of |f(x)| is a good one though, and somehow "independent" of this issue. $\endgroup$
    – Kevin Buzzard
    Commented Apr 19, 2010 at 19:30
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    $\begingroup$ Heh, hi Brian. I'm sure you put it as early as possible in the notes! Yes I appreciate that in your current incarnation you are limited to 500 chars :-) $\endgroup$
    – Kevin Buzzard
    Commented Apr 19, 2010 at 19:30
  • $\begingroup$ Dear Kevin, this answer is also very enlightening. It makes precise the vague idea I had that MaxSpec is a better object in this setting than in algebraic geometry. $\endgroup$
    – user5117
    Commented Apr 20, 2010 at 9:58
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Another point to bear in mind, in addition to those raised by Brian and Kevin, is that generic points (in the sense of non-maximal prime ideals) don't make sense in analytic geomtery.

For example, the Tate algebra $\mathbb Q_p\langle\langle x\rangle \rangle$ contains one non-maximal prime ideal, the zero ideal. Geometrically it corresponds to the closed disk $|x| \leq 1$. Where in this disk would the generic point corresponding to the zero ideal live? The point is that, unlike in algebraic geometry, in rigid analytic geometry one can find disjoint open subsets of irreducible spaces such as the closed disk.

In Berkovich's theory, one does have generic points, but they consist of more data than just a prime ideal; one must also choose a norm on the residue field. (This relates to Brian's comment.) Geometrically, this choice of norm pins down where on the rigid space the generic point lives.

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  • $\begingroup$ By taking norm to be sup-norm one gets the "Gauss point", which is "generic" in many ways. First, upon removing it the Berkovich disc disconnects into disjoint open union of residue discs (of radius 1). Thus, path-connectedness of Berkovich disc rests on that "non-classical" point; paths linking classical points on "boundary" must pass through the Gauss point. Second, under "reduction map" onto affine line over residue field its unique over that generic point (corresponds to Shilov boundary, another "genericity" property). Third, local ring at this point is a field, similar to varieties. $\endgroup$
    – BCnrd
    Commented Apr 19, 2010 at 21:59
  • $\begingroup$ Dear Brian, What I had in mind is that if we, say, include the disk of radius $|p|$ into the unit disk, this does not induce a corresponding map of generic points (although the map on algebras is an injection of one domain in another, and so maps the zero ideal of one to the zero ideal of another). $\endgroup$
    – Emerton
    Commented Apr 20, 2010 at 2:00
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    $\begingroup$ Dear Matt: Yep, I should have made clearer that my comment to your answer was mainly aimed at Artie. To illustrate the consistency of your answer with the stuff in my comment, the non-functoriality of generic points as you point out meshes well with the analogous map on "reductions" being non-dominant and so also not respecting generic points: constant map of affine line over residue field to origin of another affine line over the residue field. $\endgroup$
    – BCnrd
    Commented Apr 20, 2010 at 2:22

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