7
$\begingroup$

Affine schemes are simply the Zariski spectra of commutative rings, and commutative rings occurs as models of a first-order theory.

I would guess that general schemes do not naturally correspond to the models of any first-order theory and would like to know of any theorems that formalize this.

I would also like to know of interesting first-order theories that do capture more general schemes than merely the affine ones.

Even more interesting to me: first order theories that approximate/generalize scheme theory by allowing unintended models. For example, variables in the theory might range over partial sections of the structure sheaf, one relation might indicate when one section extends another, and another when two sections have the same domain. One would then formulate addition and multiplication as relations (with the various sections all required to have the same domain.) As above, I would guess that "locally affine" resists a first-order formulation in this language. Still one can consider the first-order theory of the models that have the form of schemes, then ask about systems of axioms for such a theory and what the non-scheme models look like.

Improve this question
$\endgroup$
3
  • $\begingroup$ This is true if and only if schemes are closed under ultraproducts. Googling "ultraproducts of schemes" produces some positive results so it appears that this might be possible. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 23, 2010 at 16:59
  • $\begingroup$ Someone who you might want to talk to is Hans Schoutens: websupport1.citytech.cuny.edu/Faculty/hschoutens/index.html He does a lot of work bordering on algebraic geometry and logic. $\endgroup$
    – Steven Gubkin
    Commented Nov 23, 2010 at 17:01
  • $\begingroup$ @Qiaochu Yuan Right, but ultraproducts do not commute with passing to the Zariski spectrum. So what one even means by "ultraproducts of schemes" may well depend upon what one takes as the primitives in one's first-order theory. Indeed seems to me that a geometer would like to work geometrically and take (even in the affine case) ultraproducts of spectra, not rings, even if the result doesn't have the form of the spectrum of a ring. But then what form does it generally take? $\endgroup$
    – David Feldman
    Commented Nov 23, 2010 at 19:57

1 Answer 1

Reset to default
2
$\begingroup$

One kind of answer is as follows. A commutative ring $R$ is local iff for all $a\in R$, either $a$ is invertible or $1-a$ is invertible. The logical form of this statement is obviously much simpler than the usual definition in terms of maximal ideals, so one can easily interpret it in a wide range of categories. The pair (sheaves on spec($R$), structure sheaf) is in some sense the free locally-ringed topos generated by the ringed topos (sets,$R$). If I remember correctly, this is explained in more detail in Johnstone's book on Stone Spaces. (He probably talks about it in his topos theory books as well.) More generally, the structure sheaf on a scheme is always a local ring object in the sense just discussed. Concretely, this means that for any section $a$ of the structure sheaf over an open set $U$ we can write $U=V\cup W$ with $V$ and $W$ open, such that $a|_V$ and $(1-a)|_W$ are invertible.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Just so I understand, you mean this to speak to my final paragraph? One can't first-order capture "closure under arbitrary unions," so one can't presuppose first-order topology (even with the open sets rather than the points as primitives). $\endgroup$
    – David Feldman
    Commented Nov 23, 2010 at 19:58
  • $\begingroup$ I don't really want to talk explicitly about open sets at all. Given a topos $\mathcal{E}$ and a sentence $\phi$ in a kind of first order language that refers to objects and morphisms of $\mathcal{E}$, there is a standard system of semantics that gives a subobject $[\phi]$ of the terminal object (or in other words, a truth value). If $\mathcal{E}$ is the category of sheaves on a scheme and $R$ is the structure sheaf and $\phi$ is the sentence $\forall a\in R (\exists b\in R ab=1)\vee (\exists b\in R (1-a)b=1)$ then $[\phi]$ is the whole terminal object (so $\phi$ is 'true'). $\endgroup$
    – Neil Strickland
    Commented Nov 23, 2010 at 20:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .