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Can someone give me an example of a non-quasi-compact morphism of schemes which arises naturally in the field of Algebraic Number Theory?

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Just a start, suppose you have such a morphism. It's domain has to be non-noetherian. Since every open subset in a noetherian space is quasi-compact. So the natural question to ask is, what are examples of non-noetherian schemes that arise naturally in algebraic number theory? I cannot think of any offhand, since usually in algebraic number theory, the focus is on Dedekind domains which are noetherian. – Jamie Weigandt Apr 16 '11 at 18:01
Yes. I agree. That is the question to ask! – Andrew Stout Apr 16 '11 at 18:14
@REX: could you explain why you're looking for such an example? For what it's worth, one non-noetherian scheme that algebraic number theorists might think about is $Spec(\overline{\mathbb Q})\otimes_{Spec(\mathbb Q)}Spec(\overline{\mathbb Q})$, the absolute galois group of $\mathbb Q$. The inclusion of the complement of the identity (diagonal) $\overline{\mathbb Q}$-point is a non-quasi-compact open immersion. However, I'd guess this doesn't arise "naturally" in algebraic number theory. – Anton Geraschenko Apr 16 '11 at 18:28
Thanks for your response. I am looking for a "nice" example of a morphism of schemes $f: X\rightarrow Y$ such that the scheme theoretic image Z is not equal to $\overline{f(X)}$, which also comes along with a nice picture. All the examples I have seen so far seem "contrived." Of course, I may have to adjoin a nilpotent to some of these examples, but that's not a big deal. – Andrew Stout Apr 16 '11 at 18:59
How about the adele ring? – Kevin Ventullo Apr 16 '11 at 21:00
up vote 10 down vote accepted

A typical non-Noetherian ring that would arise in algebraic number theory would be the ring $\mathbb Z_p \otimes_{\mathbb Z_{(p)}} \mathbb Z_p$, where I am writing $\mathbb Z_{(p)}$ to denote the localization of $\mathbb Z$ at the prime ideal $(p)$, and $\mathbb Z_p$ to denote its completion (the usual ring of $p$-adic integers).

Such tensor products come up in considerations related to faithfully flat descent, and can arise for example in justfiying (in certain situations) the passage from working over the localization $\mathbb Z_{(p)}$ to its completion $\mathbb Z_p$. (The kind of thing I have in mind is studying finite flat group schemes over $\mathbb Z$, say, by working over $\mathbb Z[1/p]$ and $\mathbb Z_p$ separately. It is not hard to justify working over $\mathbb Z[1/p]$ and $\mathbb Z_{(p)}$ separately, but to justify the replacement of $\mathbb Z_{(p)}$ by $\mathbb Z_p$, one needs to make (or at least, might naturally find oneself making) a descent argument, in which the tensor product written above could play a role.)

Another (perhaps simpler) example of a non-Noetherian ring that naturally appears in algebraic number theory is the ring of all algebraic integers.

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Just want to point out that M. Artin has a wonderful theorem ("formal glueing of module categories") that says: given a ring R and an element f, specifying an R-module M is equivalent to specifying an R_f-module M_f, a \hat{R}-module \hat{M} (where \hat{R} is the f-adic completion of R), and an isomorphism between M_f and \hat{M} over \hat{R}_f; this procedure preserves tensor products, and so passes to algebras, quasi-projective (group) schemes, etc. In particular, one can quite often avoid contemplating big rings like \hat{R} (x)_R \hat{R} for making descent arguments. – Bhargav Apr 16 '11 at 18:56
Dear Bhargav, Thanks for this comment. It was because of these sorts of results that I put in my weasely parenthetical remarks "or at least ..."! Best wishes, Matt – Emerton Apr 16 '11 at 19:10
About this "formal glueing" theorem: it was also proved by Ferrand and Raynaud (appendix to "Fibres formelles..., Ann. Sci. ENS 1970). Artin's version is in "Algebraization of formal moduli II" (also 1970) and, if I remember correctly, Artin attributes it to Grothendieck. In any case the result is not very hard to prove, and would deserve to be better known. – Laurent Moret-Bailly Apr 17 '11 at 7:25
Dear Laurent, Thanks for adding these details, and for the additional references. The reference I knew for this was a paper of Beauville (perhaps with a coauthor?), and I'm glad to learn of some other (earlier) sources. Best wishes, Matthew – Emerton Apr 17 '11 at 21:10
The paper of Ferrand and Raynaud is on Numdam, – ACL Apr 18 '11 at 12:30

Rex, about your scheme-theoretic image problem, I don't know how "contrived" the following example is (I am afraid it is not particularly related to number theory).
Notations: $R$ is a discrete valuation ring, $t$ a uniformizer, $R_n:=R/(t^{n+1})$ ($n\in\mathbb{N}$), $X_n=\mathrm{Spec}\,R_n$, $A=\prod_n R_n$.
Take $X:=\coprod_n X_n$ and $Y:=\mathrm{Spec}\,A$. There is a natural open immersion $f:X\to Y$ since each $X_n$ embeds in $Y$ as an open and closed subscheme.
The scheme-theoretic image of $f$ is $Y$: since $Y$ is affine, it just means that each $x\in A$ vanishing on each $X_n$ is zero, which is obvious. (In fact, $A=\Gamma(X,\mathcal{O}_X)$).
But $X$ is not topologically dense in $Y$: indeed, consider $x=(t,t,\dots)\in A$. Then $x$ is locally nilpotent on $X$ but not nilpotent on $Y$, hence the open set $D(x)\subset Y$ is nonempty and disjoint from $X$.

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I really like this example. Again, $A$ is not Noetherian which prevents us from applying Krull's Intersection Theorem. You have exhibited elements of A contained in $inf(A) = \cap_{n=1}^{\infty} A/\mathfrak{m}^n$. It seems that these are exactly the elements in the topological closure of Y. It is nice to see the interaction between the m-adic topology of A and the zariski topology on X. I might add, now that you have given this great example, that if I choose an ultrafilter on $\mathbb{N}$, Then ultraproduct $B = A/\sim$ would still be non-Noetherian, and the same problem would arise. – Andrew Stout Apr 18 '11 at 13:53
@Laurent Actually, something needs to add to your set up. We need to make R is not artinian; or else, x will be nilpotent on $Y$. – Andrew Stout Apr 18 '11 at 19:58
$R$ is a DVR, hence not artinian. – Laurent Moret-Bailly Apr 19 '11 at 6:36
Dear Laurent, I think in your second last line, you mean "But $X$ is not topologically dense in $Y$". Best wishes, Matthew – Emerton Apr 19 '11 at 6:48
OK, but be careful with non-noetherian rings: it may happen that $\mathfrak{m}^2=\mathfrak{m}\neq0$. – Laurent Moret-Bailly Apr 19 '11 at 18:33

You might see non-quasi-compact maps in the context of universal covers of maximally degenerate pointed curves, and depending on who you ask, this might be called algebraic number theory. Specifically, a maximally degenerate positive genus (proper) curve $X$ has the form of a connected graph made out of finitely many projective lines intersecting transversely, where each line has exactly 3 special points (namely intersections and markings). If $x$ is a marked point, then $\pi_1^{geom}(X,x)$ is a finitely generated free group. One then has a universal cover $(\tilde{X},\tilde{x}) \to (X,x)$, where $\tilde{X}$ is a tree of projective lines. The covering map is étale but not quasi-compact.

Gerritzen and van der Put wrote a book (Schottky Groups and Mumford Curves, Springer LNM 817) describing some number-theoretic data like theta functions on these objects. Some brief web searching suggests that there seem to be some more modern treatments using rigid analytic techniques (that I don't really understand).

I'm afraid this doesn't answer the more focused question about scheme-theoretic image that you posed in the comments.

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I appreciate the answer anyway. I will have to look into this more. – Andrew Stout Apr 18 '11 at 13:58

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