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

6$\begingroup$ Just a start, suppose you have such a morphism. It's domain has to be nonnoetherian. Since every open subset in a noetherian space is quasicompact. So the natural question to ask is, what are examples of nonnoetherian 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. $\endgroup$– James WeigandtApr 16 '11 at 18:01

$\begingroup$ Yes. I agree. That is the question to ask! $\endgroup$– Andrew StoutApr 16 '11 at 18:14

4$\begingroup$ @REX: could you explain why you're looking for such an example? For what it's worth, one nonnoetherian 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 nonquasicompact open immersion. However, I'd guess this doesn't arise "naturally" in algebraic number theory. $\endgroup$– Anton GeraschenkoApr 16 '11 at 18:28

$\begingroup$ 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. $\endgroup$– Andrew StoutApr 16 '11 at 18:59

3$\begingroup$ How about the adele ring? $\endgroup$– Kevin VentulloApr 16 '11 at 21:00
A typical nonNoetherian 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 nonNoetherian ring that naturally appears in algebraic number theory is the ring of all algebraic integers.

5$\begingroup$ 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 Rmodule M is equivalent to specifying an R_fmodule M_f, a \hat{R}module \hat{M} (where \hat{R} is the fadic 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, quasiprojective (group) schemes, etc. In particular, one can quite often avoid contemplating big rings like \hat{R} (x)_R \hat{R} for making descent arguments. $\endgroup$– BhargavApr 16 '11 at 18:56

1$\begingroup$ 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 $\endgroup$– EmertonApr 16 '11 at 19:10

2$\begingroup$ 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. $\endgroup$ Apr 17 '11 at 7:25

$\begingroup$ 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 $\endgroup$– EmertonApr 17 '11 at 21:10

$\begingroup$ The paper of Ferrand and Raynaud is on Numdam, numdam.org/item?id=ASENS_1970_4_3_3_295_0 $\endgroup$– ACLApr 18 '11 at 12:30
Rex, about your schemetheoretic 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 schemetheoretic 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 $X$.

$\begingroup$ 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 madic 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 nonNoetherian, and the same problem would arise. $\endgroup$ Apr 18 '11 at 13:53

$\begingroup$ @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 $Y$. $\endgroup$ Apr 18 '11 at 19:58

2$\begingroup$ $R$ is a DVR, hence not artinian. $\endgroup$ Apr 19 '11 at 6:36

$\begingroup$ Dear Laurent, I think in your second last line, you mean "But $X$ is not topologically dense in $Y$". Best wishes, Matthew $\endgroup$– EmertonApr 19 '11 at 6:48

2$\begingroup$ OK, but be careful with nonnoetherian rings: it may happen that $\mathfrak{m}^2=\mathfrak{m}\neq0$. $\endgroup$ Apr 19 '11 at 18:33
You might see nonquasicompact 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 quasicompact.
Gerritzen and van der Put wrote a book (Schottky Groups and Mumford Curves, Springer LNM 817) describing some numbertheoretic 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 schemetheoretic image that you posed in the comments.

$\begingroup$ I appreciate the answer anyway. I will have to look into this more. $\endgroup$ Apr 18 '11 at 13:58