Lemma 110.80.1. There exists a ring A and an infinite family of flat ring maps \{ A \to A_ i\} _{i \in I} such that for every A-module M
but there is no finite subfamily where the same thing is true.
Of course the title of this section doesn't refer to the spectrum of the integers as a topological space, because any spectrum is quasi-compact as a topological space (Algebra, Lemma 10.17.8). No, it refers to the spectrum of the integers in the canonical topology on the category of schemes, and the definition of a quasi-compact object in a site (Sites, Definition 7.17.1).
Let U be a nonprincipal ultrafilter on the set P of prime numbers. For a subset T \subset P we denote T^ c = P \setminus T the complement. For A \in U let S_ A \subset \mathbf{Z} be the multiplicative subset generated by p \in A. Set
Observe that \mathop{\mathrm{Spec}}(\mathbf{Z}_ A) = \{ (0)\} \cup A^ c \subset \mathop{\mathrm{Spec}}(\mathbf{Z}) if we think of P as the set of closed points of \mathop{\mathrm{Spec}}(\mathbf{Z}). If A, B \in U, then A \cap B \in U and A \cup B \in U and we have
(fibre product of rings). In particular, for any integer n and elements A_1, \ldots , A_ n \in U the morphisms
factors through \mathop{\mathrm{Spec}}(\mathbf{Z}[1/p]) for some p (namely for any p \in A_1 \cap \ldots \cap A_ n). We conclude that the family of flat morphisms \{ \mathop{\mathrm{Spec}}(\mathbf{Z}_ A) \to \mathop{\mathrm{Spec}}(\mathbf{Z})\} _{A \in U} is jointly surjective, but no finite subset is.
For a \mathbf{Z}-module M we set
Claim I: for every \mathbf{Z}-module M we have
First, assume M is torsion free. Then M_ A \subset M_ P for all A \in U. Hence we see that we have to prove
Namely, since U is nonprincipal, for any prime p we have \{ p\} ^ c \in U. Also, M_{\{ p\} ^ c} = M_{(p)} is equal to the localization at the prime (p). Thus the above is clear because already M_{(2)} \cap M_{(3)} = M. Next, assume M is torsion. Then we have
and correspondingly we have
because we are localizing at the primes in A. Suppose that (x_ A) \in \prod M_ A is in the equalizer. Denote x_ p = x_{\{ p\} ^ c} \in M[p^{\infty }]. Then the equalizer property says
and in particular it says that x_ p is zero for all but a finite number of p \not\in A. To finish the proof in the torsion case it suffices to show that x_ p is zero for all but a finite number of primes p. If not write \{ p \in P \mid x_ p \not= 0\} = T \amalg T' as the disjoint union of two infinite sets. Then either T \not\in U or T' \not\in U because U is an ultrafilter (namely if both T, T' are in U then U contains T \cap T' = \emptyset which is not allowed). Say T \not\in U. Then T = A^ c and this contradicts the finiteness mentioned above. Finally, suppose that M is a general module. Then we look at the short exact sequence
and we look at the following large diagram
Doing a diagram chase using exactness of the columns and the result for the torsion module M_{tors} and the torsion free module M/M_{tors} proving Claim I for M. This gives an example of the phenomenon in the following lemma.
Lemma 110.80.1. There exists a ring A and an infinite family of flat ring maps \{ A \to A_ i\} _{i \in I} such that for every A-module M
but there is no finite subfamily where the same thing is true.
Proof. See discussion above. \square
We continue working with our nonprincipal ultrafilter U on the set P of prime numbers. Let R be a ring. Denote R_ A = S_ A^{-1}R = R \otimes \mathbf{Z}_ A for A \in U. Claim II: given closed subsets T_ A \subset \mathop{\mathrm{Spec}}(R_ A), A \in U such that
for all A, B \in U, there is a closed subset T \subset \mathop{\mathrm{Spec}}(R) with T_ A = (\mathop{\mathrm{Spec}}(R_ A) \to \mathop{\mathrm{Spec}}(R))^{-1}(T) for all A \in U. Let I_ A \subset R_ A for A \in U be the radical ideal cutting out T_ A. Then the glueing condition implies S_{A \cup B}^{-1}I_ A = S_{A \cup B}^{-1}I_ B in R_{A \cup B} for all A, B \in U (because localization preserves being a radical ideal). Let I' \subset R be the set of elements mapping into I_ P \subset R_ P = R \otimes \mathbf{Q}. Then we see for A \in U that
I_ A \subset I'_ A = S_ A^{-1}I', and
M_ A = I'_ A/I_ A is a torsion module.
Of course we obtain canonical identifications S_{A \cup B}^{-1}M_ A = S_{A \cup B}^{-1}M_ B for A, B \in U. Decomposing the torsion modules M_ A into their p-primary components, the reader easily shows that there exist p-power torsion R-modules M_ p such that
compatible with the canonical identifications given above. Setting M = \bigoplus _{p \in P} M_ p we find canonical isomorphisms M_ A = S_ A^{-1}M compatible with the above canonical identifications. Then we get a canonical map
of R-modules which recovers the map I_ A \to M_ A for all A \in U. This is true by all the compatibilities mentioned above and the claim proved previously that M is the equalizer of the two maps from \prod _{A \in U} M_ A to \prod _{A, B \in U} M_{A \cup B}. Let I = \mathop{\mathrm{Ker}}(I' \to M). Then I is an ideal and T = V(I) is a closed subset which recovers the closed subsets T_ A for all A \in U. This proves Claim II.
Lemma 110.80.2. The scheme \mathop{\mathrm{Spec}}(\mathbf{Z}) is not quasi-compact in the canonical topology on the category of schemes.
Proof. With notation as above consider the family of morphisms
By Descent, Lemma 35.13.5 and the two claims proved above this is a universal effective epimorphism. In any category with fibre products, the universal effective epimorphisms give \mathcal{C} the structure of a site (modulo some set theoretical issues which are easy to fix) defining the canonical topology. Thus \mathcal{W} is a covering for the canonical topology. On the other hand, we have seen above that any finite subfamily
factors through \mathop{\mathrm{Spec}}(\mathbf{Z}[1/p]) for some p. Hence this finite family cannot be a universal effective epimorphism and more generally no universal effective epimorphism \{ g_ j : T_ j \to \mathop{\mathrm{Spec}}(\mathbf{Z})\} can refine \{ \mathop{\mathrm{Spec}}(\mathbf{Z}_{A_ i}) \to \mathop{\mathrm{Spec}}(\mathbf{Z})\} _{i = 1, \ldots , n}. By Sites, Definition 7.17.1 this means that \mathop{\mathrm{Spec}}(\mathbf{Z}) is not quasi-compact in the canonical topology. To see that our notion of quasi-compactness agrees with the usual topos theoretic definition, see Sites, Lemma 7.17.3. \square
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Comment #3569 by David Roberts on
Comment #3693 by Johan on