The Stacks project

110.79 The spectrum of the integers is not quasi-compact

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 10.17.10). 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.

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

1. $I_ A \subset I'_ A = S_ A^{-1}I'$, and

2. $M_ A = I'_ A/I_ A$ 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 wich 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.