Lemma 34.9.14. Let $R$ be a nonzero ring. There does not exist a set $A$ of fpqc-coverings of $\mathop{\mathrm{Spec}}(R)$ such that every fpqc-covering can be refined by an element of $A$.

Proof. Let us first explain this when $R = k$ is a field. For any set $I$ consider the purely transcendental field extension $k_ I = k(\{ t_ i\} _{i \in I})/k$. Since $k \to k_ I$ is faithfully flat we see that $\{ \mathop{\mathrm{Spec}}(k_ I) \to \mathop{\mathrm{Spec}}(k)\}$ is an fpqc covering. Let $A$ be a set and for each $\alpha \in A$ let $\mathcal{U}_\alpha = \{ S_{\alpha , j} \to \mathop{\mathrm{Spec}}(k)\} _{j \in J_\alpha }$ be an fpqc covering. If $\mathcal{U}_\alpha$ refines $\{ \mathop{\mathrm{Spec}}(k_ I) \to \mathop{\mathrm{Spec}}(k)\}$ then the morphisms $S_{\alpha , j} \to \mathop{\mathrm{Spec}}(k)$ factor through $\mathop{\mathrm{Spec}}(k_ I)$. Since $\mathcal{U}_\alpha$ is a covering, at least some $S_{\alpha , j}$ is nonempty. Pick a point $s \in S_{\alpha , j}$. Since we have the factorization $S_{\alpha , j} \to \mathop{\mathrm{Spec}}(k_ I) \to \mathop{\mathrm{Spec}}(k)$ we obtain a homomorphism of fields $k_ I \to \kappa (s)$. In particular, we see that the cardinality of $\kappa (s)$ is at least the cardinality of $I$. Thus if we take $I$ to be a set of cardinality bigger than the cardinalities of the residue fields of all the schemes $S_{\alpha , j}$, then such a factorization does not exist and the lemma holds for $R = k$.

General case. Since $R$ is nonzero it has a maximal prime ideal $\mathfrak m$ with residue field $\kappa$. Let $I$ be a set and consider $R_ I = S_ I^{-1} R[\{ t_ i\} _{i \in I}]$ where $S_ I \subset R[\{ t_ i\} _{i \in I}]$ is the multiplicative subset of $f \in R[\{ t_ i\} _{i \in I}]$ such that $f$ maps to a nonzero element of $R/\mathfrak p[\{ t_ i\} _{i \in I}]$ for all primes $\mathfrak p$ of $R$. Then $R_ I$ is a faithfully flat $R$-algebra and $\{ \mathop{\mathrm{Spec}}(R_ I) \to \mathop{\mathrm{Spec}}(R)\}$ is an fpqc covering. We leave it as an exercise to the reader to show that $R_ I \otimes _ R \kappa \cong \kappa (\{ t_ i\} _{i \in I}) = \kappa _ I$ with notation as above (hint: use that $R \to \kappa$ is surjective and that any $f \in R[\{ t_ i\} _{i \in I}]$ one of whose monomials occurs with coefficient $1$ is an element of $S_ I$). Let $A$ be a set and for each $\alpha \in A$ let $\mathcal{U}_\alpha = \{ S_{\alpha , j} \to \mathop{\mathrm{Spec}}(R)\} _{j \in J_\alpha }$ be an fpqc covering. If $\mathcal{U}_\alpha$ refines $\{ \mathop{\mathrm{Spec}}(R_ I) \to \mathop{\mathrm{Spec}}(R)\}$, then by base change we conclude that $\{ S_{\alpha , j} \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(\kappa ) \to \mathop{\mathrm{Spec}}(\kappa )\}$ refines $\{ \mathop{\mathrm{Spec}}(\kappa _ I) \to \mathop{\mathrm{Spec}}(\kappa )\}$. Hence by the result of the previous paragraph, there exists an $I$ such that this is not the case and the lemma is proved. $\square$

There are also:

• 7 comment(s) on Section 34.9: The fpqc topology

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).