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 $k \subset k_ I = k(\{ t_ i\} _{i \in I})$. 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. $\square$

## Comments (0)

There are also:

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

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BBK. Beware of the difference between the letter 'O' and the digit '0'.