The Stacks project

Lemma 18.40.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. The following are equivalent

  1. For every object $U$ of $\mathcal{C}$ and $f \in \mathcal{O}(U)$ there exists a covering $\{ U_ j \to U\} $ such that for each $j$ either $f|_{U_ j}$ is invertible or $(1 - f)|_{U_ j}$ is invertible.

  2. For $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $n \geq 1$, and $f_1, \ldots , f_ n \in \mathcal{O}(U)$ which generate the unit ideal in $\mathcal{O}(U)$ there exists a covering $\{ U_ j \to U\} $ such that for each $j$ there exists an $i$ such that $f_ i|_{U_ j}$ is invertible.

  3. The map of sheaves of sets

    \[ (\mathcal{O} \times \mathcal{O}) \amalg (\mathcal{O} \times \mathcal{O}) \longrightarrow \mathcal{O} \times \mathcal{O} \]

    which maps $(f, a)$ in the first component to $(f, af)$ and $(f, b)$ in the second component to $(f, b(1 - f))$ is surjective.

Proof. It is clear that (2) implies (1). To show that (1) implies (2) we argue by induction on $n$. The first case is $n = 2$ (since $n = 1$ is trivial). In this case we have $a_1f_1 + a_2f_2 = 1$ for some $a_1, a_2 \in \mathcal{O}(U)$. By assumption we can find a covering $\{ U_ j \to U\} $ such that for each $j$ either $a_1f_1|_{U_ j}$ is invertible or $a_2f_2|_{U_ j}$ is invertible. Hence either $f_1|_{U_ j}$ is invertible or $f_2|_{U_ j}$ is invertible as desired. For $n > 2$ we have $a_1f_1 + \ldots + a_ nf_ n = 1$ for some $a_1, \ldots , a_ n \in \mathcal{O}(U)$. By the case $n = 2$ we see that we have some covering $\{ U_ j \to U\} _{j \in J}$ such that for each $j$ either $f_ n|_{U_ j}$ is invertible or $a_1f_1 + \ldots + a_{n - 1}f_{n - 1}|_{U_ j}$ is invertible. Say the first case happens for $j \in J_ n$. Set $J' = J \setminus J_ n$. By induction hypothesis, for each $j \in J'$ we can find a covering $\{ U_{jk} \to U_ j\} _{k \in K_ j}$ such that for each $k \in K_ j$ there exists an $i \in \{ 1, \ldots , n - 1\} $ such that $f_ i|_{U_{jk}}$ is invertible. By the axioms of a site the family of morphisms $\{ U_ j \to U\} _{j \in J_ n} \cup \{ U_{jk} \to U\} _{j \in J', k \in K_ j}$ is a covering which has the desired property.

Assume (1). To see that the map in (3) is surjective, let $(f, c)$ be a section of $\mathcal{O} \times \mathcal{O}$ over $U$. By assumption there exists a covering $\{ U_ j \to U\} $ such that for each $j$ either $f$ or $1 - f$ restricts to an invertible section. In the first case we can take $a = c|_{U_ j} (f|_{U_ j})^{-1}$, and in the second case we can take $b = c|_{U_ j} (1 - f|_{U_ j})^{-1}$. Hence $(f, c)$ is in the image of the map on each of the members. Conversely, assume (3) holds. For any $U$ and $f \in \mathcal{O}(U)$ there exists a covering $\{ U_ j \to U\} $ of $U$ such that the section $(f, 1)|_{U_ j}$ is in the image of the map in (3) on sections over $U_ j$. This means precisely that either $f$ or $1 - f$ restricts to an invertible section over $U_ j$, and we see that (1) holds. $\square$

Comments (0)

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 04ES. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 18.40.1 as pdf