The Stacks project

Lemma 18.40.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Consider the following conditions

  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 every point $p$ of $\mathcal{C}$ the stalk $\mathcal{O}_ p$ is either the zero ring or a local ring.

We always have (1) $\Rightarrow $ (2). If $\mathcal{C}$ has enough points then (1) and (2) are equivalent.

Proof. Assume (1). Let $p$ be a point of $\mathcal{C}$ given by a functor $u : \mathcal{C} \to \textit{Sets}$. Let $f_ p \in \mathcal{O}_ p$. Since $\mathcal{O}_ p$ is computed by Sites, Equation (7.32.1.1) we may represent $f_ p$ by a triple $(U, x, f)$ where $x \in U(U)$ and $f \in \mathcal{O}(U)$. By assumption there exists a covering $\{ U_ i \to U\} $ such that for each $i$ either $f$ or $1 - f$ is invertible on $U_ i$. Because $u$ defines a point of the site we see that for some $i$ there exists an $x_ i \in u(U_ i)$ which maps to $x \in u(U)$. By the discussion surrounding Sites, Equation (7.32.1.1) we see that $(U, x, f)$ and $(U_ i, x_ i, f|_{U_ i})$ define the same element of $\mathcal{O}_ p$. Hence we conclude that either $f_ p$ or $1 - f_ p$ is invertible. Thus $\mathcal{O}_ p$ is a ring such that for every element $a$ either $a$ or $1 - a$ is invertible. This means that $\mathcal{O}_ p$ is either zero or a local ring, see Algebra, Lemma 10.18.3.

Assume (2) and assume that $\mathcal{C}$ has enough points. Consider the map of sheaves of sets

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

of Lemma 18.40.1 part (3). For any local ring $R$ the corresponding map $(R \times R) \amalg (R \times R) \to R \times R$ is surjective, see for example Algebra, Lemma 10.18.3. Since each $\mathcal{O}_ p$ is a local ring or zero the map is surjective on stalks. Hence, by our assumption that $\mathcal{C}$ has enough points it is surjective and we win. $\square$

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Your email address will not be published.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 04ET. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 18.40.2 as pdf