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 ( 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 ( 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. $\square$

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