In Hakim's book "Topos annelés et schémas relatifs", Chap. III, Def. 2.3 states that a ringed topos $(X,A)$ is a locally ringed topos when two equivalent conditions are satisfied:


(i) For each $U \in X$ and each section $s \in A(U)$ one has $U = U\_s \cup U\_{1-s}$.


(ii) For each $U \in X$ and each family $(s\_i)\_{i \in I}$ in $A(U)$ generating the unit ideal one has $U = \bigcup\_{i \in I} U\_{s\_i}$.


Here $ U\_s \subseteq U$ is the largest subobject on which $s$ is invertible.


But I don't think that (i) is equivalent to (ii), since (i) is satisfied for $A=0$, right? Notice that (ii) implies that $A(U)=0 \implies U=0$ (take $I=\emptyset$, cf. [MO/45951](https://mathoverflow.net/questions/45951)), which I would expect from a local ring object (see also [here](https://ncatlab.org/nlab/show/too+simple+to+be+simple)).


Am I missing something?



Hakim's definition of a locally ringed topos
Agreed. $(ii)$ is equivalent to "$(i)$ and $( 0 = 1$ in $A(U) ) \Rightarrow U= \emptyset$".


*Remark:* I haven't looked at Hakim's convention, by $U= \emptyset$ I mean "as a sheaf", that is, if we are talking about an object of a site and not an object of a topos it means that the empty sieve is a covering of $U$.


This corresponds to the fact that (constructively) a Ring is local if either:


$(i)$ $0 \neq 1$ and $\forall s \in A(U)$, either $s$ or $1-s$ is invertible.


$(ii)$ If $\sum s\_i =1$ then $\exists i$ such that $s\_i$ is invertible.


The interpretration of these claim ni the sheaf semantics corresponds exactly to the proposition given in Hakim's these, except the missing condition $0 \neq 1$ which interprets as $( 0 = 1$ in $A(U) ) \Rightarrow U= \emptyset$.


One can also say (but I don't think that it is the intended meaning) that $(i)$ (without $0 \neq 1$) is equivalent to $(ii)$ restricted to familly with at least one elements.


