Ariyan Javanpeykar said [here](https://mathoverflow.net/questions/308474/atlas-of-gerbe-over-stack?noredirect=1#comment789174_308474) in comments that,



> 
> $X\times\_{\mathcal{X}}X$ being a scheme is equivalent to representability of $X\rightarrow \mathcal{X}$.
> 
> 
> 


Context is as in [this question](https://mathoverflow.net/questions/316484/diagonal-is-representable-then-any-morphism-is-representable?noredirect=1&lq=1).


Suppose $p:X\rightarrow \mathcal{X}$ is representable, then, for any scheme $M$ with a map of stacks $M\rightarrow \mathcal{X}$, the $2$-fiber product $X\times\_{\mathcal{X}}M$ is a scheme. In particular, we can take $M=X$ then, $X\times\_{\mathcal{X}}X$ is a scheme. Thus, $p:X\rightarrow \mathcal{X}$ is representable implies $X\times\_{\mathcal{X}}X$ is a scheme.


Suppose it is given that $X\times\_{\mathcal{X}}X$ is a scheme. Then, I want to see that $X\rightarrow \mathcal{X}$ is representable i.e., hen, for any scheme $M$ with a map of stacks $M\rightarrow \mathcal{X}$, the $2$-fiber product $X\times\_{\mathcal{X}}M$ is a scheme. I wanted to write something like $X\times\_{\mathcal{X}}M=(X\times\_{\mathcal{X}}X)\times\_X M$. But it does not make sense as there is no map $M\rightarrow X$.


any comments on how to see this are welcome.



$2$-fiber product is a scheme then map of stacks is representable
This is not true even if $\mathcal X$ is an Artin stack. For example, let $G$ be a smooth group scheme over the base $T$, and let $\mathbf BG$ be its classifying stack (the category of $G$-torsors fibered over $Sch/T$). Then $T\times\_{\mathbf BG}T=G$ is a scheme. However, the atlas $T\to \mathbf BG$ is not representable by schemes in general. An example where $G$ is an elliptic curve over a normal local scheme $T$ of dimension 2 can be found in the article *[Ineffective descent of genus one curves](https://arxiv.org/abs/1501.04304)* by Wouter Zomervrucht (arXiv:1501.04304).


What is true is that if $X$ is an algebraic space, $f:X\to \mathcal X$ is an effective epimorphism of fppf sheaves of groupoids on $Sch/T$, and $X\times\_{\mathcal X}X$ is an algebraic space, then $f$ is representable by algebraic spaces. Indeed if $U$ is any algebraic space and $U\to\mathcal X$ any map, then $U\times\_{\mathcal X}X\to U$ is representable by algebraic spaces fppf-locally on $U$ (namely after base change to $U\times\_{\mathcal X}X$, where it becomes a pullback of $X\times\_{\mathcal X}X\to X$), hence it is representable by algebraic spaces because algebraic spaces satisfy fppf descent (which is a somewhat nontrivial fact, see <https://stacks.math.columbia.edu/tag/04SJ>).


