Let $S$ be a scheme and let $$0 \to A \to B \to C \to 0$$ be an exact sequence of abelian sheaves on $(\mathrm{Sch}/S)_\text{fppf}$. Assume that $A$ and $C$ are representable by flat algebraic spaces. Why is $B$ representable by an algebraic space?

I've seen this statement several times but never seen a proof. I also wonder whether this result has an analogue for the étale topology.


1 Answer 1


The main fact here is that, since $B$ fits into this short exact sequence, we have that $B$ is an $A$-torsor over $C$. In particular, fppf-locally on $C$, $B$ is a product of $A$ and $C$. But being an algebraic space is a local condition [Stacks Project, Tag 04SK], and we win.

If you have a similar short exact sequence of étale sheaves, then [Stacks Project, Tag 076L] implies that $A$ and $C$ are also "fppf-algebraic spaces". The same idea as before also works here.


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