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Let $i : Z \to X$ be a closed immersion of schemes. Is $i_* : Ab((Sch/Z)_{fppf}) \to Ab((Sch/X)_{fppf})$ an exact functor?

The answer is yes in the \'etale or syntomic topology. It seems likely the answer is no in the fppf case however.

In terms of algebra, the problem is related to the following question. Suppose that $A$ is a ring and $I$ is an ideal. Let $A/I \to C$ be a faithfully flat, finitely presented ring map. Is there a flat, finitely presented ring map $A \to B$ with $B/IB$ nonzero such that $A/I \to B/IB$ factors through $C$?

In fact, if one finds an example $(A, I, C)$ for which there does not exist any $B$, then one has a counter example to exactness of $i_*$.

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Just because it is the first thing that occurs to me, what happens if $A$ is an algebra over $\mathbb{Z}_{p\mathbb{Z}}$, $I$ equals $pA$, $C$ equals $A/pA$, and the map $A/I\to C$ is the Frobenius map $A/pA \to A/pA$. Does $B$ always exist in that case? – Jason Starr Jan 21 '14 at 12:45
@JasonStarr : For the Frobenius map to be flat, you need $A/pA$ to be regular, by Kunz's regularity criterion. But then, the Frobenius $A/pA\to A/pA$ is lci, and it should be easy to extend it as a flat lci morphism to an $A$-algebra $B$ (this is question_bot's remark about syntomic topology). – Olivier Benoist Jan 22 '14 at 9:42
@OlivierBenoist: Thanks for the explanation. – Jason Starr Jan 22 '14 at 12:33

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