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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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  • $\begingroup$ 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? $\endgroup$ Jan 21, 2014 at 12:45
  • $\begingroup$ @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). $\endgroup$ Jan 22, 2014 at 9:42
  • $\begingroup$ @OlivierBenoist: Thanks for the explanation. $\endgroup$ Jan 22, 2014 at 12:33

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