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We will always work with finite-type, smooth schemes over a field $k$. Let $\pi: Y \to X$ be an etale map of $k$-schemes. Let $Z$ be another $k$-scheme admitting a morphism $f: Y \to Z$. Suppose that there exists a scheme $Y'$ and an etale map $\pi': Y' \to Y$ such that $\pi' \circ f : Y' \to Z$ descends to a map $X \to Z$. Is it necessary that $f$ also descends to a morphism $X \to Z$? If not, can one give me a counterexample?

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1 Answer 1

If I understand correctly, you have a morphism $g:X\rightarrow Z$ such that $g\circ \pi \circ \pi '=f\circ \pi '$, and you ask whether this implies $g\circ \pi =f$. You have to assume $Y$ connected (or at least that $\pi '(Y')$ hits every component of $Y$). Then since an étale map is open, $g\circ \pi $ and $f$ coincide on an dense open subset of $Y$, hence they are equal.

Calling this descent is a bit pompous, by the way...

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Dear abx: The argument via density implicitly uses that $Z$ is separated if $\pi'$ isn't surjective (even if $Y$ is connected), but the hypotheses don't impose separatedness. For the affine line $L$ over $k$, let $Z=Y$ be "$L$ with doubled origin" over $k$, $f={\rm{id}}_Y$, $X=L$, $\pi$ the natural etale surjection, $Y'=L-\{0\}$, $\pi'$ the natural dense open immersion (complementary to the two origins), and $X\rightarrow Z$ one of the two natural dense open immersions (complementary to one of the origins). By the way, the OP's name seems more pompous than anything in the question. :) –  user76758 Dec 31 '13 at 15:41
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Indeed I was assuming that the schemes are separated. –  abx Dec 31 '13 at 15:51

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