It's true that the pushforward of a coherent sheaf is coherent via a proper morphism: but do proper morphisms preserve a finite presentation? Under some assumptions perhaps? Does it change if we are working with algebraic spaces instead of just schemes?
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$\begingroup$ wait a minute: is the answer no, for trivial reasons? Take a non-finitely generated ideal $I$ in a ring $R$, then $R \to R/I$ wants to give a presentation of $R/I$ but it's not a finite presentation as $I$ is not finitely generated. $\endgroup$– Yosemite SamMar 28, 2012 at 17:42
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$\begingroup$ and we know that if $R/I$ admits a finite presentation then the kernel of any surjection $R^n \to R/I$ will be finitely generated. mathoverflow.net/questions/1788/… $\endgroup$– Yosemite SamMar 28, 2012 at 17:43
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Consider a ring $A$ and an ideal $I\subseteq A$, then $A/I$ is finitely presented as an $A/I$-module, but only finitely presented as an $A$-module if $I$ is finitely generated.
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$\begingroup$ ah, good. so my comments made sense, cheers. $\endgroup$ Mar 28, 2012 at 17:47
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$\begingroup$ Dear Y-S: The proper morphism in a-fortiori's example (= yours) is not finitely presented, so it doesn't seem reasonable to have expected an "operation" with such a morphism to preserve finite presentation. A more natural version of your question (to which the examples by you and a-fortiori are not applicable) is whether pushforward of quasi-coherent sheaves along a finitely presented proper morphism preserves finite presentation. Perhaps restate the question with this extra condition on the morphism? A counterexample (or proof!) for this refinement would be very interesting. Regards, q-c $\endgroup$ Mar 29, 2012 at 4:48
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