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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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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. –  Yosemite Sam Mar 28 '12 at 17:42
    
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/… –  Yosemite Sam Mar 28 '12 at 17:43
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up vote 1 down vote accepted

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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ah, good. so my comments made sense, cheers. –  Yosemite Sam Mar 28 '12 at 17:47
    
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 –  user22479 Mar 29 '12 at 4:48
    
I see, good point. –  Yosemite Sam Mar 29 '12 at 11:16
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