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Let $f:X \to Y$ be a proper, flat morphism of noetherian schemes. Let $\mathcal{F}$ be a coherent sheaf on $X$ non-zero on every fibers of $f$. Is it true that $\mathcal{F}^{\vee \vee}$ is going to be flat over $Y$? If not true in general is there any condition on $f$, under which this can hold true?

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    $\begingroup$ First, an example. Take $X=Y$ smooth, $f=Id$. Then, flatness means, the double dual should be locally free. Examples abound where this is false. For your second part, unless you suggest what kind of conditions on $f$ are you looking for, difficult to answer. $\endgroup$ – Mohan Mar 9 '16 at 19:13
  • $\begingroup$ For any reasonable Noetherian scheme , double dual of a coherent sheaf is coherent. $\endgroup$ – Mohan Mar 9 '16 at 19:19
  • $\begingroup$ If $X=Y$ as in my example, $F$ is coherent on $Y$ since it is so on $X$. $\endgroup$ – Mohan Mar 9 '16 at 19:20
  • $\begingroup$ @Mohan Sorry. I have removed my previous comments. The original question that I had was: I took a coherent pure sheaf $\mathcal{F}$ on $X$ flat over $Y$ of rank $n$. I wanted to ask if $(\wedge^n \mathcal{F})^{\vee \vee}$ is flat over $Y$. In my cases, $X$ is of the form $Y \times X'$ where $X'$ is integral, normal of dimension at least $1$. $\endgroup$ – Ron Mar 9 '16 at 19:25

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