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Let $R$ be a discrete valuation ring and $X$ be a regular, integral. projective $R$-scheme, flat over $R$. Let $F$ be a torsion-free coherent sheaf on $X$ of rank $n$, flat over $\mathrm{Spec}(R)$. Is there any known additional conditions on $X$ under which we can conclude that the exterior power $\wedge^n F$ is again flat over $\mathrm{Spec}(R)$?

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  • $\begingroup$ Maybe I'm missing something. Isn't it always flat? $\endgroup$
    – Piotr Achinger
    Commented Mar 5, 2016 at 11:20
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    $\begingroup$ @Ron Flatness of an $R$-integral domain $A$ over a dvr $R$ is automatic when $R\to A$ is injective. $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 5, 2016 at 11:45
  • $\begingroup$ @AriyanJavanpeykar I do not quite follow. Are you saying as well that it is always flat? Could you please elaborate a bit more. $\endgroup$
    – Ron
    Commented Mar 5, 2016 at 11:59
  • $\begingroup$ @PiotrAchinger As in your other answer, here as well the wedge product is not taken as $R$-modules. Do you still this it is always flat? $\endgroup$
    – Ron
    Commented Mar 5, 2016 at 12:00
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    $\begingroup$ Here is an example where $\Lambda^nF$ is not $R$-flat: Assume, say, $\dim X=2$ and let $x$ be a closed point (in the closed fiber). Let $I\subset\mathscr{O}_X$ be the ideal sheaf of $x$, which is torsion-free of rank $1$ on $X$. Observe that $\Lambda^2I$ is nonzero and supported at $x$. Now take $F=I\oplus\mathscr{O}_X$: then $\Lambda^2F$ contains $\Lambda^2I$ as a direct summand, hence it has $R$-torsion. $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 5, 2016 at 14:10

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