Let $\pi:X\rightarrow S$ be a smooth scheme over $S$ and let's assume for simplicity that $S=\mathrm{Spec} R$ is affine, Noetherian and regular. Let $\mathcal F$ be a coherent sheaf on $X$ and let's assume as well that $\mathcal F$ is torsion-free. Then I can take the derived global sections $R^\bullet\pi_* \mathcal F$, taking further its cohomology I get some $R$-modules $R^i\pi_*\mathcal F$.

Now the question is:

Given $R^i\pi_* \mathcal F$ can I say anything about torsion in it (as in an $R$-module)? In particular is there an open dense $U\subset S$ such that $R^i\pi_*\mathcal F$ is torsion-free restricted to $U$?

If $\pi$ is proper this is automatic since then $R^i\pi_* \mathcal F$ is finitely generated over $R$. If $\pi$ is affine then $R^\bullet\pi_* \mathcal F=\pi_*\mathcal F$ and we just consider some torsion-free module $M$ over $R'$ as a module over $R$ under an injection $R\subset R'$, so the statement holds as well. For the general case I tried to use a Cech cover of $X$ by affine $S$-schemes, but what you get is a complex whose terms are infinitely generated $R$-modules (though torsion-free) and I'm not sure what I could use to bound the torsion in the cohomology there.

Also, not sure if this makes life any simplier, but under the assumptions above $\mathcal F$ is a perfect complex on $X$ and it would be enough to take $\mathcal F=\mathcal O_X$.


1 Answer 1


No, this isn't true even when $S = \text{Spec}(\mathbf{Z})$ because of an example of Anurag Singh. Set $$ R = \mathbf{Z}[X, Y, Z, U, V, W]/(XU + YV + ZW) $$ and let $\mathfrak a \subset R$ be the ideal generated by $X, Y, Z$ in $R$. Set $$ X = \text{Spec}(R) \setminus V(\mathfrak a) $$ Then $H^2(X, \mathcal{O}_X) = H^3_\mathfrak a(R)$ has a $p$-torsion element for infinitely many primes $p$ (see link given above).

  • $\begingroup$ By the way, do you know if there is still a counterexamle for $S$ of characteristic 0? $\endgroup$
    – user42024
    Aug 31, 2019 at 19:01

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