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$\newcommand{\F}{\mathcal{F}}$ $\newcommand{\ox}{\mathcal{O}_X}$ Let $f:X \to \operatorname{Spec}(k)$ be a projective scheme of dimension one over a field $k$. The Riemann-Roch equation for such curves is given, for instance in Liu's book (Thm. 7.3.26), and it states

$$ h^0(\F) - h^1(\F) = \deg_k \F + \chi(\ox)$$

for $\F$ an invertible $\ox$-module. Moreover, we have $h^1(\F) = h^0(\omega_f \otimes_{\ox} \F^{-1})$ where $\omega_f$ denotes the $1$-dualizing sheaf of $f$.

I am looking for a reference which states the following:

In the above situation, we have $$ h^0(\F) = \deg_k \F + \chi(\ox)$$ whenever $\deg_k \F$ is greater than a bound solely depending on $X$, for instance a constant multiple of $\chi(\ox)$.

Liu does give such a statement in his book, but he needs to assume that $X$ is a local complete intersection and, moreover, integral. But I wonder whether this is true in a more general setting.

Edit If $X$ is Gorenstein, this is easily seen to be true. In particular, I am interested in the case that $X$ is not Gorenstein.

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  • $\begingroup$ If your curve is not irreducible, you are in trouble — Gorenstein or not. $\endgroup$
    – abx
    Commented Sep 14, 2018 at 15:42
  • $\begingroup$ @abx Could you please elaborate why this is the case? I am grateful for any kind of insight. $\endgroup$
    – windsheaf
    Commented Sep 14, 2018 at 15:51
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    $\begingroup$ Suppose for instance that your curve is the union of 2 smooth components $A$ and $B$, meeting transversally at a point $p$. Take a line bundle of negative degree on $A$, and very large degree on $B$, so that the total degree is as large as you want. Then there will be nonzero sections in $H^0(A,\omega_A(p) \otimes \mathcal{F}^{-1})$ vanishing at $p$. Just extend these sections by $0$ on $B$. $\endgroup$
    – abx
    Commented Sep 14, 2018 at 17:36
  • $\begingroup$ @abx Thank you, that makes sense! So what if we assume $\mathcal{F}$ to be ample? Then what you've mentioned cannot happen anymore. $\endgroup$
    – windsheaf
    Commented Sep 17, 2018 at 14:57
  • $\begingroup$ Yes it can — just take a line bundle of degree 1 on $A$, and very large on $B$, and do the same construction. $\endgroup$
    – abx
    Commented Sep 17, 2018 at 15:38

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