Let $X$ be a smooth projective curve over an Algebraically closed field $k$. Let $k(X)$ denote its function field.

If $A, B$ are Weil divisors on $X$ such that $A$ is effective (i.e. $A\ge 0$) , then is there any (in)equality between $l(A+B)$ and $ l(B), l(A)$ ?

Here, for a Weil divisor $D$ on $X$, by $l(D)$ we denote the $k$-vector space dimension of the Riemann-Roch space $L(D):=\{f\in k(X)^*: D+ div(f)\ge 0\}\cup \{0\}$.

For a divisor $D$ on $X$, the complete linear system $|D|$ be the collection of all effective divisors which are linearly equivalent with $D$. $|D|$ can be given the structure of a projective space by identifying it with $( L(D)\setminus \{0\})/k^*$ and by that structure, $\dim |D|=l(D)-1$. Now it is known (Hartshorne, Chapter IV, Lemma 5.5) that if $D,E$ are both effective divisors, then $\dim |D|+\dim |E|\le \dim |D+E|$ i.e. $l(D)+l(E)\le l(D+E)+1$ . What I'm basically asking is that if something similar holds if we assume only one of the divisors is effective...


1 Answer 1


If $D$ and $E$ are linearly equivalent to effective divisors, this is OK from what's in Hartshorne, as both sides are invariant under linear equivalence.

If $E$, say, is not linearly equivalent to an effective divisor, then $l(E)=0$, so your desired inequality is $l(D) \leq l(D+E) +1$. It is easy to produce counterexamples to this. We can take $E$ to be a very negative divisor, or, alternately, we can take $D$ to be $k$ times the hyperplane class on a hyperelliptic curve of genus $g>2k$ and $E$ to be a generic divisor of degree $0$, which will give $l(D) = k+1$ and $l(D+E)=0$. One can even make examples with $E$ of positive degree.

  • $\begingroup$ Agh yes of-course I should have added that both my divisors have positive degree (otherwise, as you say, it's very easy to get counterexamples) ... what would be a counterexample with a divisor of positive degree ? $\endgroup$
    – Louis
    Jun 21, 2020 at 2:00
  • $\begingroup$ @Louis The simplest example would be $D$ the hyperelliptic divisor on a hyperelliptic curve of genus $>3$, so that $l(D)=2$, and $E$ a generic divisor of degree $1$, so that $D+E$ is a generic divisor of degree $3$, which means $l(D+E)=0$ since $3<g$. $\endgroup$
    – Will Sawin
    Jun 21, 2020 at 2:02

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