# On relating $l(A), l(B)$ and $l(A+B)$ for Weil divisors on a smooth projective curve where one of the divisors is effective

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

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.

• 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 ? – Louis Jun 21 '20 at 2:00
• @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 $3<g$. – Will Sawin Jun 21 '20 at 2:02