I'm studying Fulton's algebraic curves book and I have the following question:

Clifford's theorem says that if $D$ is a divisor and $W$ is a canonical divisor with $l(D)\gt 0$ and $l(W-D)\gt 0$, then $$l(D)\le\frac{1}{2}\deg(D)+1$$

What I didn't understand is why if the curve is hyperelliptic, then we have $l(D)=\frac{1}{2}\deg(D)+1$.

The definition of hyperelliptic I know is $C$ is hyperelliptic if $2$ is not a gap (see page 111 for the definition of gap). The problem is I can't connect the equality of the Clifford's theorem and the gap concept. According to my knowledge it seems two completely different notions without any relation.

Any suggestion using Fulton's book level is very welcome!

I need help!

Thanks a lot.


1 Answer 1


In page 111 it is also written that $C$ is hyperelliptic iff there is a degree 2 map from $C$ to $\mathbb{P}^1$. Having such degree 2 map to $\mathbb{P}^1$ is the same to have a divisor $D$ of degree $2$ with $l(D) = 2$. In the classical language people use to say a $g^1_2$. Now if you have an special divisor $D$ different from $W$ and $0$ such that $l(D) = \frac{\deg{D}}{2} + 1$ then it is possible to show the existence of another effective divisor $D'$ of degree $2$ and $l(D') = 2$ hence $C$ is hyperelliptic. A nice proof of this is by induction and it is written in the last two paragraphs at page 344 of Hartshorne "Algebraic Geometry".

Here it is a Fulton's level proof: First notice that Noether's reduction Lemma has a kind of converse (I'am taking 'almost' literally it from page 186 of Walker's book):

Theorem 6.12 for any effective divisor $A$ and any point $P \in C$ at most one of the following holds: $$ \begin{cases} l(A + P) = l(A) \\ l(W - A - P) = l(W - A)\end{cases}$$

Now let $D$ the special divisor $D$ different from $W$ and $0$ such that $l(D) = \frac{\deg{D}}{2} + 1$. As Hartshorne we proceed by induction on $l(D)$. So if $l(D) = 2$ then $C$ is hyperelliptic by the very definition and we have nothing to prove. So assume $l(D) - 1 \geq 4$. Then there is an effective canonical divisor $W = D + E$. The effective divisor $E$ is a sum of a finite number of points of $C$. So there is a point $Q \in C$ which is not in $E$. Take now $P \in D$ a point of the divisor $D$. Now we apply twice Theorem 6.12 to $D - P - Q$. Namely, we want to compute $l(D - P - Q)$. One possibility is that $$l(D - P - Q) = l(D) - 2$$. In this case we have two times equality in the other one (just one of two must hold!). Namely, we get $$l(W - D - P - Q) = l(W - D) \, .$$ But the you can check that the point $Q$ belongs to $E$ contradicting our choise of it i.e. we chose $Q \notin E$. Thus, we get $$l(D - P - Q) = l(D) -1 \, .$$ Setting $D' = D-P-Q$ we can apply the induction hypothesis to any effective divisor in $|D'|$. Indeed, if $\tilde{D} \in |D'|$ is an effective divisor you can check that $\deg(\tilde{D}) = \deg(D) - 2$, $l(\tilde{D}) = l(D) - 1$ and of course $\tilde{D}$ is neither a canonical divisor nor the zero one. QED

  • $\begingroup$ Thank you for your answer. Do you know where I can find a proof more readable by someone who has only Fulton's book as background? $\endgroup$
    – user42912
    Jun 13, 2015 at 22:00
  • $\begingroup$ You said: "Having such degree $2$ map to $\mathbb P^1$ is the same to have a divisor $D$ of degree $2$ with $l(D)=2$", do you know where I can find this result? thank you again. $\endgroup$
    – user42912
    Jun 13, 2015 at 22:47
  • 1
    $\begingroup$ This is not just a result for divisors of degree 2. Without exageration I can say that this is one of the more important ideas of algebraic geometry. Unfortunately, Fulton tells you this in page 110 at the end of exercise 8.33. I suggest to you to have a look to Walker's "Algebraic curves" page 140 paragraph "Rational Transformations Associated with a Linear Series" or also Rick Miranda's "Algebraic curves and Riemann surfaces" page 153 'Divisors and maps to projective spaces' $\endgroup$
    – Holonomia
    Jun 14, 2015 at 8:44
  • $\begingroup$ I'm trying to understand your answer. Thank you very much for your help! $\endgroup$
    – user42912
    Jun 14, 2015 at 17:40

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