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Let $C$ be an irreducible projective cubic on $k[X,Y,Z]$, $k$ algebraically closed. Let $x = \frac{X}{Z}$, $y = \frac{Y}{Z}$, $z = x^{-1} = \frac{Z}{X}$. Also define the divisor of zeros of $z$: $(z)_0 = \sum_{\text{ord}_P (z)>0} \text{ord}_P (z)P$ , and the vector space of functions $L(D) = \{f\in k(C)\mid \text{ord}_P (f)\geq -n_P\text{ for all }P\in C\}$ with dimension $l(D)$, for any divisor $D = \sum n_P P$. I understand this is known as the Riemann-Roch space.

Problem 8.10 in Fulton's algebraic curves requires us to prove, for an integer $r$:

  1. Show that $L(r(z)_0)\subset k[x,y]$
  2. Show that $l(r(z)_0) = 2r$ if $r > 0$

I have proven (1) like so. Given $\phi\in L(r(z)_0)$, the pole set is $V(S)$ where $S = \{f\in \Gamma(C)\mid f\phi\in\Gamma(C)\}$, the set of possible denominator representatives. But $\text{ord}_P (z) > 0$ only when $P\in V(Z) $, and therefore $\text{ord}_P(\phi)$ must be at least zero outside of $V(Z)$. Hence, $V(S)\subset V(Z)$ and so $Z^n\in S$ for some $n$ by the Nullstellensatz. Thus, $\phi = \frac{g}{Z^n}\in k[x,y]$.

For (2), I am not sure how to begin. I have tried to consider a concrete example using the curve $V(Y^2Z - X(X - Z)(X - \lambda Z))$. If I have not computed wrongly, $(z)_0 = 2P$ where $P = [0:1:0]$. How would one compute an explicit basis for $L((z)_0)$? Does this lead to a proof of (2)?

Houndoom
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  • Ah, apologies, I believe $r$ is an arbitrary integer, positive in the case of (2). – Houndoom Mar 02 '16 at 08:57
  • $r((z)_0)$ is the divisor where each coefficient of $(z)_0$ is multiplied by the integer $r$. – Houndoom Mar 02 '16 at 09:25
  • Dear Houndoom, I don't think your description for (1) is correct: do you realize that $\Gamma (C)=k$ ? – Georges Elencwajg Mar 02 '16 at 11:44
  • On the other hand your calculation $(z)_0=2P$ is correct: bravo! This leads to a proof of (2) if you are allowed to use Riemann-Roch, but I suppose Fulton has something else in mind at that stage. – Georges Elencwajg Mar 02 '16 at 12:09
  • Ah! I'm sorry, my understanding is probably flawed. By $\Gamma(C)$ I refer to the coordinate ring of $C$, which is $\frac{k[X,Y,Z]}{\mathcal{I}(C)}$, whose field of fractions is $k(C)$, the field of rational functions. My understanding is that the pole set should be the vanishing set of "possible denominators", which is an ideal in $\Gamma(C)$. How should I correct this? Also, thank you for confirming my calculation for (2)! Is there a relatively simple way of computing an explicit basis then? – Houndoom Mar 02 '16 at 12:39
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    Dear Houndoom, the field of fraction of $Q=\frac {k[X,Y,Z]}{\mathcal I(C)}$ is not $k(C)$. You have to take the subfield of homogeneous fractions of degree zero of $Q$ to obtain $k(C)$ You are confirming my suspicion that it is virtually impossible for a beginner to learn the content of the last chapters of Fulton's book. – Georges Elencwajg Mar 02 '16 at 17:23

1 Answers1

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We have in obvious notation $$\operatorname {div}(x)=2P_0-2P_\infty, \quad \operatorname {div}(y)=P_0+P_1+P_\lambda-3P_\infty $$ so that $\operatorname {div}(z)=\operatorname {div}(x^{-1})=2P_\infty-2P_0$ from which we get our divisor $$D_r:=r(\operatorname {div}(z))_0=2rP_\infty$$ Then it is clear that $$Vect(1,x,\dots,x^r;y,xy,\dots,x^{r-2}y)\subset L(D_r)$$ and the left-hand side has dimension $2r$ because the displayed vectors are linearly independent.
On the other hand we have $$l(D_r)\leq \operatorname {deg}(D_r)=2r$$(a slight improvement for curves not isomorphic to $\mathbb P^1$ of Proposition 3(3) ).
Putting the above considerations together we obtain what was required: $$L(D_r)= Vect(1,x,\dots,x^r;y,xy,\dots,x^{r-2}y)\subset k[x,y]$$ and thus $l(D_r)=2r$.

Georges Elencwajg
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  • Dear Georges, this is a very enlightening answer, showing an explicit basis (I see that the power of $y$ doesn't go higher than 1 in the basis because the curve equation shows $y^2 = x(x - 1)(x - \lambda)$). And you are perfectly right about $k(C)$; I am missing some statement of homogeneity. Thank you so much for your time and patience in answering the question and clearing up my misconceptions! I must confess that this has been a very daunting subject to pick up. – Houndoom Mar 03 '16 at 01:22
  • Dear Houndoom, I'm glad I could help you. Fulton's book gets very technical at the end and I find his exercises very difficult in that he does not do any preparatory solved computations. Was that book assigned to you or did you choose it yourself? – Georges Elencwajg Mar 03 '16 at 07:22
  • I chose it myself as an introduction, before trying to attempt Hartshorne and the material heavy notes of Vakil. But as I am nearing the end of Fulton, I will start on other sources soon, which will hopefully give me different perspectives and good examples to work with, which will improve my understanding. – Houndoom Mar 03 '16 at 08:10
  • Dear Houndoom, I warmly recommend Shafarevich's book for a general introduction to algebraic geometry and Miranda's for algebraic curves and Riemann surfaces (there are easily accessible PDF's of Miranda's book floating online...) Also: here is a relevant thread on algebraic geometry books. – Georges Elencwajg Mar 03 '16 at 08:26
  • Thanks Georges! I have heard good things about Shafarevich as well, I shall certainly add it to my reading list. You have been very helpful indeed! – Houndoom Mar 03 '16 at 09:00
  • Good luck in your study and welcome to the enchanted Kingdom of Algebraicgeometryland :-) – Georges Elencwajg Mar 03 '16 at 11:49