This problem arose in an algebraic geometry course I'm taking, and my understanding of it comes from Shavarevich's "Basic Algebraic Geometry." The question is this: Given a projective variety $$X = V(y^2z - x^3 - xz^2)$$ and a divisor $D = 2[0:0:1]$, find a basis for $\mathcal{L} (D)$. That is, I'm trying to find a rational map corresponding to $D$ that let's me construct any other map from $X$ corresponding to $D$ by composing it with various projection maps. I'm working over $\mathbb{C}$.

This is my approach:

A basis of rational functions $\{f_0, \cdots , f_r\}$ for $\mathcal{L} (D)$ must satisfy $div(f_i) + D \geq 0$. Thus it is necessary for any such $f$ to have a pole at infinity with multiplicity 2. If I write any such $f$ as a quotient of 1-forms, Bézout's theorem is going to give that we'll have another pole of $f$, so this has to cancel with a zero in the numerator.

I then found the divisors of different 1-forms in x, y, and z, which really amounted to finding $div(x), div(y), div(z)$ then using properties of valuations to conclude that, if the denominator is a 1-form, it has to be $x$, as only $div(x)$ has the point at infinity with multiplicity 2. That is, $$div(x) = 2[0:0:1] + [0:1:0].$$ I then found that the only rational functions in $k(X)$ that are in $\mathcal{L} (D)$ are of the form $$f = \frac{az + by}{x}$$ and so a basis for $\mathcal{L} (-D)$ is $\{ f_1 = \frac{z}{x}, f_2 = 1\}$, whence we find our rational map from $X$ associated to $D$ is $$\phi = [f_1: f_2].$$ We can clear the denominator to then have $$\phi = [z:x].$$

Is this correct? This is the first time I've attempted to construct such a rational map from a divisor, so I'm not sure the process was correct. I'm concerned because this isn't defined at $[0:1:0]$, and I was expecting it to be defined everywhere on X.



EDIT: I'll show how I found $div(x)$ using Bézout's theorem. The points of intersection of $V(y^2z - x^3 - xz^2)$ with $V(x)$ are easily seen to be $[0:0:1]$ and $[0:1:0]$. Using Bézout's theorem, I know that X and $V(x)$ have to intersect in three points with multiplicity. I'll find the intersection multiplicity of $[0:1:0]$:

To do this, I'll intersect both X and $V(x)$ with the open set $U_y = \{[x:1:z]\}$ and work in affine space. In particular, $$X \cap U_y = V(z-x^3-xz^2)$$ and $V(x) \cap U_y$ is just $V(x)$; I then consider the point $p = (0,0)$.

Because the tangent to $X \cap U_y$ is given by the line $z=0$ and $X \cap U_y$ is nonsingular at $(0,0)$, I conclude that the maximal ideal of the local ring of $X \cap U_y$ at p is $$\mathfrak{M}_p(X\cap U_y) = (x)$$. From this it's easy to see that the intersection multiplicity of $X\cap U_y$ with $V(x)$ at $(0,0)$ is just 1, whence the intersection multiplicity of $X$ with $V(x)$ at $[0:1:0]$ is 1. By Bézout's theorem, I can immediately conclude that the intersection multiplicity of $X$ with $V(x)$ at $[0:0:1]$ is 2, so $$div(x) = 2[0:0:1] + [0:1:0]$$.

A similar calculation shows that $div(z) = 3[0:1:0]$, and so $$div(\frac{z}{x}) = div(z) - div(x) = 2[0:1:0] - 2[0:0:1].$$

  • $\begingroup$ Dear Garnet, a better notation would be to define $E=2[0:0:1]$ and then ask for a basis of $L(E)$.Why all these minus signs? $\endgroup$ – Georges Elencwajg Sep 14 '13 at 23:53
  • $\begingroup$ Also, the notation $div(x)$ does not make sense because $x$ is not a rational function. And forget about Bézout. $\endgroup$ – Georges Elencwajg Sep 14 '13 at 23:54
  • $\begingroup$ Thanks so much for the response! You're right - I don't know why I included the negative signs. I was getting caught up with the differences in approach used in my textbook and that used in my lecture. I don't understand why $div(x)$ does not make sense? It was my understanding that the divisor of a rational function $\frac{a}{b}$ is $div(a) - div(b)$, where $a$ and $b$ are in the coordinate ring of the variety $X$. I've shown how I've used Bézout's theorem in my edit above, and I'm not sure how I would find the divisor otherwise. $\endgroup$ – Garnet Sep 15 '13 at 11:01
  • $\begingroup$ Dear Garnet: let me repeat that it makes no sense to mention $div(x)$ or $div(z)$ because $x$ and $z$ are not rational functions. $\endgroup$ – Georges Elencwajg Sep 15 '13 at 11:55

Try to show that $\operatorname {div}(\frac zx)=2\cdot [0:1:0]-2\cdot[0:0:1]$ and that a basis of $L(2\cdot[0:0:1])$ is $(1,\frac zx)$.

Conclude by using the following result from Miranda's book, Problem V.3.H, page 153:
If $X$ is a compact Riemann surface not isomorphic to $\mathbb P^1$, then for every divisor $D$ of positive degree we have: $$\operatorname {dim} L(D)\leq \operatorname {deg}(D)$$

  • $\begingroup$ I can show that the divisor of $\frac{z}{x}$ is as given, but I'm less certain of how to show that a basis of $\mathcal{L}(D)$ is $(1, \frac{z}{x})$. Certainly $\frac{z}{x}$ and $1$ are contained in $\mathcal{L}(D)$, and I know it is finite dimensional. If I could show that $\mathcal{L}(D)$ has dimension 2 I would be done. $\endgroup$ – Garnet Sep 15 '13 at 11:07
  • $\begingroup$ Dear Garnet, I have written an edit showing you how to conclude. $\endgroup$ – Georges Elencwajg Sep 15 '13 at 11:50
  • $\begingroup$ Ooops, my edit was not there: I have rewritten it (I must have forgotten to save it!) $\endgroup$ – Georges Elencwajg Sep 15 '13 at 13:50
  • $\begingroup$ Thank you so much! I'll have to pick up Miranda's book at my school's library, as I was not aware of this result. I'm still a bit confused as to your comment on divisors. In [Fulton's Book] (f3.tiera.ru/2/M_Mathematics/MA_Algebra/MAg_Algebraic%20geometry/…) on page 97, the divisor of a plane curve is defined, not just a rational function. The divisor of a rational function is then just the difference of the divisor of the numerator with the divisor of the denominator, no? $\endgroup$ – Garnet Sep 15 '13 at 19:05
  • $\begingroup$ Dear Garnet, Fulton defines on page 97 divisors for elements of $K$, the field of rational functions. I think the difficulty is that for Fulton $z$ is a rational function and for you a coordinate. In case this is still not clear, you might ask a new question. $\endgroup$ – Georges Elencwajg Sep 15 '13 at 19:47

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