Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm doing a course on elliptic curves. It starts with a bit of a crash course in algebraic geometry, giving statements alone. We were given the following definition

The Riemann-Roch space of $D$ is $$\mathcal{L}(D)=\{f\in K(C)^*: \mbox{div}(f)+D\geq 0\}\cup \{0\}$$ i.e. the $K$-vector space of rational functions on $C$ with "poles no worse than specified by D".

Does the asterisk mean dual or the units? Later we have

If $f \in K(C)^*$ we define $$\mbox{div}(f)=\sum_{P\in C}\mbox{ord}_P(f)P$$

and I don't understand how order is defined for the dual (the definition he gave was for elements of the function field). However on http://planetmath.org/encyclopedia/SpaceOfFunctionsAssociatedToADivisor.html it says we should be using the dual.

I am also unclear as to whether the definition should instead be

The Riemann-Roch space of $D$ is $$\mathcal{L}(D)=\{f\in \overline{K}(C)^*: \mbox{div}(f)+D\geq 0\}\cup \{0\}$$ i.e. the $\overline{K}$-vector space of rational functions on $C$ with "poles no worse than specified by D".

since at the time we were working in an algebraically closed field, whereas later, we have the remark

If $D$ is defined over $K$, i.e. it is fixed by the natural action of Gal$(\overline{K}/K)$, then $\mathcal{L}(D)$ has a basis in $K(E)$. (Not just in $\overline{K}(E))$.

Should this basis be a $K$-basis or a $\overline{K}$-basis? Later we get a bunch of things living in $\mathcal{L}(\mathcal{O}_E)$, where $\mathcal{O}_E$ is a rational point, and they have to be linearly dependent: $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ and it is claimed that $a_i \in K$ becuase of the remark. I don't see how this follows from us being able to find a $\overline{K}$-basis.

share|improve this question
5  
$K(U)^*$ means units. (I prefer to write $K(U)^\times$ to avoid confusion.) –  Zhen Lin Jan 27 '13 at 18:47
    
Do you have a PlanetMath account? I don't and it's not letting me set one up - it would be nice to correct that page. –  porkramen Jan 27 '13 at 18:54
1  
$\bar{K}$-basis, but the basis elements lie in $K(E)$. You can then show that if an element in $K(E)$ is written as linear sum of this basis, then the coefficients all lie in $K$. –  Sanchez Jan 29 '13 at 7:30
    
Anyway, I don't see the interest to define $L(D)$ in $\bar{K}(C)$. Your first definition is better. –  user18119 Jan 29 '13 at 15:30
1  
@porkramen, write down such a linear combination, and let $Gal(\bar{K}/K)$ acts on it. All the $K(E)$ elements are fixed, so you would get a different linear combination unless all the coefficients are in $K$ already. If there is a different linear combination, take the difference with the original combination, we get linear dependence between the basis elements, contradiction. –  Sanchez Jan 29 '13 at 19:53
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.