4
$\begingroup$

I was trying to understand better the definition of the group law for an elliptic curve given in Katz and Mazur's book (http://books.google.com.br/books/about/Arithmetic_Moduli_of_Elliptic_Curves.html?id=M1IT0J_sPr8C&redir_esc=y).

They define an elliptic curve $E$ to be an $S$-scheme with geometrically connected fibers of genus 1, together with a section called $0$. Furthermore, for each section $P$, they consider the ideal sheaf $\mathcal{I}(P)$ viewed as an effective Cartier divisor of degree $1$ over $E$.

For each $S$-scheme $T$, three section of $E(T)$ (now viewed as a $T$-scheme) are defined to satisfy $P + Q = R$ iff there exists an line bundle $\mathcal{L}_0$ over $T$ such that $$\mathcal{I}^{-1}(P) \otimes \mathcal{I}^{-1}(Q) \otimes \mathcal{I}(0) \cong \mathcal{I}^{-1}(R) \otimes f_T^{*}\mathcal{L}_0 $$ in $E(T)$, where $f_T$ is the structure morphism.

Well , I could not understand exactly the relation between this definition and the classical one. Furthermore, the "$^{-1}$" sign appears to makes no sense at all (what is the problem in changing everything to $\mathcal{I}$ instead of the dual?).

I tried to recover the classical group law, using $S =\text{Spec} (k)$ for an field $k$ and $T$ equals some field extension $L$. In this context $\mathcal{I} (P)$ is just a point and $\mathcal{L}_0$ a vector space over $L$ of dimension $1$, the the pullback looks like the trivial bundle, so things appears to make no sense at all. Where is the line connecting the points $P$, $Q$ and $-R$? How can I recover this classical group law?

Thanks in advance.

$\endgroup$
  • $\begingroup$ In the case $S = spec(k)$ this just reduces to saying that $(P) + (Q) = (R) + (0)$ in the Picard group, which is the usual definition of elliptic curve addition. The point is that when $S$ is some complicated base scheme you don't want the Picard group of $S$ to be involved, which is why you require the equality to hold modulo pullbacks from $S$ rather than exactly. $\endgroup$ – David Loeffler Jun 4 '14 at 22:54
  • $\begingroup$ @DavidLoeffler Thanks for the comment, but how can I see the line that is connecting the three points? $\endgroup$ – user40276 Jun 4 '14 at 22:59
  • 2
    $\begingroup$ If you don't know that the group law on an elliptic curve can be expressed in terms of the Picard group, then I'm afraid you probably don't have any hope of getting much out of Katz--Mazur. You should probably read a more basic elliptic curves textbook (e.g. Silverman) first. $\endgroup$ – David Loeffler Jun 5 '14 at 1:16
  • $\begingroup$ @DavidLoeffler I´m not hoping to understand Katz and Mazur book entirely, I was just interested in seeing a scheme theoretical approach to elliptic curves (and I´m study Silverman too), which seems to be done only in this book (do you know other reference with this approach?). Could you explain me or cite a reference for the proof of the equivalence between the group law in terms of the Picard group and the classical one (in terms of a line in the projetive space connecting three points)? $\endgroup$ – user40276 Jun 5 '14 at 12:20
  • 1
    $\begingroup$ This is Prop III.3.4 of Silverman. $\endgroup$ – David Loeffler Jun 5 '14 at 17:14
3
$\begingroup$

(Composite of comments above, reposted as an answer)

In the case $S=Spec(k)$ this just reduces to saying that $(P)+(Q)=(R)+(0)$ in the Picard group, which is the usual definition of elliptic curve addition. The point is that when $S$ is some complicated base scheme you don't want the Picard group of $S$ to be involved, which is why you require the equality to hold modulo pullbacks from $S$ rather than exactly.

To see why this definition of addition on points of an elliptic curve over a field, using the Picard group, coincides with the more elementary "chord-and-tangent" definition of addition, see Prop III.3.4 of Silverman.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.