Let $E$ be an elliptic curve and $k$ a field. It is well know that $E(k)$ has an (additive) group structure and indeed there are a lot of sources describing what geometrically there is going on.

The point of my interest is to find a derivation of this group law in language of divisors - especially using properties of the divisor class group.

Indeed the divisors $Div(E)$ are formal sums $\sum_{P \in E(k)} n_P (P)$ with $n_P \in \mathbb{Z}$ and the principal divisors $div(f) = \sum_P ord_P(f) (P)$ form a subgroup of $Div(E)$; denote it by $PrDiv(E)$.

The divisor class group is the quotient $Cl(E)= Div(E)/PrDiv(E)$.

Obviosly we can define canonically a map $E(k) \to Cl(E), p \to (P)-(O)$ where $O$ is the special point (=neutral element).

How to show that this map determine the group law on $E(k)$.

  • 2
    $\begingroup$ $(O)$ isn't a principal divisor. Principal divisors have degree zero. $\endgroup$ – Lord Shark the Unknown Jan 11 at 21:27
  • $\begingroup$ @LordSharktheUnknown: ah sure sure, $(O)-(O)$ should be the prinipal one and is obviously one. Thank you. But the point is why in $Cl(E)$ the group law is already determined. $\endgroup$ – KarlPeter Jan 11 at 21:31
  • $\begingroup$ This is Proposition III.3.4 (p. 61) in Silverman. Have you looked there? $\endgroup$ – André 3000 Jan 11 at 21:33
  • $\begingroup$ Every degree zero divisor is congruent, modulo principal divisors, to exactly one of the form $(P)-(O)$. $\endgroup$ – Lord Shark the Unknown Jan 11 at 21:33
  • $\begingroup$ It suffices to show (or define) that $P+Q-R-O$ is a principal divisor iff $P + Q= R$ in the group law. From this then $\sum_j P_j-Q_j = R-O+div(f)$ where $R = \sum_j P_j-Q_j$ in the group law and $div(f) = O-R+\sum_j P_j-Q_j$ where $f$ is meromorphic $\endgroup$ – reuns Jan 11 at 22:05

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