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I'm trying t find good references to understand "why" the elliptic curves have a group structure. The canonical answer seems to be that:

  • Every curve has a group associated to it, the picard group
  • the elliptic curves' set of solutions in projective $\mathbb Q$ space is in set theoretic bijection with the Picard group.
  • Hence, we pullback the group structure of the "correct object" (the Picard group) onto the elliptic curve itself, giving the "weird geometric group law" of:

draw a line between $P$ and $Q$, find third point of intersection $R$, reflect $R$ to $R'$. Define $P \cdot Q \equiv R'$.

Can I find an elementary exposition of this somewhere? Silverman expects one to know sophisticated algebraic geometry as far as I can tell. I found picard groups in Hartshorne as well, but it once again seems like quite a lot of effort to get to the idea of a picard group in the book.

Is there an elementary (read: undergrad who knows a first course in varieties/diffgeo/algebra/number theory/topology) source to understand the structure of the Picard group of the elliptic curve, and how it relates to the elementary group law one is taught?

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    $\begingroup$ “Rational points on elliptic curves” by J. Silverman and J. Tate $\endgroup$ – marwalix Dec 27 '20 at 12:24
  • $\begingroup$ @marwalix It doesn't explain the picard group, best as I recall? I'd seen some of the book in a crypto course I had taken. $\endgroup$ – Siddharth Bhat Dec 27 '20 at 12:32
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Is this clear to you ?

For an affine elliptic curve defined by $C:y^2=x^3+Ax+B$ then the line passing through $P,Q$ is defined by some equation $ax+by+c=0$ (assuming $P,Q$ are distinct otherwise we are considering the tangent line at $P$)

$ax+by+c$ is a rational function on the curve and its divisor (zeros & poles) on the projective closure (ie. $E=C\cup O$ where $O$ is the point at infinity) is $P+Q+R-3O$,

where $R$ is the 3rd point of the line.

So $P+ Q+R=3O$ in the Picard group $Pic(E)$, choosing $O$ as the neutral element gives $P+Q=-R=R'$ in $Pic(E)/\langle O\rangle$.

Where $R'=(x_R,-y_R)$ because the divisor of the rational function $x-x_R$ is $R+R'-2O$.

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  • $\begingroup$ The problem is that I don't know what the picard group is. If you could expand the answer with a handwavy-but-morally-correct explanation of the picard group, this would be perfect! You can assume I know homology and cohomology to draw analogies. I know that hand-wavily the divisor group counts "roots with multiplicity", and the picard group is some sort of quotient (?) of the divisor group? $\endgroup$ – Siddharth Bhat Dec 27 '20 at 15:25
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    $\begingroup$ The Picard group is the formal sums of points on the curve quotiented by the subgroup of divisors (zeros/poles counted with multiplicity) of rational functions, ie. $Pic(E)=Div(E)/Prin(E)$. Then we quotient by the subgroup $\{ nO,n\in \Bbb{Z}\}$ to get the group law. $\endgroup$ – reuns Dec 27 '20 at 15:28
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    $\begingroup$ Well since we are going to quotient by $\langle O\rangle$ we don't really care of the pole. It is of order 3 because rational functions have the same number of zeros/poles. For a complex elliptic curve you can say that $x/y$ is a chart at $\infty$ (ie. it has a simple zero), $x^3/y^2\to 1$ at $\infty$, so $ax+by+c\sim by\sim b(y/x)^3 x^3/y^2\sim b (y/x)^3 $ has a pole of order 3. Over arbitrary fields we need to go to the projective curve $ZY^2=X^3+AXZ^2+BZ^3$ and say that $X/Y$ is an uniformizer of the local ring of rational functions with no pole at $[0:1:0]$. $\endgroup$ – reuns Dec 27 '20 at 15:51
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    $\begingroup$ Because $P+Q=R'$ so any element $\sum_j S_j-\sum_i T_i\in Div(E)$ is represented by at most one point in $Pic(E)/ \langle O \rangle$ $\endgroup$ – reuns Dec 27 '20 at 16:07
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    $\begingroup$ $\sum_{j=1}^n S_j=\sum_{j=1}^{n-2} S_j+ W$, doing so until $n=0$ we get that our divisor $\sum_j S_j-\sum_i T_i$ is $ = A-B=A+B' = C$ $\endgroup$ – reuns Dec 27 '20 at 16:32

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