# Elementary exposition of the Picard group?

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?

• “Rational points on elliptic curves” by J. Silverman and J. Tate Dec 27, 2020 at 12:24
• @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. Dec 27, 2020 at 12:32

## 1 Answer

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 $$R+R'-2O$$.

• 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? Dec 27, 2020 at 15:25
• 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. Dec 27, 2020 at 15:28
• 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 $[0:1:0]$. Dec 27, 2020 at 15:51
• 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 $Pic(E)/ \langle O \rangle$ Dec 27, 2020 at 16:07
• $\sum_{j=1}^n S_j=\sum_{j=1}^{n-2} S_j+ W$, doing so until $n=0$ we get that our divisor $\sum_j S_j-\sum_i T_i$ is $= A-B=A+B' = C$ Dec 27, 2020 at 16:32