# Elements of the Picard group on elliptic curves

I was wondering about the elements of a picard group on elliptic curves.

The Picard group of degree 0 divisors on an elliptic curve $$E$$ is defined as $$Pic^0(E)= \frac{Div^0(E)}{Princ(E)}$$, where $$Princ(E)$$ denotes the principal divisors of $$E$$.

The Picard group is used to prove the group law on elliptic curves, where a bijection $$\sigma$$ is defined as

$$\sigma: E \rightarrow Pic^0(E)$$ and $$P \mapsto (P)-(O)$$,

with O being the point at infinity. It is clear to me that $$(P)-(O)$$ is a degree 0 divisor, but why is it also an element of the Picard group $$Pic^0(E)$$?

• $(P)-(O)\in\text{Div}^0(E)$. Strictly speaking, $(P)-(O)$ is not in $\text{Pic}(E)$ but the coset $(P)-(O)+\text{Princ}(E)$ is. Jul 19 '20 at 12:08

To elaborate on Angina Seng's comment. What you have run into here is just a typical abuse of notation, you've definitely done it before yourself!

We write "$$m$$" (which strictly speaking just an element of $$\mathbb{Z}$$) for an element of $$\mathbb{Z}/n\mathbb{Z}$$ when we really mean the coset $$m + n\mathbb{Z}$$. Similarly we write "$$(P) - (O)$$" (which again is just an element of $$\operatorname{Div}^0(E)$$) when we really mean the element $$(P) - (O) + \operatorname{Princ}(E)$$, its image in $$\operatorname{Pic}^0(E)$$.

In your comment you say that

Why is it than allowed to say that because of the bijection of $$P \mapsto (P)−(O)$$ the map $$P \mapsto (P)−(O)+\operatorname{Princ}(E)$$ is also a bijection?

The map $$E \to \operatorname{Div}^0(E) : P \mapsto (P) - (O)$$ is not a bijection (since, for example, $$2(P) - 2(O)$$ does not get hit for any $$P \neq O$$). But we can show that the map $$E \to \operatorname{Pic}^0(E) : P \mapsto (P) - (O)$$ is a bijection when again we are using the abuse of notation above (this is e.g., Silverman, The Arithmetic of Elliptic Curves, Prop 3.4).

A final tiny point, at this level of abstraction $$O$$ is your given $$K$$-point of $$E$$, it doesn't have to be the point at infinity - i.e., you don't need to choose a Weierstrass equation.

• @AnujJamadagni maybe it will be clear if you think about the simpler situation. Consider the map $\phi : \mathbb{Z} \to C_n$ taking $1$ to a generator of $C_n$. Then this induces a map $\mathbb{Z}/n\mathbb{Z} \to C_n$ (choose a representative and send it there) which is well defined since $n\mathbb{Z} \subset \ker(\phi)$, which turns out to be a bijection. The same phenomenon is happening here. Jul 21 '20 at 0:55