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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 $Pic^0(E)$?

Thanks for your help! :)

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

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  • $\begingroup$ @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. $\endgroup$ Jul 21 '20 at 0:55

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