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A divisor on an elliptic curve E is a formal sum of points $$D=\sum_{P\in E}n_P(P)$$ where the $n_P$ are integers only a finite number of which are nonzero.

Could anyone please explain what is the difference between $(P)$ and $P$ ? I know $P\in E$ is a point on $E$. And what is the use of formal sum here ?

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  • $\begingroup$ Typically one works with rational points on the elliptic curve, which can be given a group operation "+" related to the geometry of the curve, and it is with respect to this operation that one can take integer multiples and finite sums of points. $\endgroup$ – hardmath Apr 15 at 20:27
  • $\begingroup$ For an elliptic curve $E : \{(x,y) \in \overline{k}^2, y^2=x^3+x\} \cup \{\infty\}$ let $f(x,y) \in k(E)=k(x)[y]/(y^2-x^3-x)$ the function field then $Div(f)$ is the list of zeros and poles of $f$ weighted by multiplicity. The point is that $Div(fg) = Div(f)+Div(g)$ and $Div(f) = \emptyset \implies f \in k^*$ and $Div(f)$ always has degree $0$ (same number of zeros/poles). From there the big question is which (degree $0$) divisors come from some function $\in k(E)$. Exercice : factorize $x^3+x$ and deduce $Div(x),Div(y),Div(x-a)$ $\endgroup$ – reuns Apr 16 at 1:11
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A divisor is a gadget that we use to keep track of values attached to points on a curve. The notation $(P)$ is just to indicate what values are attached to what points. For instance, the divisor $D=3(P)+5(Q)$ is meant to indicate that $n_P=3$ and $n_Q=5$. If you are confused about this notation, you could think of a divisor as a vector, such that each coordinate is associated with a point on the curve, i.e., $D=(n_P)_{P\in C}$, but you can easily see that this notation is inefficient because if I write now $(\ldots,0,0,0,3,0,\ldots,0,5,0,0,\ldots)$, it is not clear what point is attached to $3$ and what point is attached to the value $5$, unless I used some notation like $(\ldots,0,0,0,3_P,0,\ldots,0,5_Q,0,0,\ldots)$ of something like that. Moreover, this vector notation has all those unnecessary zeroes, and, of course, it is also misleading because there may be an uncountable number of points on $C$. Thus, we prefer to write $D=3(P)+5(Q)$ to indicate a divisor with $n_P=3$, and $n_Q=5$, and $n_R=0$ for all $R\neq P,Q$.

If you want to think about a concrete divisor, think about the divisor of a function. Let $f(x)= (x-P)^3(x-Q)^5$, for some real numbers $P,Q$, and define $D=\sum_{x\in \mathbb{R}} \operatorname{ord}_x(f)\cdot (x)$, there $\operatorname{ord}_x(f)$ is the order of vanishing of $f$ at $x$. Then, $D=3(P)+5(Q)$.

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  • $\begingroup$ Thank you for explaining briefly. $\endgroup$ – Laba Sa Apr 16 at 6:28
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    $\begingroup$ Thanks. Totally save my day! $\endgroup$ – lkahtz May 13 at 10:01
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I don't think there should be a difference between $(P)$ and $P$. Seeing it as a "formal sum" is probably a little bit confusing, since it doesn't really tell you anything. It might be helpful to think of a divisor $D$ as an element of the free abelian group on the points of your elliptic curve. As far as the usage goes, they are basically everywhere. It might help to just go through an algebraic geometry book and see it in action.

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Well, I think the author wants to distinguish formally between the point $P$ on the curve and the symbol $(P)$ (that stands for $P$) in a formal linear combination of points. This distinction is not necessary. It should be clear from the context.

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