Help understanding geometric proof (using Cayley - Bacharach) that group law on elliptic curves is associative I have a specific question about a geometric proof that the group law on elliptic curves is associative, as carried out in Fulton's Algebraic Curves (Section 5.6) or here.
The main thrust is to begin with a curve $\mathcal{C}$, and then to find two sets of 9 points (with 8 points in common) on 3 lines on this curve $\mathcal{C}$, to posit that each of these sets of points determines some elliptic curve that is not $\mathcal{C}$ (let us call these curves $\mathcal{C}'$ and $\mathcal{C}''$), and then to apply the Cayley-Bacharach Theorem to show that the 9th point on both $\mathcal{C}'$ and $\mathcal{C}''$ must be the same.
What I do not understand - surely Cayley - Bacharach applies only if  $\mathcal{C}'$ and $\mathcal{C}''$ are distinct curves, and I do not understand why we are able to say that there MUST be another curve going through 9 points of $\mathcal{C}$ that is not equal to $\mathcal{C}$. Also, how is the proof valid once we recognize that the "different" curves $\mathcal{C}'$ and $\mathcal{C}''$, for which the only things we know is that they have the same nine points, share these same nine points in common?
 A: I will try to isolate the idea of the proof of the associativity, hope this answers the unclear points. First, we are doing the following, when we use the Cayley-Bacharach link in the OP.
The definition of the sum:
We start with $O$, fixed (rational) point on some fixed cubic curve
$$
E\subset \Bbb P^2\ ,
$$ 
(all spaces defined over a fixed field,)
then consider two other points, $P,Q$, and use a specific receipt to define the point $P+Q$. The notations in loc. cit are rather irritating, and i will never use something like $PQ$ (for a point). The point P+Q is defined uniquely as in the "picture":
 O
 |
 |
P+Q
 |
 |
 |
 *---- P------Q

Here, points in triple joined in the picture through a line correspond to points on a line in the geometry (of the affine space where the elliptic curve also lives in). Now consider a further point $R$, we want to show:
$$(P+Q)+R = P+(Q+R)\ .$$
This means the equality of the following points "in the middle" X and X' of the diagrams:
   O
   |
   |
  P+Q --- X --------- R
   |                  |
   |                  |
   |                  |
   A ---- P --------- Q

and 
   O ----- Q+R -------- B
            |           |
            |           |
            X' -------- R
            |           |
            |           |
            |           |
            P --------- Q

(Above, starting from X, and respectively X', we have to intersect 
the lines OX and OX' with the curve, the "third point" is then 
$(P+Q)+R$, and respectively $P+(Q+R)$.)
So it is natural to show that both diagrams fit in the same picture:
   O ----- Q+R -------- B ------- Line L3
   |        |           |
   |        |           |
  P+Q ----- * --------- R ------- Line L2
   |        |           |
   |        |           |
   |        |           |
   A ------ P --------- Q ------- Line L1
   |        |           |
   |        |           |
   |        |           |
  Line     Line        Line
   M1       M2          M3

The question is explicitly, if the points * = X and respectively * = X', 
constructed as follows starting with the eight points "on the margin", $A,P,A,R,B,Q+R,O,P+Q$ do coincide:


*

*Consider the lines $L_1,L_2,L_3$, then on $L_1,L_3$ we already have by  construction three points, let $X$ be the third point on $L_2$. Let $l_1,l_2,l_3$ be degree one homogeneous polynomials, so that the equations $l_1=0$, $l_2=0$, $l_3=0$, describe the lines $L_1,L_2,L_3$. Then the degree three polynomial $l_1l_2l_3$ defines a (degenerated) cubic.

*Consider the lines $M_1,M_2,M_3$, then on $M_1,M_3$ we already have by  construction three points, let $X'$ be the third point on $M_2$.


The proof forgets now everything about $X,X'$, introduces a new point, $Y$, defined as the intersection of the lines $L_2$ and $M_2$. (A priori, this point may or may not lie on the elliptic curve. In the end, all three points $X,X',Y$ coincide.) We are now in the position now to apply in the generic case (eight distinct points) the theorem Cayley-Bacharach for the (degenerated) cubic curves 
$$
\begin{aligned}
C_l &:& l_1l_2l_3&=0\ ,\\
C_m &:& m_1m_2m_3&=0\ ,
\end{aligned}
$$
and the given elliptic curve $E$.The $8+1$ points are 
$A,P,A,R,B,Q+R,O,P+Q$ plus $Y$. It follows, that $Y$ is also on the cubic $E$. We get by construction $X=Y=X'$, the relation we wanted.
In case some of the points coincide, we have to use multiplicities, this leads to the solution from [Fulton, Algebraic Curves], a sort of intersection number (as part of an intersection theory) is needed.
