Algebraic Geometry – Understanding Terence Tao’s Proof of Cayley-Bacharach’s Theorem

algebraic-curvesalgebraic-geometrycurvesplane-curvesproof-explanation

I am studying Cayley-Bacharach's theorem, which Tao states as (here)

Theorem (Cayley-Bacharach). Let $\gamma_0 = \{ P_0(x,y) = 0\}$ and $\gamma_\infty = \{ P_\infty(x,y) =0 \}$ be two cubic curves that
intersect (over an algebrically closed field $k$) in precisely nine
distinct points $A_1,\dots,A_9.$ Let $P$ be a cubic polynomial that
vanishes on eight of those points (say $A_1,\dots,A_8).$ Then $P$ is a
linear combination of $P_0,P_\infty$ and in particular vanishes on the
ninth point $A_9.$

After this, Tao presents a proof of this theorem which I believe I understand almost totally, there is just one small detail that I am not getting. Below I will show an outline of the proof presented in the link above (just the topics) and be more detailed in the part that I didn't get.

Outline of proof. Essentialy, we can split the proof Tao did in the following steps:

Let $P$ be a cubic polynomial that vanishes on the points $A_1,\dots,A_8$ such that $P$ is not a linear combination of $P_0,P_\infty.$
Some observation on the $9$ original points: no four of these can be colinear and no seven of these can lie on the same quadric.
A consequence of the observations above: any five points from $\{A_1,\dots,A_9\}$ determine a unique quadric curve $\sigma$.
Some further conclusions: no three points from $\{ A_1,\dots,A_8\}$ are colinear and no six points from $\{A_1,\dots,A_8\}$ lie on the same quadric curve.

After this, Tao continues: Finally, let $l$ be the line that goes through the points $A_1,A_2$ and let $\sigma$ be the quadric curve that goes through $A_3,\dots,A_7.$ Note that the quadric curve must be a conic section based on $4.$ Even more, from $4.$ It also follows that $A_8 \notin l$ and $A_8 \notin \sigma.$ Now let us consider two points $B,C$ that are in $l$ but not in $\sigma.$ We can construct $Q = aP + bP_0 + cP_\infty$ such that $Q$ vanishes in $B$ and $C$. Furthermore, $Q$ vanishes in $4$ points of $l$ $(B,C,A_1,A_2)$ and thus $l$ must be a component of $Q$ (Bézout). The other component must be a quadric curve (degree comparasion). Also, this quadric curve must contain the points $A_3,\dots,A_7$ and thus it must be $\sigma$, based on $3.$ But then $Q$ doesn't vanish at $A_8$, which is a contradiction, because $A_8$ vanishes at the three polynomials $P$, $P_0$ and $P_\infty$.

My doubt. I believe I understand almost every argument that Tao used: I just don't see where we use the fact that $P$ is not a linear combination of $P_0,P_\infty.$ Basically, I understand that we find a contradiction and how we find it, I just don't see how the fact that $P$ is not a linear combination of $P_0$ and $P_\infty$ influences this same contradiction. As far as my mind can think, we would reach this same contradiction if $P$ was a linear combination of $P_0,P_\infty$. I don't see what I am missing but it must be something really trivial.

Thanks for any help in advance.

Best Answer

As $P$ is not a linear combination of $P_0$ and $P_\infty$, $\{P,P_0,P_\infty\}$ is a linearly independent set of degree $3$ polynomials. Let $B$ and $C$ be two points as above. We need to find a nonzero $Q=aP+bP_0+cP_\infty$ such that $Q$ vanishes at both $B$ and $C$. This is the same as solving the system $$\begin{align} \begin{bmatrix} P(B)& P_0(B)&P_\infty(B)\\ P(C)& P_0(C)&P_\infty(C) \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix} \end{align},$$ where $P(B)$ denotes $P$ evaluated at $B$ etc. A nontrivial solution $(a,b,c)$ exists in this case and hence a nonzero $Q$.

If $P$ were a linear combination of $P_0$ and $P_\infty$, we would only have a $2\times 2$ matrix, and there is no guarantee of a nontrivial solution to the system.

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Algebraic Geometry – Projective Cubic Curve Passing Through 9 Points

If you pick smooth cubic $C\subset \mathbb{P}^2$ and 8 general points $p_1,\ldots,p_8\in C$ and pick the 9th point $p_9\in C$ such that $p_1+\cdots+p_9\sim 3L$, where $L$ is the class of a line, then this should be an example. The fact that $p_1+\cdots+p_9\sim 3L$ means exactly that there is another cubic passing through these 9 points.

We know such a point $p_9$ exists and is unique by Riemann-Roch for curves (or fix a basepoint and use the group law on an elliptic curve). To see that the points satisfy the stated conditions, I claim that in general no 3 lie on a line and no 6 lie on a conic.

We should be able to pick the 8 points $p_1,\ldots,p_8\subset C\times\cdots \times C$ within some Zariski dense subset, because having 3 lying on a line or 6 on a conic is a closed condition (for example by using the group law on an elliptic curve).

Put more concretely, after choosing a base point on $C$, we have a map from $\mathbb{C}/\Lambda$ to the complex points of $C$ that respects the group law on $C$. Then, the condition that, for example, $p_1+p_2+p_3\in \Lambda$ cuts out a codimension 1 locus in the complex manifold $(\mathbb{C}/\Lambda)^8$. We repeat this for all subsets of 3 and 6 points of $p_1,\ldots,p_9$ (where $p_9=-p_1-\cdots-p_8$), generate a finite number of bad loci, and throw them out.

Help understanding geometric proof (using Cayley – Bacharach) that group law on elliptic curves is associative

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.

Best Answer

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Algebraic Geometry – Projective Cubic Curve Passing Through 9 Points

Help understanding geometric proof (using Cayley – Bacharach) that group law on elliptic curves is associative

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