In this post we will talk about rational cubic curves. Before starting, we should know what a group is. Here is a brief definition of what a group is (which would be sufficient for this post, but to learn it in detail, you can view the group theory lectures conducted in OMC!)

A group is basically an algebraic structure with a few properties. If $G$ is a set, and $\ast$ is a binary function defined over $F$ then we say that $(G, \ast)$ is a group iff

- $a, b \in G \implies a \ast b \in G$.
- $a \ast (b \ast c) = (a \ast b) \ast c$.$\exists e \in G$ such that $a \ast e = e \ast a = a$. $e$ is said to be the
**identity element**of $G$. - $\forall a \in G, \ \exists a^{-1} \in G$ such that $a \ast a^{-1} = e$

Note that not all groups follow the commutative property, those which follow are known to be

*abelian groups*. That is, if $a \ast b = b \ast a$, then we say that $G$ is an abelian group.INTRODUCTION

Let $\mathcal{C}$ be a three degree curve. The goal of this post is to prove that all the rational points lying on $\mathcal{C}$ form a group. Before we begin, here are few terms :-

**Rational Point**- A point $(x,y)$ is said to be a rational point if $x,y \in \mathbb{Q}$**Rational Line**- We call a line a rational line if the equation of the line can be written with rational numbers, that is, if it has an equation $ax + by + c = 0$ with $a,b,c \in \mathbb{Q}$**Rational Curve**- We will say that the curve is rational if the coefficients of its equation are rational numbers.

Here's an important result - Two rational lines always intersect at a rational point. (The proof of this is quite straightforward and the readers should try to prove it.)

SOME IMPORTANT THEOREMS

We first state some useful results before and then move on to the actual part.*Bezout's theorem :-*A curve of degree $m$ and a curve of degree $n$ intersect in exactly $mn$ points.

For the above theorem to be valid we need to keep a few things in mind - we use projective plane, to have the "extra" points i.e. Points at infinity, the intersections are being counted with multiplicities, for example counting points of tangency as intersections of multiplicities greater than one. And finally, we should allow complex numbers for coordinates.

The above theorem isn't directly used, I just stated it for sanity purposes. The only use of it is to note the $m = n = 3$ case, because we are dealing with cubics in this post.

*Cayley Bacharach :-*Let $C,C_1, C_2$ be cubic curves. Suppose that $C$ goes through eight of the nine intersection points of $C_1$ and $C_2$. Then $C$ goes through the ninth intersection point too.

I am just giving an outline of proof as of now, which might be intuitive enough and omitting the details, you can skip the proof if you want to, it's not relevant for what our aim of the post is.

*Proof of Cayley Bacharach :*

First of all note that the equation of a cubic curve is of the form$$ax^3 + bx^2y + cxy^2 + dy^3 + ex^2 + fxy + gy^2 + hx + iy + j = 0$$So basically we need $10$ coefficients to define a cubic curve, but do we actually need 10?? The answer is NO. If $j \ne 0$ then divide the whole expression by $j$ and consider $\frac{a}{j}$ as a single variable. This reduces the number of variables required to $9$ and hence a cube is "$9$-dimensional".

Let $C_1$ and $C_2$ intersect at $P_1, P_2, \ldots, P_9$. Suppose $C$ passes through $P_1, \ldots, P_8$ and we need to show $P_9 \in C$.

Now we note that whenever we say that a particular point passes through the curve, we are basically imposing a linear condition on the curve. We have that $C$ passes through $8$ fixed points ($P_1,\ldots, P_8$), which means the family defining $C$ is linear/one dimensional.

If $F_1(x,y) = 0$ and $F_2(x,y) = 0$ are the equations of curves $C_1, C_2$ then due to $C$ being linear, the family defining $C$ is precisely$$\lambda_1 \cdot F_1 + \lambda_2 \cdot F_2$$Which means for all $\lambda_1, \lambda_2$,$$\lambda_1 \cdot F_1 + \lambda_2 \cdot F_2$$is the equation of some cubic curve passing through $P_1, \ldots, P_8$.

Since $F_1$ and $F_2$ both vanish at the $P_9$ too, we get that $\lambda_1 \cdot C_1 + \lambda_2 \cdot F_2 = 0$, at $P_9$.

Which means $C$ passes through $P_9$ too.

Let $C_1$ and $C_2$ intersect at $P_1, P_2, \ldots, P_9$. Suppose $C$ passes through $P_1, \ldots, P_8$ and we need to show $P_9 \in C$.

Now we note that whenever we say that a particular point passes through the curve, we are basically imposing a linear condition on the curve. We have that $C$ passes through $8$ fixed points ($P_1,\ldots, P_8$), which means the family defining $C$ is linear/one dimensional.

If $F_1(x,y) = 0$ and $F_2(x,y) = 0$ are the equations of curves $C_1, C_2$ then due to $C$ being linear, the family defining $C$ is precisely$$\lambda_1 \cdot F_1 + \lambda_2 \cdot F_2$$Which means for all $\lambda_1, \lambda_2$,$$\lambda_1 \cdot F_1 + \lambda_2 \cdot F_2$$is the equation of some cubic curve passing through $P_1, \ldots, P_8$.

Since $F_1$ and $F_2$ both vanish at the $P_9$ too, we get that $\lambda_1 \cdot C_1 + \lambda_2 \cdot F_2 = 0$, at $P_9$.

Which means $C$ passes through $P_9$ too.

MAIN PART

Now, we move onto proving that all the rational points lying on a rational cubic form a group. (everywhere below, $\mathcal{C}$ refers to a cubic). We assume that there exists atleast one rational point on $\mathcal{C}$ which is known to us. Also we have assumed that the curve $\mathcal{C}$ is non-singular, which just means at any given point on $\mathcal{C}$, you can construct a tangent.

First of all, note that if $P$ and $Q$ are rational points lying on $\mathcal{C}$, then we can find another point rational point, $P \ast Q$ lying on $\mathcal{C}$.

To do so, Construct the line $PQ$ and let it intersect $\mathcal{C}$ again at $R$. Since $P,Q$ are rational, this line is a rational line. Writing the equation of line, and doing some manipulations we can get that the points that lie both on this line and on $\mathcal{C}$ are solutions to a cubic equation, coefficients of which are rational. Suppose $R$ was irrational, then there must exist one more irrational root (conjugate of $R$ ) that satisfies this cubic equation. But since there can be only $3$ solutions to a cubic equation, two of which ($P,Q$) are rational, this is not possible. Which means $R$ has to be rational.

We denote $R$ by $P \ast Q$ basically. So, given any two rational points we can construct another rational point lying on the curve. Definition of $P \ast Q \rightarrow$ Suppose $P,Q \in \mathcal{C}$ then $P \ast Q$ is the third intersection of the line joining $P,Q$ with $\mathcal{C}$.

There is a more interesting thing to it!! Suppose $P$ is a rational point lying on $\mathcal{C}$, then what about considering $P \ast P$ ? It is the tangent to $\mathcal{C}$ at $P$. Repeating the same argument as above, we get that $P \ast P$ is also a rational point, and obviously lying on $\mathcal{C}$.

Till now, we have that, given some rational points lying on a cubic curve, we can construct more rational points on it using the existing ones. But unfortunately, we can't form a group with the function $\ast$ because we can't pick an identity element here. So what do we do now?

We consider a fixed rational point lying on $\mathcal{C}$, say $\mathcal{O}$ and make it our identity element, note that $\mathcal{O}$ can be any rational point lying on $\mathcal{C}$. We denote the group law by $+$ where,

$$P + Q = \mathcal{O} \ast (P \ast Q)$$In simpler words, first construct $R = P \ast Q$ and then take the third intersection of line joining $\mathcal{O}, R$ with $\mathcal{C}$. This point is $P + Q$.

Proving that $(\mathcal{C}, +)$ is a group.

We claim that $(\mathcal{C}, +)$ is an abelian group.We first prove that it's commutative.

*Proof :-*It's clear that $P \ast Q = Q \ast P$, which means

$$P + Q = \mathcal{O} \ast (P \ast Q) = \mathcal{O} \ast (Q \ast P) = Q + P$$

Now, we show that $\mathcal{O}$ is indeed the identity element.

*Proof :-*$$P + \mathcal{O} = \mathcal{O} \ast (P \ast \mathcal{O})$$Note that $P \ast \mathcal{O}$ is the third intersection of line $P\mathcal{O}$ with $C$, say $P'$. Now $\mathcal{O} \ast P'$ is the third intersection of line $P'\mathcal{O}$ with $C$ which is just $P$. Hence,

$$P + \mathcal{O} = \mathcal{O} + P = P$$

Inverses are a bit hard to think of directly, but it's very cute :-

Draw the tangent at $O$ and let it intersect the curve at $T$. We claim that inverse of a point $S$ is just the third intersection of $ST$ with the curve. The proof is left as an exercise to the reader, if you face any difficulty you can just ask in the comments :D

Now we are just left with proving associativity. If you actually tried to prove it by yourself, and couldn't do so (because it's not that intuitive like the other $3$, atleast wasn't for me :( ) here are a few nudges :-

*Nudge 1 :*Consider two cubics (basically degenerate cubics) in a clever way.

*Nudge 2 :*Try to use cayley bacharach

*Proof :-*We need to prove that if $P,Q,R \in \mathcal{C}$ are three rational points, then$$(P + Q) + R = P + (Q + R)$$Clearly, this is equivalent to proving$$(P + Q) \ast R = P \ast (Q + R)$$To get $P + Q$, we form $P \ast Q$ and take the third point of intersection of the line connecting $P \ast Q$ to $\mathcal{O}$. Now to add $P +Q$ to $R$, we draw the line through $P +Q$, which meets the curve at $(P + Q) \ast R$

To form $P \ast (Q + R)$ we have to find $Q \ast R$, join that to $\mathcal{O}$, and take the third intersection, which

is $Q+R$. Then we must join $Q+R$ to $P$, which gives the point $P \ast (Q+R)$.

We have $8$ points as of now :-

$$\mathcal{O}, P, Q, R, P \ast Q, P + Q, Q \ast R, Q + R$$each of which lies on one dashed line and one solid line.

Let us consider the dashed line through $P+Q$ and $R$, say $\ell_1$ and the solid line through $P$ and $Q+R$, say $\ell_2$. If we show that their intersection lies on $\mathcal{C}$, then we will be done. because this intersection will be then $(P+Q) \ast R$ and $P \ast (Q+R)$ respectively, and thus both will be same.

We have $9$ points with us, the eight points stated above and $\ell_1 \cap \ell_2$. Since a line has a linear equation, if we multiply the equations of $3$ lines we get a cubic equation. The set of solutions to that cubic is just the three lines taken together.

Now we consider two degenerate cubics. $C_1$ is a cubic formed by the three dashed lines and $C_2$ is the cubic formed by the three solid lines. Note that the nine intersection points of $C_1$ and $C_2$ are the nine points stated above. Also, it's evident that $\mathcal{C}$ passes through eight of these nine points, basically all except $\ell_1 \cap \ell_2$.

Now we consider two degenerate cubics. $C_1$ is a cubic formed by the three dashed lines and $C_2$ is the cubic formed by the three solid lines. Note that the nine intersection points of $C_1$ and $C_2$ are the nine points stated above. Also, it's evident that $\mathcal{C}$ passes through eight of these nine points, basically all except $\ell_1 \cap \ell_2$.

Using cayley bacharach, which I stated at the start of this post, we get that $\ell_1 \cap \ell_2 \in \mathcal{C}$. And this is what we wanted! Hence the associativity holds too.

So now we have proved that $(\mathcal{C}, +)$ is a group. yayyy!

Just to mention, there is nothing special about our choice of $\mathcal{O}$. If we choose another rational point say $\mathcal{O}'$ to be the identity element, then we get a group with the same structure. Formally, the map$$P \rightarrow P + \mathcal{O}'$$is an isomorphism from the group $(\mathcal{C},\mathcal{O}, +)$ to the group $(\mathcal{C}, \mathcal{O}', +')$, where the new addition law is defined by$$P +' Q = P + Q - \mathcal{O}'$$I won't get into much details of this, I just stated it for fun and if someone was curious about $\mathcal{O}$ :P

Also, all the things covered in this post has dodged some things. Like, If line through $P,Q$ is tangent to $\mathcal{C}$ then their intersection is $P$ and you don't get a new point for that. Secondly, if $P$ is a point of inflection on $\mathcal{C}$, then the tangent to $P$ meets the curve three times at $P$. In simpler words, if $P$ is a point of inflection then$$P \ast P = P.$$ Despite that, I hope you enjoyed the post and this cute result :D

Pranav Choudhary

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