Unique group law on cubic

Question

I'm looking at the following problem from Pg. 47 of Miles Reid's Undergraduate Algebraic Geometry (https://homepages.warwick.ac.uk/staff/Miles.Reid/MA4A5/UAG.pdf):

2.11 (Group law on cuspidal cubic.) Consider the curve $$C:(z=x^3)\subset k^2;$$ $C$ is the image of the bijective map $\varphi\colon k\to C$ by $t\mapsto (t,t^3)$, so it inherits a group law from the additive group $k$. Prove that this is the unique group on $C$ such that $(0,0)$ is the neutral element and $$P+Q+R=0\iff P,Q,R\text{ are collinear}$$ for $P,Q,R\in C$. [Hint: You might find useful the identity $$\left|\begin{array}{ccc} 1 & a & a^3\\ 1 & b & b^3\\ 1&c&c^3\end{array}\right| = (a-b)(b-c)(c-a)(a+b+c).]$$ In projective terms, $C$ is the curve $(Y^2Z=X^3)$, our old friend with a cusp at the origin and an inflexion point at $(0,1,0)$, and the point of the question is that the usual construction gives a group law on the complement of the singular point.

There is a typo in the equation defining $C$; it should be $y=x^3$. The group operation on $C$ should be $$ (a,a^3)+(b,b^3):=(a+b, (a+b)^3), $$ if I understand correctly. Showing that this operation satisfies the above biconditional is, I think, fairly straightforward. We can use the fact that the above determinant is zero iff the points $P$, $Q$, and $R$ are collinear. But I'm having trouble with the uniqueness part. Here's what I have so far: Suppose $\odot$ is a group operation on $C$ satisfying the relevant properties. We wish to show that this operation equals $+$ as defined above. Take $(s,s^3)$ and $(t,t^3)$ on the curve, and write $$ (s,s^3)\odot(t,t^3)=(x_{s,t},x_{s,t}^3). $$ We want $x_{s,t}$ to equal $s+t$. If these points are distinct, then the line connecting them intersects $C$ at the point $(-s-t,(-s-t)^3)$, so, using the collinearity condition, we get $$ (x_{s,t},x_{s,t}^3)\odot(-s-t,(-s-t)^3)=(0,0). $$ So we're almost there! This implies that $(-s-t,(-s-t)^3)$ is the inverse of $(x_{s,t},x_{s,t}^3)$ in the operation $\odot$. But can we conclude that $$ (x_{s,t},x_{s,t}^3)=(s+t,(s+t)^3)? $$ Thank you.

Answer 1

Let me try and resolve the confusion here. The projective variety $C=V(Y^2Z=X^3)$ is the cuspidal cubic, and it has a group law on its nonsingular $k$-rational points, which are exactly the $k$-rational points of $D(Y)\subset C$. But $D(Y)\subset C$ is $V(z=x^3)\subset\Bbb A^2_{x,z}$, so that's why Reid is telling you to work with that - in particular, $z=x^3$ is correct and not a typo.

Now we'll show the equivalence of $P+Q+R=0$ and $P,Q,R$ collinear. We'll need one piece of linear algebra:

Lemma. If $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ are points in $k^2$, then they are collinear iff the matrix $$\begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{pmatrix}$$ has determinant zero.

Proof. Suppose the points lie on the line $ax+by+c=0$ (with not all $a,b,c$ zero). Then $(a,b,c)$ is in the kernel of the matrix, so it has determinant zero. Conversely, if the determinant is zero, there is a nonzero vector $(a,b,c)$ in the kernel, and so $ax+by+c$ vanishes on each point. $\blacksquare$

We'll need to split up the cases when $P,Q,R$ are distinct or not. Here's the equivalence for distinct $P,Q,R$:

If $P=(p,p^3)$, $Q=(q,q^3)$, $R=(r,r^3)$, and the group operation is $(a,a^3)+(b,b^3)\mapsto (a+b,a^3+b^3)$, then when $P+Q+R=0$, we have $p+q+r=0$ and hence the determinant of $$\begin{pmatrix} p & p^3 & 1 \\ q & q^3 & 1 \\ r & r^3 & 1 \end{pmatrix}$$ vanishes from the hint, so $P,Q,R$ are collinear by the lemma.

If $P,Q,R$ are points on $z=x^3$ which are collinear, then the determinant vanishes. If $P,Q,R$ are distinct, then $(p-q)(q-r)(r-p)\neq 0$, implying $p+q+r=0$ and therefore $P+Q+R=0$.

In the case when $P,Q,R$ are not distinct, we'll need a slight re-imagining of what we mean by collinear. What we really want here is that there is a line $L$ so that the three points of intersection $L$ with $C$ (when counting points with multiplicity) are exactly $P$, $Q$, and $R$.

Without loss of generality, assume $P=Q$. If $P,Q,R$ are collinear in our upgraded definition, then the line $L$ must be the tangent line to $P=Q$, since the tangent line is the unique line which has a multiple intersection with a smooth point on a curve. So the line $L$ is $y-p^3=3p^2(x-p)$, and if $(r,r^3)$ satisfies this equation as well, we must have $r^3-p^3=3p^2(r-p)$, or $(p-r)^2(2p+r)=0$. So either $p=r$, implying that $P=Q=R$ is an inflection point, or $2p+r=0$. The only inflection point on $(t,t^3)$ is $t=0$, so either way $p+q+r=0$ and therefore we've proven collinear implies $P+Q+R=0$ for non-distinct $P,Q,R$.

The reverse direction is essentially the same argument. If $P+Q+R=0$ and $P=Q$, then we get $2p+r=0$, hence $(p-r)^2(2p+r)=0$, hence $(r,r^3)$ is on the tangent line to $P=Q$ and $P,Q,R$ are collinear in our upgraded definition.

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For uniqueness, suppose $(0,0)$ is the identity and $P+Q+R=0$ is equivalent to $P,Q,R$ collinear. First we'll observer that for $P\neq 0$, we have $-P=(-p,-p^3)$, as $P,0,-P$ are collinear via the line $y=p^2x$.

Now let $P,Q$ be arbitrary and $R$ the third point of intersection of $L=\overline{PQ}$ with $C$. As $L$ is given by $z-q^3=(q^2+qp+p^2)(x-q)$, substituting this expression in for $z$ in $z=x^3$, we see that the $x$-coordinate of $R$ must satisfy $p+q+r=0$ by Vieta, so $R$ has first coordinate $-p-q$ and therefore $R=(-p-q,(-p-q)^3)$, so $P+Q=(p+q,(p+q)^3)$.