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I'm working on a problem,

Show that in an elliptic curve $E/\mathbb{Q}$ the sum of three colinear rational points of it is equal to $O$ exactly when the neutral element of the group $E(\mathbb{Q})$, $O$ is an inflection point of the curve.

I found the following in my notes.

Let $C$ be a cubic curve that is defined over a field $k$ and let $O \in C(k)$. $O$ is an inflection point if and only if $P+Q+R=O$, where $P, Q, R$ are three intersection points of $C$ with a line.

Proof: If $P, Q, R$ are three intersection points of $C$ with a line, then $R=PQ$. Now, $-R=P+Q$ if and only if $-R=(OO)R=O(PQ)=P+Q$. This holds exactly when $(OO)R=OR$ or $O=OO$, i.e. exactly when $O$ is an inflection point.

Could you explain this to me? I understand that if $P, Q, R$ are three intersection points of $C$ with a line, then $R=PQ$, but I am facing difficulties with the remaining.

EDIT:

Could we say it as follows?

Let $R$ be a point of the curve.

We have that $R+(-R)=O$. To add the points $R,-R$ we find the third intersection point of the line $(R,-R)$ and the curve, which is $R(-R)$.

Then we take the line $(R(-R),O)$. The third intersection point is $R+(-R)=O$.

That means that the line $(R(-R),O)$ is the tangent at $O$. Since $R(-R)$ is the third intersection point of the tangent at $O$, it should be $R(-R)=OO$.

If we take the line $(OO,R)$ the third intersection point is $(OO)R$. So $(OO)R=-R$.

So we have that the following relation holds: $P+Q=(PQ)O$, where $P,Q$ are points of the elliptic curve and $-R=(OO)R$.

$P,Q,R$ are collinear, so let $R=PQ$. Then \begin{align}P+Q+R=O \iff &P+Q=-R \\ \iff &(PQ)O=(OO)R \\ \implies &RO=(OO)R\end{align} so $(OO,R)=(R,O),$

Since the lines are the same it has to hold $OO=O$. Does this mean that $O$ is an inflection point?

Have we shown in this way both directions?

mjc
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evinda
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  • $P+Q$ according to the addition law in the elliptic C, is the symmetrical (with respect to Ox-axes) of $R$, say $R'.$ Thus $R+R'=O$ – 111 Jan 13 '15 at 00:34
  • @111 What does this mean:

    (OO)R=O(PQ) ?

     – evinda Jan 13 '15 at 00:40
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    I think the symbol AB means the (third) point in the intersection of the cubic C and the line defined by the points A,B. Probable OO is the tangent line at O.... – 111 Jan 13 '15 at 00:52
  • So $O$ is the tangent line at $O$. But what is $(OO)R$? Is it the third intersection point of the cubic $C$ and the line that is defined by the points $OO$, $R$? $$$$ And at $O(PQ)$, is $PQ$ the third intersection point of the cubic $C$ and the line that is defined by the points $P$, $Q$? And is $O(PQ)$ the third intersection point of the cubic $C$ and the line that is defined by the points $O$ and $PQ$ ? $$$$ Why does the equality $(OO)R=O(PQ)$ hold? – evinda Jan 13 '15 at 11:22
  • @111 I edited my post. Could you take a look at it? – evinda Feb 05 '15 at 00:03

1 Answers1

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Let us write $A*B$ for the third point in the intersection of the cubic with the line through $A$ and $B$. If we fix a point $O$ on the cubic we define $A+B=O*(P*Q)$.

Remark

For each point $P$ of $C$, $-P=(O*O)*P$.

Infact $P+X=O$ is equivalent to $O*(P*X)=O$, implying that $O*O=P*X$, that gives (applying $*P$ to both sides) $(O*O)*P=(P*X)*P$. And the righthandside has to give $X$.

i) Assume $P,Q,R$ collinear points of $C$ such that $P+Q+R=O$. This equations leads to $P+Q=-R=(O*O)*R$.

Now $P+Q=O*(P*Q)$ and collinearity tells us that $P*Q=R$, so $$O*R=(O*O)*R.$$ This can be seen as $$R*(O*R)=O*O.$$ Now observe that $(R*(O*R))=O$ and that then $O*O=O$, that is exactly saying that $O$ is an inflection point.

ii) Assume that $O$ is an inflection point, i.e. $O*O=O$ and let us take three collinear point $P,Q,R$, i.e. $P*Q=R$.

Then $P+Q=O*(P*Q)=O*R=(O*O)*R=-R$.

Pgatti
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