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