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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.

Ifound 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.

The proof is the following:

If $P, Q, R$ are three intersection points of $C$ with a line, then $R=PQ$. $-R=P+Q$ if and only if $-R=(OO)R=O(PQ)=P+Q$. This hols exactly when $(OO)R=OR$ or otherwise $O=OO$ that means that exactly when $O$ is an inflection point.

Could you explain it 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 points of the elliptic curve and -R=(OO)R.

P,Q,R are collinear, so let R=PQ. P+Q+R=O iff P+Q=-R iff (PQ)O=(OO)R => RO=(OO)R. 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?

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  • $\begingroup$ $P+Q$ according to the addition law in the elliptic C, is the symmetrical (with respect to Ox-axes) of $R$, say $R'.$ Thus $R+R'=O$ $\endgroup$
    – 111
    Jan 13 '15 at 0:34
  • $\begingroup$ @111 What does this mean: (OO)R=O(PQ) ? $\endgroup$
    – evinda
    Jan 13 '15 at 0:40
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    $\begingroup$ 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.... $\endgroup$
    – 111
    Jan 13 '15 at 0:52
  • $\begingroup$ 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? $\endgroup$
    – evinda
    Jan 13 '15 at 11:22
  • $\begingroup$ @111 I edited my post. Could you take a look at it? $\endgroup$
    – evinda
    Feb 5 '15 at 0:03
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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 $P+Q=O*(P*Q)=O*R=(O*O)*R=-R$.

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  • $\begingroup$ I edited my post.. Could you take a look at it? $\endgroup$
    – evinda
    Feb 5 '15 at 0:03

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