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My question is about geometric interpretation of addition. It is known that if you have two points $A,B$ on elliptic curve $E$, draw straight line passing through $A$,$B$ and thanks to the Bezout theorem, curve and points intersect at the third point $C$. $A+B$ turns out to be reflection of third point $C$ with respect to $x$-axis.

Now, let $A=(x,y)$ is on the curve then the inverse of point $A$ is $-A=(x,-y)$. When you draw straight line passing through $A$ and $-A$, there is no third intersection point. Of course, I know that $A + (-A)= O$ where $O$ is point at infinity. Even I say that let say third intersection point is $O$ then how can I take its reflection w.r.t the $x$-axis?

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3 Answers 3

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COMMENT.-When the elliptic curve is referred to its Weirstrass form you have trivially symmetry respect to $x$-axis, for example $E$, defined by $y^2=x^3-5x+4$. You have written that $(x,-y)$ is the opposite of $(x,y)$ because you know the identity element is the point at infinity which have been choosen by comfort.

But the zero of the group law, it is well known that, can be any point. For example take as zero in $E(\Bbb Q)$ the point $(3,4)$ in the attached figure. If $A$ and $B$ are rational points of $E$ then $C$ is the sum $A+B$ and the opposite of $C$ is $D$ but these opposite points have distinct $x$-coordinate.

All this is quite different but, does this change the answer to your question? Try to see that it doesn't.

enter image description here

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  • $\begingroup$ While it's true that any point on an abstract elliptic curve can function as the identity element, the only points on an elliptic curve embedded in $\Bbb P^2$ which can function as the identity under the standard geometric construction of the group law are inflection points. In particular, in your example, the tangent line through $(3,4)$ intersects your curve at $(\frac{25}{16},\frac{3}{64})$, and therefore in your example $O+O\neq O$. $\endgroup$
    – KReiser
    Apr 13 at 20:30
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    $\begingroup$ .@KReiser. Thanks dear friend for your comment. I think you had a distraction because your point $(\frac{25}{16},\frac{3}{64})$ is not $2((3,4))$ but $-2((3,4))$. Regards. $\endgroup$
    – Piquito
    Apr 13 at 20:59
  • $\begingroup$ I don't think there are any problems with the prior comment: if a point $P$ could function as the identity under the standard geometric construction of the group law, $nP$ would equal to $P$ for any $n\in\Bbb Z$. Since that is not the case for your point $(3,4)$ (as $-2((3,4))=(\frac{25}{16},\frac{3}{64})\neq (3,4)$ since the tangent line to $(3,4)$ intersects the curve at $(\frac{25}{16},\frac{3}{64})$), it cannot function as the identity under the standard geometric construction of the group law. $\endgroup$
    – KReiser
    Apr 13 at 21:11
  • $\begingroup$ I have learned in my first lessons on elliptic curves that the point at infinity is chosen for comfort as zero but that another point can equally be chosen. After these first notions I learned a lot more about this topic but since almost everyone uses the point at infinity, I used it too and the arbitrary point thing was left in the past. I still think the opposite of you but I don't discard I were wrong. $\endgroup$
    – Piquito
    Apr 13 at 21:30
  • $\begingroup$ @Piquito You are correct, I think this is a good answer and don't think it should be deleted. $\endgroup$
    – Merosity
    Apr 13 at 21:33
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The Point at Infinity is the identity element on the group: $A+O=A$ for all$A$. In particular, $O+O=O$; in other words, $-O=O$. So the reflection of $O$ about the $x$-axis is $O$ itself.

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  • $\begingroup$ I have no problem about the theory. I was trying to figure out that even the intersection point is not in $\mathbb R^2$, how is that possible to take its reflection in $x$-axis? Is it something like all elements of empty set satisfies everything. I mean all elements of empty set is $2$ or $3$ otherwise there is an element which is not $2$ or $3$. So, contrad. $\endgroup$
    – Fuat Ray
    Apr 13 at 18:18
  • $\begingroup$ Well, $O$ is obviously a special case, brought into the picture so that (i) the group has an identity element and (ii) $A-A$ is well-defined (these are really the same thing). So it's not surprising that it doesn't behave like a regular point in $\Bbb R^2$. If you understand the theory, you can forget about reflections in the $x$-axis. $\endgroup$
    – TonyK
    Apr 13 at 18:38
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Let's cast everything in $\Bbb P^2$. Then $A+B+C=0$ in the group law iff there is a line $L\subset\Bbb P^2$ such that $L\cap E$ is exactly $A,B,C$ (where we're counting points with multiplicity). So the line through $A$ and $B$ meets $E$ in a third point which must be $-(A+B)$, and the line through $O$ and $-(A+B)$ meets $E$ in a third point, which must be $(A+B)$.

The process of "reflecting a point $P\neq O$ over the $x$-axis" is exactly constructing the third point of intersection of the line $OP$ with $E$: the projective line through $P=[a:b:1]$ and $O=[0:1:0]$ is cut out by $X-aZ=0$, which when dehomogenized is the line $x=a$, the vertical line through $P$, which intersects the curve $E$ (cut out by $y^2=x^3+c_4x+c_6$) in the points $(a,\pm b)$.

In the case when we're trying to construct the third point of intersection of a line $L$ with $E$ when $L$ meets $E$ with multiplicity 2 at $O$, we have that $L$ is the tangent line to $O$, which is the projective line $Z=0$, and this line intersects $E$ with multiplicity 3 at $O$, so the third point of intersection is just $O$.

(Of course, there's also a much more low-tech way to reflect points over the $x$-axis which even gives the right answer when applied to the point at infinity: say $[a:b:c]$ is a point, then $[a:-b:c]$ is its reflection over the $x$-axis, and sure enough, $[0:1:0]\mapsto [0:-1:0]=[0:1:0]$.)

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