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If I understand correctly, the identity point on all elliptic curves is the point at infinity.

But on the Edwards curve, this can be written in Affine form?

  • Does this have something to do with the fact that Edwards curve formulas are complete?

  • What would the point (0,1) be on the Weierstrass curve signify? If b=1, then we can represent this point on the Weierstrass curve?

  • It still has not clicked with me, the fact that the point at infinity is in the group, yet it cannot be represented? But somehow it can b represented on the Projected form, which is another form of the same curve? Whereas Montgomery is a different curve entirely?

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In mathematics, every group need and identity element. The points on the elliptic curves forms a group. Therefore they need an identity element, too.

The identity element is determined by the addition law of the curve. Some curves need a point at infinity $\mathcal{O}$ as the identity element (neutral element).

In the Edwards curve, the neutral element is chosen as $(0,1)$ and this can be represented in the affine coordinates. Other coordinate like $(0,-1)$ can be identity, however, this requires different formulas.

If I understand correctly, the identity point on all elliptic curves is the point at infinity.

No not necessarily. Edwards curves doesn't need a point at infinity.

But on the Edwards curve, this can be written in Affine form?

Yes, it is $(0,1)$

Does this have something to do with the fact that Edwards curve formulas are complete?

No, it is just by the equation of the curve and the defined addition law. This is very similar to the analogous to clock

What would the point (0,1) be on the Weierstrass curve signify? If b=1, then we can represent this point on the Weierstrass curve?

Converting formulas exist like this one

it still has not clicked with me, the fact that the point at infinity is in the group, yet it cannot be represented? But somehow it can b represented on the Projected form, which is another form of the same curve? Whereas Montgomery is a different curve entirely?

An birationally equivalence doesn't say about the representation of the elements. Consider the isomorphism of two group, they are same but they can be completely defined over different sets.

The the neutral element can be defined but you force it to be defined over the affine coordinates. Some curves need the protective coordinates so that every element can be represented with some coordinates.

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  • $\begingroup$ Thank you. - Is the identity point on all weierstrass curves the point at infinity? - Can you link me to a post to see how to find out what the identity point is on an edwards and weierstrass curve? $\endgroup$ – WeCanBeFriends Sep 22 at 8:33
  • $\begingroup$ With a birational equivalence, my confusion is that the identity element is in the set for Weierstrass curves, but it cannot be represented in affine form, while in edwards curves it can be represented $\endgroup$ – WeCanBeFriends Sep 22 at 8:35
  • $\begingroup$ So for Edwards curve, (0,1) is the identity element but not the point at infinity? Whereas for all Weierstrass curves, the identity element is the point at infinity, which cannot be represented as (x, y) ? $\endgroup$ – WeCanBeFriends Sep 22 at 8:37
  • $\begingroup$ It seems that there is a contradiction in your answer; First it says "The point at infinity is the identity element (neutral element) of the additive group" and then you say "No not necessarily" to the statement on the identity point being the point at infinity on elliptic curves. Could you clarify? $\endgroup$ – WeCanBeFriends Sep 22 at 8:39
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    $\begingroup$ I'll update give me some time. $\endgroup$ – kelalaka Sep 25 at 11:31

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