in the Montgomery Curve using Affine coordinates, we have points that lie on either the (0,0) coordinate or have a y-coordinate of 0. The question here is: Can we use these points as the Generator Point (Base Point)? Because if we did, then the denominator of the slope in the point doubling process would be zero, and we know there's no multiplicative inverse for zero. So, How to solve this?
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1$\begingroup$ Any member of a group is a generator of a subgroup, thus any point on a Montgomery curve is a "Generator Point (Base Point)" for some subgroup of the curve. Depending on the characteristics we want for that subgroup, we can or we cannot use certain points as generator. Hence, the question is lacking a characterization of the required properties of the generated subgroup. That could be e.g. "is the whole group", "has large prime order", "has order more than 1", or "none". And that changes the answer. $\endgroup$– fgrieu ♦Commented Mar 21 at 9:30
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$\begingroup$ @fgrieu , the equation of the slope in the point doubling for Montgomery curve in the affine coordinate is (((3*X^2)+(2*Ax)+1) * Multiplicative invers of (2*By)) according to (en.wikipedia.org/wiki/Montgomery_curve#Doubling) and since the y-coordinate for the point (0,0) is zero, then there is no multiplicative inverse. how to make the calculations in this situation ? $\endgroup$– Nawras HusseinCommented Mar 21 at 10:12
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Any member of a group is a generator of a subgroup, thus any point on a Montgomery curve is a "Generator Point (Base Point)" for some subgroup of the curve.
If in a Montgomery curve group we choose as generator a point $P$ with y-coordinate of $0$, then doubling of this point yields $P+P=\infty$ (that is the group's unity, aka neutral, aka point at infinity). That can not be computed by the formulas that apply for doubling of other points, much like in point addition $Q+(-Q)=\infty$ requires special treatment. The resulting subgroup is $\{\infty,P\}$, which has order $2$.
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$\begingroup$ The first paragraph is problematic since the $\mathcal {O} := \infty$ must be excluded from generators. $\endgroup$– kelalakaCommented Mar 21 at 19:54
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$\begingroup$ @kelalaka: $\{\infty\}$ is a proper subgroup, and it's a semantic issue/decision if $\infty$ is a proper generator for that group. $\endgroup$– fgrieu ♦Commented Mar 22 at 6:30