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It is not possible to draw tangent at all the points of a singular curve. What is the specialty of this and how it is related to cryptography and elliptic discriminant?

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  • $\begingroup$ In ECC we use elliptic curves, where by definition are non-singular. But you can use varieties in cryptography. For instance, you can use Jacobian varieties, because they form a group where dlog is hard. If it occurs non-singular curves to form a group, whee dlog is hard, then they will also used such curves. $\endgroup$ – 111 Jan 25 '17 at 0:29
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Not-so-useful answer: An elliptic curve is by definition a non-singular curve. Therefore by definition we use non-singular curves in elliptic-curve cryptography.

Why not use singular curves? It turns out that the group structure on those curves is isomorphic to the multiplicative group of a (quadratic extension of) a field. Therefore the discrete logarithm problem is not harder than of that in a finite field, so there is little reason to use the curve in the first place.

The discriminant characterizes all curves which are non-singular. In other words, the curves that are non-singular are exactly those for which the discriminant $\Delta$ is non-zero.

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