I'm trying to understand the process of ECDSA signing. I already figured out the process of generating a public key out of a private key by multiplying the private key with G.

The equation for ECDSA signing is as follows: $S = k^{-1} (\text{hash} + dA * R) \pmod p$. What I'm trying to understand is the process of multiplying $dA$ (my private key) by $R$ ($x$ of public key generated from random private key). Is it a "simple" multiplication (such as 5*5 = 25) or the same way we multiply $k$ by $G$ to generate a public key?

  • $\begingroup$ The public key is G added to itself repeatedly, with the number of G in the sum defined by the private key (and a speedup method for large number), which IMHO is more multiplying G by the private key than it is "multiplying the private key with G". $\endgroup$ – fgrieu Jan 11 '18 at 11:11
  • $\begingroup$ @fgrieu: point multiplication is used for Q and R, but not S. $\endgroup$ – dave_thompson_085 Jan 12 '18 at 4:37
  • $\begingroup$ @dave_thompson_085: yes; I was criticizing the vocabulary used in the question's first sentence. $\endgroup$ – fgrieu Jan 12 '18 at 8:57

Is it a "simple" multiplication (such as 5*5 = 10) or the same way we multiply k by G to generate a public key?

It's multiplication modulo the curve order. For example, if dA = 9, R = 7 and p = 13, then dA * R = 9 * 7 mod 13 = 63 mod 13 = 11.

  • $\begingroup$ Conventionally we use n for the order of the group = order of G on the curve, and p for the modulus of the underlying field if $F_p$ (and occasionally the reducing polynomial if $F_{2^m}$). These are not the same, although for cofactor 1 (common for ECDSA as used, like Suite B and Bitcoin) they are close due to Hasse. $\endgroup$ – dave_thompson_085 Jan 12 '18 at 4:37

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