# ECDSA signing vs generating public key

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?

• 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".
– fgrieu
Commented Jan 11, 2018 at 11:11
• @fgrieu: point multiplication is used for Q and R, but not S. Commented Jan 12, 2018 at 4:37
• @dave_thompson_085: yes; I was criticizing the vocabulary used in the question's first sentence.
– fgrieu
Commented Jan 12, 2018 at 8:57

• 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. Commented Jan 12, 2018 at 4:37