I'm working my way through some papers and ran across what seems to be division of two points that produce a third point. I'm new to ECC and am having a terrible time trying to figure out what this notation means, any thoughts?

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This is from the BLS paper: https://crypto.stanford.edu/~dabo/pubs/papers/aggreg.pdf

Point division appears on pages

  • 6 (A potential attack on aggregate signatures)
  • 18 (Ring Signing equation)
  • 1
    $\begingroup$ It is not ECC notation/ it is multiplicative group notation where the division exist! $\endgroup$
    – kelalaka
    Commented Oct 26, 2021 at 15:09
  • $\begingroup$ @kelalaka I think I see! so does division here actually correspond to subtraction in the exponent? i.e. if A = g^a, B = g^b then A / B = g^(a-b) ? $\endgroup$
    – David Rusu
    Commented Oct 26, 2021 at 15:21
  • 1
    $\begingroup$ As I observed elsewhere, if we assume additive notation, then 'point division' corresponds to 'discrete log'. That is, if we set up the equation $x = A/B$, then this should be equivalent to $xB = A$, which is the discrete log of A to the base B - hence, it is well defined (if somewhat intractable to compute). Of course, this has nothing to do with the paper, which is written in multiplicative notation... $\endgroup$
    – poncho
    Commented Oct 26, 2021 at 18:38

1 Answer 1


I don't see a point division.

I read $v_B=v'_B/v'_A$ as $v_B=v'_B\cdot({v'_A}^{-1})$ where $\cdot$ is the group law, and ${v'_A}^{-1}$ is the inverse of $v'_A$ in that group, that is such that ${v'_A}\cdot{v'_A}^{-1}={v'_A}^{-1}\cdot{v'_A}=1$, the group's neutral.

If the group was noted additively, that would be $v_B=v'_B-v'_A$, read as $v_B=v'_B+(-v'_A)$ where $-v'_A$ is the opposite of $v'_A$ in that group, that is such that ${v'_A}+(-v'_A)=(-v'_A)+{v'_A}=0$, the group's neutral.

  • $\begingroup$ Thanks! Yes this makes more sense. I'm a bit out of practice, but I'm starting to remember this now more clearly from university. $\endgroup$
    – David Rusu
    Commented Oct 26, 2021 at 19:22

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