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Math Help - Finite field subtraction, i.e. additive inverse of a geometric point

  1. #1
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    Finite field subtraction, i.e. additive inverse of a geometric point

    Hello

    I hope someone can help me with this.

    Let's imagine I want to perform the following subtraction operation -

    (7, 1) - (2, 3)

    Both are geometric points on an elliptic curve.

    As far as I know, finite field subtraction is simply finite field addition using the additive inverse of the bit I want to subtract, i.e. (7, 1) + [additiveInverse(2, 3)]

    Is this correct?

    Does anyone know how I can calculate the additive inverse of a point?

    Is it simply (-2, -3)?

    Any help greatly appreciated.

    Thanks!
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  2. #2
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    I think in order to answer this equation you need to say which finite field you are working in.
    And also what elliptic curve you are dealing with.

    I guess that if y^2 = x^3 + ax+b is the elliptic curve and (7,1),(2,3) \in E(\mathbb{F}_p) then 1 = 343 + 7a+b \mbox{ and }8=9+2x+b. This gives us infromation to solve for a,b and determine the curve. Now we just need to know what p is.
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  3. #3
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    Thanks for your reply.

    I think there was some misunderstanding though.

    All I'm asking is this: is the additive inverse of (2, 3) simply (-2, -3)?

    Thanks.

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  4. #4
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    Hmmm. I'm after finding this on the internet:

    J = (Xj, Yj)
    K = (Xk, Yk)

    J - K = J + (-K)
    where -K = (Xk, Xk + Yk)

    Does this ring any bells with anyone?

    Thanks
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  5. #5
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    That looks correct. I been just gazing for curves over \mathbb{R} for motivation. If you have (x,y) on the curve then the line joining (x,y) and (x,-y) passes through \infty. Thus, the inverse of (x,y) is (x,-y). Therefore, if we have a point (x,y) over a finite field \mathbb{F}_p then it seems -(x,y) = (x,-y) = (x,p-y).
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  6. #6
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    Perfect! Thank you.
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