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

• June 15th 2008, 09:04 AM
MrSteve
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!
• June 15th 2008, 09:06 AM
ThePerfectHacker
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.
• June 15th 2008, 09:34 AM
MrSteve

I think there was some misunderstanding though.

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

Thanks.

:)
• June 15th 2008, 10:44 AM
MrSteve
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
• June 15th 2008, 11:02 AM
ThePerfectHacker
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 $-(x,y) = (x,-y) = (x,p-y)$.
• June 15th 2008, 12:43 PM
MrSteve
Perfect! Thank you. :)