finite field question

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qubits1
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finite field question

Postby qubits1 » Thu Mar 10, 2016 11:09 pm UTC

Wikipedia article on Finite Field Arithmetic says addition and subtraction are equivalent operations a+b=a^b and a-b=a^b.
I'm trying to implement multiplication and division in GF(256) by using the identity a*b=antilog(log(a)+log(b)) mod 255 and a/b =anitlog(log(a)-log(b)) mod 255
Does this mean that the addition and subtraction in the multiplication and division identities above are equal to a^b?
Wouldn't this give you the same result for multiplication and division?

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jaap
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Re: finite field question

Postby jaap » Fri Mar 11, 2016 1:10 am UTC

You seem to be assuming that log in this context is a function from GF(2n) to GF(2n), but it is not.
The multiplicative group of GF(2n) is cyclic of order 2n-1. If g is a generator of that group, then you are multiplying two elements a=ge and b=gf as a*b = gegf = ge+f = c. The addition of these exponents is not done in the field GF(2^n) itself but in (an additive version of) this cyclic group. So log is a function from GF(2n) to Z_(2n-1).


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