From: paul@fatmans.demon.co.uk
To: cypherpunks@toad.com
Message Hash: 76fa5e275a9d8ef3e9246b0c0db5a399418dac262fca39f16b4e4617e8bdbdd8
Message ID: <842896368.27767.0@fatmans.demon.co.uk>
Reply To: N/A
UTC Datetime: 1996-09-17 02:40:03 UTC
Raw Date: Tue, 17 Sep 1996 10:40:03 +0800
From: paul@fatmans.demon.co.uk
Date: Tue, 17 Sep 1996 10:40:03 +0800
To: cypherpunks@toad.com
Subject: Diffie Hellman - logs in Galois fields
Message-ID: <842896368.27767.0@fatmans.demon.co.uk>
MIME-Version: 1.0
Content-Type: text/plain
Hi all,
A question for the matematicians out there:
I am looking at the Diffie Hellman public key exchange protocol and
am trying to find out why it is computationally hard to take logs in
a finite (Galois) field.
My maths tutor has told me a bit about the construction of Galois
fields (If I`m correct the construction is Z mod N, N some integer,
then a transformation F(x) on the residue classes already in the
field) I know also the definition is that there are P**k elements, p
a prime.
My questions are as follows:
1. How can a field be finite, as by definition it has to be closed
under addition, subtraction, multiplication and division???? (sorry
if this one is a bit of a no brainer, maybe the definition is
different but I can`t seem to see how)
2. Why is taking logs in a finite field computationally hard? - Me
and Alec (My maths tutor at college) guessed that it is because
exponentiation and logs are each others inverse functions, and
somehow this becomes a one way function in a finite field.
3. Are the Galois fields used in Diffie Hellman specially constructed
in any way or are they just normal GF????
Thanks for any help anyone can give me....
Datacomms Technologies web authoring and data security
Paul Bradley, Paul@fatmans.demon.co.uk
Http://www.fatmans.demon.co.uk/crypt/
"Don`t forget to mount a scratch monkey"
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