Trying to wrap my brain around finite fields. I get how one can construct a finite field with an order of a prime number, but I don't get how it works with powers of primes. Everything I try to read on the subject eventually ends up getting into notation that I don't know how to read.
I think I get that a GF(p^n) has something to do with converting the field into a polynomial where all the coefficients are of GF(p), but that's where my understanding starts to fall apart.
Can anyone point me at something that will help me to better understand this?
Jonathan Lamothe
in reply to Jonathan Lamothe • •I found this article, which brings me a little closer to understanding, but:
How in the hell did they arrive at those values?
Jonathan Lamothe
in reply to Jonathan Lamothe • •I mean, I get why it would need to contain an x^3, but is finding a modulus that works just a matter of trial and error until one lands on one that works, or is there a way to calculate this?
If the former, how did Galois work out that it's always possible to do for any power of a prime?
Matthew Skala
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I think the polynomials that work are irreducible polynomials - the ones that can't be factored into smaller pieces. In that way they are analogous to prime numbers.
So just like GF(p) works for p=2, 7, 23, other prime numbers, but not 12 (which isn't a prime number), GF(2^3) works for a polynomial that is like a prime in the sense it can't be factored, and not for a polynomial that can be factored over GF(2).
There is some trial and error involved just as with finding prime numbers.
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Jonathan Lamothe
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