Since you're in characteristic 2, another cool way to calculate (α+1)(α+1) is through the Frobenius map. I've really been enjoying your content, especially since you go in so many different interesting directions. And you've been on a tear, putting out new videos!
I heard that every finite field has a size that is prime or a power of a prime I could understand fields of prime order: Zp where p is prime, but I couldn’t understand fields with job prime order
I don't get the part at <a href="#" class="seekto" data-time="198">3:18</a> - <a href="#" class="seekto" data-time="235">3:55</a> . Can someone explain to me?
Can someone help explain why AES , in GF(2^8), uses the irreducible polynomial x^8 + x^4 + x^3 + x +1 while the Twofish algorithm, in GF(2^8), uses a different irreducible polynomial x^8 +x^6 +x^3 +x^2 +1 ? Wont both polynomials produce the same values if there is only one finite field of order 2^8? I was told never to mix polynomials -- pick one polynomial and stick with it.
They're not "mixing" them. AES defined the field in one (legal) way; Twofish defined them in a different (also legal) way. They're incompatible, but that doesn't matter, because no one is "hotwiring" or "mixing" internal pieces of AES implementations into Twofish implementations.
The different versions of GF(2^8) are _isomorphic_ (8:33), which means “the same, if you squint”-you can map between the elements of the different versions in a way that preserves multiplication Map(a)Map(b)=Map(ab) and vice versa-but the different versions of the field look obviously different from each other, if you're just looking at the polynomials. If Wikipedia is to be believed, GF(2^2) is the ONLY case where you have just 1 option of irreducible polynomial.
I think you can take any polynomial p that’s degree at least 2 and use Euclidean division to express it as p = (x² + x + 1)q + r. So then modulo x² + x + 1, this just looks like r which is degree strictly less than 2.
