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!
Me neither. I always just thought Z2/P(x) is just notation to mean a+b alpha. I don’t understand, why he calls x +p(x) -> alpha. Or why alpha^2=alpha+1 then implies Z2/P(x) ={a + b alpha| a,b in Z2}.
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
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.
Fantastic channel ... I don't know how I could have missed this type of content on this platform. Many thanks Professor. Greetings
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!
Also note that the finite field of order 4 is NOT isomorphic to the integers mod 4. That's a somewhat common misconception.
The integers mod 4 isn’t a field because it’s not a division ring
At 5:58 Z alpha (alpha) is actually Z2 (alpha)
I've rarely ever had less doubt that someone completely understands something...
Very nice explanation, I really thanks for this video Sir.
I don't get the part at 3:18 - 3:55 . Can someone explain to me?
Me neither. I always just thought Z2/P(x) is just notation to mean a+b alpha. I don’t understand, why he calls x +p(x) -> alpha. Or why alpha^2=alpha+1 then implies Z2/P(x) ={a + b alpha| a,b in Z2}.
thank you king
Sir can u make some more videos on finite field?
Take the product and use synthetic division keep the remainder.
thank you sir Please I want to solve practical exercises in fields and groups
Thank you!!
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
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.
Nice !
Why x^2+x^2 become zero?
Coefficients of polynomials are in Z_2. 1+1=0
من فضلك أريد حل تمارين عملية في حقول ومجموعات👍👍👍👍👍👍
Any standard abstract algebra textbook have ton of exercises. Try it on your own and seek help from online forums If you’re stuck
why is 2x^2=0?
why are you screaming?
I think i can't understand this type of math
but how x³ , x⁴ are not needed ???
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.