Abstract Algebra | Constructing a field of order 4.

We use the standard strategy involving a quotient of the polynomial ring Z2[x] by a maximal ideal in order to construct a field of order 4.
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[@davidescobar7726]

Fantastic channel ... I don't know how I could have missed this type of content on this platform. Many thanks Professor. Greetings

[@jordancourtemanche562]

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!

[@josh34578]

Also note that the finite field of order 4 is NOT isomorphic to the integers mod 4. That's a somewhat common misconception.

[@cameronspalding9792]

The integers mod 4 isn’t a field because it’s not a division ring

[@kevinsuryapranata]

Very nice explanation, I really thanks for this video Sir.

[@paul21353]

At 5:58 Z alpha (alpha) is actually Z2 (alpha)

[@Adityarm.08]

Thank you!!

[@eh0394]

thank you king

[@mac2105]

I've rarely ever had less doubt that someone completely understands something...

[@user-uh9se1ks9j]

thank you sir Please I want to solve practical exercises in fields and groups

[@julien31415]

Nice !

[@tLearningTree]

I don't get the part at 3:18 - 3:55 . Can someone explain to me?

[@cameronspalding9792]

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

[@howardcheung8304]

Sir can u make some more videos on finite field?

[@mark_tilltill6664]

Take the product and use synthetic division keep the remainder.

[@Jim-vr2lx]

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.

[@codegeek98]

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.

[@codegeek98]

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.

[@darkguild8624]

Why x^2+x^2 become zero?

[@henk7747]

Coefficients of polynomials are in Z_2. 1+1=0

[@user-uh9se1ks9j]

من فضلك أريد حل تمارين عملية في حقول ومجموعات👍👍👍👍👍👍

[@ibrahimn628]

Any standard abstract algebra textbook have ton of exercises. Try it on your own and seek help from online forums If you’re stuck

[@vanessamichaels9512]

why is 2x^2=0? why are you screaming?

[@maxamedaxmedn6380]

I think i can't understand this type of math

[@KurdaHussein]

but how x³ , x⁴ are not needed ???

[@praneeshkumar7341]

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.