I need to solve few similar exercises. One of them is:
Add the polynomials (bytes) in the ring $Z_2 [x]/(x^8+x^4+x^3+x+1)=GF(2^8)$
a) ‘57’+’02’
b) ‘03’+’03’
c) ‘FF’+’0F’
I try to learn it by myself. I bought a book with some exercises and solutions, but I can't find anything similar to that. I searched whole YouTube, and check other topics on this site. I'm not asking to solve this one, but maybe someone can send a link to similar exercises and tutorials on how it supposed to be solved? Or can write what are the steps to do so on other examples?
Thank you for any hints in that matter!
UPDATE: I think I understood how to calculate it. I was lookin through the Internet with wrong keywords. I calculated each answer. Could someone tell me is it right?
a) $001010101, x^6+x^4+x^2+1$
b) $0$
c) $11110000, x^7 + x^6 + x^5 + x^4$
Welcome to MSE!
The field has 256 elements and are all of the form $\sum_{i=0}^7 a_ix^i$.
Addition is easy as in the usual polynomial ring and is done component-wise:
$\sum_{i=0}^7 a_ix^i + b\sum_{i=0}^7 a_ix^i = \sum_{i=0}^7 (a_i+b_i)x^i$
where the coefficients $a_i,b_i$ are added modulo 2.
‘57’+’02’ or in binary '00101111 + 00000010'
Here $a(x) = x^5+x^3+x^2+x+1$ and $b(x)=x$.
So the addition yields '00101101' or
$a(x)+b(x) = x^5+x^3+x+1$.