0

As I know all arithmetic operations in AES are done over $GF(2^8)$ so I want to know which field generator AES uses? is it 0x11B? and choosing another field generator would it give a different results?

mark
  • 101
  • Related Design properties of the Rijndael finite field? and Replacing the Rijndael S-Box? Pancho's answer : Actually, the choice of irreducible polynomial is unimportant in AES; for any polynomial representation of$GF(2^8)$ – kelalaka Jan 04 '19 at 17:25
  • @kelalaka sorry for the following stupid question but would you kindly answer it : is irreducible polynomial is same as the field generator? if not what is the relation? and is $x^{8}+x^{3}+x+1$ is the same 0x11b? – mark Jan 04 '19 at 17:38
  • See Why generator polynomial of $GF(2^m)$ are irreducible? for a complete answer – kelalaka Jan 04 '19 at 17:43
  • 1
    The binary representation of $0x11b$ is $100011011$, which corresponds to the polynomial $x^8+x^4+x^3+x+1$ (you forgot the $x^4$ term, and yes, this is the field generator used by AES. You should use it, if you care not just about the theoretical security, but also about getting the correct result. – j.p. Jan 04 '19 at 17:50
  • 1
    A different generator would result in a different incompatible cipher. The choice of generator shouldn't influence the security, but you shouldn't use a generator which hasn't received as much scrutiny at the one in the standard. The generator in the standard had to be a nothing-up-my-sleeve number. – kasperd Jan 04 '19 at 17:54

0 Answers0