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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?

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  • $\begingroup$ 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 $GF(2^8)$ $\endgroup$
    – kelalaka
    Jan 4, 2019 at 17:25
  • $\begingroup$ @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? $\endgroup$
    – mark
    Jan 4, 2019 at 17:38
  • $\begingroup$ See Why generator polynomial of $GF(2^m)$ are irreducible? for a complete answer $\endgroup$
    – kelalaka
    Jan 4, 2019 at 17:43
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    $\begingroup$ 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. $\endgroup$
    – j.p.
    Jan 4, 2019 at 17:50
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    $\begingroup$ 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. $\endgroup$
    – kasperd
    Jan 4, 2019 at 17:54

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