Timeline for In AES is 0x11b is the specified field generator? [duplicate]
Current License: CC BY-SA 4.0
9 events
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| Jan 4, 2019 at 18:08 | history | closed |
kelalaka Ella Rose |
Duplicate of Design properties of the Rijndael finite field? | |
| Jan 4, 2019 at 17:55 | review | Close votes | |||
| Jan 4, 2019 at 18:08 | |||||
| Jan 4, 2019 at 17:54 | comment | added | kasperd | 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. | |
| Jan 4, 2019 at 17:50 | comment | added | j.p. | 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. | |
| Jan 4, 2019 at 17:43 | comment | added | kelalaka | See Why generator polynomial of $GF(2^m)$ are irreducible? for a complete answer | |
| Jan 4, 2019 at 17:38 | comment | added | mark | @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? | |
| Jan 4, 2019 at 17:25 | review | First posts | |||
| Jan 4, 2019 at 18:08 | |||||
| Jan 4, 2019 at 17:25 | comment | added | kelalaka |
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 $GF(2^8)$
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| Jan 4, 2019 at 17:21 | history | asked | mark | CC BY-SA 4.0 |