1
$\begingroup$

If irreducible polynomial $m(x) = x^8+x^4+x^3+x+1$ is chosen, or even for any other value, the multiplicative inverse will not exist for $01$, as $0000 0001$ will perfectly divide $m(x) = 100011011$ leaving remainder zero.

So then how it the value for $01$ mapped to $7c$ in AES S-box?

$\endgroup$
  • 4
    $\begingroup$ The inverse of $1$ is $1$ (in any ring). $\endgroup$ – yyyyyyy Mar 10 '18 at 8:09
1
$\begingroup$

$00000001$ is its own inverse in the Rijndael field, because polynomial multiplication by itself gets $00000001$ and is unchanged by the modulo operation. This will always be true in any ring, as noted in a comment above.

However, the specification of the Rijndael S-box is that after taking the inverse of the number in the Galois field, you have to multiply the result by a matrix then add (XOR) a constant: the "affine transformation". That is what makes 0x01 map to 0x7C.

Note that for the purposes of the Rijndael S-box, zero's inverse in the field is considered to be zero. The affine transformation is applied to zero, getting the expected 0x63.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.