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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?

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    $\begingroup$ The inverse of $1$ is $1$ (in any ring). $\endgroup$ – yyyyyyy Mar 10 '18 at 8:09
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$00000001$ 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.

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