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The S-box is defined as the generalised inverse function $S:\mathbb{F}_{2^n}\rightarrow \mathbb{F}_{2^n}$,in quotient ring $\mathcal{R}:=\mathbb{F}_{2^n}[X]/(X^{2^n}-X)$ with $S(x)=x^{-1}$, is correct $S(X):=X^{2^n-2}$. But the Euler's theorem says $x^{\varphi(n)}\equiv1\pmod{n}$,so the answer is $x^{\varphi(n)-1}=x^{2^{n-1}-1}\equiv x^{-1}\pmod{n}$,why is $S(X):=X^{2^n-2}$

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Euler's theorem is a special case of Lagrange's theorem applied to the group $(\mathbb Z/m\mathbb Z)^\times$. It can be applied in the case $m=2^n$ where $|(\mathbb Z/m\mathbb Z)^\times|=2^{n-1}$ to deduce that for any odd integer $x$ $x^{2^{n-1}-1}\equiv x^{-1}\pmod{2^n}$. However, this is different to the group $\mathbb F_{2^n}^\times$. In this case $|\mathbb F_{2^n}^\times|=2^n-1$ and we can use this to deduce that for any element $x\in\mathbb F_{2^n}^\times$ we have $x^{2^n-2}\equiv x^{-1}$. Elements of $\mathbb F_{2^n}$ are often written as polynomials in some variable, say $X$, over $\mathbb F_2$ modulo some irreducible polynomial, say $f(X)$ of degree $n$. Thus another way to express this is to say that for any polynomial $g(X)$ coprime to $f(X)$ over $\mathbb F_2$ we have $$g(X)^{2^n-2}\equiv g(X)^{-1}\pmod{\langle 2,f(X)\rangle}.$$

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  • $\begingroup$ Can you explain the notation $\langle 2,f(X)\rangle$ ? I know from context that the outcome is the reduction polynomial, but I can't figure the logic behind the notation. Ah, or is it that coefficients of the polynomial $\langle m,f(X)\rangle$ are in $\mathbb Z/m\mathbb Z$? $\endgroup$
    – fgrieu
    Dec 6, 2022 at 16:57
  • $\begingroup$ generally $\langle x_1,x_2,\dots,x_k\rangle$ is notation for the (ring-theoretic) ideal generated by the elements $x_1, x_2,\dots, x_k$. So concretely $\langle 2, f(X)\rangle = \{2r_1 + r_2f(X)\mid r_i\in\mathbb{F}_2[X]\}.$ $\endgroup$
    – Mark Schultz-Wu
    Dec 6, 2022 at 21:11

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