Chiffrement de Hill (1929)

soit $m \in \mathbb{Z}_{> 0},\; \EuScript{P}$= $\EuScript{C}$= $\frac{\mathbb{Z}}{26\mathbb{Z}}^{m}$

$\EuScript{K} = GL_{m}(\frac{\mathbb{Z}}{26\mathbb{Z}}) = \{A \in M_{m}(\frac{\mathbb{Z}}{26\mathbb{Z}}) \mid A \; inversible\}$

$K = (k_{1}, k_{2}, ... ,k_{m}) \in \EuScript{K},\; x=(x_{1},x_{2}, ..., x_{m}) \in \EuScript{P},\; y=(y_{1},y_{2}, ..., y_{m}) \in \EuScript{C}$

$\mid \EuScript{K} \mid\; = 26^{m^{2}}(1- \frac{1}{2})(1- \frac{1}{13})(1- \frac{1}{2^{2}})(1- \frac{1}{13^{2}})...(1- \frac{1}{2^{m}})(1- \frac{1}{13^{m}})$

Remarque : en général $\mid GL_{m}(\frac{\mathbb{Z}}{n\mathbb{Z}}) \mid\; = n^{m^2} \prod_{p\vert n}^{} ((p-\frac{1}{p})(p-\frac{1}{p^{2}})...(p-\frac{1}{p^{m}}))$ avec $p$ premier.

$E_{K}(x) = xK = y$

$D_{K}(y) = yK^{-1} = x$

vérifions (1) : $D_{K}oE_{K}(x) = (xK)K^{-1} = x$