# Le cryptosystème Kyber
## STA Flashback
$$
\begin{align}
&a = 4X^2 + 7X + 1 \\
&b = 3X^2 + 2
\end{align}
$$
### a)
- $a+b = 7X^2 + 7X + 3$
- $a \cdot b = 12X^4 + 8X^2 + 21X^3 + 14X + 3X^2 + 2 = 12X^4 + 21X^3 + 11X^2 + 14X + 2$
### b)
$X^3 = -1, \, X^4 = -X$
- $a+b = 7X^2 + 7X + 3$
- $a\cdot b = -12X -21 + 11X^2 + 14X + 2 = 11X^2 + 2X -19$
### c)
$X^3 = -1, \, X^4 = -X$
- $a+b = 7X^2 + 7X + 3$
- $a\cdot b = 2X^2 + 2X + 8$

## Kyber
### a)
$R_{7,5} = \mathbb{Z}/7 \mathbb{Z}[X]/(X^5 + 1)$
Donc $X^5 = -1, \, X^6 = -X, \, X^7 = -X^2$
$$
\begin{align}
t &= a \cdot s + e \\
&= (X^4 + 3X^2 + 6X + 2) \cdot (X^3 - X - 1) + X^3 \\
&= -X^2 + 1 -X^4 -3 - 3X^3 - 3X^2 + 6X^4 - 6X^2 - 6X + 2X^3 - 2X - 2  + X^3\\
&= 5X^4 + 4X^2 + 6X + 3
\end{align}
$$
### b)
$$
\begin{align}
m' &= v - s\cdot u \\
&= 5 + 5X + X^2 + 2X^3 + 2X^4 - (-1 -X + X^3) \cdot (6 + 4X + 5X^2 + 4X^3 + 4X^4) \\
&= 5 + 5X + X^2 + 2X^3 + 2X^4 - (-6 - 4X - 5X^2 - 4X^3 - 4X^4 -6X - 4X^2 - 5X^3 -4X^4 + 4 + 6X^3 + 4X^4 - 5 - 4X - 4X^2) \\
&= 5 + 5X + X^2 + 2X^3 + 2X^4 - (X^2 - 3X^3 +3X^4) \\
&= 5 + 5X + 5X^3 + 6X^4 \\
m &= 1 + X + X^3
\end{align}
$$

## 1-3
### a)