# An Application of Algebraic Geometry to Encryption: Tame Transformation Method

### T.T. Moh

Purdue University, West Lafayette, USA

## Abstract

Let $K$ be a finite field of $2_{ℓ}$ elements. Let $ϕ_{4},ϕ_{3},ϕ_{2},ϕ_{1}$ be tame mappings of the $n+r$-dimensional affine space $K_{n+r}$. Let the composition $ϕ_{4}ϕ_{3}ϕ_{2}ϕ_{1}$ be $π$. The mapping $π$ and the $ϕ_{i}$'s will be hidden. Let the component expression of $π$ be $(π_{1}(x_{1},…,x_{n+r}),…π_{n+r}(x_{1},…,x_{n+r}))$. Let the restriction of $π$ to a subspace be $π^$ as $π^=(π_{1}(x_{1},…,x_{n},0,…,0),…,π_{n+r}(x_{1},…,x_{n},0,…,0))=(f_{1},…,f_{n+r}):K_{n}mapstoK_{n+r}$. The field $K$ and the polynomial map ($f_{1},…,f_{n+r}$) will be announced as the public key. Given a plaintext $(x_{1},…,x_{n})∈K_{n}$, let $y_{i}=f_{i}(x_{1},…,x_{n})$, then the ciphertext will be $(y_{1},…,y_{n+r})∈K_{n+r}$. Given $ϕ_{i}$ and ($y_{1},…,y_{n+r}$), it is easy to find $ϕ_{i}(y_{1},…,y_{n+r})$. Therefore the plaintext can be recovered by $(x_{1},…,x_{n},0,…,0)=ϕ_{1}ϕ_{2}ϕ_{3}ϕ_{4}π^(x_{1},…,x_{n})=ϕ_{1}ϕ_{2}ϕ_{3}ϕ_{4}(y_{1},…,y_{n+r})$. The private key will be the set of maps ${ϕ_{1},ϕ_{2},ϕ_{3},ϕ_{4}}$. The security of the system rests in part on the difficulty of finding the map $π$ from the partial informations provided by the map $π^$ and the factorization of the map $π$ into a product (i.e., composition) of tame transformations $ϕ_{i}$'s.

## Cite this article

T.T. Moh, An Application of Algebraic Geometry to Encryption: Tame Transformation Method. Rev. Mat. Iberoam. 19 (2003), no. 2, pp. 667–685

DOI 10.4171/RMI/364