User:Vahedi.vahid

Let $(F,+,\cdot)$ be a field and $G$ be a subgroup of $(F^{*}, \cdot),$ where $F^{*}=F\smallsetminus\{0\}$ and $(\frac{F}{G}, \oplus, \odot)$ where $\frac{F}{G}=\{aG\mid a\in F\}$ with the hyperoperation and the multiplication defined by: i) $aG\oplus bG=\{cG| c\in aG+bG \},$ ii) $aG\odot bG=abG,$ for all $aG,bG\in \frac{F}{G}$. Then $(\frac{F}{G}, \oplus, \odot)$ is a hyperfield. From now on, we denote $\bar{a}=aG,$ for all $aG\in \frac{F}{G}$ and the constructed hyperfield $(\frac{F}{G}, \oplus, \odot)$ by $\bar{F},$ and call it the Krasner hyperfield.

Now let $\bar F$ be a Krasner hyperfield and $ (\bar A,\bar B)\in \bar F^{2} $ Then the

relation $\bar y^{2}\in \bar x^{3}\oplus \bar A\bar x\oplus \bar B$; for all $ (\bar x,\bar y)\in \bar F^{2} $; is called the generalized Weierstrass (reduced)

equation. Moreover, the set

$$ E_{\bar A,\bar B}=\{(\bar x,\bar y)\in \bar F^{2}|\bar y^{2}\in \bar x^{3}\oplus \bar A\bar x\oplus \bar B\}$$

is called the Weierstrass hypercurve on the Krasner hyperfield $\bar F$

If $ 0\notin \vartriangle_{\bar A,\bar B} $; where $\vartriangle_{\bar A,\bar B}=\{4x^{3}+27y^{2}|(x,y)\in \bar A \times\bar B\}$ and the following implication holds:

$$E_{a,b}\cap E_{c,d}\neq\emptyset\Rightarrow E_{a,b}= E_{c,d}, \forall a,c\in \bar A,b,d\in \bar B,$$;

where$E_{m,n}=\{(x,y)\in F^{2}|y^{2}=x^{3}+mx+n\},$; for all $(m,n)\in F^{2},$; then we say that $E_{\bar A,\bar B}(F)=E_{\bar A,\bar B}\cup{\infty}$ is an

elliptic hypercurve.

Without losing the generality, we assume that $E_{\bar A,\bar

B}\neq\emptyset$. We can do this because there exists a

field extension of $F$ where the Weierstrass equation has a solution.

By selecting the appropriate $G$, $E_{\bar A,\bar B}(F)$ has a regular hypergroup structure.

see for more information.