What is the map $\xi$ over $E/F_q : y^2 = x^3 + b$

Question

I have read the following in Pairings for Beginners:

Consider $E/\mathbb{F}_q : y^2 = x^3 + b $. The map, $\xi$, defined by $\xi : (x, y) \rightarrow (\xi_3x,y)$ with $\xi_3^3 = 1$ and $\xi_3 \neq 1$ is a non-trivial endomorphism on $E$

What does the $\xi$ mean in this context?

Answer 1

$\xi$ is simply a dummy variable. The meaning of this variable is given directly in the sentence that you quoted:

• $\xi$ is an "endomorphism", meaning that it is a function from $E$ to $E$.
• $\xi$ maps an input point $(x,y)$ on $E$ to the output point $(\xi_3 x, y)$.

You may be wondering what $\xi_3$ is. This meaning is also given in the sentence that you quoted: $\xi_3$ is an element of $\mathbb{F}_q$ satisfying

• $\xi_3^3 = 1$ (i.e.: $\xi_3 \cdot \xi_3 \cdot \xi_3 = 1$), and
• $\xi_3 \neq 1$.

The above quantity is not strictly speaking well-defined. There is more than one element of $\mathbb{F}_q$ satisfying the stated properties. In particular, if $\xi_3$ is one such element, then $\xi_3^2 = \xi_3 \cdot \xi_3$ is a different such element.