$x+y=x-y$ in $\mathbb F_q$ when $q$ is even.

Question

I had the following doubt while reading a proof :

If we have a finite field $\mathbb F_q$ where $q$ is even, then show that $x+y=x-y$ in $\mathbb F_q$ where $x,y \in \mathbb F_q$.

kindly help..

Answer 1

If $q$ is even, $q=2^n$ and the characteristic of $F_q$ is $2$, that is $-1=1$ and $x+y=x-y$ since $-y=y$.

Answer 2

It is easy to see that you only need to show that for all $x\in F_q$ we have $x+x=0$.

First, you need to note that since $F_q$ is a field, it is in particular a domain^$\dagger$, so its characteristic must be a prime number $p$ (as it can't be zero, since it is finite).

Second, note that $(F_q,+)$ is an abelian group of exponent $p$ (by definition of characteristic). By classification of finitely generated abelian groups, the only finite abelian groups of exponent $p$ are $({\bf Z}/p{\bf Z})^n$, which have order $q=p^n$. Since we had assumed that $q$ is even...

($\dagger$ It is a nice exercise to show that a finite domain is always a field.)