Why does $\alpha = -\alpha$, where $\alpha \in F = \{0, 1, \alpha, \alpha^2\}$? $F$ is a field with 4 elements, $\{0, 1, \alpha, \alpha^2\}$, where $\alpha \neq 0$ and $\alpha \neq 1$
This is the setup for a previous exam paper question. The question is of little importance, as the solution isn't too difficult for me to understand. However one of the fast conclusions my lecturer uses is the following: (the sentence may seem out of context, but it shouldn't be that important)
"Since $(\alpha^2)^2=\alpha^4=\alpha=-\alpha$ because $\alpha^3=1$ ...." And the explanation continues to talk about the exercise.
However the confusion for me comes from how we know that $\alpha = - \alpha$? This is written in such an off-hand manner that it makes me think it must be very easy to see. But I cannot see it. I know that $\alpha^3 = 1$, and it therefore makes sense that $\alpha^4 = \alpha$. But my understanding falters thereafter.
Thanks

As mentioned in the comments, $\alpha = - \alpha$ follows at once because a field with $4$ elements has characteristic $2$ and so $2x=0$ for all $x$. This follows from Lagrange's theorem applied to the additive group of the field. Indeed, $0 = 4 \cdot 1 = (2 \cdot 1)(2 \cdot 1)$ implies $2 \cdot 1
=0$, because a field has no zero divisors.
If you want to argue from first principles, then:


*

*$-\alpha \ne 0$ because otherwise $\alpha = 0$ 

*$-\alpha \ne 1$ because otherwise $\alpha^2=1$ 

*$-\alpha \ne \alpha^2$ because otherwise because otherwise $\alpha^2=\alpha^4=\alpha$
The only possibility left is $-\alpha=\alpha$.
