The problem I'm working with:
As mentioned in the book, the quadratic polynomial $P(x) = x^2+x+1$ is irreducible over $\mathbb{Z_2}$ Hence, we can form the Galois field $$F = \mathbb Z/2\mathbb Z$$
Let $\alpha$ be the "Galois imaginary" associated for this field - that is, an element such that $$\alpha^2+\alpha+1 = 0$$
Find all the roots of the equation $x^2+\alpha = 0$ in $F$.
Here is my thought process:
Since $P(x)$ is a quadratic equation, it can have at most $2$ roots and let those $2$ roots be called $r_1$ and $r_2$. So we need to find these two roots such that $$x^2+\alpha+1 = (x-r_1)(x-r_2)$$ So since we are working in $\mathbb{Z_2}$, I can write $$\alpha^2 = \alpha + 1$$ Then we have
$\alpha^3 = \alpha^2\cdot \alpha = (\alpha+1)(\alpha) = \alpha^2+\alpha = (\alpha+1)+ \alpha = 1$
Therefore, $\alpha$ has order 3 and we have $x = \pm (1+\alpha)$. However, my friend told me that I didn't specify which element of the field F is $-(1+\alpha)$ What does he mean by this? Also, is my thought process correct?