order in $\mathbb{F}_5^*$ I have a confusion with the following problem.
Let $\alpha$ a root of $f(x)=x^2-x+2\in\mathbb{F}_5[x]$ and $\mathbb{F}=\mathbb{F}_5{(\alpha)}$. Compute the order of $\alpha$ in $\mathbb{F}^*$.
My attempt.
If $\alpha$ is a root of $f$, then
\begin{align*}
\alpha^2-\alpha+2&=0\\
\alpha^2-\alpha&=-2\\
\alpha(\alpha-1)&=3
\end{align*}
but there's no $\alpha$ that satisfies this (so I don't have a clear conclussion).
What I'm missing? or I'm doing this in a wrong way?
Thanks for your help
 A: When you say "there's no $\alpha$ that satisfies this", keep in mind that you're working in $\Bbb F_5(\alpha)$, which is a field specifically constructed to make $\alpha$ a root of that polynomial. So there is an $\alpha$ that satisfies that, namely $\alpha$!
It's a bit hard to say what $\alpha$ "looks like", since it isn't an element of $\Bbb F_5$. It's a symbol that lives in a world outside $\Bbb F_5$ and is a root of $f(x)$ in that world. (Compare to defining $i$ as a root of $x^2+1$ over the reals, giving us $\Bbb C = \Bbb R(i)$. It's hard to say what $i$ "looks like" from within the reals.)
Given that $\alpha^2 - \alpha + 2 = 0$, it follows that $\alpha^2 = \alpha - 2$. Use this to simplify the progressive powers of $\alpha$:
$$
\begin{align}
\alpha^2 &= &\alpha -2\\
\alpha^3 &= &\alpha(\alpha-2)\\
&= &\alpha^2 - 2\alpha\\
&= &(\alpha-2) - 2\alpha\\
&= &-\alpha - 2\\
\alpha^4 &= &\alpha(-\alpha-2)\\
&= &\ldots
\end{align}$$
Keep on going until you get $\alpha^n = 1$ for some $n > 1$.
