In the field $F_{3^2} = F_3[x]/(2x^2 + x + 1)$, let $\alpha$ be the root and primitive element. The elements of $F_{3^2}$ can be represented in 2 ways:

*

*As polynomials (degree $\leq 1$) in $\alpha$ with coefficients in $F_3$

*As $0, 1, \alpha, \alpha^2, \dots, \alpha^{3^2 - 2}$.

And the $2$ ways can be related by using the fact that
$2\alpha^2 + \alpha + 1 = 0$ in $F_3[\alpha]$.
But that is true only in $F_3[\alpha]$?
For e.g. I know $\alpha^5 = 2\alpha$ in $F_3[\alpha]$. But if I want to find $\alpha + \alpha$ in $F_{3^2}$, that will be $2\alpha$, where $2$ is just an integer, that will not be $\alpha^5$ right?

First of all, your skepticism about $\alpha^5 = 2\alpha$ is misplaced. You are absolutely correct about that equation.
We can simplify the minimal polynomial for $\alpha$ over $F_3$. Since $2 \equiv -1 \pmod{3}$, the equation $2\alpha^2 + \alpha + 1 = 0$ in $F_3[\alpha]$ is equivalent to the equation
$$
\alpha^2 = \alpha + 1.
$$
We can treat this as a power-reducing identity to inductively express each power of $\alpha$ (your second representation) as an $F_3$-linear combination of $1$ and $\alpha$ (your first representation).
Here's what this looks like:
$$
\alpha^3 = \alpha \, \alpha^2 = \alpha (\alpha + 1) 
= \alpha^2 + \alpha = (\alpha + 1) + \alpha = 2\alpha + 1
$$
and
$$
\alpha^4 = \alpha \, \alpha^3 = \alpha (2\alpha + 1) 
= 2\alpha^2 + \alpha = 2(\alpha + 1) + \alpha = 2
$$
etc. It's interesting to recalculate $\alpha^4$ as $\alpha^2 \, \alpha^2$ instead:
$$
\alpha^2 \, \alpha^2 = (\alpha + 1) (\alpha + 1) = \alpha^2 + 2\alpha + 1 
= (\alpha + 1) + 2\alpha + 1 = 2
$$
Repeating this process produces the dictionary:
\begin{array}{l|r}
\alpha^n & b\alpha + c \\
\hline 
0 & 0 \\
1 & 1 \\
\alpha & \alpha \phantom{{}+1} \\
\alpha^2 & \alpha + 1 \\ 
\alpha^3 & 2\alpha + 1 \\
\alpha^4 & 2 \\
\alpha^5 & 2\alpha \phantom{{}+1} \\
\alpha^6 & 2\alpha + 2 \\
\alpha^7 & \alpha + 2
\end{array}
Of course, $\alpha^8 = 1$, showing that all of the nonzero elements of the field form the cyclic group of order $8 = 3^2 - 1$. This is true in general.
