Find the Galois group of $x^3-2$ over the field $\mathbb F_5$ If the problem is "Find the Galois group of $x^3-2$ over the field $\mathbb Q$", I may know the answer confidently. But in this case I am a little uncertain about my answer as follows:
According to my calculation, $x^3-2$ is also irreducible over $\mathbb F_5$, and so the Galois extension is just $\mathbb F_{5^3}$. Hence, the Galois group is simply $Gal(\mathbb F_{5^3}/\mathbb F_5)\cong\mathbb Z_3$.
Is my argument right?
 A: We have that any irreducible polynomial $f(X)\in\mathbb{F}_q[X]$ of degree $n$ has the Galois group of $f$ over $\mathbb{F}_q$ cyclic of degree $n$, and the Galois group is generated by the Frobenius map $\phi(x)=x^q$. Succinctly
$$\text{Gal}(\mathbb{F}_q(\alpha)/\mathbb{F}_q)=\langle\phi\rangle\cong C_n$$
where the field extension $\mathbb{F}_q(\alpha)/\mathbb{F}_q$ is of degree $n$, and $\mathbb{F}_q(\alpha)$ is the splitting field for $f$ over $\mathbb{F}_q$; in particular any finite extension of finite fields has a cyclic Galois group.

So for $f(X)=X^3-2$ over $\mathbb{F}_5$ we have: as $0^3=0$, $(\pm1)^3=\pm1$, $(-2)^3=-8= 2$ then $-2=3$ is a root. Factorize:
$$ X^3-2=(X+2)(X^2-2X+4)=(X+2)(X^2-2X-1)$$
Now $-2$ is not a root of $g(X)=X^2-2X-1$ so $g$ is irreducible over $\mathbb{F}_5$, and it follows the splitting field $E$ for $X^3-2$ over $\mathbb{F}_5$ is the splitting field for $X^2-2X-1$ over $\mathbb{F}_5$ which has degree $2$ and is $\mathbb{F}_5(\alpha)$ for some root $\alpha$ of $X^2-2X-1$. The Galois group is isomorphic to the cyclic group $C_2$, $\text{Gal}(\mathbb{F}_5(\alpha)/\mathbb{F}_5)=\langle\phi\rangle\cong C_2$, where $\phi\colon x\mapsto x^5$ is the Frobenius map which generates the group. The roots of $X^3-2$ are then $-2$, $\alpha$ and $\beta=2-\alpha$ (since the sum of the roots of $X^2-2X-1$ is $2$ and so $\alpha+\beta-2=0$, or we could use the decomposition of the cubic in its splitting field:
\begin{align*}
X^3-2&=(X-\alpha)(X-\beta)(X-\gamma)\\
&=X^3-(\alpha+\beta+\gamma)X^2+(\alpha\beta+\alpha\gamma+\beta\gamma)X-\alpha\beta\gamma
\end{align*}
where $\alpha$, $\beta$, $\gamma$, with $\gamma=-2$, are the three roots). The Frobenius map $\phi$ then acts by
$$\phi\colon -2\mapsto -2,\quad\alpha\mapsto\beta\mapsto\alpha$$
So numbering the roots $1$, $2$, $3$ we have $\phi$ as the permutation $(1)(2\,3)$.

To illustrate the case where $X^3-2$ is irreducible over a finite field consider it over $\mathbb{F}_7$, then trying $0$, $\pm1$, $\pm2$, $\pm3$ as roots shows $f$ has none in $\mathbb{F}_7$ and is thus irreducible (here we can immediately see the Galois group $G$ of $f$ over $\mathbb{F}_7$ is $C_3$ and that since $f$ is irreducible this implies $G$ acts transitively on the roots). Hence its roots lie in $\mathbb{F}_{7^3}$. If $\alpha$ is some root of $X^3-2$ in $\mathbb{F}_{7^3}$, then the roots are generated by the Frobenius map, $\phi(x)=x^7$, and are 
$$\phi^0(\alpha)=\alpha^{7^0},\quad\phi^1(\alpha)=\alpha^{7^1},\quad\phi^2(\alpha)=\alpha^{7^2},\quad\phi^3(\alpha)=\alpha^{7^3}=\alpha$$
Working these out explicitly:
$$\alpha,\quad\alpha^{7}=(\alpha^3)^2\alpha=2^2\alpha=4\alpha,\quad\alpha^{49}=(\alpha^3)^{16}\alpha=2^{16}\alpha=65536\alpha=2\alpha$$
then the Frobenius map takes us back to $\alpha$ since $\alpha^{7^3}=(2\alpha)^7=2^7\cdot4\alpha=512\alpha=\alpha$, and thus acts by cyclically permuting the roots:
$$\phi\colon \alpha\mapsto4\alpha\mapsto2\alpha\mapsto\alpha$$
(Here we note that $\alpha\in\mathbb{F}_{7^3}$ is a primitive element for the finite separable and simple field extension $\mathbb{F}_{7^3}/\mathbb{F}_7$, since $\mathbb{F}_{7^3}=\mathbb{F}_7(\alpha)$.) Hence $\text{Gal}(\mathbb{F}_7(\alpha)/\mathbb{F}_7)=\langle\phi\rangle\cong C_3$. So numbering the roots $1$, $2$, $3$ we have $\phi$ as the permutation $(1\,2\,3)$.
