$ [F_p(\sqrt[3]{a}): F_p] = 3$ if and only if $3$ divides $p-1$ Prove that for any prime $p$, the field $F_p$ with $p$ elements contains an element $a$ such that $ [F_p(\sqrt[3]{a}): F_p] = 3$ if and only if $3$ divides $p-1$
Hint: Use the fact that for any finite field $F$, multiplicative group is cyclic.
My attempt:
Since $F_p$ is finite I know that it is algebraic, so $x^3-a$ is the min polynomial with $\sqrt[3]{a}$ as a root. with using the hint $ |F^{*}|=p-1$ but I don't know how to connect those. Can someone please help me with both if and only if parts?
 A: I've tried to write some hints of important properties to assemble to get the proof :
Hint1 : a cyclic group of cardinal $p-1$ has a generator, which is of order exactly $p-1$
Hint2 : a polynomial of degree $3$ is irreducible on a field if and only if it has no roots
Hint3 : use arithmetic ! what does it mean for $3$ to divide/not divide $p-1$ ? note that $3$ is prime
The full proof if you want to see it :
Assume $3$ divides $p-1$. If for every $a \in \mathbb F_p^{\times}$ the polynomial $x^3 - a$ is reducible, then every element of $\mathbb F_p^{\times}$ has a third-root (a polynomial of degree $3$ is irreducible on a field if and only if it has no roots). Since $\mathbb F_p^{\times}$ is cyclic of order $p-1$, it has a generator $\alpha$. Let $\beta$ be a third-root of $\alpha$.
Since $3$ divides $p-1$, there exists an integer $d$ such that $3d = p-1$. We get $\alpha^d = (\beta^3)^d = \beta^{p-1} = 1$, which is absurd since $\alpha$ has order exactly $p-1$. Hence, there exists $a \in \mathbb F_p^{\times}$ such that $x^3 - a$ is irreducible over $\mathbb F_p$. Thus, $[\mathbb F_p(\sqrt[3]{a}): \mathbb F_p] = 3$.
Now, assume $3$ doesn't divide $p-1$. Let $a \in \mathbb F_p^{\times}$. Again, let $\alpha$ be a generator of $\mathbb F_p^{\times}$. There exists an integer $k$ such that $a = \alpha^k$. Since $3$ is prime and doesn't divide $p-1$, $3$ and $p-1$ are coprime. Bézout's identity gives us integers $u, v$ such that $3u + (p-1)v = k$. So $(\alpha^u)^3 = \alpha^k = a$ and $a$ has a third root in $\mathbb F_p^{\times}$, that is $\alpha^u$. Thus, $[\mathbb F_p(\sqrt[3]{a}): \mathbb F_p] = 1$.
A: $\newcommand{\Fp}{\mathbb{F}_p}$Note that you want to find exactly those $p$ for which there exists a non-cube-root $a \in \Fp$. Since $0$ is always a cube, this is equivalent to the existence of a non-cube-root $a \in \Fp^{\times}$.
Note that $\Fp^{\times}$ is a cyclic group of order $p - 1$. Let $g$ be a generator of this group.
Case 1. $3$ divides $p - 1$.
In this case, $g$ cannot be a cube. Indeed, if $h^3 = g$, then $g^{(p - 1)/3} = h^{p-1} = 1$, showing that $g$ does not have the correct order.
Case 2. $3$ does not divide $p - 1$.
We will show that $g$ itself is a cube. In this case, every element is then a cube as well.
Write $1 = 3a + (p - 1)b$ for some integers $a, b \in \Bbb Z$.
Then, $g = g^{3a} \cdot (g^{p - 1})^b = g^{3a}$. Thus, $h = g^{a}$ is a cube root.
