Irreducibility of $x^3 − a$ over the field of $q$ elements

Question

I have done much research on this specific question. I have come across many different theorems and definitions on this topic. However, I am having difficulties piecing them together to create a nice looking proof.

I have found a similar, general (slightly altered proof), however, there are some slightly different properties. Any help would be so great.

Let q be a prime power and let $a\in \Bbb{F}_q^*$ . Prove that $x^3 − a$ is irreducible over $\Bbb{F}_q$ if and only if $3|(q−1)$ but $3$ doesn't divide $(q−1)/ord(a)$.

Bonus: Verify that $x^3 −2$ is irreducible over $\Bbb{F}_{13}$ but not over $\Bbb{F}_{17}$. Moreover, show that $x^3 − 5$ is reducible over $\Bbb{F}_{13}$.

Once I prove the above statement, I will be able to use it to verify the "bonus" question.

Thank you for all your help!!

Answer 1

Hint: Consider the contrapositive:

$x^3 − a$ is reducible over $F_q$ if and only if $3\not\mid (q−1)$ or $3\mid(q−1)/ord(a)$.

Note that $x^3 − a$ is reducible over $F_q$ iff it has a root in $F_q$.

Answer 2

I'm assuming $q$ is prime, not a "prime power". Note $x^3-a$ is reducible over $F_q$ iff $x^3\equiv a\pmod{q}$ is solvable. You can prove the contrapositive:

If $a\not\equiv 0\pmod{p}$, then $x^3\equiv a\pmod{p}$ is solvable if and only if either $p\equiv 2\pmod{3}$ or $3\mid\frac{p-1}{\text{ord}_p(a)}$.

Proof: If $p\equiv 2\pmod{3}$, then $x^3\equiv a\pmod{p}$ is solvable, because $x\equiv a^{2k+1}\pmod{p}$ is a solution, where $p=3k+2$.

If $p\equiv 1\pmod{3}$, let $p=3k+1$.

$$3\mid \frac{p-1}{\text{ord}_p(a)}\iff 3\text{ord}_p(a)\mid 3k\iff \text{ord}_p(a)\mid k$$

$$\iff a^k\equiv 1\pmod{p}\iff a^{\frac{p-1}{3}}\equiv 1\pmod{p}$$

$$\stackrel{(1)}\iff \exists x\in\Bbb Z\left(x^3\equiv a\pmod{p}\right)$$

I'll elaborate on $(1)$. If $x^3\equiv a\pmod{p}$, then clearly $a^{\frac{p-1}{3}}\equiv \left(x^3\right)^{\frac{p-1}{3}}\equiv x^{p-1}\pmod{p}$ by Fermat's Little theorem.

If $a^{\frac{p-1}{3}}\equiv 1\pmod{p}$, then let $a\equiv g^t\pmod{p}$ for some $t\in\Bbb Z^+$, where $g$ is the primitive root mod $p$. Then $g^{\frac{t(p-1)}{3}}\equiv 1\pmod{p}\iff p-1\mid \frac{t(p-1)}{3}\iff t=3l$, so $a\equiv \left(g^l\right)^3\pmod{p}$.