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!!

  • $\begingroup$ Is $F\times q$ meant to be $F_q^{\times}$ ? $\endgroup$ – lhf Nov 10 '15 at 1:02
  • $\begingroup$ Yes! sorry, I'm new to this and I am not really sure how to properly input symbols, etc. $\endgroup$ – queence Nov 10 '15 at 1:16
  • $\begingroup$ You mean $q$ is prime, not a prime power. A prime power would be $p^{\alpha}$ for some prime $p$ and $\alpha\ge 2$. $\endgroup$ – user236182 Nov 10 '15 at 1:46
  • $\begingroup$ That is the exact question I found. I'm not very familiar with prime powers unfortunately $\endgroup$ – queence Nov 10 '15 at 1:53
  • $\begingroup$ A similar question was asked on November 6 by user286213. Sorry, I seem to have lost the capacity to copy links. $\endgroup$ – André Nicolas Nov 10 '15 at 2:03

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 $F_q$.

  • $\begingroup$ I know several definitions that I can work with, I'm just not sure how to incorporate them all together. • Say x^3 - a is reducible, then for say root b, (b^3 - a) = 0 (mod q). • Also, since 3 and q are prime, then for 3 to be able to divide (q-1)/ord(a), then ord(a) must even (hence not prime). • For 3 to divide (q-1) => 3y = q-1, for some y in Fq => q= 3y+1 I'm just really having a hard time piecing it together $\endgroup$ – queence Nov 10 '15 at 1:32
  • $\begingroup$ In all honestly, I feel like i'm having a hard time with this question because I'm not very strong when it comes to group orders $\endgroup$ – queence Nov 10 '15 at 1:39
  • 1
    $\begingroup$ @queence This exercise cannot really be done without knowing the basics of orders and powers of elements of a cyclic group. Anyway, following up on lhf's (+1) hint. The polynomial is reducible iff $a$ has a cube root in $\Bbb{F}_q$, i.e. iff $a$ is a cube of another element. If $q-1$ is not divisible by 3, then all the elements are cubes. If it is divisible, then only one third of them are. But you do need to review everything you know about cyclic groups. $\endgroup$ – Jyrki Lahtonen Nov 10 '15 at 9:11
  • $\begingroup$ @JyrkiLahtonen i know for squares you can use fermats to prove whether or not its a square in Fq, however, is there a similar proof for cubics... (or would we do the same the, but divide by 3 instead: a^(q-1)/2) $\endgroup$ – queence Nov 10 '15 at 16:44

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 $a\equiv \left(g^l\right)^3\pmod{p}$.

  • $\begingroup$ No reason to assume that $q$ is a prime. Because the multiplicative group of $\Bbb{F}_q$ is always cyclic of order $q-1$ you can do more or less the same. But IMVHO it would look better if you used equalities inside the finite field instead of all those congruences $\endgroup$ – Jyrki Lahtonen Nov 10 '15 at 8:46

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