Let $q=p^n$.

  1. For what $q$ does $\mathbb F_q$ contain a primitive cube root of unity?
  2. Deduce for which $q$ the polynomial $x^2+x+1$ splits into linear factors in $\mathbb F_q[x]$. Use the quadratic formula when appropriate to find these factors. Over which fields is it inappropriate to use the quadratic formula?
  3. Which finite fields contain a square root of $-3$ and which do not?

If $x$ is a primitive cube root of unity in $\mathbb F_q$, then $x\ne 1, x^2\ne 1$ but $x^3=1$ in $\mathbb F_q$. Equivalently, $x^3-1=0$ or $(x-1)(x^2+x+1)=0$. Since $x$ is primitive and $\mathbb F_q$ has no zero divisors, $x^2+x+1=0$. Now I believe the fields containing a primitive cube root of 1 are the fields over which this polynomial splits. So are the first two parts asking the same question? Anyway, I don't know how to describe the fields having either property.

For the quadratic formula part, $x=\frac{-1\pm \sqrt{-3}}{2}$; this is valid provided $2\ne 0$ i.e. provided the characteristic isn't 2. So the polynomial splits iff $\sqrt{-3}$ lies in the field (provided the characteristic isn't 2). So is the third part asking the same as the first two?

Let $F=\mathbb F_q$.

$F$ contains a primitive cube root of unity $\iff$ $x^2+x+1$ has a root over $F$ that isn't equal to $1$ $\iff$$x^3-1$ has a root over $F$ that isn't equal to $1$ $\iff$ $F^\ast $ contains an element of order $3$ $\iff$ $3$ divides $p^n-1$ $\iff$ the conditions on $p,n$ from the answer of @lhf are met. So those conditions is the answer to 1.

To get an answer for 2, we need to add to those $p$ and $n$ the values of $p,n$ for which $x^2+x+1$ has a root that is equal to $1$. Such values are $p=3, n$ is arbitrary. So the answer to 2 is the values from lhf's answer as well as $p=3, n$ arbitrary.

  • $\begingroup$ Yes, all the parts are asking "the same question" in the sense you figured out. $\endgroup$ Jun 19, 2018 at 4:43
  • $\begingroup$ @JyrkiLahtonen Seems it's not quite the same. I think there will be more values of $q$ for part 2 (see my edit). Or am I wrong? $\endgroup$
    – user557
    Jun 20, 2018 at 1:12
  • $\begingroup$ Yeah, But $3$ is just a single prime whereas there are infinitely choices of $p$ such that $3\mid q-1$: A) $p\equiv1\pmod3$, $n$ anything, or B) $p\equiv-1\pmod n$, $n$ even. For the sake of completeness, sure $p=3$ counts in parts 2) and 3), but not in part 1. I only made a quick comment to A) acknowledge your good work, B) I spent a while looking for a duplicate (this sounded familiar), but I didn't find one yet. $\endgroup$ Jun 20, 2018 at 6:55

2 Answers 2



A primitive cube root of unity in $\mathbb F_q$ is an element of order $3$ in $\mathbb F_q^\times$.

Because $\mathbb F_q^\times$ is a cyclic group, $\mathbb F_q$ contains a primitive cube root of unity iff $3$ divides $q-1$.

If $p \equiv 0 \bmod 3$, then $q \equiv 0 \bmod 3$ and so $3$ never divides $q-1$.

If $p \equiv 1 \bmod 3$, then $q \equiv 1 \bmod 3$ and so $3$ always divides $q-1$.

If $p \equiv -1 \bmod 3$, then $q \equiv (-1)^n \bmod 3$ and so $3$ divides $q-1$ iff $n$ is even.

  • $\begingroup$ I've edited my question. Could you take a look and say whether my understanding of 1 and 2 is correct? $\endgroup$
    – user557
    Jun 19, 2018 at 15:19

If $p=2$, the polynomial $x^2+x+1$ is irreducible over $\mathbb{F}_2$; so the primitive cube root exists if and only if $n$ is even.

If $p$ is an odd prime, then we need that $3$ divides $q-1$.

For $p\ne2$, if a primitive cube root $w$ exists, it is not restrictive to assume it is $$ \frac{-1+\sqrt{-3}}{2} $$ In particular, $(2w+1)^2=-3$, which can also be proved directly: $w^2=-w-1$, so $$ (2w+1)^2=4w^2+4w+1=-4w-4+4w+1=-3 $$

For $p=2$, we have $-3=1$, so the square root of $-3$ always exists.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .