# Let $\mathbb{F}$ be any field. Show that the number of cube-roots of unity in $\mathbb{F}$ is either $1$ or $3$.

Let $\mathbb{F}$ be any field.

Show that the number of cube-roots of unity in $\Bbb F$ is either $1$ or $3$.

Show that if $\mathbb{F}$ has characteristic $3$ then it has only one cube-root of unity.

Show that if $\mathbb{F}$ has any characteristic other that $2$ or $3$ then there are $3$ cube-roots of unity in $\mathbb{F}$ if and only if the element $-3$ (in $\mathbb{F}$) has a square-root in $\mathbb{F}$.

• What did you try ? – Belgi Feb 19 '13 at 16:38
• Hint: It is clear that there is at least one cube root of $1$ and at most $3$. Can you show that there can't be $2$? – Thomas Andrews Feb 19 '13 at 16:39

First of all there must be at most 3 cube roots of unity because there are at most 3 solutions to the degree 3 polynomial $x^3-1$.

Clearly one is one, so it remains to show that there cannot be two:

Suppose there are at least two cube roots of unity: 1 (necessarily) and $\omega$ then $\omega^2$ is another distinct cube root of unity since $(\omega^2)^3-1 = (\omega^3-1)(\omega^3+1) = 0$ and $\omega^2 \not =1$ since otherwise $\omega = 1$.

Hint $\$ Over a field (or domain), a cubic has at most $3$ roots. If it has two roots then it has three, since the field contains the coefficient of $\rm\,x^2,\:$ which is minus the sum of the roots.

A cube root of unity is some $\zeta\in\Bbb F$ such that $\zeta^3-1=0$. It's clear that the unity, itself, satisfies this, so there's always at least one. Since $\zeta^3-1=(\zeta-1)(\zeta^2+\zeta+1)$, then for $\zeta$ to be any other cube root of unity, it must satisfy $\zeta^2+\zeta+1=0.$ But since $\zeta^3=1$, then $$(\zeta^2)^2+(\zeta^2)+1=\zeta^4+\zeta^2+1=\zeta+\zeta^2+1=0,$$ so $\zeta^2$ is another such root of unity (if $\zeta$ is). I leave it to you to show that $\zeta^2$ is distinct from $\zeta$ and $1$ (so long as $\zeta\neq 1$).

Let's consider the complex case for a moment, to give us an idea of how to deal with the rest of it. By the quadratic formula, a non-trivial cube root of unity is $$z=-\frac12(1+\sqrt{-3}),$$ from which it follows that $$(-2z-1)^2=-3.$$

Now let's transition back to our general field, and see what happens there. If $\zeta^2+\zeta+1=0$, then $$(-2\zeta-1)^2=4\zeta^2+4\zeta+1=4(\zeta^2+\zeta+1)-3=-3.\tag{1}$$ If we aren't in characteristic $2$ or $3$ fields, then $(1)$ shows us that $-3$ has a square root in the field if there is a non-trivial cube root of unity. The other direction shouldn't be too hard to figure out.

In characteristic $3$ fields, we have $$\zeta^3-1=\zeta^3-0+0-1=\zeta^3-3\zeta^2+3\zeta+1=(\zeta-1)^3,$$ so there's only one root of $\zeta^3-1$.

Clearly, $1$ is a cube root of unity. In characteristic $3$ we simply have $X^3-1=(X-1)^3$, i.e. $1$ is a triple (and thus the only) root.

Assume from now on $\operatorname{char}(F)\ne3$. If $a$ is another cube root of unity, then the polynomial $\frac{X^3-1}{(X-1)(X-a)}$ is linear, i.e. of the form $X-b$ making $b$ a third cube root of unity - unless we have a double root $b=1$ or $b=a$. But a double root of $X^3-1$ is also a root of the derivative $3X^2$, which can't happen, i.e. there is no double root and thus either one or three roots.

Note that $\frac{X^3-1}{X-1}=X^2+X+1$ has solutions $\frac{-1\pm\sqrt{-3}}{2}$ if that makes sense, i.e. if $\operatorname{char}(F)\ne2$ and $-3$ is a square. On the other hand, if $a,b$ are the two solutions of $X^2+X+1=X^2-(a+b)X+ab=0$ then $-3$ is a square because $$(a-b)^2=(a+b)^2-4ab =(-1)^2-4\cdot1= -3.$$ (We don't make direct use of $\operatorname{char}(F)\ne3$ in this last part directly, but we need it so that $\frac{-1\pm\sqrt{-3}}2\ne1$)