Roots of unity in a general field $k$.

Question

Let $k$ be an algebraically closed field of characteristic zero. Then $x^4-1$ factors linearly in $k$. Usually, I would consider the roots of $x^4-1=0$, the fourth roots of unity. Which in the case $k = \mathbb{C}$ would be $\{1, -1, i, -i\}$.

However $k$ is not necessarily $\mathbb{C}$. Do I still recognize the roots of the equation $x^4-1=0$ as the fourth roots of unity?

These roots of unity in $k$ would not be the set $\{1, -1, i, -i\}$.

Answer 1

Yes; if $\zeta\in k$ is a root of $x^4-1$ then by definition it satisfies $\zeta^4=1$, so it is a fourth root of unity.

If $\operatorname{char}(k)\neq2$ then $x^4-1$ splits into distinct linear factors: Because $k$ is algebraically closed we have a factorisation $$x^4-1=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta),$$ with $\alpha,\beta,\gamma,\delta\in k$. It is clear that $1,-1\in k$ are two distinct roots of $x^4-1$, so we get $$x^4-1=(x-1)(x+1)(x-\gamma)(x-\delta).$$ Dividing both sides by $(x-1)(x+1)=x^2-1$ we find that $$x^2+1=(x-\gamma)(x-\delta)=x^2-(\gamma+\delta)x+\gamma\delta.$$ This shows that $\gamma=-\delta$ and $\delta\neq0\neq\gamma$. It is also clear that $\gamma$ and $\delta$ do not equal $1$ or $-1$, so indeed we have four distinct roots. It is customary to denote these latter two roots $\gamma$ and $\delta$ by $i$ and $-i$, in analogy to the case $k=\Bbb{C}$. They are however not elements of $\Bbb{C}$ unless $k=\Bbb{C}$, or at least $k$ contains $\Bbb{Q}(i)\subset\Bbb{C}$ as a subfield.

Note that if $\operatorname{char}(k)=2$ there is a problem because $$x^4-1=(x-1)^4,$$ so we don't get four distinct fourth roots of unity as roots of $x^4-1$. In fact this shows that $1$ is then the unique fourth root of unity in $k$.

Answer 2

The roots won't be $\pm 1, \pm i$ if you're working with a finite field, because not every finite field is a subfield of $\Bbb C$.

For example, every element of $\Bbb F_5$ satisfies the equation $x^4=1$, because as a group with multiplication $\Bbb F_5^\times$ has order $4$, and every element raised to the order of the group is the identity.

It's easy to see that $\Bbb F_5$ is not a subfield of $\Bbb C$, because the elements of $\Bbb C$ are numbers, while the elements of $\Bbb F_5$ are equivalence classes of numbers.