In Exercise 13.6.4 of Dummit's & Foote's Abstract Algebra,
it is asked to:
Prove that if $n=p^km$ where $p$ is a prime and $m$ is relatively prime to $p$
then there are precisely $m$ distinct $n$th roots of unity over a field $F$ of
characteristic $p$.
My question: Where exactly should the roots of $x^n-1$ be taken?
They can't be taken only in $F$.
For example, if $F=\mathbb{F}_2$ and $n=2\cdot 3$,
then there certainly do not exist 3 distinct roots of $x^6-1$ in $F$,
which only has 2 elements.
I've read a solution to this exercise by positrón0802.
It roughly goes like this:
Since $\text{char}\ F=p$, we have $x^n-1=(x^m-1)^{p^k}$,
so the roots of $x^n-1$ are precisely the roots of $x^m-1$,
which does not have any multiple root by considering its derivative.
Hence, the $\boldsymbol{m}$ distinct roots of $\boldsymbol{x^m-1}$ are precisely the $\boldsymbol{m}$ distinct $\boldsymbol{n}$th roots of unity over $\boldsymbol{F}$.
The part in boldface seems to suggest that the roots are taken in the splitting field for $x^n-1$ over $F$.
Am I correct?
I don't manage to find a reference in the textbook about this.
Best Answer
Yes, I believe we have:
Roots of unity over a field $F$ are elements of some large enough algebraic extension of $F$.
Roots of unity in a field $F$ are elements of the field itself.
So you can say that $\mathbb{F}_5$ 'has' a $4$th root of unity (Because $X^4-1=(X-1)(X+1)(X-2)(X+2)$ splits completely), and there 'exists' a $6$th root of unity over $\mathbb{F}_5$ (Because $X^6-1$ does not split completely).
There is no $6$th root of unity in $\mathbb{F}_5$, but there is a $6$th root of unity over $\mathbb{F}_5$.
I hope this makes sense.