roots of a polynomial in closure of a finite field I'm having trouble with the following practice qual problem:
$(a)$ Determine the number of distinct roots of the polynomial $x^
n−1$ in an algebraic closure
of $F_p = \mathbb{Z}/p\mathbb{Z}$, where $p$ is a prime number and $n > 0$.
$(b)$ Let $K/\mathbb{Q}$ be a finite extension of fields, and let $α ∈ K$. Suppose that there is a monic
polynomial $f ∈ \mathbb{Z}[x]$ so that $f(α) = 0$. Show that the minimal polynomial $mα,\mathbb{Q}(x)$ of
$α$ over $\mathbb{Q}$ lies in $\mathbb{Z}[x]$.
 A: For a: The polynomial $X^n -1$ decomposes into linear factors over the algebraic closure (by definition of the algebraic closure). Thus, counted with multiplicity there are $n$ roots. Yet this is not what you want. However, it shows that you only need to care about understanding repeated roots. 
Now, if a root is repeated then it is also a root of the deriviative $nX^{n-1}$ and $\gcd(X^n-1, nX^{n-1})$ is non-constant. 
If $p \nmid n$, then $\gcd(X^n-1, nX^{n-1})=1$, so all the roots are distinct and you have $n$ distinct roots. (One says the polynomial is separable.)
If $p\mid n$, then proceed as in the other answer. That is,  write $n = p^vn'$ with $p\nmid n'$ and observe that $(X^{n'} -1)^{p^v} = X^n -1$. 
The roots of $X^n -1$ (ignoring multiplicity) are the same as those of
$X^{n'} -1$. And for $X^{n'}-1$ we know that there are $n'$ distinct roots by the  argument just given. 
A: Hints:
(a) If $\;n=pk\;$ , then 
$$\forall\,a\in\overline{\Bbb F_p}\;,\;\;a^{pk}-1=(a^k-1)^p$$
or what is more or less the same (depends...), $\;(x^n-1)'=0\;$ in $\;\Bbb F_p[x]\;$
(b) Either develop this yourself, or read about  Gauss Lemma
