Splitting field of $x^n-a$ contains all $n$ roots of unity This statement is suggested as a correction to this question:
If $K$ is the splitting field of the polynomial $P(x)=x^n-a$ over $\mathbb{Q}$, prove that $K$ contains all the $n$th roots of unity.
How to prove it?
 A: If $K$ is the splitting field for that polynomial then it contains all $n$th roots of $a$, as you know.  So if $\omega$ is an primitive $n$th root of unity, then the roots of $P(x)=x^n-a$ are $\sqrt[n]{a},\sqrt[n]{a}\omega,...,\sqrt[n]{a}\omega^{n-1}$ just as JJ Beck said.  We can multiply $\sqrt[n]{a}\omega$ by $\frac{1}{\sqrt[n]{a}}$, since inverses exist in $K$.  This give us $\omega \in K$.  Therefore, all powers of $\omega$ are in $K$.  Since $\omega$ was primitive, we get all $n$th roots of unity.
A: Let $F$ be an arbitrary field, $P(x) = x^n - a \in F[x]$ with $p = \text{char} \; F \not\mid n$ and $a \ne 0$, and let $K$ be the splitting field of $P(x)$ over $F$.  Then $P(x)$ splits into linear factors in $K$; that is, there exist $\beta_j \in K$, $1 \le j \le n$, with
$P(x) = x^n - a = \prod_1^n (x - \beta_j) \tag{1}$
in $K$.  Furthermore, since
$P'(x) = nx^{n - 1}, \tag{2}$
we see that $P(x)$ and $P'(x)$ have no common root;  since $p \not\mid n$, $n \ne 0$ in $F$, so the only zero of $P'(x)$ is $0$, which is not a root of $P(x)$.  This implies the $\beta_j$ are in fact all distinct, each from the others.  Now setting $\omega_i = \beta_1^{-1}\beta_i \in K$, we see that $\omega_i^n = \beta_1^{-n}\beta_i^n = a^{-n}a^n = 1$, since each $\beta_i$ satisfies $\beta_i^n = a$.  Furthermore, the $\omega_i$ are distinct, since the $\beta_i$ are.  Thus the $n$ $n$-th roots of unity $\omega_i$ are all contained in $K$.  QED.
Applying the above result to the case $F = \Bbb Q$, we have that the splitting field $K$ of
$x^n - a$ contains all the $n$-th roots of unity for any $n$, since $0 = \text{char} \; \Bbb Q \not\mid n$.
It should be observed that the above argument makes no assumptions about the roots of unity lying in $K$; rather, it deduces their existence from that of the distinct zeroes $\beta_i$ of $P(x)$, hence no explicit reference to the complex field $\Bbb C$ is required.  Nor is the primitivity of any of the $\omega_i$ an issue, since we obtain $n$ distinct $\omega_i$ via the formula  $\omega_i = \beta_1^{-1}\beta_i$.  And apparently, the essential result is true for many fields other than $\Bbb Q$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
