Roots of unity in splitting field of all polynomials of given degree Let $K$ be the splitting field of all polynomials of degree $4$ in $\mathbb{Q}[x]$. For which $n\in \mathbb{N}$ the $n$-th primitive root of unity $\xi_n\in K$ ?
I've shown that $K$ is an algebraic, normal and separable extension of $\mathbb{Q}$. Clearly $\xi_n\in K$ for $n=1, 2, 3, 4, 5, 6, 8, 10, 12$  (the minimal polynomial of $\xi_n$ over $\mathbb{Q}$ has degree $\varphi(n)$). Any hints to continue?
 A: 
$\xi_n$ is in $K$ iff the group $(\mathbb Z/n\mathbb Z)^*$ (of invertible elements of $\mathbb Z/n\mathbb Z$) is isomorphic to a product of cyclic groups which are of order $\leq 4$ $\ \ \ \ \ \ \ \ \ (*)$

to prove it: 
one direction: if $L$ is the splitting field of $k$ polynomials of degree $4$ in $\mathbb Q[x]$ then $G=Gal(L/\mathbb Q)$ is a subgroup of $(S_4)^k$ (as it permutes the roots of those polynomials). If $\xi_n\in L$ then we have a surjection $r:G\to Gal(\mathbb Q(\xi_n)/\mathbb Q)=(\mathbb Z/n\mathbb Z)^*$. Now $(\mathbb Z/k\mathbb Z)^*$ (being a finite Abelian group) is isomorphic to a product of $C_{p^m}$'s (cyclic groups of prime-power order). If $p^m>4$ then the generator $c$ of $C_{p^m}$ can't be in the image of $r$ (as there is no element in $G\subset(S_4)^k$ whose order is divisible by $p^m$). 
the other direction: if $Gal(\mathbb Q(\xi_n)/\mathbb Q)=(\mathbb Z/n\mathbb Z)^*\cong\prod_i C_{s_i}$ with $s_i\leq4$ then let $F_i\subset\mathbb Q(\xi_n)$ be the fixed field of $\prod_{j\neq i} C_{s_j}$, so that $Gal(F_i/\mathbb Q)=C_{s_i}$. If $\alpha_i\in F_i$ is such that $F_i=\mathbb Q(\alpha_i)$ and $f_i$ is the minimal polynomial of $\alpha_i$ then we see that $\mathbb Q(\xi_n)$ is the splitting field of the $f_i$'s.

Now for which $n$'s is the condition $(*)$ actually satisfied: if $n=\prod p^a$ (prime powers with different $p$'s) then $(\mathbb Z/n\mathbb Z)^*=\prod(\mathbb Z/p^a\mathbb Z)^*$, so let's suppose that $n$ is a prime power. If $p>2$ then $(\mathbb Z/p^a\mathbb Z)^*$ is cyclic $C_{p^{a-1}(p-1)}\cong C_{p^{a-1}}\times C_{p-1}$. So $(*)$ fails if $a>1$ and $p>3$, and also for $a>2$ and $p=3$. When $a=1$ then permissible $p>3$'s are  $p=5$ ($C_4$), $p=7$ ($C_2\times C_3$), $p=13$ ($C_4\times C_3$). For $p=2$ it's uglier, and there $2^a$ works for $a\leq4$

So the final answer is (modulo mistakes): $n$ has to be a product of $2^a,3^b,5,7,13$ with $a\leq4$, $b\leq2$, using each number at most once.

