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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?

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  • $\begingroup$ $n=7$ is also there. The condition is that the group $(\mathbb Z/n\mathbb Z)^*$ (of invertible elements) should be isomorphic to a product of cyclic groups of order $\leq 4$. I'm not quite sure how to characterize such $n$'s (it's enough to consider $n$'s which are prime powers, and the prime must be either $\leq 4$ or the power must be $1$, but after thatit needs some thinking) $\endgroup$ – user8268 Jan 23 '15 at 12:19
  • $\begingroup$ Dear @user 8268: I didn't check the result in your comment but I have no objection any more. $\endgroup$ – Georges Elencwajg Jan 23 '15 at 12:53
  • $\begingroup$ @user8268: how do you show, for example, that $\xi_{11}\not\in K$ ? $\endgroup$ – user72870 Jan 23 '15 at 13:13
  • $\begingroup$ if $L$ is the splitting field of $k$ polynomials of deg $4$ then the Galois of $L$ over $\mathbb Q$ is a subgroup of $(S_4)^k$, so $10$ doesn't divide the degree $[L:\mathbb Q]$. However $10$ is the degree of $\mathbb Q(\xi_{11})$, so it can't be a subfield of $L$. $\endgroup$ – user8268 Jan 23 '15 at 13:30
  • $\begingroup$ @user8268: Ok, you should post your comments as an answer. $\endgroup$ – user72870 Jan 23 '15 at 13:36
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$\xi_n$ 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.

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