1
$\begingroup$

i have a question on a decomposition of the set of all roots of unity $\mu_{\infty}$ in $\overline{\mathbb{Q}}$.

Let $\zeta \in \mu_{\infty}$ be any root of unity and let $p$ be any prime number in $\mathbb{Z}$. How can i find two others root of unity $\zeta_1,\zeta_2 \in \mu_{\infty}$ such that

$$\zeta = \zeta_1\zeta_2$$

and the order of $\zeta_1$ is a power of $p$ and the order of $\zeta_2$ is not divisible by $p$?

In other words i want to decompose

$$\mu_{\infty} = \{\alpha \in \overline{\mathbb{Q}}: \alpha^{p^n}=1, \mbox{ for some } n \geq 0\}\cdot\{\alpha \in \overline{\mathbb{Q}}: \alpha^m=1, \mbox{ for some } m \mbox{ such that } p\nmid m\}.$$

My attempt:

If $n$ is the order of $\zeta$ we can write $n = p^ab$ with $p \nmid b$ and $a,b \geq 0$. Define $\zeta_1 = \zeta^b$ and $\zeta_2 = \zeta^{p^a}$. Then the order of $\zeta_1$ is $p^a$ and the order of $\zeta_2$ is $b$, however the multiplication $\zeta_1\zeta_2 = \zeta^{p^a + b}$, but why should this also equals to $\zeta$??

Thanks in advance!

$\endgroup$
3
  • 1
    $\begingroup$ Every finite Abelian group is the direct product of its Sylow subgroups. $\endgroup$ Apr 29, 2018 at 17:03
  • $\begingroup$ I'm sorry but i do not see why can i use this, since all these three groups are infinite. $\endgroup$ Apr 30, 2018 at 6:51
  • $\begingroup$ But $\zeta$ generates a finite subgroup. $\endgroup$ Apr 30, 2018 at 16:53

2 Answers 2

1
$\begingroup$

You have $(p^a,b)=1$, so you know that $1$ is a linear combination of them by elementary number theory: $1$=$cp^a+db$. Now let $\zeta_1=\zeta^{db}$ and $\zeta_2=\zeta^{cp^a}$, and you have what you were looking for.

$\endgroup$
2
  • $\begingroup$ It is true that $\zeta_1^{p^a} = 1$ and $\zeta_2^b = 1$ too, but why are these two exactly the order of $\zeta_1, \zeta_2$? I mean, i cannot really see why these numbers are the minimal with this property. $\endgroup$ Apr 30, 2018 at 8:03
  • $\begingroup$ They are minimal with this property since otherwise $\zeta$, which is their product, would have order less than $p^ab$. $\endgroup$
    – C Monsour
    Apr 30, 2018 at 11:52
0
$\begingroup$

Let $n$ be the order of $\zeta$. Take the prime power factorization: $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$.

Case 1: For the moment assume $$\zeta = \prod_j \alpha_j\quad \mbox{where }\alpha_j=\exp{\frac {2\pi i}{p_j^{a_j}}}$$

One of the $p_j$'s must be the $p$ in your question. The $\zeta_1=\alpha_j$, and $\zeta_2$ must be the product of other $\alpha_r$'s for $r\neq j$.

Case 2: But a general primitive $n$ th root of unity may not be like the $\zeta$ above. It has to be a power of the above $\zeta$, say $\zeta^t$ with $\gcd(n,t)=1$. Now modify the definition of $\zeta_1$ and $\zeta_2$ in case 1 by taking their $t$ th powers now.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .