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Let $R$ be a commutative ring and $\zeta$ be a root of unity in $R$ such that $\mathrm{ord}(\zeta) = n$. If $n = p_1^{e_1} \dots p_r^{e_r}$ is the decomposition in prime factors of $n$, is it true that there exists $r$ roots of unity $\zeta_i \in R$ such that $\mathrm{ord}(\zeta_i) = p_i^{e_i}$? In other words, if we denote by $Z^R$ the set of roots of unity in $R$ and $Z_n = \{ \zeta \in Z^R \mid \mathrm{ord}(\zeta) = n \}$, then

Is the map $Z_{p_1^{e_1}} \times \cdots \times Z_{p_r^{e_r}} \to Z_n$, $(\zeta_1, \dots , \zeta_r) \mapsto \zeta_1 \cdots \zeta_r$ surjective?

This comes from trying to prove that the map is bijective. For the injectivity part, I believe it's true. If $\zeta_1 \dots \zeta_r = \zeta_1' \dots \zeta_r'$, then $$\mathrm{ord}(\zeta_1)\cdots \mathrm{ord}(\zeta_r) = \mathrm{ord}(\zeta'_1) \cdots \mathrm{ord}(\zeta_r') = p_1^{e_1} \cdots p_r^{e_r}$$ and since the factorization is unique up to the other of the factors, we can assume that $\mathrm{ord}(\zeta_i) = \mathrm{ord}(\zeta'_i) = p_i^{e_i}$, and hence $(\zeta_1, \dots , \zeta_r) = (\zeta_1' , \dots , \zeta_r')$. So it's the subjectivity part I'm having trouble with. Any hint would be appreciated. Thank you.

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    $\begingroup$ This is really just the Chinese Remainder Theorem. $\endgroup$ – Lord Shark the Unknown Dec 29 '17 at 12:46
  • $\begingroup$ CRT : $\sum_{i=1}^j \frac{a_i}{p_i^{e_i}} = \frac{\sum_{i=1}^j a_i \frac{n}{p_i^{e_i}}}{n}$. You want to find $a_i$ such that $\sum_{i=1}^j a_i \frac{n}{p_i^{e_i}} \equiv 1 \bmod n$. For this take $a_i \frac{n}{p_i^{e_i}} \equiv 1 \bmod p_i^{e_i}$, you know it exists because .. $\endgroup$ – reuns Dec 29 '17 at 13:13
  • $\begingroup$ @LordSharktheUnknown thanks! Didn't think about it $\endgroup$ – user313212 Dec 30 '17 at 9:39

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