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PS: Oops! This is somewhat duplicate. Thanks for those who made interest for this post. I will delete this post in a few hours. Please cite the following link Which cyclotomic fields are different?

PS2: Now this post has a one correct answer. Should I delete this post?

Let $\zeta_n \in \mathbb{C}$ be a primitive $n^{\textrm{th}}$ root of unity and call $\mathbb{Q}[\zeta_n]$ the $n^{\textrm{th}}$ cyclotomic extension of $\mathbb{Q}$.

Let $m$ and $n$ be positive integers. Describe a simple relation of $m$ and $n$ which is equivalent to $\mathbb{Q}[\zeta_n] = \mathbb{Q}[\zeta_m]$

A student solving exercises on Galois theory made the following argument. Consider $\zeta_{19}$. She could see that $-\zeta_{19}$ is one of primitive $38$th root of unity. So she conclude $\mathbb{Q}[\zeta_{19}] = \mathbb{Q}[\zeta_{38}]$. That was right. But it seems that she thought those two fields are not same. I have told her the fact. However I cannot tell a simple criterion amounts to the above question.

Would you please help me?

seoneo
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1 Answers1

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Assuming that $n \ge m$ WLOG, $\mathbb{Q}[\zeta_n] = \mathbb{Q}[\zeta_m]$ (note that $=$ is unambiguous here only because these extensions are Galois) iff either $n = m$ or $m$ is odd and $n = 2m$.

Probably there are a couple of different ways to do this but here's a relatively straightforward one. $\mathbb{Q}[\zeta_n]$ and $\mathbb{Q}[\zeta_m]$ both embed into $K = \mathbb{Q}[\zeta_{\ell}]$ where $\ell = \text{lcm}(n, m)$, and as subfields of $K$ they are determined by the Galois correspondence: they are the fixed fields of the subgroups of the Galois group $G = U(\ell)$ given by the kernels of the quotient maps $U(\ell) \to U(n), U(\ell) \to U(m)$, and $\mathbb{Q}[\zeta_n] = \mathbb{Q}[\zeta_m]$ (again, this is only unambiguous because they're Galois) iff these subgroups are equal.

By the Chinese remainder theorem, to understand these kernels it suffices to work one prime at a time and understand the kernels of the quotient maps $U(p^a) \to U(p^b), a \ge b$. If $p$ is odd this is a surjective homomorphism between cyclic groups so its kernel is the unique subgroup of $U(p^a)$ of index $p^{b-a}$ and in particular, for fixed $a$ its size uniquely determines the value of $b$. It follows that $\nu_p(n) = \nu_p(m)$ for all odd primes $p$.

So the only possible discrepancy between $n$ and $m$ is at the prime $2$. At this point we don't have to keep analyzing the group of units and we can argue as follows: the condition $\varphi(n) = \varphi(m)$ implies, after dividing out the contribution of the odd primes, that either

$$2^{\nu_2(n) - 1} = 2^{\nu_2(m) - 1}$$

(and hence that $n = m$) if $\nu_2(n), \nu_2(m) \ge 1$, or that $\nu_2(m) = 0$ and $\nu_2(n) = 0, 1$. So either $n = m$, or $m$ is odd and $n = 2m$, as stated. In this second case we have that $- \zeta_m$ is a primitive $n^{th}$ root of unity, generalizing your example.

Qiaochu Yuan
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  • Can you please explain your point about the = being unambiguous? Also what does the notation $v_p(n)$ mean? – gen Feb 13 '23 at 01:25
  • @gen: well, it depends on your definitions. If by $\mathbb{Q}(\zeta_n)$ you mean the subfield of $\mathbb{C}$ generated by $e^{\frac{2 \pi i}{n}}$, say, then there's already no ambiguity. But if you just mean any splitting field of $x^n - 1$ over $\mathbb{Q}$ then a priori it only makes sense to ask whether two such splitting fields are isomorphic, not literally equal. What makes equality unambiguous here is that splitting fields are normal, so fixing an algebraically closed field containing $\mathbb{Q}$ (such as $\mathbb{C})$, there is a unique splitting field of $x^n - 1$ in it... – Qiaochu Yuan Feb 13 '23 at 03:47
  • ...namely the one generated by any primitive $n^{th}$ root of unity, so it makes sense and is unambiguous to ask whether these subfields are equal on the nose.

    $\nu_p(n)$ means the exponent of the prime $p$ in the unique prime factorization of $n$.

     – Qiaochu Yuan Feb 13 '23 at 03:47