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My question is that Identify the least integer $n\ge 1$ such that there is a field $K$ and a primitive $n$-th root of unity $\xi$ within the algebraic closure $\bar{k}$ of $K$, where the Galois extension with group $Gal(K(\xi)/K)$ is not cyclic.

Here is one useful result in cyclotomic extension of field $K$

Suppose that $char(F)$ does not divide $n$, and let $K$ be a splitting field of $x^n-1$ over $F$. Then $K/F$ is Galois, $K=F(\xi)$ is generated by any primitive $n$-th root of unity. And $Gal(K/F)$ is isomorphic to a subgroup of $(\mathbb{Z}/n \mathbb{Z})^\times$.Thus, $Gal(K/F)$ is abelian and $[K:F]$ divides $\phi(n)$.

We know that $Gal(K(\xi)/K)\cong H<(\mathbb{Z}/n \mathbb{Z})^\times$. Clearly, the smallest $n\ge 1$ so that $(\mathbb{Z}/n \mathbb{Z})^\times$ not cyclic is $n=8$. However, although $(\mathbb{Z}/8 \mathbb{Z})^\times$ is not cyclic, we cannot say subgroup $H$ is also not cyclic... How do we find such $n$?

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  • $\begingroup$ @K02 Yes, that is $n=8$. But how to deal with that? $\endgroup$
    – H.Y Duan
    Nov 27, 2023 at 5:05

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In the case of $K = \mathbb{Q}$, you can show that the Galois group is precisely $(\mathbb{Z}/n\mathbb{Z})^{\times}$. This follows from the irreducibility of cyclotomic polynomials over $\mathbb{Q}$ (this is not true over all fields, which is why the Galois group is only a subgroup of $(\mathbb{Z}/n\mathbb{Z})^{\times}$ in general). Thus, the answer is $n = 8$.

P.S. Note that this is not the least $n$ for all choices of $K$. For instance, if $K = \mathbb{F}_q$ is a finite field, then all finite extensions of $K$ are cyclic (this follows from the classification of finite fields: the finite extensions of $K$ are $\mathbb{F}_{q^t}$ and $\mathrm{Gal}(\mathbb{\mathbb{F}}_{q^t}/\mathbb{F}_q) \cong \mathbb{Z}/t\mathbb{Z}$). This means that such a value $n$ does not exist for finite $K$.

On the other hand, if $K = \mathbb{Q}(\zeta_r)$ for some primitive $r$th root of unity, then adjoining a primitive root of unity to $K$ will give a field $L = \mathbb{Q}(\zeta_n)$ for $r \mid n$. We know $\mathrm{Gal}(L/\mathbb{Q}) = (\mathbb{Z}/n\mathbb{Z})^{\times}$ and $\mathrm{Gal}(K/\mathbb{Q}) = (\mathbb{Z}/r\mathbb{Z})^{\times}$, so $\mathrm{Gal}(L/K)$ is the kernel of the map $(\mathbb{Z}/n\mathbb{Z})^{\times} \to (\mathbb{Z}/r\mathbb{Z})^{\times}$. By the Chinese Remainder Theorem, if $r = \prod p^{\alpha}$ and $n = \prod p^{\beta}$ where $\alpha \leqslant \beta$, then $\mathrm{Gal}(L/K)$ is the product of the kernels of the maps $(\mathbb{Z}/p^{\beta}\mathbb{Z})^{\times} \to (\mathbb{Z}/p^{\alpha}\mathbb{Z})^{\times}$. This can be worked out in general using the structure of $(\mathbb{Z}/p^{\alpha}\mathbb{Z})^{\times}$ as an abelian group.

One specific case we can think about is $r = 4$. In this case, $n \neq 8$ since the kernel of the map has order $\tfrac{n}{r} = 2$, so the kernel must be cyclic. In this case, you can work out that $n = 24$ is the least value (this is clearly a lower bound since we need $\tfrac{n}{r} \geqslant 6$ and we can see $(\mathbb{Z}/24\mathbb{Z})^{\times} \to (\mathbb{Z}/4\mathbb{Z})^{\times}$ has kernel isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ using the Chinese Remainder Theorem).

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    $\begingroup$ Ok... So we can just take $K=\mathbb{Q}$? $\endgroup$
    – H.Y Duan
    Nov 27, 2023 at 0:00
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    $\begingroup$ Oh, sorry, one easy question: why $(Z/6Z)^\times$ is not cyclic? I am confused because $(Z/6Z)^\times=\{-1, +1\}$ and then $(-1)^2=1$. So it is cyclic? $\endgroup$
    – H.Y Duan
    Nov 27, 2023 at 1:30
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    $\begingroup$ So it seems that $n=8$. $\endgroup$
    – H.Y Duan
    Nov 27, 2023 at 1:35
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    $\begingroup$ But I still have a question. When $K=\mathbb{Q}$, we take the smallest $n=8.$ But if we choose other number fields $K$, can we also ensure that $n=8$ is the smallest? I am a little bit Not sure about this. Or do we not need to consider other fields not $\mathbb{Q}$? $\endgroup$
    – H.Y Duan
    Nov 27, 2023 at 1:38
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    $\begingroup$ Whoops, my bad! I just took what you wrote for granted. It is indeed $n = 8$. For other fields, this may not be the case. For instance, if $K$ is a finite field, then every finite extension of $K$ will be Galois with cyclic Galois group, so no such $n$ would exist. You should try thinking about what will happen when you take $K$ to be $\mathbb{Q}(\omega)$ for some root of unity $\omega$. $\endgroup$
    – Haran
    Nov 27, 2023 at 2:51

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