# Find smallest $n$ so that $Gal(K(\xi)/K)$ is not cyclic

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 $$n$$?

• @K02 Yes, that is $n=8$. But how to deal with that? Nov 27, 2023 at 5:05

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).

• Ok... So we can just take $K=\mathbb{Q}$? Nov 27, 2023 at 0:00
• 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? Nov 27, 2023 at 1:30
• So it seems that $n=8$. Nov 27, 2023 at 1:35
• 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 $\mathbb{Q}$? Nov 27, 2023 at 1:38
• 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 $\omega$. Nov 27, 2023 at 2:51