Considerations on cyclotomic extensions

Question

I stopped in front of some issues regarding certain passages of the theorem 9.4 of Algebra of Serge Lang, in which we suppose to have k a field such that $[k (\mu_n):k]=\phi (n)$ where $\mu_n$ is the set of n-th units of k and $\phi $ is the Euler function and n and char (k) are coprime and n is odd. Why have we immediately that $[k (\mu_m):k]=\phi (m)$ for any integer m that divides n ? Further, how can I deduce again for any integer m that divides n that if $[k (\beta,\mu_m):k (\mu_m) ]=m$ for some $\beta $ root of $ X^m-a $ for some a in k then $[k (\beta,\mu_n):k (\mu_n) ]=m$? I think the relations between different cyclotomic extensions of a same field k aren't clear to me ... thank in advance

Answer 1

Lemma: Assume $m \mid n$. The homomorphism $$(\mathbb{Z}/n\mathbb{Z})^{\ast} \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\ast}$$ given by $t + n\mathbb{Z} \mapsto t + m\mathbb{Z}$ is surjective.

Proof: Write $n = p_1^{e_1} \cdots p_j^{e_j}$, and $m = p_1^{d_1} \cdots p_j^{d_j}$ with $d_i \leq e_i$. We have a commutative diagram

$$\begin{matrix} \mathbb{Z}/n\mathbb{Z} & \rightarrow & \mathbb{Z}/p_1^{e_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_j^{e_j} \mathbb{Z} \\ \downarrow & & \downarrow \\ \mathbb{Z}/m\mathbb{Z} & \rightarrow & \mathbb{Z}/p_1^{d_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_j^{d_j} \mathbb{Z} \end{matrix}$$

with horizontal isomorphisms, from which you can see it suffices to prove the lemma when $n$ is a power of a prime number. But in this case it's obvious. $\blacksquare$

Let $\zeta$ be a primitive $n$th root of unity, and $L = K(\zeta)$. Then $L$ is a Galois extension of $K$, since it is a splitting field and the polynomial $X^n -1$ is separable over $K$. Let $G = \textrm{Gal}(L/K)$. There is an injection $$G \rightarrow (\mathbb{Z}/n\mathbb{Z})^{\ast}$$which by your hypothesis is an isomorphism. This is equivalent to saying that for every $1 \leq t \leq n$ which is relatively prime to $n$, there is a $\sigma \in G$ such that $\sigma \zeta = \zeta^t$.

Now if $m \mid n$, then $\xi := \zeta^{n/m}$ is a primitive $m$th root of unity. Again there is an injection of groups $$\textrm{Gal}(K(\xi)/K) \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\ast}$$ To say this map is surjective is to say that for $1 \leq s \leq m$, relatively prime to $m$, there is a $\tau \in \textrm{Gal}(K(\xi)/K)$ such that $\tau(\xi) = \xi^s$.

And this is the case: since $s$ is relatively prime to $m$, by the lemma there is a $1 \leq t \leq n$, relatively prime to $n$, with $t \equiv s \pmod{m}$. Let $\sigma$ be the element of $G$ which sends $\zeta$ to $\zeta^t$. It sends $\zeta^{n/m} = \xi$ to $\zeta^{t(n/m)} = \xi^t$. Let $\tau$ be the restriction of $\sigma$ to $K(\xi)$. Then $\tau$ does what you want:

$$\tau \xi = \xi^t = \xi^s$$