On subfields of the cyclotomic field $\mathbb{Q}(\zeta_p)$

Question

Let $p$ be an odd prime. Let $\zeta_p=e^{2\pi{\bf i}/p}$ and let $1\le k\le p-1$ be a divisor of $p-1$. Recently, when I learnt algebraic number theory, I met the following problem.

If we let $$U_k=\{y\in\mathbb{Z}/p\mathbb{Z}:\ y^k\equiv 1\pmod p\}$$ be the subgroup of all $k$-th root of unity over the finite field $\mathbb{Z}/p\mathbb{Z}$, then we can define an element $$x_k=\sum_{y\in U_k}\zeta_p^y.$$ Using the Galois theory, it is easy to see that $x_k$ is indeed contained in the unique subfield $M_k$ of $\mathbb{Q}(\zeta_p)$ with $[M_k:\mathbb{Q}]=(p-1)/k$.

I want to know if $x_k$ is exactly the generator of the subfield $M_k$, i.e., $\mathbb{Q}(x_k)=M_k$. For example, when $k=1$ or $2$, the answer is positive.

Thank you for you comments!

Answer 1

This is the same argument as Olivier's, but I would tell the story as follows. By the explicit knowledge of $\operatorname{Gal}(\mathbb Q(\zeta_p)/\mathbb Q)$, the Galois conjugates of $x_k$ are the sums $$\sum_{y\in U_k}\zeta_p^{ay}=\sum_{v\in aU_k}\zeta_p^v,$$ where $a\in(\mathbb Z/p\mathbb Z)^{\times}$. By the linear independence of the primitive $p$-adic roots of unity, the above sums correspond bijectively to the cosets of $U_k$ in $(\mathbb Z/p\mathbb Z)^{\times}$. Hence the degree of $x_k$ over $\mathbb{Q}$ is the index of $U_k$ in $(\mathbb Z/p\mathbb Z)^{\times}$, which is $(p-1)/k$.

Answer 2

By definition, $U_s$ is the only subgroup of the cyclic group $(\mathbb Z/p\mathbb Z)^{\times}$ of cardinality $s$. The cyclotomic character $\chi_p:\operatorname{Gal}(\mathbb Q(\zeta_p)/\mathbb Q)\longrightarrow(\mathbb Z/p\mathbb Z)^{\times}$ thus identifies $U_s$ with $\operatorname{Gal}(\mathbb Q(\zeta_p)/M_s)$. The question of whether $x_s$ belongs to a sub-extension of $M_s$ is thus equivalent to the question of whether $x_s$ is fixed by a subgroup of $\operatorname{Gal}(\mathbb Q(\zeta_p)/\mathbb Q)$ larger than $U_s$. By linear independence of the $\zeta_p^{y}$, an element $\sigma_a:\zeta_p\longmapsto\zeta_p^{a}$ of the Galois group fixes $x_s$ if and only if multiplication by $a\in(\mathbb Z/p\mathbb Z)^{\times}$ induces a permutation of $U_s$. Indeed, for $y\in U_s$, $(ay)^s=1$ if and only if $a^s=1$, that is to say if and only if $a\in U_s$.

Answer 3

Here is a broader context for your question. In every finite Galois extension $L/K$ there is an $\alpha \in L$ such that the $K$-conjugates $\{\sigma(\alpha) : \sigma \in G\}$ form a $K$-basis of $L$, where $G$ is the Galois group. This is called a normal basis. As an example, in $\mathbf Q(i)/\mathbf Q$ the sets $\{1,i\}$ and $\{i,-i\}$ are not normal bases ($\{1,i\}$ is a $\mathbf Q$-basis but its elements are not $\mathbf Q$-conjugates, and $\{i,-i\}$ is a set of $\mathbf Q$-conjugates of the necessary size but it is not linearly independent), while $\{1+i,1-i\}$ is a normal basis. Here is the key result: when $H$ is a subgroup of $G$, its fixed field $L^H$ is $K(\alpha_H)$ where $\alpha_H = {\rm Tr}_{L/L^H}(\alpha) = \sum_{\tau \in H} \tau(\alpha)$. The proof of this uses linear independence of characters, as other answers have already done in the particular case you ask about. See Theorem 3.8 here.

You use $K = \mathbf Q$, $L = \mathbf Q(\zeta_p)$, $G = (\mathbf Z/p\mathbf Z)^\times$, $\alpha = \zeta_p$, and $H = \{y \bmod p : y^k \equiv 1 \bmod p\}$. That $\zeta_p$ and its $\mathbf Q$-conjugates are a normal basis of $\mathbf Q(\zeta_p)/\mathbf Q$ is Example 3.2 here. Watch out: when $n \geq 1$, $\zeta_n$ and its $\mathbf Q$-conjugates are not always a normal basis of $\mathbf Q(\zeta_n)/\mathbf Q$: they are if and only if $n$ is squarefree and that is shown in the top answer here.