Strengthened Dirichlet's Unit Theorem for Cyclotomic Fields If $p$ is an odd prime, $\xi$ is a $p$th root of unity, and $\mu_k = \frac{1-\xi^k}{1-\xi}$, then $\mu_2, \mu_3, \ldots, \mu_{\frac{p-1}{2}}$ are multiplicatively independent. I would greatly appreciate a reference for this fact. I am also interested in the analogous fact when $p$ is replaced with $p^k$.
Thanks.
 A: I believe that this follows from the following two theorems of Dirichlet (seems like everything in this post is named after him):
Theorem: Let $\chi$ be a nontrivial primitive Dirichlet character. Then $0<|L(1, \chi)| < \infty$.
Theorem: Let $\chi$ be a nontrivial primitive Dirichlet character of conductor $D>1$. Then
$$L(1, \chi) = \frac{1}{\tau(\chi)} \sum_{a=0}^{D-1} \chi^{-1}(a) \log(1-\zeta_D^a).$$
If $D=p$ is prime, then the matrix $(\chi^{-1}(a))$, where $\chi$ runs over all characters of conductor dividing $p$ and $a$ runs over the nonzero residue classes mod $p$, is invertible, by the linear independence of characters. It follows that the values $\log(1-\zeta^a)$ are linearly independent over $\mathbf Q$; I think that this should do it.
A: This is basically the answer of Marie, now deleted, except that my answer is considerably longer and so I like it much less.
It's enough to show that the vectors $v_k=(\log|\phi(\mu_k)|)_{\phi}\in \mathbb R^{p-1}$, $k=2,3,\dots,(p-1)/2$, where $\phi$ runs over all the embeddings of $\mathbb Q(\xi)$ to $\mathbb C$, are linearly independent (over $\mathbb R$). This would  follow from the fact that the vectors $w_k=(\log\phi(1-\xi^k))_\phi=(\log(1-\xi^k),\log(1-\xi^{2k}),\dots,\log(1-\xi^{(p-1)k}))\in\mathbb C^{p-1}$, $k=1,2,\dots,p-1$, are linearly independent (over $\mathbb C$). Indeed, $v_k=((w_k-w_1)+(w_{p-k}-w_{p-1}))/2$.
To prove independence of $w_k$'s, let us view them as functions on $G:=(\mathbb Z/p\mathbb Z)^\times$, with $w_k(a)=\log(1-\xi^{ka})$, or equivalently as elements of $\mathbb C[G]$. Let $W$ be the vector space spanned by $w_k$'s. $W$ is invariant under the action of $G$ on $\mathbb C[G]$, as $w_k=k\cdot w_1$. But $\mathbb C[G]$ splits to 1-dim irreducible (and pairwise inequivalent) representations of $G$, namely $\mathbb C[G]=\bigoplus_{\chi\in G^*}\mathbb C\chi$, where $\chi$ runs over the characters of $G$ ($G^*$ is the dual group). As $W$ is a $G$-invariant, it must be $W=\bigoplus_{\chi\in X}\mathbb C\chi$ for some subset $X\subset G^*$. By orthogonality of characters, if we show that $(\chi,w_1)\neq0$ for every $\chi\in G^*$ then $X=G^*$ and we are done.
Showing $(\chi,w_1)\neq0$ is the actual work. For the trivial character $\chi=1$ it is obvious, $\sum_{a\in G}w_1(a)=\log((1-\xi)(1-\xi^2)\dots(1-\xi^{p-1}))=\log p\neq0$. For a non-trivial $\chi$ let us use the following:
$$(w_1,\chi)=\sum_{a\in G}\overline{\chi(a)}\log(1-\xi^a)=-\sum_{n=1}^\infty\sum_{a\in G}\overline{\chi(a)}\xi^{na}/n=$$
$$=-\tau(\chi)\sum_n\chi(n)/n,$$
where $\tau(\chi)=\sum_{a\in G}\overline{\chi(a)}\xi^a\neq 0$ is the Gauss sum and $\chi$ was extended to all integers by setting $\chi(n)=0$ if $p|n$. And finally, by the celebrate theorem of Dirichlet,
$$L(1,\chi):=\sum_n\chi(n)/n\neq0$$
(which follows from $\zeta_{\mathbb Q(\xi)}(s)=\prod_{\chi\in G^*}L(s,\chi)$ and from looking at the residue at $s=1$; there is a purely analytic (and very simple) proof in Serre's Course in Arithmetics).
