a certain finite index subgroup of an unit group of a number field I don't understand the proposition 2.3 of Gerald J.Janusz "Algebraic Number Fields" in Chapter V. It states as follow (changed some expressions based on Daileda's note p.39);

Proposition. Let $L / K$ be Galois extension of number fields with Galois group $G$ and let $S_{L, \infty}$ denote the set of infinite primes of $L$ (which consists of $r$ real primes $\mathfrak{P}_1\dots\mathfrak{P}_r$ and $s$ complex primes $\mathfrak{P}_{r+1}\dots\mathfrak{P}_{r+s}$). $S_{K, \infty}$ is similar. There exist units $w_{\mathfrak{P}} \in U_{L}$ ($U_{L}$ is an unit group of $L$) indexed by the primes $\mathfrak{P} \in S_{L, \infty}$ such that
(a)$\tau (w_{\mathfrak{P}})=w_{\tau (\mathfrak{P})} \text { for all } \tau \in G \text { and } \mathfrak{P} \in S_{L, \infty}$
(b)$\prod_{\mathfrak{P} \in S_{L, \infty}} w_{\mathfrak{P}}=1,$ and this is the only relation among these units
(c) $W=\left\langle w_{\mathfrak{P}} \mid \mathfrak{P} \in S_{L, \infty}\right\rangle$ has finite index in $U_{L}$

Before a proof:


*

*$\tau (\mathfrak{P})$ means $\mathfrak{P}\circ\tau^{-1}$ (identifying infinite prime with corresponding embedding). So the decomposition group $G(\mathfrak{P})$ of $\mathfrak{P}$ is defined.

*Is "only relation" a term of some kind of theory? I think this means $\prod_{\mathfrak{P} \in S_{L, \infty}} w_{\mathfrak{P}}^{v_{\mathfrak{P}}}=1\Rightarrow \text{all $v_{\mathfrak{P}}$ are equal ($v_{\mathfrak{P}}$ are integers)}$ as far as seeing following argument.


Proof.
Let $\mathfrak{P}\mid\mathfrak{p}$ (where $\mathfrak{P}=\mathfrak{P}_{i_0}\in S_{L,\infty},\mathfrak{p}\in S_{K,\infty}$).
Take an unit $w_{\mathfrak{P}}\in U_L$ such that $|w_{\mathfrak{P}}|_{\mathfrak{P}_i}>1 (i=i_0), |w_{\mathfrak{P}}|_{\mathfrak{P}_i}<1 (i\neq i_0)$ using Lemma2 of a proof of Dirichlet's unit theorem ($|\cdot|_{\mathfrak{P}}$ is a valuation in $\mathfrak{P}$).
Let $w_{\mathfrak{P}}'\colon=\prod_{\tau\in G(\mathfrak{P})}\tau (w_{\mathfrak{P}})(\in U_L)$, then for any $i=1\dots r+s$,
$$
|w_{\mathfrak{P}}'|_{\mathfrak{P}_i}
=\prod_{\tau\in G(\mathfrak{P})}|\tau (w_{\mathfrak{P}})|_{\mathfrak{P}_i}
=\prod_{\tau\in G(\mathfrak{P})}|\tau^{-1} (w_{\mathfrak{P}})|_{\mathfrak{P}_i}
=\prod_{\tau\in G(\mathfrak{P})}|\mathfrak{P}_i(\tau^{-1} (w_{\mathfrak{P}}))|
=\prod_{\tau\in G(\mathfrak{P})}|w_{\mathfrak{P}}|_{\tau (\mathfrak{P}_i)}.
$$
For $\tau\in G(\mathfrak{P})$ and $\mathfrak{P}\neq\mathfrak{P}_i$, it follows that $\tau(\mathfrak{P}_i)\neq\mathfrak{P}$ and so $|w'_{\mathfrak{P}}|_{\mathfrak{P}_i}>1 (i=i_0), |w'_{\mathfrak{P}}|_{\mathfrak{P}_i}<1 (i\neq i_0)$.
Now if $\mathfrak{Q}\in S_{L,\infty}$ also extending $\mathfrak{p}$ and if $\tau(\mathfrak{P})=\mathfrak{Q}$ then we set $w_{\mathfrak{Q}}\colon=\tau(w_{\mathfrak{P}})$.
Let $W'\colon=\left\langle w_{\mathfrak{P}}' \mid \mathfrak{P} \in S_{L, \infty}\right\rangle$, then $\ell(W')\cong \mathbb{Z}^{r+s-1}$ and any $r+s-1$ elements of $\{\ell(w'_{\mathfrak{P}})\}$ give a basis (see a proof of Dirichlet's unit theorem).

For each $\mathfrak{p}\in S_{K,\infty}$, let $v_{\mathfrak{p}}\colon=\prod_{\mathfrak{P}\mid\mathfrak{p}}w_{\mathfrak{P}}' (\in U_K)$. Let $|S_{K,\infty}|\colon=r'+s'$ then any $r'+s'-1$ elements of $\{v_{\mathfrak{p}}\}_{\mathfrak{p}\in S_{K,\infty}}$ generate a free abelian group.
There exist integers $a_{\mathfrak{p}}$ (not all zero) such that
  $$
1=\prod_{\mathfrak{p}\in S_{K,\infty}} {v_{\mathfrak{p}}}^{a_{\mathfrak{p}}}.
$$
In fact none of the $a_{\mathfrak{p}}$ can be zero because there cannot be a relation between $r+s-1$ of the generators $\{\ell(w'_{\mathfrak{P}})\}$. 
Finally replaceing $w_{\mathfrak{P}_i}$ with $(w'_{\mathfrak{P}_i})^{a_{\mathfrak{p}}}$ if $\mathfrak{P}_i\mid\mathfrak{p}$. Then the product of the $w_{\mathfrak{P}_i}$ is $1$ and there is no other relation between them and $\ell(W)\cong \mathbb{Z}^{r+s-1}$(where $W$ be the group generated by $\{w_{\mathfrak{P}_i}\}$).

So $(\ell(U_L):\ell(W))$ is finite (because of this) and $\operatorname{Ker}{\ell}$ is also finite, it follows that $(U_L:W)$ is finite.
Question.
I don't understand the part of block quoted. Could you explain in details?
 A: *

*First paragraph


First we prepare easy lemma.

lemma.
    Let $L/K$ be a Galois extension of number fields with degree $n$ and assume that $d$ real primes of $K$ ramify in $L/K$. Let $r_L,s_L$ be the number of real primes and pairs of complex primes of $K$. $r_K,s_K$ are similarly. Then the following formulas hold:
    $$
    r_L=(r_K-d)n,\quad s_L=ns_K+\frac{n}{2}d
  $$

Indexing element of $S_K$ as $\mathfrak{p}_1,\dots,\mathfrak{p}_{r'+s'}$, then asuume that there are $\mathfrak{P}_{i,1},\dots,\mathfrak{P}_{i,g_{\mathfrak{p}_i}}$ primes of $L$ on $\mathfrak{p}_i$.
Now
$$
  g_{\mathfrak{p}_i} =\frac{n}{e_{\mathfrak{p}_i}f_{\mathfrak{p}_i}}=
  \begin{cases}
    \dfrac{n}{2} & (\text{$\mathfrak{p}_i$ ramifies}) \\
    n & (\text{otherwise}).
  \end{cases}
$$
Let $D:=\sum_{i=1}^{r'+s'-1} g_{\mathfrak{p}_i}$ be the number of $w_{\mathfrak{P}_{i,j}}'$ which consist $v_{\mathfrak{p}_1},\dots ,v_{\mathfrak{p}_{r'+s'-1}}$.
Then $r+s=n(r'+s')-\dfrac{n}{2}d$ by the lemma. 
(i) $\mathfrak{p}_{r'+s'}$ ramifies
\begin{align*}
  D
  =& \dfrac{n}{2}(d-1)+n(r'+s'-d)\\
  =& n(r'+s')-\dfrac{n}{2}d-\dfrac{n}{2}\\
  =& r+s-\dfrac{n}{2}\\
  \leq & r+s-1.
\end{align*}
(ii) $\mathfrak{p}_{r'+s'}$ unramifies
\begin{align*}
  D
  =& \dfrac{n}{2}d+n(r'+s'-1-d)\\
  =& n(r'+s')-\dfrac{n}{2}d-n\\
  =& r+s-n\\
  \leq & r+s-1.
\end{align*}
So $D\leq r+s-1$ and $\{v_{\mathfrak{p}_i}\}_{i=1,\dots, r'+s'-1}$ is linearly independent(consider image by $\ell$).


*

*Second paragraph


The existence of $a_{\mathfrak{p}}$ follows by following fact:
"Let $M$ be a free $R$ module with rank $n$(where $R$ is commutative ring), then any $m (\geq n)$ elements of $M$ is linearly dependent on $R$".
If $a_{\mathfrak{p}}=0$ for some $\mathfrak{p}$, then $a_{\mathfrak{p}}=0$ for all $\mathfrak{p}$ since linearly independence of remaining $\{v_{\mathfrak{p}}\}$ (at most $r'+s'-1$ elements). But this is contradiction to ''(not all zero)''


*

*Third paragraph


We assume $\prod_{\mathfrak{P}\in S_{L,\infty}} w_{\mathfrak{P}}^{v_{\mathfrak{P}}}=1$ ($v_{\mathfrak{P}}$ are integers) . 
Then $\prod_{\mathfrak{P}\in S_{L,\infty}} w_{\mathfrak{P}}^{v_{\mathfrak{P}}-v}=1$ (where $v:= \min_{\mathfrak{P}\in S_{L,\infty}}{v_{\mathfrak{P}}}$) and since this is a product of $r+s-1$ elements and $a_{\mathfrak{p}}\neq 0$ for all $\mathfrak{p}\in S_{K,\infty}$, $\{\ell(w_{\mathfrak{P}})\}$ is linearly independent. Therefore $v_{\mathfrak{P}}=v$ for all $\mathfrak{P}\in S_{L,\infty}$, and $\ell(W)\cong \mathbb{Z}^{r+s-1}$.
