In its Algebraic number theory (second edition), Lang uses in the proof of Theorem 1, chapter XI, paragraph 2, page 215-216, this group index identity:
$\displaystyle{\prod\_{v\in S}(k^\ast\_v:k^{\ast n}\_v)=n^{2s}}$
in which:


1. $n>2$ is an integer;
2. $k$ is a number field containing $n$-th roots of unity;
3. $S$ is a finite set of $s$ absolute values of $k$ containing all the archimedean ones and all primes $\mathfrak p$ of $k$ such that $\mathfrak p\mid n$.


I'll try to prove it.
We have:


* $(k\_v^\ast:k\_v^{\ast n})=\frac{n^2}{\|n\|\_v}$ for $v$ non-archimedean (corollary page 48);
* $(k\_v^\ast:k\_v^{\ast n})=(\mathbb R^\ast:\mathbb R^{\ast n})=\frac{3+(-1)^n}2$ for $v$ real;
* $(k\_v^\ast:k\_v^{\ast n})=(\mathbb C^\ast:\mathbb C^{\ast n})=1$ for $v$ complex.


By product formula, we have $\prod\_{v\in S}\|n\|\_v=1$.


Let $S\_\infty\subseteq S$ denote the set of archimedean absolute value of $k$, $r\_1$ the number of real absolute values and $r\_2$ the number of complex absolute values.
Then


$\displaystyle{\prod\_{v\in S}(k^\ast\_v:k^{\ast n}\_v)=(\mathbb R^\ast:\mathbb R^{\ast n})^{r\_1}(\mathbb C^\ast:\mathbb C^{\ast n})^{r\_2}\prod\_{v\in S\setminus S\_\infty}\frac{n^2}{\|n\|\_v}=}$


$\displaystyle{=\Bigl(\frac{3+(-1)^n}2\Bigr)^{r\_1}\prod\_{v\in S}\frac{n^2}{\|n\|\_v}\prod\_{v\in S\_\infty}\frac{\|n\|\_v}{n^2}=\Bigl(\frac{3+(-1)^n}2\Bigr)^{r\_1}n^{2s-r\_1}}$.


Where I get wrong?



Group index formula in Lang-Algebraic number theory
If there is a real embedding then $n=2$, because all the solutions to the equation $x^n-1$ in 
$\mathbb R$ are $\pm 1$ and the field contains the $n$-th roots of unity. Hence, there are 2 possibilities $r\_1=0$ and $n$ arbitrary and the formula is ok, or $r\_1>0$, $n=2$ and the formula is also ok.


