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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?

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up vote 2 down vote accepted

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.

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