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Let $L/\mathbb{Q}$ be a cyclic galois extension of degree $ 2n $ and $\sigma $ be a generator of $\operatorname{Gal}(L/\mathbb Q)$. Let $ U $ be the collection of all norm $ 1 $ elements of $L^\times$. Then $ U $ is a subgroup of $ L^{\times} $ = the set of all non zero elements in $L$. Now what is the index of the subgroup $ U $ of $ L^{\times} $?

Note: By Hilbert's theorem 90 we have $ U =\{\sigma(a)/a : a\in L^{\times}\} $. So one can consider the map $ \phi : L^{\times} \rightarrow L^{\times} $ by $ \phi(a) = \sigma(a)/a $. Then $ L^{\times}/ \mathbb{Q}^\times \cong U $.

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    $\begingroup$ Actually $U = \operatorname{Im}(\phi)$. In fact, $L^\times/U$ is naturally isomorphic to $\operatorname{Im}(\operatorname N_{L/\mathbb Q})$, not, as far as I can see, to $\operatorname{Im}(\phi)$. Or did you mean something else? \\ Also, this index will always be infinite in your setting, so are you sure that's what you mean to ask about? $\endgroup$
    – LSpice
    Commented Oct 9, 2021 at 18:29
  • $\begingroup$ Yes @Lspice you are right, I have edited my question. $\endgroup$
    – Sky
    Commented Oct 9, 2021 at 18:45
  • $\begingroup$ Yes @Lspice if we consider norm map then image of this map will be required quotient group. But does the map subjective? $\endgroup$
    – Sky
    Commented Oct 9, 2021 at 18:58
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    $\begingroup$ @LSpice, dunno. My reaction was/is just a charades-sort-of-thing "sounds like" [tugs ear]. There are certainly versions of such a question that have no sensible answers, and some that have trivial... and more interesting ones in-between. Possibly the original asker can refine the question... $\endgroup$
    – paul garrett
    Commented Oct 9, 2021 at 22:46
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    $\begingroup$ I voted to close, because the index of $U$ in $L^\times$ is infinite ($\omega$) for obvious reasons (as LSpice explained). If the OP meant to ask something else, he should ask it in a separate question. $\endgroup$
    – GH from MO
    Commented Oct 10, 2021 at 1:40

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