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let $L=\mathbb{Q}(\sqrt[5]{n},\zeta_5)$ and $K=\mathbb{Q}(\zeta_5)$ the $5^{th}$ cyclotomic fields, we now that $[L:K] = 5$ and $ GAl(L/K) =\langle\sigma\rangle$ so we call $\mathcal{A}$ an ambigous ideal class of the extension $L/K$ if and only if $\mathcal{A}^{\sigma}= \mathcal{A}$.

My question is how to prove using that $\sigma^4+\sigma^2+\sigma^2+\sigma+1 =0$ that it exist an ambiguous ideal class??

For example, in the case of $\mathbb{Q}(\sqrt[3]{n},\zeta_3)/\mathbb{Q}(\zeta_3)$ we have $\sigma^3=1$ and $ \sigma^2+\sigma+1=0$ and we have $\mathcal{A}^{3}=\mathcal{A}$, so we prouve that $\mathcal{A}^{\sigma-1}$ is ambigous. I need to do the same for the case of 5

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The standard argument (see the early articles by George Gras) is the following: Let $L/K$ be a cyclic extension of degree $\ell$ and let $\sigma$ denote a generator of the Galois group. If $c$ is a nontrivial ideal class in $K$ whose order is $\ell$ and which is killed by the relative norm, then $c^{(1 - \sigma)^{\ell-1}} = 1$ (binomial expansion). Thus the sequence $$ c, c^{1-\sigma}, c^{(1-\sigma)^2}, \ldots, c^{(1 - \sigma)^{\ell-1}} $$ must contain an ideal class whose $(1-\sigma)$th power is trivial; this ideal class is clearly ambiguous and hs order $\ell$.

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Sorry, my first answer was too hasty. However, I maintain that you should look up the notion of Lagrange resolvent, which can be found in many algebraic number theory books. In your case it is $A^{s^3+2s^2+3s+4}$ which is ambiguous.

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  • $\begingroup$ what do you mean by s in your answer ?? $\endgroup$ – Fouad El Feb 5 at 10:31
  • $\begingroup$ sorry, of course s is $\sigma$. $\endgroup$ – Henri Cohen Feb 5 at 11:32
  • $\begingroup$ yes is verified for $A^{\sigma^3+2\sigma^2+3\sigma+4}$ that is ambigous, but how you get that ?? $\endgroup$ – Fouad El Feb 5 at 12:21

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