# On the determination of ambiguous ideal class of the extension $\mathbb{Q}(\zeta_5,\sqrt{m})/\mathbb{Q}(\zeta_5))$

let $$L=\mathbb{Q}(\sqrt{n},\zeta_5)$$ 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 $$\mathbb{Q}(\sqrt{n},\zeta_3)/\mathbb{Q}(\zeta_3)$$ 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

• – Martin Sleziak Dec 11 '19 at 17:04
• As a side note, the tag (abstract-algebra) is deprecated on MathOverflow, see the tag-info. Perhaps some other tags can be chosen instead. – Martin Sleziak Dec 11 '19 at 17:05

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 $$\ell$$.
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.
• sorry, of course s is $\sigma$. – Henri Cohen Feb 5 at 11:32
• yes is verified for $A^{\sigma^3+2\sigma^2+3\sigma+4}$ that is ambigous, but how you get that ?? – Fouad El Feb 5 at 12:21