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I have tried to solve exercise 1.7.4 in Neukirch's Algebraic Number Theory which states that $1+\zeta $ is a fundamental unit of $\mathbb Z [\zeta]$ when $\zeta$ is a primitive $5$th root of unity.

I have troubles proving that $1+\zeta$ is actually a fundamental unit. I have no idea of proving it. I have assumed that this is false, and let $x^k = \mu (1 + \zeta ) $ for some $1<k\in \mathbb Z$ . I have tried to prove that there is no such $x\in \mathbb Z[\zeta]$ with its minimal polynomial, which led me to my previous question, Proving the irreducibility of a specific family of polynomials, but I have troubles in solving both questions. I believe that I am not familiar with dealing with units. Can anyone give me, at least, a hint?

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Gheehyun Nahm
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1 Answers1

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Let $K=\mathbb{Q}(\zeta)$ with $\zeta=\zeta_5$. Then Dirichlet's unit theorem gives, with $(r,s)=(0,\phi(5)/2)=(0,2)$, that the unit group of $K$ is given by $\langle u \rangle \times \langle -\zeta \rangle \simeq \mathbb{Z}\times \mathbb{Z}/10\mathbb{Z}$. Here $u$ is a fundamental unit which generates the infinite cyclic group. Note that $\mathbb{Q}(\zeta+\zeta^{-1})=\mathbb{Q}(\sqrt{5})$. We can choose a fundamental unit $\frac{1-\zeta ^2}{1-\zeta}=1+\zeta$, see here. So we have $$ \mathbb{Z}=\langle 1+\zeta\rangle =\langle \frac{1+\sqrt{5}}{2}\rangle. $$

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Dietrich Burde
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  • Could you please detail why we can choose $\frac{1-\zeta ^2}{1-\zeta}=1+\zeta$ as a fundamental unit (without using the fact that cyclotomic units have index 1(the class number of $\Bbb Z(\zeta)$)) in the unit group of $\Bbb Z[\zeta_5]$? It seems that your remark that $\mathbb{Q}(\zeta+\zeta^{-1})=\mathbb{Q}(\sqrt{5})$ leads to that but I don't see how without computing the index of the group of cyclotomic units – Conjecture Nov 28 '19 at 17:17