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I am trying to prove the following result, which states that any unit in $\Bbb Z[\zeta_p]$ can be (multiplicatively) decomposed into a power of $\zeta_p$ and a real unit in $\Bbb Z[\zeta_p]$.

Let $K = \Bbb Q(\zeta_p)$, where $p$ is an odd prime. For any unit $\varepsilon \in \mathscr O_K^\times$, we can write $\varepsilon = \zeta_p^k u$ for some real unit $u \in \mathscr O_K^\times \cap \Bbb R$, and $k \in \Bbb Z/p\Bbb Z$.

My work:

Let $W_K$ denote the group of roots of unity in $K$. Since $p$ is odd, we have $W_K \cong \Bbb Z/2p\Bbb Z$ and $W_K$ consists of the $2p^{\text{th}}$ roots of unity. By Dirichlet's unit theorem, we have $$\mathscr O_K^\times \cong W_K \times \Bbb Z^r \cong \Bbb Z/2p\Bbb Z \times \Bbb Z^r$$ where $r = r_1 + r_2 - 1$, $r_1$ denotes the number of real embeddings $K \hookrightarrow \Bbb C$ and $r_2$ denotes the number of pairs of complex embeddings $K \hookrightarrow \Bbb C$. Furthermore, $$p-1 = \varphi(p) = [K:\Bbb Q] = r_1 + 2r_2$$ Now, choose any $\varepsilon \in \mathscr O_K^\times$. Let $\varepsilon^{(i)} := \sigma_i(\varepsilon)$, where $\sigma_1, \ldots,\sigma_n: K \hookrightarrow \Bbb C$ are all the embeddings of $K$, and $1\le i\le n$. Observe that for each $1\le i\le n$, $\frac{\varepsilon^{(i)}}{\overline{\varepsilon^{(i)}}}$ lies on the unit circle, i.e., $\left| \frac{\varepsilon^{(i)}}{\overline{\varepsilon^{(i)}}} \right| = \left|\left(\frac{\varepsilon}{\overline\varepsilon} \right)^{(i)} \right| = 1$. By a classical result due to Kronecker, we know that any algebraic integer whose all conjugates lie on the unit circle must be a root of unity. Thus, $\varepsilon/\overline{\varepsilon}$ is a root of unity in $K$. By the description of $W_K$ above, we get $$\frac{\varepsilon}{\overline \varepsilon} = \exp\left({\frac{2\pi i k}{2p}}\right) = \zeta_p^{k/2}$$ for some $1\le k\le 2p-1$. How should I proceed? I'm stuck here.

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2 Answers 2

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You showed that for any unit $\epsilon$ there is a root of unity $\alpha$ such that $\bar\epsilon = \alpha \epsilon$. We can then find another root of unity $\beta$ in $K$ such that $\alpha = \pm\beta^2$ (stealing from the answer of @reuns). In the $+$-case we are done, as then $\beta\epsilon = \overline{\beta\epsilon}$ is real. In the $-$-case $\beta\epsilon$ is purely imaginary, and we need to exclude this option.

One purely imaginary element of $K$ is $x:=\zeta_p -\zeta_p^{-1}$, so any other element $z$ of $K\cap i\mathbb R$ is of the form $z=xy$ for some $y\in L:=K\cap\mathbb R$. Now $N_{K/\mathbb Q}y = (N_{L/\mathbb Q}y)^2$ is a square of a rational number, whereas $N_{K/\mathbb Q}(x)=p$, so $N_{K/\mathbb Q}(z)=N_{K/\mathbb Q}(x)N_{K/\mathbb Q}(y)$ is not a square, in particular it is not $1$, so $z$ cannot be a unit.

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    $\begingroup$ Great solution ${}{}$ $\endgroup$
    – reuns
    May 3, 2023 at 9:39
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Let $\phi:O_K^\times \to \Bbb{C}^\times, \phi(z)= |z|^2$.

If $\phi(z)=1$ then $\phi(\sigma(z))= \sigma(\phi(z))= 1$ for all $\sigma\in Gal(K/\Bbb{Q})$.

This implies that $z$ is a root of unity, ie. $z = s\zeta_p^r$ with $s=\pm 1$ and $r\in \Bbb{Z}/(p)$.

Now take an arbitrary element $a\in O_K^\times$.

$\phi(a/\overline{a})=1$ so $a/\overline{a} = s\zeta_p^r$.

Assume that $s=1$. So $$a \zeta_{2p}^{-r} = \overline{a} \zeta_{2p}^r \in O_K^\times \cap \Bbb{R}$$

Writing $\zeta_{2p}^r = \pm \zeta_p^{r'}$ you get the claim $$a \zeta_p^{-r'} \in O_K^\times \cap \Bbb{R}$$

See @user8268's answer for excluding the case $s=-1$ (that is $a\zeta_p^{-r'}$ purely imaginary).

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  • $\begingroup$ Why is $\phi(\sigma(z))= \sigma(\phi(z))= 1$? $\endgroup$ May 1, 2023 at 15:40
  • $\begingroup$ The Galois group is abelian. @esoteric-elliptic $\endgroup$
    – reuns
    May 1, 2023 at 16:29
  • $\begingroup$ I'm not too familiar with Galois theory, but $\phi$ doesn't look like an automorphism of $K$ fixing $\Bbb Q$ pointwise. Is there any other way you could argue? @reuns $\endgroup$ May 1, 2023 at 16:36
  • $\begingroup$ $\phi(z)=z \,\rho(z)$ where $\rho$ is the complex conjugaison. $\rho$ commutes with $\sigma$. $\endgroup$
    – reuns
    May 1, 2023 at 16:49
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    $\begingroup$ I realize I made a mistake with $s=1$, let me find a way to repair it. $\endgroup$
    – reuns
    May 1, 2023 at 16:53

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