# Any unit in $\Bbb Z[\zeta_p]$ can be decomposed into a power of $\zeta_p$ and a real unit in $\Bbb Z[\zeta_p]$

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.

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.

• Great solution ${}{}$ May 3, 2023 at 9:39

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).

• Why is $\phi(\sigma(z))= \sigma(\phi(z))= 1$? May 1, 2023 at 15:40
• The Galois group is abelian. @esoteric-elliptic May 1, 2023 at 16:29
• 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 May 1, 2023 at 16:36
• $\phi(z)=z \,\rho(z)$ where $\rho$ is the complex conjugaison. $\rho$ commutes with $\sigma$. May 1, 2023 at 16:49
• I realize I made a mistake with $s=1$, let me find a way to repair it. May 1, 2023 at 16:53