Question about proof that an algebraic integer is a unit if and only if it has norm $\pm 1$.

Question

Let $K/\mathbb{Q}$ be a number field and let $\mathcal{O}_K$ be its ring of integers. I have a question about the proof that $\mathcal{O}_K^\times =\{\alpha\in \mathcal{O}_K\mid N_{K/\mathbb{Q}}(\alpha)=\pm1 \}$. I understand that if $\alpha \in \mathcal{O}_K^\times$ then $N_{K/\mathbb{Q}}(\alpha)=\pm1$ but the converse is where I'm stuck. The proof I'm following (Neukirch `Algebraic Number Theory', page 12) argues that if $N_{K/\mathbb{Q}}(\alpha)= \pm1$ then $aN_{K/\mathbb{Q}}(\alpha)=1$ for some $a\in \{\pm1\}$, whence $$ 1=aN_{K/\mathbb{Q}}(\alpha)=a\prod_{\sigma:K\rightarrow\overline{ \mathbb{Q}}}\sigma\alpha=a\alpha\prod_{\sigma\neq 1}\sigma\alpha=\alpha\beta $$ for some $\beta\in \mathcal{O}_K$. So in this case, $\beta= a\prod_{\sigma\neq 1}\sigma\alpha$. Here is my issue:

How do we know that $\prod_{\sigma\neq 1}\sigma\alpha$ is in $\mathcal{O}_K$?

I understand that all the conjugates $\sigma\alpha$ of $\alpha$ are integral over $\Bbb{Z}$, but this does not, a priori, mean that each conjugate $\sigma\alpha$ lies in $\mathcal{O}_K:=\{x\in K\mid \text{$x$ integral over $\Bbb{Z}$}\}$ since each $\sigma \alpha$ may not be in $K$ itself.

For example, if $K=\mathbb{Q}(\sqrt[3]{2})$ then we know $\mathcal{O}_K=\Bbb{Z}[\sqrt[3]{2}]$. The conjugates of, say, $\alpha=\sqrt[3]{2}$ are $$\sqrt[3]{2},\,\, \zeta\sqrt[3]{2},\,\,\zeta^2\sqrt[3]{2},$$ where $\zeta$ is a primitive third root of unity. So in this case $\sigma\alpha\notin \mathcal{O}_K=\Bbb{Z}[\sqrt[3]{2}]$ for $\sigma\neq 1$, even though $\prod_{\sigma\neq 1}\sigma\alpha=\sqrt[3]{4}\in \mathcal{O}_K$.

Answer 1

The product $\prod_{\sigma\neq 1}\sigma\alpha$ lies in $K$ because it is $N(\alpha)/\alpha$. It is an algebraic integer because each $\sigma\alpha \in \overline{\mathbb Z}$.

We conclude using $\overline{\mathbb Z} \cap K = \mathcal O_K$, by definition of integral closure.

Answer 2

Consider the Galois closure $K'$ of $K/\Bbb Q$.

The element $u := \prod_{\sigma\neq 1}\sigma(\alpha)$ is invariant under any embedding $\tau : K' \to \overline{\Bbb Q}$ fixing $K$ (i.e. any element of the Galois group of $K' / K$), since $\alpha u = \mathrm{N}_{K / \Bbb Q}(\alpha) \in \Bbb Q$ and $\alpha \in K$. Thus $u$ must lie in $K$, by Galois theory.

Since all the elements $\sigma(\alpha)$ are integral over $\Bbb Z$, as you noted, this implies that $u$ is integral over $\Bbb Z$. Finally, this means that $u \in \mathcal O_K$.

Answer 3

Another way of looking at the same thing: If you recognize that $\text{Norm}^{\Bbb Q(\alpha)}_{\Bbb Q}(\alpha)$ is, up to sign, the constant coefficient of $\alpha$’s minimal polynomial $\text{Irr}(\alpha,\Bbb Q[X])$, then since $\alpha$ is an algebraic integer, the Irr is actually in $\Bbb Z[X]$.

We then have $\alpha^n+c_{n-1}\alpha^{n-1}+\cdots+c_1\alpha\pm1=0$ with all $c_i\in\Bbb Z$; divide by $\alpha$ and get $\mp1/\alpha=\alpha^{n-1}+c_{n-1}\alpha^{n-2}+\cdots+c_1$, where the right-hand side is an algebraic integer.

(You need only a minor adjustment in case $K\ne\Bbb Q(\alpha)$.)

Answer 4

Denote by $N$ the norm of $K/\mathbf Q$. If $x$ is a unit of $K$, then $N(x)$ and $N(x^{-1})$ are rational integers whose product is $1$, hence $N(x)=\pm 1$. Conversely, if $x\in O_K$ has norm $\pm 1$, its characteristic polynomial reads $X^n+a_{n-1}x^{n-1}+...+a_1x\pm 1\in\mathbf Z[X]$. Then $\pm (x^{n-1}+...+a_1)$ is an algebraic integer which is the inverse of $x$.

NB. I have just noticed that this is almost exactly the answer given by Lubin ! The characteristic polynomial of $x$ is by definition the characteristic polynomial (in the sense of linear algebra) of the multiplication by $x$. Its roots are exactly those of the minimal polynomial repeated $[K:\mathbf Q(x)]$ times.