Is a totally positive unit necessarily a square?

Question

Let $F$ be a totally real number field with narrow class number one which means every non-zero ideal has a totally positive generator.

Let $\mathfrak{o}$ be the ring of integers, $\mathfrak{o}_+$ the subset of totally positive integers, $\mathfrak{o}^\times$ the group of units, $\mathfrak{o}^\times_+$ the group of totally positive units.

In "Automorphic forms and representations" by Daniel Bump, p82 line 4, it is written: "as $\eta$ runs through $\mathfrak{o}^\times$, $\eta^2$ runs through $\mathfrak{o}_+^\times$ twice."

My question is why every totally positive unit is a square?

Answer 1

Let $\Sigma$ denote the real places of $F$, and let $\Sigma = \Sigma^{+} \cup \Sigma^{-1}$ denote any partition of $\Sigma$.

It is certainly easy enough to find an integral element $\alpha \in F$ which is positive for places in $\Sigma^{+}$ and negative for places in $\Sigma^{-}$. Now consider the ideal $I = (\alpha)$. Since the narrow class group is trivial, the ideal $I$ must be generated by a totally positive element, hence $I = (\beta)$. But then if $\varepsilon = \alpha/\beta$, then $\varepsilon$ is a unit which is positive in $\Sigma^{+}$ and negative in $\Sigma^{-}$.

It follows (with $d = [F:\mathbf{Q}]$ that the map

$$\mathcal{O}^{\times}_F \rightarrow \bigoplus_{\Sigma} F^{\times}_v/F^{\times 2}_v \simeq (\mathbf{Z}/2 \mathbf{Z})^d,$$

which sends each unit to its sign at each real place is surjective, and the kernel is the group of totally positive units. But by Dirichlet's theorem, the unit group is isomorphic to $\mathbf{Z}^{d-1} \oplus \mu_2$, so the kernel is also the group of square units.

One variation:

If $C$ is the class group and $N$ the narrow class group (which surjects onto $C$) you can replace the condition $|N|=1$ by the condition that $N=C$. This is because the ideal $I = (\alpha)$ is already trivial in $C$, so if $N=C$ it is trivial in $N$. Actually, this argument really shows that there is an exact sequence:

$$\mathcal{O}^{\times}/\mathcal{O}^{\times 2} \rightarrow \bigoplus_{\Sigma} F^{\times}_v/F^{\times 2}_v \rightarrow N \rightarrow C.$$