Multiplicative group of number field mod field norms of quadratic extension

Question

I'm reading some notes(*) about arithmetic lattices in $\operatorname{SU}(n,1)$. I'm trying to understand the data that classifies (up to commensurability) the arithmetic lattices of the "first type" in this group. Fix a totally real number field $F$ with a totally imaginary quadratic extension $E$. Then these lattices are defined as the integer points of a unitary group preserving a Hermitian form $H$ of signature $(n,1)$ defined over $E$, i.e. $\operatorname{SU}(H;\mathcal O_E)$, which is positive- or negative-definite at all other infinite places.

One invariant you might try to pick up from such a construction is the determinant of $H$ (thought of as an $(n+1) \times (n+1)$ Hermitian matrix), which lies in $F^\times$. But this determinant is not actually invariant under a change of basis. It turns out that the relevant invariant lies in $F^\times/N_{E/F}(E^\times)$. Now here's my confusion: in section 6.6 ("Parity and describing commensurability classes") of the cited notes, the author seems to be claiming that the group $F^\times/N_{E/F}(E^\times)$ is a group with only two elements. He refers to the "trivial and nontrivial class[es]". I can see that all elements of this group have order 2, but I don't see why the group itself has order 2.

Is it true that $F^\times/N_{E/F}(E^\times)$ is a group of order 2?

(*) D.B. McReynolds. Arithmetic lattices in $\operatorname{SU}(n,1)$.

(I should point out that these notes are clearly unfinished and may not have been intended for publishing. I don't want to call to question the author's work.)

Answer 1

The index is definitely not finite. If $F$ is an arbitrary number field and $E$ is a finite extension with $[E:F] > 1$, the norm subgroup ${\rm N}_{E/F}(E^\times)$ has infinite index in $F^\times$. A proof in general is in a recent stackexchange post here.

Let's consider the simplest case relevant to what you are reading: $F = \mathbf Q$ and $E = \mathbf Q(i)$. Then $F^\times/{\rm N}_{E/F}(E^\times) = \mathbf Q^\times/{\rm N}_{\mathbf Q(i)/\mathbf Q}(\mathbf Q(i)^\times)$, and for primes $p$ and $q$ such that $p, q \equiv 3 \bmod 4$, the ratio $p/q$ is not of the form $a^2 + b^2$ for $a, b \in \mathbf Q$. Thus the infinitely many primes $p$ such that $p \equiv 3 \bmod 4$ are all inequivalent in $\mathbf Q^\times/{\rm N}_{\mathbf Q(i)/\mathbf Q}(\mathbf Q(i)^\times)$, so they show that quotient group not only size has bigger than $2$, but is in fact infinite.

The standard setting where a quotient group $F^\times/{\rm N}_{E/F}(E^\times)$ has order $2$ when $[E:F] = 2$ is for a quadratic extension of local fields. (In case of local fields of characteristic $2$, I should say quadratic Galois extension of local fields, which would be redundant outside of characteristic $2$.) In that case it is true that ${\rm N}_{E/F}(E^\times)$ has index $2$ in $F^\times$. This is a special case of local class field theory: if $E/F$ is an abelian extension of local fields then ${\rm Gal}(E/F) \cong F^\times/{\rm N}_{E/F}(E^\times)$. Here "local field" includes archimedean completions, e.g., ${\mathbf R^\times}/{\rm N}_{\mathbf C/\mathbf R}(\mathbf C^\times) = {\mathbf R}^\times/{\mathbf R_{>0}} = \{\pm 1\}$ has order $2$. Maybe there is an implicit use of archimedean completions when you refer to quotient groups with norms, since compatible archimedean completions of a totally real number field and a totally imaginary quadratic extension will look like the extension $\mathbf C/\mathbf R$.