# Multiplicative group of number field mod field norms of quadratic extension

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.)

• There is some variant of that assertion which is correct, and is the assertion of global classfield theory over your (totally real) basefield. But the quotient would need to include much more stuff, like the connected component of the multiplicative group of the archimedean part $F^\times_\infty$ of the ideles of the the base, and so on. So there is some true assertion there, and could conceivably actually suffice for the issues there. But, you're right... as in @KConrad's answer... the assertion is not literally correct, etc. Feb 8, 2022 at 23:20

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 $$\mathbf C/\mathbf R$$.