Question

Let $K/Q$ be a finite Galois extension with Galois group $G$. Let $U\subset K^\times$ be the group of units. I am interested in any available information about $H^1(G,U)$.

Motivation: in the theory of fusion categories one is interested in "d-numbers": an algebraic number $\alpha$ is a d-number if for any Galois conjugate $\beta$ of $\alpha$ the ratio $\frac{\alpha}{\beta}$ is a unit. Let us look at d-numbers contained in the number field $K$. It is clear that d-numbers form a group under multiplication; this group contains two obvious subgroups: units and rational numbers. An exact sequence $1\to U\to K^\times\to K^\times/U \to 1$ and Hilbert theorem 90 show that the quotient of d-numbers in $K$ by the units and rational numbers is precisely $H^1(G,U)$.

In the theory of fusion categories one is mainly concerned with the case when $K/Q$ is abelian and totally real. Using the properties of Herbrand quotient one shows that if $K/Q$ is cyclic (and real) of degree $n$ then the order of $H^1(G,U)$ is $n$ if $K$ contains a unit of norm $-1$ (this is always the case if $n$ is odd) and $2n$ otherwise. I suspect that group $H^1(G,U)$ is cyclic or direct sum of two cyclics in these cases but I don't see how to prove this. I don't know how to extend this computation to more general extensions (say, to biquadratic).

Finally, the computation of norm of a d-number gives a map from $H^1(G,U)$ to positive rationals modulo $|G|-$th powers. What can be said about image of this map? This seems to be nontrivial even for quadratic fields not containing a unit of negative norm.

Answer 1

I think it's actually easier to work out the invariants of $K^\times/U^\times$. $K^\times/U^\times$, is a subgroup of the group of fractional ideals, the kernel of the map to the ideal class group. Fractional ideals have a decomposition into powers of primes. To be Galois-invariant, the powers of conjugate primes must be equal.

Thus, the group of invariants in the group of fractional ideals is generated by, for each prime $p$ of $\mathbb Q$, the product $p_1\dots p_r$ where $p_1,\dots,p_r$ are the primes of $K$ lying over it. We need to quotient by the image of $H^0(K^\times)=\mathbb Q^\times$, which is generated by the primes $p$ of $\mathbb Q$.

We have the ideal factorization $(p_1\dots p_r)^e_p=p$, where $e_p$ is the ramification index of $p$ in $K$. Thus, the group of invariant fractional ideals, modulo the contribution of $\mathbb Q$, is $\prod_{p\in \mathbb Q} \mathbb Z/e_p$. The invariants in $K^\times/U^\times$ mod the contribution of $\mathbb Q$ are just the kernel of the map from this group to the ideal class group.

The norm map sends $p_1\dots p_r$ to $p^{|G|/e_p}$, so the image of the norm map is nontrivial and generated by powers of the ramified primes, as long as the class group doesn't get in the way.

This makes it clear that if a counterexample to your suspicion exists, it is a field ramified at three or more primes and probably with small class number.