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.

  • $\begingroup$ Just a thought, but for the local case $K/\mathbb{Q}_p$, I think it's not so hard to describe $H^1(U)$ using the exact sequence $1\to U\to K^*\to\mathbb{Z}\to0$ coming from the discrete valuation, the exact sequence $1\to U_1\to U\to\mathbb{F}^*\to1$ coming from the reduction map, and the fact that $U_1$ is the formal group, so more-or-less isomorphic to the additive group (in any case, has a subgroup of finite index isomorphic to the additive group). Now you can map the global $H^1(U)$ into the product of the local ones and (?) apply a local-global principle. $\endgroup$ Jul 28, 2012 at 1:02
  • $\begingroup$ Dear Joe, thanks for your comment! I do hope that something local-global will work here.. $\endgroup$ Jul 30, 2012 at 4:40
  • $\begingroup$ Dear Effulgent, thanks again for your help. I need more time to come up with a more precise question.. $\endgroup$ Aug 22, 2012 at 23:20

2 Answers 2


$\newcommand\Q{\mathbf{Q}} \newcommand\OL{\mathcal{O}} \newcommand\I{\mathcal{I}} \newcommand\Z{\mathbf{Z}} \newcommand\eps{\epsilon} \newcommand\p{\mathfrak{p}} \newcommand\PP{\mathfrak{P}} \newcommand\Hom{\mathrm{Hom}} \newcommand\R{\mathbf{R}} \newcommand\q{\mathfrak{q}} \newcommand\Gal{\mathrm{Gal}}$ Summary: I think that it's difficult problem and that there won't be a "nice" answer in general.

Let $\I_K$ denote the group of invertible fractional ideals. There is a tautological exact sequence $$1 \rightarrow K^{\times}/\OL^{\times} \rightarrow \I_K \rightarrow C_K \rightarrow 0.$$ Taking cohomology gives the following sequence: $$0 \rightarrow (K^{\times}/\OL^{\times})^{G} \rightarrow \I^{G}_K \rightarrow C^{G}_K \rightarrow H^1(G,K^{\times}/\OL^{\times}),$$ which can naturally be modified to yeild the sequence: $$0 \rightarrow (K^{\times}/\OL^{\times})^{G}/\Q^{ > 0} \rightarrow \I^{G}_K/\Q^{ > 0} \rightarrow C^{G}_K \rightarrow H^1(G,K^{\times}/\OL^{\times}).$$ You are interested in the order of $ (K^{\times}/\OL^{\times})^{G}/\Q^{ > 0}$. As Will Sawin noted, the second term is simply $\prod \Z/e_p$. Thus one wants to understand the classes $I \in C^G_K$ (that is, the so-called ambiguous classes) which are actually strongly ambiguous , that is, $\sigma I = I$ as ideals , not just ideal classes. If one defines $S_K \subset C^G_K \subset C_K$ to denote the subset of strongly ambiguous classes, then one "formally" has an answer to your question, namely, $$\frac{1}{|S_K|} \cdot \prod e_p.$$ OTOH, this is really a proof by definition, so much content so far, although it gives a "name" to some of the objects to connect you with the literature.

The strongly ambiguous classes can also be described (given the exact sequence above) as the kernel of the map $$C^G_K \rightarrow H^1(G,K^{\times}/\OL^{\times}) \hookrightarrow H^2(G,\OL^{\times})$$

What makes things much easier when $K$ is cyclic is that one can essentially determine $C^G_K$ (by the ambiguous class number formula, which only exists for cyclic extensions), and also $S_K$. In fact, for a cyclic extension $S_K$ and $C^G_K$ are (almost) the same group. Let $C^{+}_K$ denote the group of invertible fractional ideals $I_K$ modulo the following relation: $[I] \sim [J]$ if and only if $I = (\alpha) J$ for some $\alpha$ with $N(\alpha) > 0$. In particular, $C^{+}_K$ is a quotient of the narrow class group, and surjects onto the class group. For example, $C^{+}_K = C_K$ if there exist units of norm $-1$. Suppose that $K/\Q$ is cyclic. I claim that the image of $(C^{+}_K)^G$ in $C^{G}_K$ lands in $S_K$. By assumption, $[\sigma I] \sim [I]$ in $C^{+}_K$, so $\sigma I = (\alpha) I$ for some $\alpha$ with positive norm. Clearly $N(\alpha)$ is a unit, so $N(\alpha) = 1$. Since $K/\Q$ is cyclic, by Hilbert 90 there exists a $\beta \in K^{\times}$ such that $\alpha = \beta/\sigma \beta$. (This version of Hilbert 90 only makes sense for cyclic extensions, which is one of the difficulties in the general case.) Replacing $I$ by $J = (\beta) I$ we deduce that $\sigma J = J$ and $[I] = [J]$ in $C_K$. When $K/\Q$ is not cyclic, however, then it's much trickier to get a handle on the ambiguous classes --- I don't think that there will be a nice answer in general.

The simplest possible non-cyclic case is the case of biquadratic extensions. Suppose that $K$ is totally real. Then $K$ contains three subfields $K_1$, $K_2$, and $K_3$. The unit group of $K$ is, up to finite index, generated by $U:=\{\eps_1, \eps_2, \eps_3\}$. Kubota (Uber den bizyklischen biquadratischen Zahlkörper) classified the possible $\Gal(K/\Q)$-module structure of $\OL^{\times}_K$ (there are $8$ or so different types, of indices ranging from $2^0$ to $2^3$). One could simply compute the corresponding cohomology groups of each type, and see what one gets. It's not clear that the answer will be any more precise than a list of cases. (Even in the case of real quadratic fields the answer depends on the existence of unit of norm $-1$, which is itself a notoriously fickle condition.)

Yet another way to explain why the cyclic case is not typical is that, since (at least for $p$ odd) the units tensor $\Z_p$ are annihilated by the norm map, they form a module under $$\Z_p[x]/(x^n-1,1+x+x^2+ \ldots + x^{n-1}) = \Z_p[\zeta_n],$$ which essentially a direct sum of PIDs.

I guess it depends on exactly what you are interested in doing, but it might be useful to consider the following approach. Assuming $K/\Q$ is abelian, one can easily compute the genus class field $L/K$. The field $L$ is the largest field such that $L/\Q$ is totally real, abelian, and $L/K$ is unramified everwhere. The point of working with $L$ is that one knows that the strongly ambiguous classes in $\OL_K$ (and in $\OL_L$) become (are) principal in $\OL_L$, so the transgression map is injective, and $$(L^{\times}/\OL^{\times}_{L})^{G_L}/\Q^{ > 0} \simeq \I^{G}_L/\Q^{ > 0} \simeq \prod \Z/e_p.$$ (The $e_p$ are the same for $L$ and $K$ since $L/K$ is everywhere unramified.) If $L = \Q(\zeta_N)$, then one has explicit generators for this group, namely, $1 - \zeta$ for each $p^n$th root of unity for $p^n \| N$. Note that, in some sense, this gives a "complete" description of the $d$-integers in $\Q^{\mathrm{ab}}$: The $d$-numbers in $K$ are the $G_K$-invariants of the finitely generated group $$\Z[\zeta_N]^{\times} \times \prod_{p^n \| N} (1 - \zeta)^{\Z}.$$

Concerning the image of the norm map, I think you are again out of luck. To explain why, consider what is close to the simplest possible non-trivial example, namely, $K = \Q(\sqrt{6p})$, where $p \equiv 1 \mod 4$ is prime. Write $(2) = \p^2_2$, $(3) = \p^2_3$, and $(p) = \p^2_p$ respectively. The field $K$ does not contain a unit of norm $-1$, so we know that the $d$-numbers generate the group $(\Z/2\Z)^2$. Since $\p_2 \p_3 \p_p$ is principal, it follows that exactly one of $\p_2$, $\p_3$, and $\p_p$ will be principal.

The genus class field of $K$ is $$F = \Q(\sqrt{6},\sqrt{p}).$$ It follows that if $C_K$ is the class group of $K$, then $C_K/2C_K$ is cyclic, and hence the $2$-part of the class group is cyclic. Via the Artin map, we can identify the primes ideals which lie in $C_K[2]$ as exactly the primes which split completely in the genus field. Since the genus field is given explicitly, we may compute that

  1. $\p_2$ splits in $F/K$ if and only if $p \equiv 1,17 \mod 24$,
  2. $\p_3$ splits in $F/K$ if and only if $p \equiv 1,13 \mod 24$,
  3. $\p_p$ splits in $F/K$ if and only if $p \equiv 1,5 \mod 24$.

In particular, if $p$ lies outside one of these equivalence classes, then the image of the corresponding $\p$ is non-trivial in $C_K/2 C_K$, and hence $\p$ is not principal. It follows that:

  1. If $p \equiv 17 \mod 24$, then the $d$-numbers are generated by $2$ and $6p$,
  2. If $p \equiv 13 \mod 24$, then the $d$-numbers are generated by $3$ and $6p$,
  3. If $p \equiv 5 \mod 24$, then the $d$-numbers are generated by $6$ and $6p$,

This leaves open the case when $q \equiv 1 \mod 24$. The computation above merely shows that $\p_2$, $\p_3$, and $\p_p$ all become trivial in $C_K/2C_K$. Since they cannot all be principal, it follows that when $p \equiv 1 \mod 4$, there must be a surjection $C_K \rightarrow \Z/4\Z$. By class field theory, this corresponds to the existence of an unramified extension $E/K$ with $\Gal(E/K) = \Z/4\Z$. We may construct $E$ explicitly as follows. Since $\Z[\sqrt{6}]$ has class number one, and $p \equiv 1 \mod 24$ splits this field, there exists a $\pi \in \Z[\sqrt{6}]$ with $N(\pi) = p$. (For local reasons the sign will be positive.) The choice of $\pi$ will be unique up to a sign and the fundamental unit $\eps = 5 + 2 \sqrt{6}$. If we explicitly write $\pi = A + B \sqrt{6}$, then we have $$A^2 - 6 B^2 = p.$$ The congruence on $p$ forces $B$ to be even and $A$ to be odd. After possibly multiplying by a unit and by $-1$, we may assume that $A \equiv 1 \mod 4$ and $B \equiv 0 \mod 4$. This determines $\pi$ up to squares, and $E$ is identified with the Galois closure of $\Q(\sqrt{\pi}) = \Q(\sqrt{A + B \sqrt{6}})$. In particular, $E$ is the splitting field of $$X^4 - 2 A X^2 + p.$$ A necessary condition (and sufficient if the $2$-part of the class group has order $4$) for $\p_p$ to be prime is that the residue degree of $p$ in $E$ is $1$, or equivalently that $(2 A/p) = 1$. Since $p \equiv 1 \mod 24$, this is the same as saying that $(A/p) = 1$. In fact, (proof omitted, because this answer is already too long and the argument is a somewhat tedious calculation of ring class fields and Kummer extensions) this is equvalent to $(6/p)_4 = 1$. So this leads to the following criterion:

  • If $p \equiv 1 \mod 24$, and the quartic residue $(6/p)_4 = -1$, then there are no $d$-numbers of norm $p$.

  • If $p \equiv 1 \mod 24$, and $8 \nmid h_K$, then there do exist $d$-numbers of norm $p$.

One could go on, giving similar criteria for $\p_2$ and $\p_3$, but it will just get worse (at least for $4 \| h_K$ one can give some sort of classical criteria due to the existence of governing fields, probably for $8 \| h_K$ and certainly for $16 \| h_K$ there won't be any non-tautological criterion.)

All in all, the "best" cases are when $K$ is its own genus field, or at least, for all $p$ dividing $[K:\Q]$, the $p$-class field of $K$ is the $p$-genus field of $K$.

If this doesn't help, perhaps you could say more precisely you want to prove about dimensions (or otherwise) of fusion categories?

  • $\begingroup$ Dear Effulgent, thanks a lot for thoughtful and useful answer! $\endgroup$ Aug 20, 2012 at 22:47

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.

  • $\begingroup$ Dear Will, thanks for your answer! I think that the image of the norm is contained in the subgroup generated by ramified primes but it is often much smaller (say for quadratic fields the image is of order $\le 4$ but we can have arbitrarily many ramified primes. $\endgroup$ Jul 30, 2012 at 4:35
  • $\begingroup$ in fact this gives you a lower bound on the class number of a quadratic field, I think. You can also see it's contained in the subgroup generated by certain powers of the ramified primes which makes sense since you can be less and more ramified. But I don't know how to deal with the class number problem $\endgroup$
    – Will Sawin
    Jul 30, 2012 at 15:18

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