most recent 30 from http://mathoverflow.net 2013-05-23T08:09:17Z http://mathoverflow.net/feeds/question/103345 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103345/what-is-known-about-first-cohomology-of-the-units-in-a-number-field Victor Ostrik 2012-07-27T20:53:13Z 2012-08-18T07:14:53Z

<p>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 $H^1(G,U)$.</p> <p>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 $H^1(G,U)$.</p> <p>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).</p> <p>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.</p>

http://mathoverflow.net/questions/103345/what-is-known-about-first-cohomology-of-the-units-in-a-number-field/103348#103348 Will Sawin 2012-07-27T22:02:17Z 2012-07-27T22:02:17Z

<p>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. </p> <p>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 $\mathbb Q$.</p> <p>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.</p> <p>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.</p> <p>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.</p>