Galois cohomology of imaginary abelian number fields

Question

Let $K$ be an imaginary abelian number field of degree $[K:\mathbb{Q}] \geq 4$. Let $G=\text{Gal}(K/\mathbb{Q})$, and denote by $U_K$ the unit group of $K$. Can we show that the order of the first cohomology group $H^1(G,U_K)$ is divisible by $[K:\mathbb{Q}]$? The statement is true for imaginary biquadratic fields as it was shown in Lemma 4.3 in Zantema's paper, but how about the "general case", say for all the "imaginary abelian number fields"?

Answer 1

As a partial answer, I think the statement holds for any abelian number field $K$ with class number $h(K)=1.$

Corollary 2.3 in [Class Number and Ramification in Number Fields, Armand Brumer and Michael Rosen Nagoya Mathematical Journal , Volume 23 , December 1963 , pp. 97 - 101] (here), shows that for such $K,$ $$|H^1(G,U_K)|=e_1\cdots e_r$$ where $e_{1-r}$ are the ramification indices of the rational primes which ramify in $K.$ So it's a case of showing the product is divisible by $|K:\mathbb Q|.$ This I think holds for any abelian number field (and I assume is well known really, unless I have made a mistake).

Suppose $K$ has conductor $n$ and let $p_1,\dots,p_s$ with $s\ge r$ be all the primes dividing $n$ i.e. the primes ramifying in $L=\mathbb Q(\zeta_n),$ where the $p_i$ for $1\le i\le r$ are as above. Let $e_{L/\mathbb Q}(p_i)$ be the ramification indices for these primes in $L.$ Choose a prime $P_i$ of $K$ lying over $p_i.$ Then $e_{L/\mathbb Q}(p_i)=e_{L/K}(P_i)e_i$ and $e_i=1$ for any $i>r.$ Since $L$ is cyclotomic, $$|L:\mathbb Q|=\prod_ie_{L/\mathbb Q}(p_i)$$ Hence to show $|K:\mathbb Q|$ divides $e_1\cdots e_r=e_1\cdots e_s,$ it's equivalent to show that $\prod_ie_{L/K}(P_i)$ divides $|L:K|.$

If $L=K$ it's clear, so assume not and choose $i$ such that $\zeta_q\notin K$ where $q$ is the power of $p_i$ dividing $n.$ Let $N=K(\zeta_q)$ and let $Q_j$ be a prime of $N$ lying over $P_j$ for $1\le j\le s.$ By induction $\prod_je_{L/N}(Q_j)$ divides $|L:N|.$ Also, $P_j$ for $j\neq i$ is unramified in $N,$ so $\prod_j e_{N/K}(P_j)=e_{N/K}(P_i)$ divides $|N:K|.$ Hence $e_{L/K}(P_j)=e_{L/N}(Q_j)e_{N/K}(P_j)$ divides $|L:N||N:K|=|L:K|$ as required.