Finite Galois module whose Ш¹ is nonzero?

Question

In algebraic number theory, we constantly make use of the nine-term Poitou-Tate sequence: Let $K$ be a number field and $M$ a finite $K$-Galois module. Then we have the nine-term exact sequence $$ H^0(K, M) \to \prod' H^0(K_v,M) \to H^2(K, M^\vee)^\vee \mathop{\to}\limits^\delta H^1(K, M) \mathop{\to}^{\operatorname{loc}} \prod' H^1(K_v,M) \to \cdots $$ The kernel of the localization map $\operatorname{loc}$, which is also the image of the connecting map $\delta$, is traditionally called $Ш^1(K, M)$. It is frequently remarked (e.g. in Neukirch, Schmidt, and Wingberg's Cohomology of Number Fields, Definition 8.6.2), that $Ш^1$ is finite but not necessarily $0$. However, in all the examples I've been able to compute explicitly (e.g. $M$ cyclic of prime order), $Ш^1 = 0$. Is there a ready example of a number field $K$ and a finite module $M$ such that $Ш^1(K, M) \neq 0$?

Answer 1

Wang's conterexample to Grunwald's theorem: $K=\mathbb{Q}(\sqrt{7})$ and $M=\mu_8$. Then $H^1(K,M) \cong K^\times/(K^\times)^8$. Now $16$ is not an $8$-th power in this field but locally an $8$-th power everywhere. Your group is cyclic of order $2$. See wikipedia.