Let $K$ be a number field and $G_K$ the absolute Galois group of $K$. Let $M$ be a finite $G_K$-module.
The Tate dual of $M$ is defined as follows: $M^D = \text{Hom}(M, \mu(K))$, where $\mu(K)$ is the group of all roots of unity.
Let $v$ be a fixed place of $K$.
It is well-known that $H^1(G_{K_v}, M^D) \cong \widehat{H^1(G_{K_v}, M)}$, where $\widehat{H^1(G_{K_v}, M)}$ is the Pontryagin dual of $H^1(G_{K_v}, M)$ (cf. Milne, Arithmetic Duality, Chap. $I, 2$).We call this sometimes Tate local duality.
My question is: Does this isomorphism hold when $M$ is not finite?
Thank you for your help.
Best Answer
One thing you'll learn very quickly when it comes to duality theorems in Galois cohomology that that there are about a million things called duality, all slightly different from each other.
First, let's be clear on the definitions. Let $K$ be a local field and $K^s$ its separable closure. Let $G_K=\text{Gal}(K^s/K)$ be its absolute Galois group.
For a finite $G_K$ module $M$, define $M^D=\hom_{ab}(M,\mu(K^s))$. Note that we are taking the morphisms to the Galois module of all roots of unity, not just the ones in $K$. It is in this setting only that we have a canonical isomorphism
$$H^1(G_K,M)\simeq \widehat{H^1(G_K,M^D)}$$
I claim this does not hold for finitely generated modules $M$. Let $M=\mathbb{Z}$ have trivial action. Then $M^D=\hom_{ab}(\mathbb{Z},\mu(K^s))\simeq\mu(K^s)$.
First, we compute $H^1(G_K,M)$. Since $G_K\simeq\varprojlim\text{Gal}(L/K)$, we have
$$H^1(G_K,M)\simeq \varinjlim H^1(\text{Gal}(L/K),\mathbb{Z})\simeq \varinjlim \hom(\text{Gal}(L/K),\mathbb{Z})\simeq \varinjlim 0=0.$$
Now we compute $H^1(G_K,M^D)$. Note that $M^D\simeq\mu(K_S)\simeq\varinjlim \mu_n(K_S)$. So by proposition 1.5.1 of Cohomology of Number Fields (or because direct limits are exact) we have
$$H^1(G_K,M^D)\simeq \varinjlim H^1(G_K,\mu_n(K^S))$$
By Kummer theory, we know that $H^1(G_K,\mu_n(K^S))\simeq K^\times/(K^\times)^n$. So we have
$$H^1(G_K,M^D)\simeq \varinjlim K^\times/(K^\times)^n\simeq \varinjlim K^\times \otimes (\mathbb{Z}/n\mathbb{Z})\simeq K^\times \otimes \mathbb{Q}/\mathbb{Z}.$$
Therefore $H^1(G_K,M)\not \simeq \widehat{H^1(G_K,M^D)}$.
There is, however, a duality theorem that does apply to finitely generated modules. For a discrete finitely generated $G_K$ module $M$. In this case, we define $M^d=\hom_{ab}(M,(K^s)^\times)$ (Note that Milne also uses $M^D$ here, but I am using $M^d$ for clarity). This case becomes more complicated because some topology enters the picture, but we obtain isomorphisms
$$H^1(G_K,M)\simeq \widehat{H^1(G_K,M^d)}.$$
Note that this recovers the duality for finite modules as a special case: Let $M$ be a finite $G_K$ module. Since a homomorphism cannot take a finite order element to an infinite order element, we have
$$M^d=\hom_{ab}(M,(K^s)^\times)=\hom_{ab}(M,\mu(K^s))=M^D.$$