Relation between Galois Cohomology $H^1(\text{Gal(L/K)}, L) = 0$ and $H^1(\text{Gal}(L/K), O_L)=0$

Question

Let $K$ be a number field and $O_K$be its ring of integers, and $L/K$ be a quadratic extension.

It is known that $H^1(\text{Gal}(L/K), L) = 0$ according to Hilbert's Theorem 90. However, what is known about $H^1(\text{Gal}(L/K), O_L)$?

Can this be easily deduced from Hilbert's Theorem 90, or is it essentially unrelated?

Indivisual solution for $H^1(\text{Gal}(L/K), O_L) = 0$ is also welcomed. Thank you for your help.

Answer 1

Here is an example computation, where $H^i(G,\mathcal O_L)\ne0$. Let $G:=\mathrm{Gal}(\mathbb Q(\sqrt{-1})/\mathbb Q)=\langle\sigma\rangle$. There is a homomorphism $\mathbb Z[G]\to\mathbb Z[\sqrt{-1}]:1\mapsto1+\sqrt{-1},\sigma\mapsto 1-\sqrt{-1}$, with an exact sequence: $$0\to\mathbb Z[G]\to\mathbb Z[\sqrt{-1}]\to\mathbb Z/2\to 0.$$ Since $\mathbb Z[G]$ is cohomologically trivial (by Shapiro's lemma), for any $i>0$, $$H^i(G,\mathbb Z[\sqrt{-1}])\simeq H^i(G,\mathbb Z/2)\simeq\mathbb Z/2.$$

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As a trivial remark, since $L\simeq\mathcal O_L\otimes\mathbb Q$, we have: $$H^i(G,\mathcal O_L)\otimes\mathbb Q\simeq H^i(G,L)=0,$$ i.e., $H^i(G,\mathcal O_L)$ is torsion.