Computing étale cohomology group $H^1( \text{Spec}(k), \mu_n)$ and $H^1( \text{Spec}(k), \underline{\Bbb{Z}/\mathord{n \Bbb{Z}}})$

Question

I am starting to learn about étale cohomology and would like to compute a simple example. Let $k$ be a field with a fixed separable extension $k^s.$ I want to compute $H^1( \operatorname{Spec}(k), \mu_n)$ (I am speaking about the étale cohomology groups)

Here is what I tried:

$\bullet \quad H^1( \operatorname{Spec}(k), \mu_n)$

According this question here, we have an isomorphism $H^1( \operatorname{Gal}(k^s|k),A) \simeq H^1( \operatorname{Spec}(k),\mu_n),$ where the cohomology on left hand side is Galois cohomology and $A = \varinjlim_{L|k} \mu_n(L)$ with $L$ running through the Galois extensions of $k$. Then, if I am not mistaken, we have $A= \mu_n(k^s)$ and it follows (explained in detail in Serre's book) that $H^1( \operatorname{Gal}(k^s|k),\mu_n(k^s))= (k^s)^{\times}/ \mathord{{(k^s)^{\times}}^n}.$

I think this is right, but again I am just beginning with this and perhaps I may have overlooked something and there is something wrong with my reasoning.

$\bullet \quad H^1(\operatorname{Spec}(k), \underline{\mathbb{Z}/\mathord{n \mathbb{Z}}})$

Here $\underline{\mathbb{Z}/ \mathord{n \mathbb{Z}}}$ is the constant sheaf associated to $\mathbb{Z}/ \mathord{n \mathbb{Z}}$ and $k$ is a field with characteristic prime to $n.$

Following the same line of reasoning yields $H^1( \operatorname{Gal}(k^s|k), \underline{\mathbb{Z}/ \mathord{n \mathbb{Z}}}(k^s)).$ Now, when $k$ is a perfect field, we have $k^s= \bar{k}$ and under algebraically closed fields the sheaves $\mu_n$ and $\mathbb{Z}/ \mathord{n \mathbb{Z}}$ coincide. However, if we don't have the guarantee that $\mathbb{Z}/ \mathord{n \mathbb{Z}}$ I am not sure how to proceed.

My field theory is a bit rusty, is it true that for any $n$-th root of unity of $k$ is a separable element over $k.$ If so then can I proceed as above?

What if the characteristic is not prime to $n$. Is there a general method to compute $H^1(\operatorname{Spec}(k), \underline{\mathbb{Z}/ \mathord{n \mathbb{Z}}})$

Answer 1

The computation of $H^1(\operatorname{Spec}(k), \mu_n)$ is not completely correct: in Serres book it is written the correct answer $H^1( \operatorname{Gal}(k^s|k),\mu_n(k^s))= k^{\times}/ \mathord{{k^{\times}}^n}.$

Another possible way to prove it is to consider the sequence $$0\to \mu_n \to \mathbb Gm \to \mathbb Gm \to 0,$$ where the map between the multiplicative groups is the "power-to-the-$n$". The sequence is exact for the etalé topology (but not for the Zarisky topology), hence you get a long exact sequence for the cohomology. But $ H^1(\operatorname{Spec}(k), \mathbb{G}m)=0$ for Hilbert's theorem 90, hence you get the result.

About the second case, so $H^1(\operatorname{Spec}(k), \underline{\mathbb{Z}/\mathord{n \mathbb{Z}}})$, you cannot reduce to the previous case since $\underline{\mathbb{Z}/\mathord{n \mathbb{Z}}}\not\cong\mu_n$ even if $k$ contains all $n$-roots of unity. But you can reduce to compute Galois cohomology as you did, so you want to compute $H^1(\operatorname{Gal}(k^s/k), \mathbb{Z}/\mathord{n \mathbb{Z}})$, where now we have a constant group, since the points over $k^s$ of the constant sheaf is the constant group (and over every field). But now we have group cohomology, where the group acts trivially, hence the $H^1$ is just the hom's, so $$H^1(\operatorname{Gal}(k^s/k), \mathbb{Z}/\mathord{n \mathbb{Z}})\cong \operatorname{Hom}(\operatorname{Gal}(k^s/k),\mathbb{Z}/\mathord{n \mathbb{Z}}).$$ This last group can be big in general, but it is small for example if $k$ is finite: it is isomorphic to $\mathbb{Z}/\mathord{n \mathbb{Z}}$. In down to earth terms, the elements of $H^1(\operatorname{Spec}(k), \underline{\mathbb{Z}/\mathord{n \mathbb{Z}}})$ correspond essentially to Galois extensions of $k$ with cyclic Galois group of order dividing $n$.