59.6 A computation
How do we compute the cohomology of $\mathbf{P}^1_\mathbf {C}$ with coefficients $\Lambda = \mathbf{Z}/n\mathbf{Z}$? We use Čech cohomology. A covering of $\mathbf{P}^1_\mathbf {C}$ is given by the two standard opens $U_0, U_1$, which are both isomorphic to $\mathbf{A}^1_\mathbf {C}$, and whose intersection is isomorphic to $\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} = \mathbf{G}_{m, \mathbf{C}}$. It turns out that the Mayer-Vietoris sequence holds in étale cohomology. This gives an exact sequence
To get the answer we expect, we would need to show that the direct sum in the third term vanishes. In fact, it is true that, as for the usual topology,
and
These results are already quite hard (what is an elementary proof?). Let us explain how we would compute this once the machinery of étale cohomology is at our disposal.
Higher cohomology. This is taken care of by the following general fact: if $X$ is an affine curve over $\mathbf{C}$, then
This is proved by considering the generic point of the curve and doing some Galois cohomology. So we only have to worry about the cohomology in degree 1.
Cohomology in degree 1. We use the following identifications:
The first identification is very general (it is true for any cohomology theory on a site) and has nothing to do with the étale topology. The second identification is a consequence of descent theory. The last set describes a collection of geometric objects on which we can get our hands.
The curve $\mathbf{A}^1_\mathbf {C}$ has no nontrivial finite étale covering and hence $H_{\acute{e}tale}^1 (\mathbf{A}^1_\mathbf {C}, \mathbf{Z}/n\mathbf{Z}) = 0$. This can be seen either topologically or by using the argument in the next paragraph.
Let us describe the finite étale coverings $\varphi : Y \to \mathbf{A}^1_\mathbf {C} \setminus \{ 0\} $. It suffices to consider the case where $Y$ is connected, which we assume. We are going to find out what $Y$ can be by applying the Riemann-Hurwitz formula (of course this is a bit silly, and you can go ahead and skip the next section if you like). Say that this morphism is $n$ to 1, and consider a projective compactification
Even though $\varphi $ is étale and does not ramify, $\bar{\varphi }$ may ramify at 0 and $\infty $. Say that the preimages of 0 are the points $y_1, \ldots , y_ r$ with indices of ramification $e_1, \ldots e_ r$, and that the preimages of $\infty $ are the points $y_1', \ldots , y_ s'$ with indices of ramification $d_1, \ldots d_ s$. In particular, $\sum e_ i = n = \sum d_ j$. Applying the Riemann-Hurwitz formula, we get
and therefore $g_ Y = 0$, $r = s = 1$ and $e_1 = d_1 = n$. Hence $Y \cong {\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} }$, and it is easy to see that $\varphi (z) = \lambda z^ n$ for some $\lambda \in \mathbf{C}^*$. After reparametrizing $Y$ we may assume $\lambda = 1$. Thus our covering is given by taking the $n$th root of the coordinate on $\mathbf{A}^1_{\mathbf{C}} \setminus \{ 0\} $.
Remember that we need to classify the coverings of ${\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} }$ together with free $\mathbf{Z}/n\mathbf{Z}$-actions on them. In our case any such action corresponds to an automorphism of $Y$ sending $z$ to $\zeta _ n z$, where $\zeta _ n$ is a primitive $n$th root of unity. There are $\phi (n)$ such actions (here $\phi (n)$ means the Euler function). Thus there are exactly $\phi (n)$ connected finite étale coverings with a given free $\mathbf{Z}/n\mathbf{Z}$-action, each corresponding to a primitive $n$th root of unity. We leave it to the reader to see that the disconnected finite étale degree $n$ coverings of $\mathbf{A}^1_{\mathbf{C}} \setminus \{ 0\} $ with a given free $\mathbf{Z}/n\mathbf{Z}$-action correspond one-to-one with $n$th roots of $1$ which are not primitive. In other words, this computation shows that
The first identification is canonical, the second isn't, see Remark 59.69.5. Since the proof of Riemann-Hurwitz does not use the computation of cohomology, the above actually constitutes a proof (provided we fill in the details on vanishing, etc).
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