# Punctured disks and punctured curves and cohomology, oh my! (Part III: the affine curve case)

This is a continuation of this post.

## The affine case

### Restating and applying purity

Let us now finally return to computing to proving Proposition 2 in the affine case. Let us first write it more concretely:

Proposition 31: Let $k$ be an algebraically closed field and $X$ a smooth connected projective curve over $k$ of genus $g$. Let $m\geqslant 1$ and $p_1,\ldots,p_m\in X(k)$. Set $U:=X-\{p_1,\ldots,p_m\}$. Then, for every prime $\ell$ invertible in $k$ and integer $n\geqslant 1$

$H^i_{\mathrm{et}}(U,\mathbb{Z}/\ell^n\mathbb{Z})\cong\begin{cases}\mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=0\\ (\mathbb{Z}/\ell^n\mathbb{Z})^{2g+m-1} & \mbox{if}\quad i=1\\ 0 & \mbox{if}\quad i>1\end{cases}$

Given our discussion of cohomology with supports this is fairly easy. In fact, it’s laughably easy if we take the Purity Theorem for granted. Namely, let us begin by noting that by Proposition 23 we have a long exact sequence

$\cdots\to H^i_{Z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})\to H^i_\mathrm{et}(X,\mathbb{Z}/\ell^n\mathbb{Z})\to H^i_\mathrm{et}(U,\mathbb{Z}/\ell^n\mathbb{Z})\to \cdots$

and thus from Proposition 4 it really suffices to compute $H^i_{Z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})$. If we assume the Purity Theorem (in the form of Corollary 30) this is trivial since $Z$ is a disjoint union of points. Namely, we see

$H^i_{Z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})=\begin{cases} (\mathbb{Z}/\ell^n\mathbb{Z})^m & \mbox{if}\quad i=2\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

from which deducing Proposition 31 is a triviality.

### Giving a more hands-on proof

But, the above feels a little bit like cheating to me. Namely, we don’t need the elephant gun that is the Purity Theorem to do this computation. In fact, this computation really is just a basic case of the Purity Theorem that we can do by hand. So, let’s do it.

Note that by iteration it suffices to deal with the case when $Z$ is a single point $z$ (but $X$ is now allowed to be affine—-this won’t change the computation by the Excision Theorem). Or, if this iteration argument doesn’t make you happy convince yourself of the easy fact that since $Z=\{p_1,\ldots,p_m\}$ that

$\displaystyle H^i_{Z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})=\bigoplus_i H^i_{p_i,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})$

which, either way, reduces us to the computation of $H^i_{z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})$ for a single closed point $z\in X(k)$.

Now, the key point for this computation is then that by Lemma 26 we can reduce this to a really explicit computation. Namely, Lemma 26 tells us that (since our schemes are Noetherian) that

$H^i_{z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})\cong H^i_{z,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{X,z}^h),\mathbb{Z}/\ell^n\mathbb{Z})$

This may not seem very helpful but, in fact, the scheme $\mathrm{Spec}(\mathcal{O}_{X,z}^h)$ is something incredibly explicit!

Namely, we have the following

Lemma 32: Let the notation be as above. Then, there is an isomorphism of schemes

$\mathrm{Spec}(\mathcal{O}_{X,z}^h)\cong \mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h)$

and $\mathcal{O}_{\mathbb{A}^1_k,0}^h$ can be described as the integral closure of $k[t]_{(t)}$ inside $k[[t]]$ (NB: since $k$ is algebraically closed there is no difference here between henselization and strict henselization).

Proof: The first claim follows from the fact that there exists an open subset $V$ of $z$ in $X$ and an etale map $V\to \mathbb{A}^1_k$ with $z\mapsto 0$. Indeed, just take any etale map $V\to \mathbb{A}^1_k$. Then, we know that $z$ will map to some point $p$ in $\mathbb{A}^1_k$ which, by etaleness, and the fact that $k$ is algebraically closed closed implies that $k(p)=k(z)=k$. One can then just postcompose with the translation isomorphism $x\mapsto x-p$. We now claim that this pointed etale map $(V,z)\to (\mathbb{A}^1_k,0)$ induces an isomorphism

$\mathcal{O}_{\mathbb{A}^1_k,0}^h=\mathcal{O}_{\mathbb{A}^1_k,0}^{sh}\xrightarrow{\approx}\mathcal{O}_{V,z}^{sh}=\mathcal{O}_{V,z}^{h}$

but since $(V,v)\to (\mathbb{A}^1_k,0)$ is an etale neighborhood this is almost by the definition of the henselization/strict henselization.

It remains to justify why $\mathcal{O}_{\mathbb{A}^1_k,0}^h=\mathcal{O}_{\mathbb{A}^1_k,0}^{sh}$ has the claimed form. There are several ways to justify this. For example, one can see [Mil, Example I.4.10 (a)] and/or [Mil, Example I.4.10 (b)] and the references therein. One can also see [Nag, Page 180]. $\blacksquare$

So, we see that (as hinted at in the idea of proof for the Purity Theorem) one can use the fact that our calculation is etale local nature and the fact that smooth things all locally look the same to reduce this to an explicit computation with a specific scheme. Namely, by the above it’s clear that we need to compute the group $H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})$. But, how do we do this?

Well, using Proposition 23 and the fact that strictly Henselian local rings have vanishing higher cohomology (e.g. see [Fu, Proposition 5.7.3] or [Stacks, Tag09AY]) it’s easy to see that

$H^1_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})=\mathrm{coker}(H^0_{\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})\to H^0_{\mathrm{et}}(U_0,\mathbb{Z}/\ell^n\mathbb{Z}))$

and for $i>1$

$H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})\cong H^{i-1}_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z})$

where here we are denoting by $U_0$ the scheme $\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h)-\{0\}$ (the punctured spectrum).

Now $\mathcal{O}_{\mathbb{A}^1_k,0}^h$ has pretty simple structure in that it’s a Henselian DVR (e.g. combine [Stacks, Tag06DI] with [Stacks, Tag06LK]). In particular, it’s easy to see that $U_0=\mathrm{Spec}(K_0)$ where $K_0:=\mathrm{Frac}(\mathcal{O}_{\mathbb{A}^1_k,0}^h)$. From this it’s easy to conclude that

\begin{aligned}H^1_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z}) &=\mathrm{coker}(H^0_{\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})\to H^0_{\mathrm{et}}(U_0,\mathbb{Z}/\ell^n\mathbb{Z}))\\ &=0\end{aligned}

and that

\begin{aligned}H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z}) & \cong H^{i-1}_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z})\\ &\cong H^{i-1}_{\mathrm{cont.}}(\mathrm{Gal}(K_0^\mathrm{sep}/K_0),\mathbb{Z}/\ell^n\mathbb{Z})\end{aligned}

Thus, we are reduced to understanding the Galois cohomology of $K_0$. But, this is surprisingly simple in our case:

Lemma 33: Let $R$ be a Henselian DVR of residue characteristic $p\geqslant 0$. Let $\pi$ be a uniformizer of $R$ and set $K=\mathrm{Frac}(R)$ and $k:=R/(\pi)$. Assume that $k$ is algebraically closed. Finally, let $L/K$ be a finite Galois extension of degree $m$ with $(m,p)=1$. Then, $L=K(\pi^{\frac{1}{m}})$.

Note that the reason this Lemma is of interest to us is that it evidently implies that

Corollary 34: In particular, if $G_K^{(p)}$ denotes the pro-prime-to-$p$ completion of $G_K:=\mathrm{Gal}(K^\mathrm{sep}/K)$ then (non-canonically)

$\displaystyle G_K^{(p)}\cong \prod_{\ell\ne p}\mathbb{Z}_\ell=:\widehat{\mathbb{Z}}^{(p)}$

This corollary also makes complete geometric sense. We are to think of $\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0})$ as being something like a ‘small etale disk around $0$ in $\mathbb{A}^1_k$‘ Then, of course, we should think of $\mathrm{Spec}(K_0)$ as being something like a ‘punctured disk’ around $0$ and, consequently, it would make sense that

$\pi_1^\mathrm{et}(\mathrm{Spec}(K_0))=G_{K_0}$

should be something like $\widehat{\mathbb{Z}}$. Corollary 34 validates this belief up to the ever-present concerns about wild ramification.

We also see that Corollary 34 immediately implies that

$H^p_{\mathrm{cont.}}(\mathrm{Gal}(K_0^\mathrm{sep}/K_0),\mathbb{Z}/\ell^n\mathbb{Z})=\begin{cases} \mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=0,1\\ 0 & \mbox{if}\quad i>1\end{cases}$

So, summarizing the above discussion we get that for $i>0$ that

\begin{aligned}H^i_{z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z}) &\cong H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{\mathbb{A}^1_k,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})\\ & \cong H^{i-1}_{\mathrm{et}}(U_0,\mathbb{Z}/\ell^n\mathbb{Z})\\ &= H^{i-1}_\text{cont.}(\mathrm{Gal}(K_0^\mathrm{sep}/K_0),\mathbb{Z}/\ell^n\mathbb{Z})\\ &\cong \begin{cases}\mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=2\\ 0 &\mbox{if}\quad \text{otherwise}\end{cases}\end{aligned}

which, combining with our trivial result that $H^0_{z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})=0$ gives us

$H^i_{z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})\cong \begin{cases}\mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=2\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

This is as already told to us by the Purity Theorem and which, we’ve already observed, easily implies Proposition 31.

Thus, all that remains to finish our hands-on proof of Proposition 31 is the proof of Lemma 33:

Proof (Lemma 33): Let $L/K$ be any finite Galois extension of $K$ of order $m$ which is prime $p$. Since $R$ is Henselian there is at a unique extension of the valuation $v$ on $K$ to a valuation $w$ on $L$ (e.g. see [EP, Lemma 4.1.1] and [EP, Theorem 4.1.3]). Let $\mathcal{O}$ be the valuation ring of $L$ and let $\varpi$ denote a uniformizer of $\mathcal{O}$. Note that since $e(w\mid v)f(w\mid v)=[L:K]$ (e.g. see [EP, Theorem 3.3.5]) and $f(w\mid v)=1$ (since the residue field of $R$ is algebraically closed) we must have that $\mathcal{O}/R$ is totally ramified. Thus, $\pi=\varpi^m u$ where $u\in\mathcal{O}^\times$.

But, note that $T^m-u$ has a solution in the residue field of $\mathcal{O}$ (since it’s algebraically closed) and since $m$ is coprime to the residue characteristic of $\mathcal{O}$ we see from the fact that $\mathcal{O}$ is Henselian (e.g. this follows from [EP, Lemma 4.1.1]) that $u$ has an $m^{\text{th}}$-root in $\mathcal{O}$. Thus, we see that, by replacing $\varpi$ with an $m^{\text{th}}$-root of $u$, we can assume $\pi=\varpi^m$. Then $L=K(\pi^{\frac{1}{m}})$ by standard theory (e.g. see [Sut, Theorem 11.5] and note that the proof doesn’t use completeness, but only Henselianess). $\blacksquare$

To summarize the proof in the affine case $U\subseteq X$: we used the theory of cohomology with supports to break the cohomology into two parts. The first part was the classes on $U$ that come from $X$. The second part was the classses on $U$ introduced by the punctures we made at the points $Z\subseteq X$. Then, using that cohomology with supports is insensitive to shrinking in the etale topology we were able to focus in on the loops around each puncture $z\in Z$ individually and, using the fact that smooth curves all etale locally look $\mathbb{A}^1_k$, we were able to essentially understand this as a computation involving the punctured spectrum $U_0$ of the henselian local ring at $0\in\mathbb{A}^1_k$ which, again, we intuited as a punctured disk. This space $U_0$ was concrete enough/well-behaved enough that we could compute its cohomology by hand and got (ignoring wild ramification issues) that its cohomology agrees with what we expected (a punctured disk).

## Two bonus computations

Even though our main goal (at least locally) was to prove Proposition 2 (which we have now done) our computations relied so heavily on smoothness hypotheses that one might be scared that the methods we have developed are ill-equipped to deal with singular cases. This is not at all the case. Namely, using our theory of cohomology with supports we can compute this cohomology much like we did for smooth curves, but in reverse.

Namely, let $X$ be our singular curve and let $U\subseteq X$ be its smooth locus and $Z\subseteq X$ its finite discrete singular locus. We are now interested in not in the cohomology of $U$ but in the cohomology of $X$. But, again, by Proposition 23 this comes down to studying the cohomology of $U$ (which we know by Proposition 2/Proposition 31) and the cohomology of $X$ with supports at $Z$. The latter can, as in the smooth case, be studied by studying etale local geometry of $X$ at $z$ which, again, allows us to make explicit computations involving the singularity type of $z$.

The two examples we compute arefortunately/unfortunately (depending on your bend) the obvious ones: the cuspidal and nodal cubics.

### The nodal cubic

As before let us fix an algebraically closed field $k$ of characteristic not $2$ or $3$ (for sanity’s sake) and a prime $\ell$ invertible in $k$. Let us define $X$ to be the projective nodal cubic curve given by

$X:=V(y^2z-x^3-x^2z)\subseteq \mathbb{P}^2_k$

this has a singular locus $Z:=\{0\}$ (where we write $0:=[0:0:1]$) and smooth locus $U=X-Z$. The structure of $U$ itself is quite simple. Namely, we claim that $U\cong \mathbb{G}_{m,k}$. Indeed, this follows from the fact that the normalization of $X$ is the map

$\nu:\mathbb{P}^1_k\to X$

(e.g. see [Mac]) which has the property that $\nu^{-1}(0)$ consists of two points so that $\nu$ isomorphically maps $\mathbb{G}_{m,k}\cong \mathbb{P}^1_k-\nu^{-1}(0)$ to $U$.

Now, by Proposition 23 we get a long exact sequence

$\cdots 0\to H^i_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})\to H^i_\mathrm{et}(X,\mathbb{Z}/\ell^n\mathbb{Z})\to H^i_\mathrm{et}(U,\mathbb{Z}/\ell^n\mathbb{Z})\to\cdots$

Now, since $U\cong \mathbb{G}_{m,k}$ we know from Proposition 2/Proposition 31 that

$H^1_\mathrm{et}(U,\mathbb{Z}/\ell^n\mathbb{Z})\cong \begin{cases}\mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=0,1\\ 0 & \mbox{if}\quad i>1\end{cases}$

and thus we really only need to compute $H^i_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})$.

To compute this we use Lemma 26 (since our schemes are Noetherian) to say that

$H^i_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})\cong H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{X,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})$

But, the point is that while the global geometry of $X$ is somewhat daunting the etale local geometry at $0$ is quite simple. Namely, we have the following:

Lemma 35: Consider $Y:=V(xy)\subseteq \mathbb{A}^2_k$. Then, there is an isomorphism

$\mathrm{Spec}(\mathcal{O}_{X,0}^h)\cong \mathrm{Spec}(\mathcal{O}_{Y,0}^h)$

where $0:=(0,0)\in Y$.

Proof: It suffices to find an etale neighborhood $(W,w)$ of $(X,0)$ such that $\mathcal{O}_{W,w}\cong \mathcal{O}_{Y,0}$. To do this we let $W\to X$ be the unique double cover of $X$.

Abstractly one can describe it as follows. As is well-known $\mathrm{Pic}(X)\cong k^\times\times \mathbb{Z}$ (e.g. see [Har, Exercise 6.9]). Let $\mathscr{L}$ be the line bundle on $X$ corresponding to $(-1,0)\in k^\times\times \mathbb{Z}$. Then, $\mathscr{L}^2\cong \mathcal{O}_X$ (it’s the unique such line bundle). Then, using this isomorphism one can put an algebra structure on $\mathscr{A}:=\mathcal{O}_X\oplus \mathscr{L}$ and we can set $W:=\underline{\mathrm{Spec}}(\mathscr{A})\to X$.

A less highfalutin way to build $W$ is to have it be the projectivization of the map

$\mathrm{Spec}(k[s,t]/(t^2-1,t(t^2-1)))\to \mathrm{Spec}(k[x,y]/(y^2-x^3-x^2))$

corresponding to the $k$-algebra map with

$x\mapsto s^2-1,\qquad y\mapsto st$

Then, one can show that this map is a finite etale cover (of degree $2$). Moreover, the point $w:=[0:1:0]$ of $W$ maps to $0\in X$. Moreover one can explicitly check that $\mathcal{O}_{W,w}\cong \mathcal{O}_{Y,0)}$ from where the conclusion follows. $\blacksquare$

In words, if we’re only interested in etale local geometry (which is the case due the fact that cohomology with supports is insensitive to shrinking in the etale topoogy) we can eschew the global complicated nature of the nodal cubic when focusing on its singular point and see it as, essentially, the intersection of two lines.

Thus, to compute

$H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{X,0}^h),\mathbb{Z}/ell^n\mathbb{Z})$

we need really only compute

$H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{Y,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})$

But, just as before in the smooth curve case we can use Proposition 23 and the vanishing of higher cohomology of strictly local rings to see that if we set $U_0:=\mathrm{Spec}(\mathcal{O}_{Y,0}^h)-\{0\}$ then

$H^1_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{Y,0}^h),\mathbb{Z}/\ell^n\mathbb{Z}) =\mathrm{coker}(H^0_\mathrm{et}(X,\mathbb{Z}/\ell^n\mathbb{Z})\to H^0_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z}))$

and for $i>1$

$H^i_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})\cong H^{i-1}_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z})$

But, what is $U_0$ in this case? It is no longer the spectrum of the fraction field of a Henselian DVR. Indeed, it’s actually the disjoint union of two such objects. The idea is that

$U_0=D_{U_0}(x)\bigsqcup D_{U_0}(y)$

(where $D_M(f)$ means the non-vanishing locus of $f$ in $M$) and, as one can check, $D_{U_0}(x)$ and $D_{U_0}(y)$ are both (essentially canonically) the spectra of the fraction field of $\mathcal{O}_{\mathbb{A}^1_k,0}^h$.

We can then use the results of the previous section (namely Corollary 34) to thus deduce that

\begin{aligned}H^1_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{Y,0}^h),\mathbb{Z}/\ell^n\mathbb{Z}) &= \mathrm{coker}(H^0_\mathrm{et}(X,\mathbb{Z}/\ell^n\mathbb{Z})\to H^0_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z}))\\ &\cong \mathbb{Z}/\ell^n\mathbb{Z}\end{aligned}

(don’t forget that that $U_0$ has two components!). In fact we can geometrically intepret this non-vanishing. Remember that $H^1_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})$ (roughly) classifies $\mathbb{Z}/\ell^n\mathbb{Z}$-torsors on $X$ that become trivialized on $U$. If $X$ is normal this can never hapen (as we saw by the vanishing of $H^1_{z,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})$) essentially because if you have a connected finite etale cover $C\to X$ one has that $C$ is actually integral (e.g. see [Stacks, Tag0BQL]) and so $C_U$ is still connected so that $C_U\to U$ cannot be trivial. But, for the nodal cubic $X$ the covers can fail to have this property. In fact note that, geometrically, the nodal curve $X$ looks like a pinched torus. Its covers $C$ then look like necklaces of wedges of spheres (e.g. see the discussion here and here) where its exactly the wedge points (the ‘kissing points’) that map to the node $0$. So, for such a cover one has that $C\to X$ is certainly not trivial, but $C_U\to U$ just looks like a bunch of disjoint copies of $\mathbb{G}_m$–the trivial torsor over $U$!

Let us return to the rigorous computation. For $i>1$

\begin{aligned}H^i_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z}) &\cong H^{i-1}_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z})\\ &\cong \begin{cases} (\mathbb{Z}/\ell^n\mathbb{Z})^2 & \mbox{if}\quad i=2\\ 0 & \mbox{if}\quad i>2\end{cases}\end{aligned}

(where we have used that $U_0$ is a disjoint union of the spectra of two Henselian discretely valued fields and Corollary 34). From this we then deduce that

$H^1_\mathrm{et}(X,\mathbb{Z}/\ell^n\mathbb{Z})=\begin{cases} \mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i\leqslant 2\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

as we would expect topologically.

### The cuspidal cubic

Finally, let’s flip our method on its head by computing the etale cohomology of an $X$ directly and then using this to say something non-trivial about the cohomology groups $H^i_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z})$.

First, let’s fix our set up. Again, let $k$ be an algebraically closed field of characteristic not $2$ or $3$ and let $\ell$ be a prime invertible in $k$. Let us consider the projective cuspidal cubic:

$X:=V(y^2z-x^3)\subseteq \mathbb{P}^2_k$

this has singular locus $Z=\{0\}$ (where we abbreviate $[0:0:1]$ to $0$) and smooth locus $u:=X-Z$.

The cheeky way to compute the cohomology of $X$ is then to employ the following observation:

Lemma 36: Let notation be as above. Then, the normalization of $X$ is $\mathbb{P}^1_k$ and the normalization map $\nu:\mathbb{P}^1_k\to X$ is a universal homeomorphism.

Proof: Note that $\nu$ is evidently finite (or even easier is that it’s proper) and surjective and thus universally closed and universally surjective. Thus, it suffices to check that $\nu$ is universally injective. It suffices by standard theory (e.g. see the discussion in [Stacks, Tag01S2]) to show that for every point $x\in X$ that $k(x)/k(\nu(x))$ is purely inseparably. For the generic point this is clear since $\nu$ is birational so it suffices to check for closed points of $x$ or, equivalently, for the points lying over the cusp. This is done in [Mac] explicitly for the cusp. $\blacksquare$

Now, one deduces from the topological invariance of the etale site (e.g. see [Stacks, Tag03SI]) that $X$ and $\mathbb{P}^1_k$ have the same cohomology. In particular, we see that

$H^i_\mathrm{et}(X,\mathbb{Z}/\ell^n\mathbb{Z})=\begin{cases} \mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=0,2\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Note though that from Lemma 36 we also deduce that $U\cong \mathbb{A}^1_k$ and thus its cohomology is known from Proposition 2/Proposition 31:

$H^1_\mathrm{et}(\mathbb{A}^1_k,\mathbb{Z}/\ell^n\mathbb{Z})\cong \begin{cases}\mathbb{Z}/\ell\mathbb{Z} & \mbox{if}\quad i=0\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Using Proposition 23 we deduce the following:

$H^i_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})=\begin{cases} \mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=2\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Let us now consider the objects $\mathrm{Spec}(\mathcal{O}_{X,0})$ and $U_0=\mathrm{Spec}(\mathcal{O}_{X,0})-\{0\}$ and see what the above computation tells us.

To start, interestingly enough, from the equality $H^1_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})=0$ (again we’re implicitly using Lemma 26 and thus Noetherianity, but I assume this needn’t be said at this point)we actually deduce that

$H^0_\mathrm{et}(\mathrm{Spec}(\mathcal{O}_{X,0}),\mathbb{Z}/\ell^n\mathbb{Z})\to H^0_\mathrm{et}(U_0,\mathbb{Z}/\ell^n\mathbb{Z})$.

is surjective and thus $U_0$ is connected. Surprisingly, this is essentially equivalent to Lemma 36. Namely, Lemma 36 roughly says that $X$ is (geometrically) unibranch (e.g. see [Stacks Tag0BPZ]), which since $X$ is normal away from $0$, is really saying that $\mathcal{O}_{X,0}$ is geometrically unibranch. But, $\mathcal{O}_{X,0}$ being geometrically unibranch is actually equivalent to the claim that $U_0$ is connected (e.g. see [Stacks, Tag0BQ4]).

Let us in fact notice that $U_0$ being connected implies that, in fact, $U_0$ is a field. Indeed, note that $\mathrm{Spec}(\mathcal{O}_{X,0}^h)$ is reduced (e.g. see [Stacks, Tag06DH]) and thus it will be domainif and only if it’s irreducible. But, note that $\mathrm{Spec}(\mathcal{O}_{X,0}^h)$ is dimension $1$ (e.g. see [Stacks, Tag06LK]) and thus there are no intermediary primes lying between the minimal primes of $\mathrm{Spec}(\mathcal{O}_{X,0}^h)$ and $0$. Thus, $U_0$ is a discrete set on the minimal primes of $\mathrm{Spec}(\mathcal{O}_{X,0}^h)$ and since $U_0$ is connected, this implies that there is a unique minimal prime of $\mathcal{O}_{X,0}^h$. Thus, $\mathcal{O}_{X_0,0}^h$ is a domain. Moreover, since it’s of dimension $1$ it’s easy to see that $U_0=\mathrm{Spec}(K_0)$ where $K_0:=\mathrm{Frac}(\mathcal{O}_{X_0,0}^h)$.

Note that this field $K_0$ is the fraction field of a Henselian local domain but one which is not a DVR. Thus, the results like Corollary 34 don’t apply. In fact, I don’t actually know how to compute $H^i(\mathrm{Gal}(K_0^\mathrm{sep}/K_0),\mathbb{Z}/\ell^n\mathbb{Z})$ directly. That said, from our sideways calculation of $H^i_{0,\mathrm{et}}(X,\mathbb{Z}/\ell^n\mathbb{Z})$. Indeed, using the usual isomorphism for $i>1$

$H^i_{0,\mathrm{et}}(\mathrm{Spec}(\mathcal{O}_{X,0}^h),\mathbb{Z}/\ell^n\mathbb{Z})\cong H^{i-1}_\mathrm{cont}(\mathrm{Gal}(K_0^\mathrm{sep}/K_0),\mathbb{Z}/\ell^n\mathbb{Z})$

we deduce that

$H^i_\mathrm{cont.}(\mathrm{Gal}(K_0^\mathrm{sep}/K_0),\mathbb{Z}/\ell^n\mathbb{Z})\cong \begin{cases}\mathbb{Z}/\ell\mathbb{Z} & \mbox{if}\quad i=0,1\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Again, this is quite different from our other calculation and quite strange. In our previous calculations we calculated a global group by studying the local group. Here we had a local group that, a priori, we didn’t know how to explicitly compute but once we globalized it we were able to exploit the global picture to get the desired result.

I have a gut feeling that there is some topological invariance trick that works locally for computing the cohomology of $\mathrm{Spec}(K_0)$ similar to how we computed the cohomology for the cuspidal cubic. I don’t currently see it though. Please feel free to let me know if you see an approach!

# References

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[Star] Starr, Jason. Brauer groups and Galois cohomology of function
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