# A computation a day: a pullback pushforward

In this post we compute the Galois representation $i^\ast R^m j_\ast\mathbb{Q}_\ell$ where $j:\mathbb{G}_{m,\overline{k}}\hookrightarrow\mathbb{A}^1_k$ is the natural inclusion and $i:\text{Spec}(k)\hookrightarrow \mathbb{A}^1_k$ is the inclusion of the origin.

# Motivation

Disclaimer: I forgot to write $k^{\text{sep}}$ everywhere instead of $\overline{k}$. When faced with such a deep issue there are only two methods of resolution, a) go through and change all the overlines to ^\text{sep}’s or b) just assume that $k$ is perfect. I adopt the latter course of action. So, in this post, let $k$ be perfect.

Someone that I greatly respect as an algebraic geometer made a statement to the effect of the following: “one may have read a lot about étale cohomology, but one also needs to able to compute basic examples”. An example of a ‘basic example’ he then gave was to compute the Galois representation $i^\ast R^m j_\ast\mathbb{Q}_\ell$. Here, $k$ is a field, $j:\mathbb{G}_{m,\overline{k}}\hookrightarrow\mathbb{A}^1_{\overline{k}}$ is the natural inclusion, $i:\text{Spec}(\overline{k})\hookrightarrow\mathbb{A}^1_{\overline{k}}$ is closed embedding of the origin, and $\ell$ is a prime different than $\mathrm{char}(k)$.

The answer to what this ‘should be’ (or, rather, at least what the underlying $\mathbb{Q}_\ell$-space should be) is fairly obvious. Namely, let us consider the topological analogue. So, consider $j$ the inclusion of $\mathbb{C}^\times$ into $\mathbb{C}$ and $i$ the inclusion of the origin into $\mathbb{C}$. We then want to compute $i^\ast R^m j_\ast\mathbb{Q}_\ell$ (we could really replace $\mathbb{Q}_\ell$ here by any field). In other words, we’re trying to compute the stalk $(R^m j_\ast\mathbb{Q}_\ell)_0$. But, as is well-known, the sheaf $R^m j_\ast\mathbb{Q}_\ell$ is the sheafification of the presheaf

$U\mapsto H^m(j^{-1}(U),\mathbb{Q}_\ell)$

and since stalks are insensitive sheafification, we see that

$\displaystyle (R^m j_\ast\mathbb{Q}_\ell)_0=\varinjlim_{U} H^m(j^{-1}(U),\mathbb{Q}_\ell)$

where $U$ travels over neighborhoods of $0\in\mathbb{C}$. That said, a cofinal system of neighborhoods of $0$ are disks around $0$ of decreasing radii. Thus, we really only need to compute

$\displaystyle \varinjlim_r H^m(j^{-1}(D_r),\mathbb{Q}_\ell)$

where $D_r$ is a disk of radius $r$ around $0$ and $r$ tends to $0$.

That said, each $j^{-1}(D_r)$ is just $D_r^\ast$–the punctured disk around $0$ of radius $r$. Moreover, it’s trivial that

$\displaystyle H^m(D_r^\ast,\mathbb{Q}_\ell)=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0\\ \mathbb{Q}_\ell & \mbox{if}\quad m=1\\ 0 & \mbox{if}\quad m>1\end{cases}$

Moreover, each map $D_r^\ast\to D_s^\ast$, for $s>r$ is a homotopy equivalence, and thus induces an isomorphism on cohomology groups. Thus, it’s easy to conclude that

$i^\ast R^m j_\ast\mathbb{Q}_\ell=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0\\ \mathbb{Q}_\ell & \mbox{if}\quad m=1\\ 0 & \mbox{if}\quad m>1\end{cases}$

as one might expect.

Now, since étale cohomology is supposed to ‘act topologically’ one would expect a similar result to hold for the case we’re actually interested in and, in fact, this is true. But, as is somewhat representative of computations in étale cohomology, it is surprisingly nuanced/technical to verify this rigorously. Specifically, depending on what one’s first exposure to étale cohomology looked like, this ‘basic computation’ could actually be exceedingly difficult even after a ‘first course in etale cohomology’–there are a lot of results, and it’s not clear which to apply for this computation.

Thus, the goal of this post is to give a full proof of why this computation in the topological case ‘holds true’ in the étale case, and what precisely the Galois representation looks like.

# Useful facts

What makes the computation of $i^\ast R^k j_\ast\mathbb{Q}_\ell$ difficult in our case is the lack of an obvious cofinal systems of neighborhoods of $0$ which have simple cohomology when pulled back by $j$. We’d like to imagine that we can consider ‘disks around $0$‘ but, of course, these are not algebraically definable objects.

What will replace these ‘simple neighborhoods’ is a ring which, in a fairly literal sense, is the ‘étale locally zoomed in neighborhood of the origin’. Namely, let $X$ be a scheme and $\overline{x}$ a geometric point of $X$. Then, recall that the strictly local ring of $X$ at $\overline{x}$, denoted $\mathcal{O}_{X,\overline{x}}$ is defined as follows:

$\displaystyle \mathcal{O}_{X,\overline{x}}=\varinjlim_{(U,\overline{u})}\mathcal{O}(U)$

as $(U,\overline{u})$ varies over étale neighborhoods of $(X,\overline{x})$. One can show that if $\mathcal{O}_{X,x}$ is the local ring of $X$ at $x$, then $\mathcal{O}_{X,\overline{x}}$ is a strict henselization of $\mathcal{O}_{X,x}$.

In essence, while one thinks of $\text{Spec}(\mathcal{O}_{X,x})$ as being the ‘space obtained by zooming in arbitrarily Zariski close on $x$‘ one thinks of $\text{Spec}(\mathcal{O}_{X,\overline{x}})$ as being ‘the space obtained by zooming in arbitrarily étale close on $x$‘.

The reason this is useful for us is that $i^\ast R^k j_\ast\mathbb{Q}_\ell$ should be (and is) taking the étale stalk of $R^k j_\ast\mathbb{Q}_\ell$. This amounts, in essence, to taking a colimit of cohomology groups over an indexing set which is, essentially, the étale neighborhoods of $(\mathbb{A}^1,i)$ (where one notes that $i$ is really a geometric point we will denote $\overline{0}$). Thus, one might imagine that the strictly local ring of $\mathbb{A}^1$ at $\overline{0}$ might be showing up since, in essence, it is a direct limit over the same indexing set.

In fact, if one looks at the direct limit

$\displaystyle (R^k j_\ast\mathbb{Q}_\ell)_{\overline{0}}=\varinjlim_{(U,\overline{u})} H^k(\mathbb{G}_{m,\overline{k}}\times_{\mathbb{A}^1_{\overline{k}}} U,\mathbb{Q}_\ell)$

one sees that if we could commute the cohomology and the direct limit we’d be getting precisely something like

$(R^k j_\ast\mathbb{Q}_\ell)_{\overline{0}}=H^k(\mathbb{G}_{m,\overline{k}}\times_{\mathbb{A}^1_{\overline{k}}}\text{Spec}(\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}),\mathbb{Q}_\ell)$

which would at least give us a place to start–turning the higher pushforward into an actual cohomology group.

Thankfully, something like the above is true:

Theorem 1: Let $f:X\to Y$ be quasi-compact quasi-separated morphism of schemes. Then, for any geometric point $\overline{y}$ of $Y$ one has that

$(R^k f_\ast\mathbb{Q}_\ell)_{\overline{y}}=H^k(X\times_Y \text{Spec}(\mathcal{O}_{Y,\overline{y}}),\mathbb{Q}_\ell)$

Proof: This follows from Tag 03Q7. $\blacksquare$

The other results that will be useful to us are some basic facts about strict Henselizations and how to compute them in some baby cases. Specifically, the following are useful facts:

Theorem 2: Let $R$ be a local ring with residue field $\kappa$ and maximal ideal $\mathfrak{m}$. Then, $R^{\text{sh}}$ is a local ring with residue field $\kappa^{\text{sep}}$ and maximal ideal $\mathfrak{m}R^{\text{sh}}$.

Proof: See Tag 04GP. $\blacksquare$

Theorem 3: Let $R$ be a DVR. Then, $R^{\text{h}}$ is the elements of $\widehat{R}$ which are algebraic over $\text{Frac}(R)$. In particular, if $R$ has separably closed residue field, then $R^{\text{h}}=R^{\text{sh}}$ and thus this also gives a description of the strict Henselization.

Proof: See this nice post of Qing Liu. $\blacksquare$

These two tools will be enough to make a rigorous computation of $i^\ast R^k j_\ast\mathbb{Q}_\ell$.

# The computation

From Theorem 1 we see that we need to compute

$H^k(\mathbb{G}_{m,\overline{k}}\times_{\mathbb{A}^1_{\overline{k}}}\text{Spec}(\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}),\mathbb{Q}_\ell)$

But, note that if we write $\mathbb{A}^1_{\overline{k}}=\text{Spec}(\overline{k}[t])$ and $\mathbf{G}_{m,\overline{k}}=\text{Spec}(\overline{k}[t][x]/(tx-1))$ then

$\mathbb{G}_{m,\overline{k}}\times_{\mathbb{A}^1_{\overline{k}}}\text{Spec}(\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}})=\text{Spec}(\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}[x]/(tx-1))$

but $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}$ is local with maximal ideal $(t)$ (this follows from Theorem 2) and thus we see that $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}[x]/(xt-1)$ is just $K:=\text{Frac}(\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}})$. Thus, we see that computing our group $i^\ast R^k j_\ast\mathbb{Q}_\ell$ largely comes down to computing the Galois cohomology groups $H^k(G_K,\mathbb{Z}/\ell^n\mathbb{Z})$

From this description we can almost immediately show that $i^\ast R^k j_\ast\mathbb{Q}_\ell=0$ for $k>1$. Indeed, note that by Theorem 3 we havethat $\text{tr.deg}(K/\overline{k})\leqslant 1$. This implies, since $\text{cd}_\ell(\overline{k})=0$ that $\text{cd}_\ell(K)\leqslant 1$ (here $\text{cd}_\ell$ denotes the $\ell$-cohomological dimension–see section 1.4 of Bjorn Poonen’s great notes for a discussion of these terms and this result). Thus, we see that the Galois cohomology groups $H^k(G_K,\mathbb{Z}/\ell^n\mathbb{Z})=0$ for $k>1$ as desired.

Since $H^0(G_K,\mathbb{Z}/\ell^n\mathbb{Z})=\mathbb{Z}/\ell^n\mathbb{Z}$ (since $\mathbb{Z}/\ell^n\mathbb{Z}$ has the trivial action!) it really remains to compute

$H^1(G_K,\mathbb{Z}/\ell^n\mathbb{Z})=\text{Hom}_{\text{cont.}}(G_K,\mathbb{Z}/\ell^n\mathbb{Z})$

which, largely, comes down to understanding what the prime-to-$p$ Galois extensions of $K$ look like.

So, let us begin by observing that $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}$ is a strictly Henselian DVR. Indeed, since $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},0}$ is Noetherian, the same is true for $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}$ (see here), and since it’s maximal ideal is $(t)$ by Theorem 2, we see that it’s a Noetherian local ring with principal maximal ideal–thus a DVR. The fact that it’s strictly Henselian is definitional (or, rather, follows from the fact that it is $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},0}^\text{sh}$).

Now, let $L/K$ be any finite Galois extension of $K$ of order prime $p$. Since $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}$ is Henselian there is at a unique extension of the valuation $v$ on $K$ to a valuation $w$ on $L$ (see the theorem on page 45 of this). Let $\mathcal{O}$ be the valuation ring of $L$ and let $\pi$ be a uniformizer of $\mathcal{O}$. Note that since $e(w\mid v)f(w\mid v)=[L:K]$ (see 3) of loc. cit.) and $f(w\mid v)=1$ (since the residue field $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}})$ is algebraically closed) we must have that $\mathcal{O}/\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}$ is totally ramified. Thus, if $m:=[L:K]$ then $t=\pi^m u$ where $u\in\mathcal{O}^\times$ (note that $t$ is a uniformizer of $\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}})$).

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 (which follows from the same Theorem of loc. cit.) that $u$ has an $m^{\text{th}}$-root in $\mathcal{O}$. Thus, we see that, by replacing $\pi$ with an $m^{\text{th}}$-root of $u$, we can assume $t=\pi^m$. Thus, clearly $L=K(t^{\frac{1}{m}})$. And thus, $\text{Gal}(L/K)=\mathbb{Z}/m\mathbb{Z}$.

From this observation it’s fairly trivial to see that

$H^1(G_K,\mathbb{Z}/\ell^n\mathbb{Z})=\text{Hom}_{\text{cont.}}(G_K,\mathbb{Z}/\ell^n\mathbb{Z})=\mathbb{Z}/\ell^n\mathbb{Z}$

completing the last portion of our computation. Well, almost, it remains to see how $G_k$ acts on this only non-trivial cohomology group–the first cohomology group. That said, note that the Galois group of $K(t^{\frac{1}{m}})/K$ is given by $t^{\frac{1}{m}}\mapsto \zeta t^{\frac{1}{m}}$ which $G_k$ acts on by the cyclotomic character. But, since the actual cohomology group is the dual of this, we see that the first cohomology group is actually $\mathbb{Q}_\ell(-1)$.

Thus, in conclusion we deduce the following:

$i^\ast R^k j_\ast\mathbb{Q}_\ell=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad k=0\\ \mathbb{Q}_\ell(-1) & \mbox{if}\quad k=1\\ 0 & \mbox{if}\quad k>1\end{cases}$

# Some intuition

The above, while technically correct, is far from intuitive, and seems miles away from the intuition garnered from the topological calculation. So, I’d like to just give some brief indication as to why the two are really not that different.

So, if one studies the above computation, one really sees that what everything came down to was essentially a computation of $G_K$ or, equivalently, the fundamental group $\text{Spec}(K)$. So, why is this at all like what happened in the topological setting? In the topological case we got punctured disks and here we’re getting some strange field.

Let us begin by noting that getting a field is not at all surprising. Namely, if we imagine that we can zoom in étale close around $0$ in $\mathbb{A}^1$ we’d expect that we shouldn’t have to worry about any closed points (of a scheme) other than those coming from $0$–if we have another closed point, it’s complement is an even further étale zooming in on $0$ ignoring that closed point. Thus, in essence, the process of étale zooming in arbitrarily close should result in an object/space/thing that has only one closed point: that corresponding to $0$. When we the pull this back to $\mathbb{G}_{m,\overline{k}}$ we are precisely removing this closed point which then, of course, should result in a field (think about what happens when you remove the closed point from a trait(=spectrum of a DVR)). This may allow one to intuit why we’re dealing ultimately with a field in this computation, but really doesn’t explain where the really nice ‘punctured disk’ comes into play.

To try and make this connection more clear, let us imagine what sort of functions we would get when we start taking smaller and smaller open disks around $0\in\mathbb{C}$. Well, each disk will have functions being the convergent power series on this disk. As we shrink this to zero (as we did when making our topological computation) we’d get a ring of convergent power series–power series which have a positive radius of convergence. Unfortunately, such a ring is not of an algebro geometric nature–how does one (in a purely algebro-geometric context) talk about ‘positive radius of convergence’. So, instead, let us think that a substitute for this ring might be $\mathbb{C}[[T]]$–the ring of ‘formally convergent’ power series. So, one might be semi-convinced to imagine that the picture of a ‘small open disk around $0$‘ (where here ‘small’ means ‘arbitrarily close’) is something like $\text{Spec}(\mathbb{C}[[T]])$. But, then, an arbitrarily small punctured open disk should be $\text{Spec}(\mathbb{C}((T))$–again we see that a small punctured disk is really the spectrum of a field, just as in our computation.

Now, it’s not true that, even when $k=\mathbb{C}$, the field $K$ we were dealing with was $\mathbb{C}((T))$–it was elements of this field algebraic over $\mathbb{C}(T)$, but one might imagine that $\mathbb{C}((T))$ serves as a reasonable approximation (see below for why this is actually reasonable). Thus, the field whose Galois cohomology we needed to compute was (integrally related to) a small formal punctured disk around the origin–precisely the type of object that were computing the cohomology of in the topological case.

This comparison is furthered bolstered by the following observation. Namely, the key part of the computation needed was that the maximal prime-to-$p$ quotient of $\pi_1^{\acute{e}\text{t}}(\text{Spec}(K))$ was $\displaystyle \prod_{\ell\ne p}\mathbb{Z}_\ell$. This should be reasonable under our analogy to small punctured disks around the origin since, there, the prime-to-$p$ profinite completion of the topological fundmental group (which is the analogy of the maximal prime-to-$p$ quotient of the étale fundamental group) is precisely $\displaystyle \prod_{\ell\ne p}\mathbb{Z}_\ell$ since the topological fundamental group of a small punctured disk is $\mathbb{Z}$.

Now, with this all being said, there is a somewhat more rigorous reason that, in the end, we were just computing the cohomology of a punctured disk. Namely, perhaps the above gave some indication as to why one would expect an analogue of a ‘punctured disk at $0$‘ in the algebraic world would be $\text{Spec}(\overline{k}((T)))$ (because, at least when $k=\mathbb{C}$, these are something like power series which can vanish at the origin). But, again, we did not litereally compute the cohomology of $\text{Spec}(\overline{k}((T)))$ we, instead, computed the cohomology of $\text{Spec}(\text{Frac}(\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}}))$. That said, as it turns out, these two spaces are ‘essentially the same’, in least as far as their cohomology goes.

Indeed, the cohomology of the spectrum of a field is computing Galois cohomology and so, perhaps, a justification of the claim we made above that ‘$\overline{k}((T))$ is a reasonable replacement for $\text{Frac}(\mathcal{O}_{\mathbb{A}^1_{\overline{k}},\overline{0}})$ is to show that they have the same Galois group.

Let us explain why this is true in a more general context:

Theorem 4: Let $R$ be a Henselian DVR with separably closed residue field and let $F=\text{Frac}(R^\text{h})$ and $K=\text{Frac}(\overline{R})$. Then, the natural map $F\hookrightarrow K$ (coming from Theorem 3) induces an isomorphism of topological groups $G_K\xrightarrow{\approx}G_F$.

Proof: To see that it’s injective we note that since $R^\text{h}$ is a Henselian DVR, and since its valuation extends uniquely to each finite separable extension $L/F$ (by loc. cit.), there is a canonical valuation on $F^\text{sep}$. Similarly, there is a canonical valuation on $K^\text{sep}$ extending that on $K$ and these agree on $F^\text{sep}$ (as a subset of $K^\text{sep}$). Moreover, since these extensions (in both the case of $F$ and $K$) were unique, we know that $G_F$ acts continuously on $F^\text{sep}$ and similarly $G_K$ acts continuously on $K^\text{sep}$.

We claim that $F^\text{sep}$ is dense in $K^\text{sep}$. This follows, essentially, from Krasner’s lemma since $\text{Frac}(R)^\text{sep}\subseteq K^\text{sep}$ is dense (this is Krasner’s lemma) and $\text{Frac}(R)^\text{sep}\subseteq F^\text{sep}$. Since the elements of $G_K$ act continuously on $K^\text{sep}$ this implies that the restriction map $G_K\to G_F$ is injective.

To show it’s surjective we note that we’re really trying to show that the map $\pi_1^{\acute{e}\text{t}}(\text{Spec}(K))\to \pi_1^{\acute{e}\text{t}}(\text{Spec}(F))$ is surjective (where we are choosing the natural geometric point $\text{Spec}(K^\text{sep})$) and there is a very well-known criterion for when a map of étale fundamental groups is surjective. Namely, if we can show that whenever $L/F$ is a finite separable extension $L\otimes_F K/K$ is a finite separable extension, we’ll be done. So, to see this note that any $L/F$ is totally ramified and thus given by an Eisenstein polynomial: $L=F(\alpha)$ where $\alpha$ is a root of an Eisenstein polynomial $f(x)$ for $F$. But, then, $L\otimes_F K=K[x]/(f(x))$. But, note that since $R^h\to \widehat{R}$ is local that $f(x)$ is still Eisenstein for $\widehat{R}$, and thus irreducible. Thus, $L\otimes_F K$ is a field, and thus we’re done. $\blacksquare$.

Thus, we see that computing $i^\ast R^m j_\ast\mathbb{Q}_\ell$ came down to computing $H^m(\text{Spec}(k((T))),\mathbb{Q}_\ell)$. Thus, in the end, we really are computing the cohomology of a small punctured disk around the origin.

Remark: One can remove the necessity that $R$ has separably closed residue field in Theorem 4 but one is forced then to give an alternate method of attack–a finite extension needn’t be totally ramified any more. One can then use a similar method of proof of surjectivity, but one will have to use Artin approximation (a slightly supped up version than that mentioned in the Wiki article) to show that $f(x)$ stays irreducible.