# Lecture 10 :::{.remark} What we've been calling a *torsor* (a sheaf with a group action plus conditions) is called by some sources a **pseudotorsor** (e.g. the Stacks Project), and what we've been calling a *locally trivial torsor* is referred to as a *torsor* instead. ::: ## Proof of Grothendieck-Hilbert 90 Recall that statement of \cref{thm:hilb90}; we'll now continue with the proof: :::{.proof title="of Hilbert 90"} :::{.observation} Let $\tau = X_{\zar}, X_\et, X_{\fppf}$, then the data of a $\GL_n\dash$torsor split by a $\tau\dash$cover $U\to X$ is the same as descent data for a vector bundle relative to $U_{/X}$. ::: \vspace{2em} This descent data comes from the following: \begin{tikzcd} U\cross_X U \ar[d, shift right=0.75ex, "\pi_1"'] \ar[d, shift left=0.75ex, "\pi_2"] \\ U \ar[d] \\ X \end{tikzcd} That $U$ trivializes our torsor means that $\pi^* T = \pi^* G$ as a $G\dash$torsor, where $G$ acts on itself by left-multiplication. We have two different ways of pulling back, and identifications with $G$ in both, yielding \begin{tikzcd} \pi_1^* \pi^* T \ar[d] \ar[r, "\sim"] & \pi_2^* \pi^* T \ar[d] \\ \pi_1^* \pi^* G \ar[r, "\sim"] & \pi_2^* \pi^* G \end{tikzcd} Both of the bottom objects are isomorphic to $\ro{G}{U\cross U}$. :::{.claim} The top horizontal map is descent data for $T$, and the bottom horizontal map is an automorphism of a $G\dash$torsor and thus is a section to $G$. I.e. a section to $\GL_n$ is an invertible matrix on double intersections (satisfying the cocycle condition) and a cover, which is precisely descent data for a vector bundle. ::: Using fppf descent, proved previously, we know that descent data for vector bundles is effective. So if we have a locally trivial $\GL_n\dash$torsor on the fppf site, it's also trivial on the other two sites, yieldings the desired maps back and forth. Thus $H^1(X_\et, \GL_n)$ is in bijection with $n\dash$dimensional vector bundles on $X$. ::: :::{.exercise title="?"} See if Hilbert 90 is true for groups other than $\GL_n$. ::: ## Representability and Local Triviality :::{.question} Suppose $G$ is an affine flat $X\dash$group scheme. Are all $G\dash$torsors representable by a $X\dash$scheme? ::: :::{.answer} Yes, by the same proof as last time, try working out the details. Idea: you can trivialize a $G\dash$torsor flat locally and use fppf descent. ::: :::{.question} Given a $G\dash$torsor $T$ that is fppf locally trivial, is it étale locally trivial? ::: :::{.answer} In general no, but yes if $G$ is smooth. ::: :::{.proof title="Sketch"} You can take an fppf local trivialization, trivialize by $p$ itself, then slice to get an étale trivialization. Given a torsor $T\to X$, we can base change it to itself: \begin{tikzcd} T\cross_X T \ar[d] \ar[r] & T \ar[d] \\ T \ar[r, "f"] \ar[u, bend right, "\exists"'] & X \end{tikzcd} The torsor $T\cross_X T\to T$ is trivial since there exists the indicated section given by the diagonal map. Another way to see this is that $T\cross T\cong T\cross G$ by the $G\dash$action map, which is equivalent to triviality here. Here $f$ is smooth map since $G$ itself was smooth and the fibers of $T$ are isomorphic to the fibers of $G$. We can thus find some $U$ such that \begin{tikzcd} & T\cross_X T \ar[d] \ar[r] & T \ar[d] \\ & T \ar[r, "f"] \ar[u, bend right, "\exists"'] & X \\ U \ar[ur, hook, "\text{closed}"] \ar[urr, bend right, "\exists \et"'] & & \end{tikzcd} Here "slicing" means finding such a $U$, and this can be done using the structure theorem for smooth morphisms. ::: :::{.example title="non-smooth group schemes"} \envlist - $\alpha_p$, the kernel of Frobenius on $\AA^1$ or $\GG_a$, - $\mu_p$ in characteristic $p$, representing $p$th roots of unity, the kernel of Frobenius on $\GG_m$, - The kernel of Frobenius on any positive dimensional affine group scheme. - $\mu_p \cross \GL_n$, etc. ::: ### What Hilbert 90 Means :::{.example title="?"} Let $X = \spec k, n=1$, so we're looking at $H^\wait(\spec k, \GG_m)$. \[ H^1 \qty{ \qty{ \spec k}_\zar, \GG_m} &= 0 \\ &= H^1 \qty{ \qty{ \spec k}_\et, \GG_m} \\ &= H^1(\Gal(k^s / k), \bar k \units) .\] The first comes from the fact that we're looking at line bundles of spec of a field, i.e. a point, which are all trivial. The last line comes from our previous discussion of the isomorphism between étale cohomology of fields and Galois cohomology. Etymology: the fact that this cohomology is zero is usually what's called **Hilbert 90**.[^why_90] [^why_90]: This is called "90" since Hilbert numbered his theorems in at least one of his books. ::: Let's generalize this observation. :::{.example title="?"} Let $X$ be any scheme and $n=1$, then $H^1(X_\et, \GG_m) = \Pic(X)$. ::: :::{.example title="?"} Let's compute $H^1(X_\et, \mu_\ell)$ where $\ell$ is an invertible function on $X$. We have a SES of étale sheaves, the **Kummer sequence**, \[ 1 \to \mu_\ell \to \GG_m \mapsvia{z\mapsto z^p} \GG_m \to 1 .\] This is exact in the étale topology since adjoining an $\ell$th power of any function gives an étale cover. We get a LES in cohomology \begin{tikzcd} & & 0 \ar[dll] \\ H^0(X_\et, \mu_\ell) \ar[r] & H^0(X_\et, \GG_m) \ar[r, "z\mapsto z^\ell"] & H^0(X_\et, \GG_m) \ar[dll] \\ {\color{NavyBlue} H^1(X_\et, \mu_\ell)} \ar[r] & \Pic(X) \ar[r, "{[\ell]}"] & \Pic(X) \ar[dll] \\ H^2(X_\et, \mu_\ell) \ar[r] & \cdots & \end{tikzcd} We know that $H^0(X_\et, \GG_m)$ are invertible functions on $X$, and the blue term is what we'd like to compute. We'll make some additional assumptions now. Suppose now $H^0(X, \OO_X) = k = \bar k$, then $H^0(X_\et,\mu_\ell) = \mu_\ell(k)$ since it is the kernel of the $\ell$th power map. We can also compute $H^1(X_\et, \mu_\ell)$, since our diagram reduces to \begin{tikzcd} & & 0 \ar[dll, hook] \\ \mu_\ell(k) \ar[r] & k\units \ar[r, "z\mapsto z^\ell"] & k\units \ar[dll, "\delta"', twoheadrightarrow] \\ {\color{NavyBlue} H^1(X_\et, \mu_\ell)} \ar[r] & \Pic(X)[\ell] \ar[r, "{[\ell]}"] & \Pic(X) \ar[dll] \\ H^2(X_\et, \mu_\ell) \ar[r] & \cdots & \end{tikzcd} where surjectivity of $\delta$ follows from the fact that $k=\bar k$ and thus every element has an $\ell$th root, making $H^1$ the kernel of $[\ell]$. ::: :::{.example title="?"} Let $X_{/k}$ with $k=\bar k$ with $\ell$ invertible in $k$, then (claim) $\ul \zlz \cong \mu_\ell$ given by sending a generator to some choice of a primitive $\ell$th root of unity. To be explicit, we have a representation $\ul \zlz = \hom \qty{ \wait, \spec k[t] / t(t-1) \cdots (t-\ell+1) }$ and $\mu_\ell = \spec k[t] / t^\ell-1$. These are both disjoint unions of points, and hence schemes of dimension zero since $\ell$ is invertible in the base and the Chinese Remainder Theorem, so one can write down the isomorphism explicitly between the schemes and hence the functors they represent. :::{.corollary title="?"} If $\mu_\ell \subseteq k$, then \[ H^i(X_\et, \ul\zlz) = H^i(X_\et, \mu_\ell) .\] ::: Since the isomorphism depends on the choice of a primitive root, this will not be Galois equivariant, which will come up when we talk about Galois actions on étale cohomology. This already happens for $H^0$, since $G\actson\zlz$ trivially but not on $\mu_\ell$. ::: ### Geometric Interpretations Let $X$ be an affine scheme, we now know \[ H^1(X_\et, \FF_p) = \cok(\OO_X \mapsvia{x^p - x} \OO_x) ,\] the Artin-Schreier map, and these are $\FF_p\dash$torsors. We also know $H^1(X_\et, \zlz)$ in terms of the LES if $k = \bar k$ and $\ch(k) = p$, and this is a $\zlz\dash$torsor. Being torsors here geometrically means they're covering spaces with those groups as Galois groups. :::{.question} How does one write down these torsors/covering spaces? ::: :::{.example title="?"} Given \[ [Y] \in H^1(X_\et, \FF_p) = \cok(\OO_X\mapsvia{x^p-x} \OO_X) \] where we write $[Y]$ to denote thinking of the torsor as some geometric object, how to we write down the covering space? Using Artin-Schreier, we can write $Y = \ts{y^p - y = a}$ for some $a\in \OO_X$, an **Artin-Schreier covering**. If $\ell \neq \ch(k)$ and $[Z] \in H^1(X_\et, \mu_\ell)$ and assume $\Pic(X) = 0$. Then we can write \[ H^1(X_\et, \mu_\ell) = \cok(\OX \mapsvia{x\mapsto x^\ell} \OX\units) \] In this case, $Z = \ts{z^\ell = f}$ where $f\in \OX\units$ is an element representing the class in cohomology, and $\mu_\ell \actson Z$ by multiplication by $z$. ::: :::{.remark} The process of explicitly writing down covers is called **explicit geometric class field theory**, which gives a recipe for writing down abelian covers of covers. In general, for $\Pic(X) \neq 0$, the Picard group plays a crucial role. ::: ## Computing the Cohomology of Curves > This is one of Daniel's favorite topics in the entire course! :::{.theorem title="?"} Let $X_{/k}$ be a smooth curve over $k=\bar k$, then \[ H^i(X_\et, \GG_m) = \begin{cases} \OO_X(X)\units & i = 0\\ \Pic(X) & i=1 \\ 0 & \text{else}, \end{cases} \] noting that $\OO_X(X)\units$ are the global sections of $\GG_m$, i.e. invertible functions on $X$. ::: The first two cases we've done, $i>1$ is the hard case. :::{.corollary title="?"} For $X$ a smooth proper connected curve of genus $g$, $k=\bar k$, and $\ell \neq \ch(k)$ is prime, \[ H^i(X_\et, \ul \zlnz) = \begin{cases} \zlnz & i = 0 \\ \Pic(X)[\ell^n] = \qty{\zlnz}^{2g} & i=1 \\ \zlnz & i=2 \\ 0 & i>2 \end{cases} .\] ::: :::{.proof title="of corollary"} We'll use some theory of abelian varieties: $\Pic^0(X) = \Jac(X)$, and we have a SES \[ 0 \to \Jac(X) \to \Pic(X) \mapsvia{\deg} \ZZ \to 0 ,\] where we identify the Néron-Severi group as $\ZZ$.[^ref_hartshorne_1] We'll use that $\Jac(X)$ is a $g\dash$dimensional abelian variety, and so $\Jac(X)[\ell^n] \cong_{\Grp} \qty{\zlnz}^{2g}$. The Kummer sequence \[ 1 \to \mu_{\ell^n}\to \GG_m \to \GG_m \to 1 \] yields a LES where we identify $\mu_{\ell^n} \cong \zlnz$: \begin{tikzcd} & & 0 \ar[dll] \\ H^1(X_\et, \ul \zlnz) \ar[r] & \Pic(X) \ar[r, "{[\ell]}"] & \Pic(X) \ar[dll] \\ H^2(X_\et, \ul \zlnz) \ar[r] & 0 \ar[r] & 0 \end{tikzcd} So we're just computing the kernel and cokernel of $[\ell]$. **Computing $H^1$**: We'll need one more fact: $\Jac(X)(\bar k)$ is a divisible group. We can identify \[ H^1(X_\et, \ul \zlnz) = \Pic(X)[\ell^n] = \Jac(X) = \qty{\zlnz}^{2g} .\] where the 2nd equality uses the fact that $\Pic(X)$ is an extension of $\ZZ$ by an abelian variety and $\ZZ$ has no torsion, and the last equality is general theory of abelian varieties. **Computing $H^2$**: Since $\Jac(X)$ is divisible, we can identify \[ \coker(\Pic(X) \mapsvia{[\ell^n]} \Pic(X) ) \cong \coker(\ZZ \mapsvia{[\ell^n]} \ZZ ) = \zlnz .\] The vanishing of higher cohomology follows from the vanishing for $\GG_m$. So assuming the theorem and the theory of abelian varieties proves this corollary. [^ref_hartshorne_1]: See Hartshorne Ch. 4, or anything that discusses cohomology of curves. ::: :::{.exercise title="?"} Check this using the snake lemma after applying multiplication by $\ell$ to the SES. ::: :::{.remark} $X$ is a scheme over $\bar k$, and if it started over some subfield $L$ then $\Gal(L_{/k}) \actson X$ and thus the corresponding functors. These isomorphisms will not be Galois equivariant, and the $\zlnz$ showing up in degree 2 cohomology will admit a Galois action via the cyclotomic character. ::: ### Outline of Proof Goal: we want to show that $H^{>1}(X_\et, \GG_m) = 0$ for $X$ a smooth curve over $k=\bar k$. Three ingredients: 1. The Leray spectral sequence, 2. The divisor exact sequence, 3. Brauer groups. ## Pushforwards and the Leray Spectral Sequence Suppose $X\mapsvia{f} Y$ is a morphism of schemes, then we get a functor $f_*:\Sh(X_\et) \to \Sh(Y_\et)$: given $\mathcal{F}\in Sh(X_\et)$, we have $f_* \mathcal{F}(U\to Y) \da \mathcal{F}(U\cross_Y X)$. This is left-exact and thus has right-derived functors $R^\wait f_*:\Sh^\Ab(X_\et) \to \Sh^\Ab(Y_\et)$. How to think about this: ![Cohomology of the fibers: but not quite!](figures/image_2020-12-28-18-24-08.png){width=350px} This is not quite true, and the obstruction is called **the base change property**, which we'll see later in the course. What's true in general is that $R^i f_* \mathcal{F}$ is the sheafification of the presheaf $V\to H^i(f^{-1}(V), \mathcal{F})$, which is not quite the cohomology of the fibers since sheafification is somewhat brutal. :::{.proposition title="Derived pushforwards for finite morphisms"} If $f$ is a finite morphism (e.g. a closed immersion) then $R^{>0} f_* = 0$. ::: :::{.exercise title="Proof, must-do!"} Prove this. The claim is that $f_*$ is right-exact, which in this case shows it is exact. Check on stalks. Compute that the stalk of $f_* \mathcal{F}$ at $\bar y\in Y$ is given by \[ f_* \mathcal{F}_{\bar y} = \bigoplus_{\bar x\in f^{-1}(\bar y)}\mathcal{F}_{\bar x} \] for $f$ a finite morphism (not necessarily unramified). ::: :::{.proposition title="technical"} $f_*$ preserves injectives. ::: :::{.exercise title="proof"} Prove this! You can do this by showing the following fact from category theory: this is true for any functor with an exact left adjoint, which here is $f^*$ and is exact since filtered colimits and sheafification are both exact, or alternatively you can check on stalks, since the stalks of $f^{-1}$ are the stalks of the original functor. ::: :::{.corollary title="The Leray Spectral Sequence"} Suppose $X \mapsvia{f} Y$ and $Y \mapsvia{g} Z$ are morphisms of schemes, then there is a spectral sequence \[ R^i g_* R^j f_* \mathcal{F} \abuts R^{i+j}(g\circ f)_* \mathcal{F} .\] As a special case, for $Z = \spec k$ with $k=\bar k$, then $g_*, f_*$ are taking global sections so we get \[ H^i(Y, R^j f_* \mathcal{F} ) \abuts H^{i+j}(X, \mathcal{F}) .\] ::: :::{.proof title="sketch"} There is a general statement (see Tohoku): given two functors between abelian functors where the first preserves injectives, you get such a spectral sequence. How to explicitly compute this: we can take an injective resolution $\mathcal{F}\to \mathcal{I}^\wait$ and compute \[ R^i f_* \mathcal{F} \mathcal{H}^i(f_* \mathcal{I}^\wait) .\] $f_* \mathcal{I}$ is a complex of injectives, and we want $\mathcal{H}^{i+j}(g_* f_* \mathcal{I}^\wait) = R^{i+j}(g\circ f)_* \mathcal{F}$, and the content here is that we don't have to take an additional injective resolution of $f_* \mathcal{I}$. Now take the spectral sequence of the filtered complex $f_* \mathcal{I}^\wait$ where the filtration is by the truncations $\tau_{\leq p}f_* \mathcal{I}^\wait$ where you replace the $p$th term with the kernel of the differential and zero beyond this point. An example of a differential is given by the following: there are SESs \[ 0 \to \tau_{\leq p} f_* \mathcal{I}^\wait \to \tau_{\leq p+1} f_* \mathcal{I}^\wait \to \mathcal{H}^{p+1}(f_* \mathcal{I}^\wait) = R^{p+1} f_* \mathcal{F} \to 0 ,\] and applying $RG_*$ yields a map \[ R^{p+1} f_* \mathcal{F} \mapsvia{\delta} R^{q+1} g_* \tau_{\leq p}f_* \mathcal{I}^\wait ,\] and after some splicing this $\delta$ will be the differential on $E_2$. ::: Next time: the Brauer group.