# 10/01/2020: Brauer Groups and $H^1$ of Curves

:::{.remark}
Recall that last time we defined $\mathrm{Br}(X)$ and determined it was a group, and given a Čech 2-cocycle representing $[\alpha]\in \mathrm{Br}(X)$ we understood it geometrically via the category of $\alpha\dash$twisted sheaves: sheaves with twisted descent data where the failure of the cocycle condition is measured by $\alpha$.
These form an abelian category resembling $\QCoh(X)$.
We had other ways of interpreting a Brauer class represented by a twisted vector bundle $\mce$:

- Severi-Brauer varieties, twisted forms of $\PP^n$ which can be obtained as $\PP(\mce)$, 
- Azumaya algebras, coherent sheaves of algebras which are étale-locally matrix algebras (endomorphisms of the trivial bundle), obtained as $\Endo(\mce)$.

Note the projectivizing $\mce$ does yield descent data, since this gets rid of the failure in the cocycle condition, and this descent data turns out to be effective.
Moreover $\alpha \dash$twisted vector bundles are étale-locally just vector bundles, hence the second interpretation.
:::

:::{.definition title="Quasi-algebraically closed fields"}
A field $k$ is **quasi-algebraically closed** if for any homogeneous polynomial $f\in \kxn$ with $\deg f < n$, $f$ has a nontrivial zero.
This is equivalently called a **$C_1\dash$field**, where a $C_m\dash$field has a similar definition but instead requires $(\deg f)^m < n$.
:::

:::{.theorem title="Tsen's Theorem"}
Suppose $k$ is a $C^1\dash$field.
Then $\mathrm{Br}(k) = 0$.
:::

:::{.theorem title="?"}
Let $k$ be either

1. The function field of a curve over an algebraically closed field, or 
2. The fraction field of a strictly Henselian DVR.

Then $k$ is quasi-algebraically closed, and thus $\mathrm{Br}(k) = 0$.
:::

:::{.remark}
Recall that we reduced our computation of $H^*_\et(X; G)$ for $G = \GG_m, \ZZ/\ell^n\ZZ, \mu_n$ to vanishing of higher cohomology, and these results give vanishing in $H^2$.
:::

:::{.definition title="Reduced norm"}
Let $\mce$ be an $\alpha\dash$twisted sheaf, so $\Endo(\mce)$ is an Azumaya algebra.
The **reduced norm** is defined as 
\[
\Norm: \Endo(\mce) &\to \Endo\qty{\Wedge^{\mathrm{top}} \mce} \cong \OO_X 
.\]
is the functorial map associated to taking wedge powers.
Note that if $\mce$ is a trivial vector bundle, this is the determinant.
:::

:::{.proposition title="?"}
If $f\in \Endo(\mce)$, then $f$ is invertible iff $\Norm(f)$ is a unit.
:::

:::{.proof title="?"}
Check locally, where $\Endo(\mce)$ is a matrix algebra.
:::

:::{.proof title="of Tsen's theorem"}
WTS: given $[\alpha]\in \mathrm{Br}(k)$ we can find an $\alpha\dash$twisted line bundle.
We have an $\alpha\dash$twisted *vector* bundle, by definition of elements in $\mathrm{Br}(k)$.
Idea: if $\rank \mce = 1$ we're done, so try to find a lower rank sub-bundle.
We can do this by finding an endomorphism with nontrivial kernel or cokernel, which will again be a vector bundle because we're working over a field and thus the rank is constant (since one only has to check rank at a single point).
So we'll look for $f$ with $\Norm(f) = 0$.

We need to use the hypothesis that $k$ is quasi-algebraically closed. Note that $\Norm: \Endo(\mce) \to k$, where $\Endo(\mce)$ is an $n^2$ dimensional affine space where $n\da \rank \mce$, so $\Norm$ is a polynomial function in $n^2$ variables -- one can check this rigorously by extending $k$ until $\mce$ splits and the norm becomes the usual determinant.
Thus $\deg \Norm = n$ as a polynomial, again checked by passing to a splitting cover.

Since $k$ is quasi-algebraically closed, $\Norm$ has a nontrivial zero when $n > 1$ so that $n < n^2$.
So this furnishes an $f$ with $\Norm(f) = 0$.
Setting $\mce' \da \ker f$, this is an $\alpha\dash$twisted bundle of lower rank and we're done.
Continuing until we reach a line bundle, $[\alpha] = 0 \in \mathrm{Br}(k)$.
:::

:::{.corollary title="?"}
If $k$ is quasi-algebraically closed, then $H^2(k; \GG_m) = 0$.
:::

:::{.proof title="?"}
A nontrivial application of field theory: for a field $k$, one has 
\[
H^2(k; \GG_m) = \mathrm{Br}(k)
.\]

:::

:::{.remark}
This is more generally true for regular rings.
This is not true for general schemes, but it's hard to construct counterexamples.
In full generality, the Brauer group here equals the cohomological Brauer group, which isn't true for all schemes but is true for affine or quasi-affine schemes.
The proof involves explicitly writing down a central simple algebra representing a cocycle, see Serre's "Local Fields".
:::

:::{.remark}
We've now produced some examples of fields with trivial Brauer group, provided we can find quasi-algebraically closed fields.

One example is $k = k(C)$ the function field of a curve over an algebraically closed field.
Note that this will imply that $H^2(k(C); \GG_m) = 0$, and the divisor exact sequence is closely related to $H^2_\et(C)$.

:::

:::{.proposition title="?"}
The function field $k = k(C)$ of a curve over an algebraically closed field is quasi-algebraically closed.
:::

:::{.proof title="Beautiful!"}
Given $f\in k(C)[x_1,\cdots, x_n]$ and $\deg f< n$, it suffices to produce a nontrivial zero of $f$ in $k(C)\units$.

A very beautiful idea: study functions on $C$ with bounded poles. Choose an ample divisor $D\in \Div(C)$ and consider 
\[
f\in H^0(\OO(mD))\cartpower{n} \to H^0(\OO(\tilde D)), \tilde D = (\deg f)\cdot mD + D'
\]
for $m\in \ZZ$, where $D'$ are the poles of the coefficients of $f$.
By Riemann-Roch, the source grows like $\bigo(mn)$ since $\OO(mD)$ grows like $m + c$ a constant (so notably this has slope 1) and taking $n$th powers multiples by $n$, and the target grows like $(\deg f)\cdot m$.
By assumption, $mn > (\deg f) n$, and so for $m\gg 0$ we get a morphism $X\to Y$ with $\dim X > \dim Y$, so the fibers are positive-dimensional.
It suffices now to show $\dim f\inv(0) > 0$, since these are maps of affine spaces over an algebraically closed field, so a nontrivial zero of $f$ is any point over $0$ whose coordinates are not all simultaneously zero, and positive dimensionality yields existence of such a point.
But $f\inv(0)$ is nonempty since $f$ is homogeneous.
:::

:::{.corollary title="?"}
\[
H^2(k(C); \GG_m) \isovia{\text{Serre}} \mathrm{Br}(k(C)) = 0
.\]
:::

:::{.proof title="?"}
Let $K$ be the function field of a strictly Henselian DVR, then this is also quasi-algebraically closed -- examples include $\OO_{X, x}$ for $X$ a curve over $\bar{\FF_q}$, or $W(\bar{ \FF_p})$ the Witt vectors.
This is shown in Lang's thesis, although it is an exercise to prove this when $K$ is equicharacteristic zero.
:::

:::{.corollary title="?"}
If $K$ is the fraction field of a strictly Henselian DVR, 
\[
\mathrm{Br}(K) = H^2(K; \GG_m) = 0
.\]
:::

:::{.theorem title="?"}
Let $K$ as above, so $K = K(C)$ or $K_{\bar x}$, then 
\[
\tau_{\geq 1} H^*(K; \GG_m) = 0
.\]
:::

:::{.remark}
Note that having $\mathrm{Br}(K) = 0$ is not quite enough, one needs that all extensions $L/K$ also have $\mathrm{Br}(L) = 0$.
This is true for quasi-algebraically closed fields, since their extensions are again quasi-algebraically closed.
What is perhaps easier to see is that any $L/K(C)$ is again the function field of a curve over an algebraically closed field.
One can also show that extensions of strictly Henselian DVRs are again strictly Henselian DVRs.
Generally, it suffices to check that $H^1$ and $H^2$ are zero for all finite extensions.
:::

:::{.remark}
Take a break! Here's the general strategy: we had a Weil divisor exact sequence relating $H^*$ of the curve to that of the pushforward from the generic point. That cohomology was related by the Leray spectral sequence to $H^*$ of $\GG_m$ on the generic point itself and $H^*$ of the strictly Henselian local rings with coefficients in $\GG_m$.
We're now showing these vanish in high degree.

Back to work!
:::

:::{.proof title="due to Tate"}
Goal: prove $H^i(L/K; \GG_m) = 0$ for all finite separable Galois extensions $L/K$, and this will suffice since Galois covers are cofinal in the system of all covers.
Equivalently, we're computing $H^*_\et(\spec k)$, and such extensions $L$ give étale covers of $k$ and we're computing the Čech cohomology of that cover.
Writing down the Čech complex will show 
\[
H^i(L/K; \GG_m) = H^i(\Gal(L/K); L\units)
.\]

**Step 1**: we know vanishing for $i=1,2$ by Hilbert 90, Tsen's theorem, and inflation-restriction.

**Step 2**: do this for $L/K$ cyclic, where the cohomology is 2-periodic and thus zero in all degrees. This comes from the 2-periodic resolution of $\ZZ\in \gmod$.

**Step 3**: do this for $L/K$ nilpotent, so iterated commutators eventually reach zero, or solvable: there is a filtration whose associated graded pieces are cyclic.
Note that solvable implies nilpotent.

Let $C \normal \Gal(L/K)$ which is normal and cyclic, which exists by solvability.
This yields a SES $C\injects \Gal(L/K) \surjects G'$ where $G'$ is nilpotent.
Since we know the required statements for $C$ by (2) and $G'$ by induction, the inflation-restriction exact sequence yields the result.

**Step 4**: do this for general $L/K$.
One can generally pass from nilpotent to general groups. Note that e.g. $p\dash$groups are nilpotent, since it always has nontrivial center by the class equation and one can keep quotienting by the center to get solvability. So for $G_p \leq \Gal(L/K)$ a $p\dash$Sylow subgroup, we have $H^i(G_p; L\units) = 0$ by part (3).
Consider the map
\[
\Phi:H^i(\Gal(L/K); L\units) \to \bigoplus _{p} H^i(G_p; L\units)
,\]
we're then done if $\Phi$ is injective.
This is true because there is a restriction/corestriction section
\[
\adjunction
{\Res}
{\cores}
{ H^i(\Gal(L/K); L\units) }
{H^i(G_p; L\units)}
,\]
see Cassels-Frohlich chapter 4.
Moreover $\cores \circ \Res = c\cdot$ is multiplication by $c=[G:G_p]$, which is prime to $p$.So $\Res$ is injective away from torsion which is prime to $p$, and this is enough to get injectivity of their direct sum: the kernel would have to be torsion prime to *all* primes $p$, which is impossible.
:::

:::{.remark}
This is a common real-life application of Sylow: reducing a statement to $p\dash$groups.
Note that we're done proving that $H^i_{\Gal}(K; \GG_m) = 0$ for these special fields, since we proved it for all extensions and the Galois cohomology is the direct limit over all such extensions.
:::

## Back to étale cohomology

:::{.corollary title="?"}
Let $C$ be a smooth curve over an algebraically closed field, then 
\[
\tau_{\geq 1} H^i(C_\et; \GG_m) \cong \tau_{\geq 1} H^i(k(C); \eta_* \GG_m)
.\]
:::

:::{.proof title="?"}
We showed $H^i(C_\et; \GG_m) \cong H^i(C_\et; \eta_* \GG_m)$ for $i > 1$ from the Weil divisor exact sequence, using vanishing for the sheaf of divisors.
The claim is that $H^i(C_\et; \eta_* \GG_m) \cong H^i(k(C); \GG_m)$.
By the Leray spectral sequence it STS that $\RR^j \eta_* \GG_m = 0$ for $i > 0$ -- however, this has stalks $H^i(K_{\bar x}; \GG_m) = 0$ since these are strictly Henselian DVRs.
:::

:::{.corollary title="?"}
\[
H^i(C_\et; \GG_m) \cong H^i(k(C) ; \GG_m) = 0
.\]
:::

:::{.proof title="?"}
Combine the previous corollary with Tate's theorem.
:::

:::{.corollary title="?"}
\[
H^*(C_\et; \GG_m) = \GG_m(C) + \Pic(C)t
.\]
and $H^i = 0$ for $i > 1$.
:::

:::{.corollary title="?"}
If $n$ is prime to $\characteristic k$, then
\[
H^i(C_\et; \mu_n) = \mu_n + \Pic(C)[n] t + C_n t^2
.\]
where the $t^2$ term is nonzero iff $C$ is proper and zero otherwise.
Note that $\mu_n \neq C_n \in \gmod$, since the action on $C_n$ is trivial.
:::

:::{.proof title="?"}
Use the Kummer sequence $\mu_n\injects \GG_m\surjects \GG_m$, where we use the assumption on the characteristic so that this is exact.
:::

:::{.remark}
The cyclic piece $C_n$ appears because $\Pic^0(C)$ is divisible if $C$ is proper, and if $C$ is not proper then $\Pic(C)$ itself is divisible by an exercise.
:::

:::{.remark}
This is a Galois-equivariant description if $C$ is proper.
The content here is that $G$ acts trivially on $H^2$, which follows from the trivial action on $\Pic^0$.
Relaxing the equivariance condition, one can write $\Pic(C)[n] = C_n^{2g}$, i.e. picking a basis for torsion yields the familiar action $G\actson C_n^{2g}$.
:::

:::{.remark}
Over $\CC$, a real surface of genus $g$, $H^2(X; C_n)$ looks the same -- this was the first evidence that $H_\et$ is a good cohomology theory. The second major step was Artin's calculation of $H_\et(\AA^2\smz)$, where $\AA^2\smz$ looks like a 3-sphere $S^3$.
:::

:::{.remark}
We've achieved our main goal of computing $H_\et(C)$ for curves.
Next time: towards Poincaré duality, cohomology with compact support, and cohomology for higher dimensional varieties.
:::