# Tuesday, September 14 :::{.remark} Goal: Severi-Brauer varieties satisfy the Hasse principle, and develop the Brauer-Manin obstruction. We have the following theorem: if $X\in \SB\slice k$, then TFAE: - $X$ has a rational point, - $X \cong \PP^n$ for some $n$, - $[X]\in \Br(k)$ is the trivial class. We'll soon prove the following theorem: ::: :::{.theorem title="Hasse principle for Severi Brauers"} For $K$ a number field, there is an injective map \[ \Br(k) \injects \bigoplus _{v\in \places{k} } \Br(k_v) ,\] which is a statement of the Hasse principle, since the previous theorem shows that if $\Br(k_v)$ is empty for all $k_v$, it will have to come from a zero class in $\Br(k)$ ::: :::{.remark} Note that the cokernel of this map is prominent in class field theory! Today we'll compute $\Br(k_v)$, or more generally $\Br(F)$ for $F$ a local field. ::: ## Cyclic Algebras :::{.remark} Setup: take $k\in \Field$, $L\slice k$ a $C_n\dash$Galois extension, which is the data of \[ \chi_L: \Gal(k^s\slice k)\to C_n .\] For $a\in K^s$, we'll consider pairs $(\chi, a) = L[x]^\chi / \gens{x^n - a}$ where commutation in $L[x]^\chi$ is given by $lx = x\sigma(\ell)$ for $l\in L$ where $C_n = \gens{\sigma}$. This is a $k\dash$vector space of dimension $n^2$, and the claim is that $(\chi, a) \in \CSA$. ::: :::{.example title="?"} Take $\chi: \Gal(\CC\slice \RR) \to C_2$ with $a=-1$, then $(\chi, a) = \HH = \RR[i, j] / \gens{i^2,j^2,[ij]}$ is the (Hamilton) quaternions. ::: :::{.fact} One can view $\chi \in H^1_\Gal(k; C_n)$ and \[ a\in H^1_\Gal(k; \mu_n) = k\units / (k\units)\powers{2} .\] In this case \[ (\chi, a) \da \chi \cupprod [a] \in H^2(k; \mu_n) \subseteq H^2(k; (k\sep)\units) .\] Note that this cup product can be computed explicitly from the product on $\Ext$ or using the standard resolution. ::: :::{.remark} Now to compute more Brauer groups! So far, we've only done relatively trivial examples. We'll start with local fields: for algebraically closed fields, Galois cohomology vanishes, so - $\Br(\CC) = 0$ - To compute $\Br(\RR) = H^2(\Gal(\CC\slice \RR); \CC\units)$, take the resolution \begin{tikzcd} && \vdots \\ {P^\bullet:} && {\ZZ[x]/\gens{x^2-1}} \\ && {\ZZ[x]/\gens{x^2-1}} \\ && {\ZZ[x]/\gens{x^2-1}} \\ && {\ZZ[x]/\gens{x^2-1}} & 1 \\ && \ZZ & 1 \arrow[from=5-3, to=6-3] \arrow["{x-1}", from=4-3, to=5-3] \arrow[from=5-4, to=6-4] \arrow["{x+1}", from=3-3, to=4-3] \arrow["{x-1}", from=2-3, to=3-3] \arrow[from=2-3, to=3-3] \arrow[from=4-3, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwzLCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMiw0LCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMiw1LCJcXFpaIl0sWzMsNCwiMSJdLFszLDUsIjEiXSxbMiwyLCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMiwxLCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMCwxLCJQXlxcYnVsbGV0OiJdLFsyLDAsIlxcdmRvdHMiXSxbMSwyXSxbMCwxLCJ4LTEiXSxbMyw0XSxbNSwwLCJ4KzEiXSxbNiw1LCJ4LTEiXSxbNiw1XSxbMCw1XV0=) Then we can take $H^*(\Hom_{\mods{\Gal}}(P^\bullet, \CC\units ))$: \begin{tikzcd} && 1 && {z\bar{z}} \\ \CC\units && \CC\units && \CC\units && \CC\units \\ z && {\bar{z}z\inv} && z && {\bar{z}z\inv} \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow[from=2-5, to=2-7] \arrow[from=3-1, to=3-3] \arrow[from=3-5, to=3-7] \arrow[from=1-3, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzAsMSwiXFxDQ1xcdW5pdHMiXSxbMiwxLCJcXENDXFx1bml0cyJdLFswLDIsInoiXSxbMiwyLCJcXGJhcnt6fXpcXGludiJdLFs0LDEsIlxcQ0NcXHVuaXRzIl0sWzQsMiwieiJdLFs2LDIsIlxcYmFye3p9elxcaW52Il0sWzYsMSwiXFxDQ1xcdW5pdHMiXSxbMiwwLCIxIl0sWzQsMCwielxcYmFye3p9Il0sWzAsMV0sWzEsNF0sWzQsN10sWzIsM10sWzUsNl0sWzgsOV1d) Check that $\bar z z\inv = 1$ then $z = \bar z$ so $z\in \RR\units$ and $\ker d = \RR\units$. Similarly, $\im d = \RR\units_{>0}$, so \[ \Br(\RR) = \RR\units/ \RR\units_{>0} = \ts{\pm 1} .\] ::: :::{.example title="?"} $\HH$ represents $-1$ in $\Br(\RR)$, as does the corresponding Severi Brauer \[ \ts{x^2 + y^2 + z^2 = 0} \subseteq \PP^2\slice \RR .\] Note that $+1$ is represented by the field itself, regarded as a $1\times 1$ matrix algebra, or projective space. ::: :::{.remark} Write $k^\unram$ for the maximal unramified extensions, where an extension is *ramified* if the degree of the residue field changes (or the valuation remains an integer?) For example, for $k = \QQpadic$, we have $k^\unram = \ff(W(\bar{\FF_p}))$ (i.e. the Witt vectors). In general, $k^\unram = k(\mu_\infty')$ where $\mu_\infty'$ is the set of roots of unity of order prime to the characteristic. As a corollary, $\Gal(k^\unram\slice k) = \bar{\FF_q}/\FF_1 = \hat \ZZ$. ::: :::{.theorem title="?"} For $k$ a nonarchimedean local field (a finite extension of $\QQpadic$), then $\Br(k) = \QQ/\ZZ$ - $H^2(k^{\unram}\slice k; (k^\unram)\units) = \QQ/\ZZ$ - $H^2(k^\unram\slice k, (k^\unram)\units) \iso H^2(k; (k^s)\units) = \Br(k)$ is an isomorphism. ::: :::{.remark} Many proofs of this are delicate! We'll follow a mix of Cassels-Frolich and Milne for this proof. ::: :::{.proof title="of 1"} Take the SES coming from the valuation map: \begin{tikzcd} 1 && {U_{k^\unram}} && {(k^\unram)\units} && \ZZ && 0 \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow["\val", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCIxIl0sWzIsMCwiVV97a15cXHVucmFtfSJdLFs0LDAsIihrXlxcdW5yYW0pXFx1bml0cyJdLFs2LDAsIlxcWloiXSxbOCwwLCIwIl0sWzAsMV0sWzEsMl0sWzIsMywiXFx2YWwiXSxbMyw0XV0=) :::{.claim} \envlist - $H^2(k^\unram\slice k; \ZZ) = \QQ/\ZZ$. - $H^*(k^\unram\slice k; U_{k^\unram}) = 0$ ::: :::{.remark} Why this implies the theorem: take the LES in cohomology to get the following: \begin{tikzcd} \textcolor{rgb,255:red,214;green,92;blue,92}{H^2(k^\unram \slice k; U_{k^\unram})=0} && {H^2(k^\unram \slice k; (k^\unram)\units )} && {H^2(k^\unram \slice k; \ZZ)} \\ \\ \textcolor{rgb,255:red,214;green,92;blue,92}{H^3(k^\unram \slice k; U_{k^\unram}) = 0} \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=3-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXjIoa15cXHVucmFtIFxcc2xpY2UgazsgVV97a15cXHVucmFtfSk9MCIsWzAsNjAsNjAsMV1dLFswLDIsIkheMyhrXlxcdW5yYW0gXFxzbGljZSBrOyBVX3trXlxcdW5yYW19KSA9IDAiLFswLDYwLDYwLDFdXSxbMiwwLCJIXjIoa15cXHVucmFtIFxcc2xpY2UgazsgKGteXFx1bnJhbSlcXHVuaXRzICkiXSxbNCwwLCJIXjIoa15cXHVucmFtIFxcc2xpY2UgazsgXFxaWikiXSxbMCwyXSxbMiwzXSxbMywxXV0=) A claim is that $H^2(k^\unram \slice k; \ZZ) = H^2(\hat \ZZ; \ZZ)$. One can compute this colimit explicitly, but there is a SES \[ 0\to \ZZ \to \QQ\to \QQ/\ZZ\to 0 .\] Now note that $H^{>0}(G; \QQ) = 0$ for profinite groups, since this is necessarily a torsion $\QQ\dash$vector space. For a full proof, use that multiplication by $n$ is an isomorphism the annihilates it. As a corollary, taking the LES above yields $H^{i}(G; \ZZ) = H^{i-1}(G; \QQ/Z)$ for $i\geq 2$. Thus \[ H^2(\hat \ZZ; \ZZ) = H^1(\hat \ZZ; \QQ/\ZZ) = \Hom_{\Top}(\hat \ZZ, \QQ/\ZZ) = \QQ/\ZZ ,\] and in fact \[ H^1(\hat \ZZ; \QQ/\ZZ) = \colim_n \Hom(C_n; \QQ/\ZZ) .\] ::: ::: :::{.proof title="of b"} Here we'll have to use the structure of $U_{k^\unram}$. It's enough to show \[ H^{>0}(k_n{} \slice k; U_{k_n}) = 0 \] for $k_n{}\slice k$ unramified of finite degree $n$, using that these are unique. We'll use the following: :::{.definition title="?"} There is a filtration $\Fil_r U_{k_n} = \ts{u\in U_{k_n} \st u = 1 \mod \pi^r}$ for $\pi$ a uniformizer. ::: :::{.fact} We can identify \[ \Fil_r/\Fil_{r+1} = \begin{cases} \kappa_n\units & r=0 \\ \kappa_n^+ & r>0. \end{cases} ,\] where $\kappa$ denotes residue fields, $\kappa_n{}\slice \kappa$ is the unique degree $n$ extension, and $\kappa^+$ is the additive group. Why: use that these look like power series, and the associated graded picks off the $r$th coefficient. Moreover, things like $1+\pi^2$ can be units by formally inverting using geometric series. ::: Thus it's enough to show for residue fields that \[ H^{>0}(\kappa_n{}\slice \kappa; \kappa_n\units) &= 0 \\ H^{>0}(\kappa_n{}\slice \kappa; \kappa_n^+) &= 0 ,\] since each graded piece of the associated grading having zero cohomology implies the entire thing has zero cohomology. For the first, - $i=1$ is Hilbert 90, - $i=2$ follows from $\Br(\kappa_n{} \slice \kappa) = 0$, - $i\geq 3$ uses that $H^* = H^*[-2]$, since the resolution used for cohomology of a cyclic group was 2-periodic. For the second, to compute the cohomology of a cyclic group we take the 2-periodic resolution: \begin{tikzcd} x && {x^{q}-x} \\ {k_n} && {k_n} && {k_n} && \cdots \\ && x && {\sum x^{q^i}} \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow["{\Frob - 1}", from=1-1, to=1-3] \arrow["\trace_ { {\kappa_n} {}\slice{\kappa} }", from=3-3, to=3-5] \arrow[from=2-5, to=2-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwxLCJrX24iXSxbMCwwLCJ4Il0sWzIsMCwieF57cX0teCJdLFsyLDEsImtfbiJdLFs0LDEsImtfbiJdLFsyLDIsIngiXSxbNCwyLCJcXHN1bSB4XntxXml9Il0sWzYsMSwiXFxjZG90cyJdLFswLDNdLFszLDRdLFsxLDIsIlxcRnJvYiAtIDEiXSxbNSw2LCJcXHRleHR7XCJBZGRpdGl2ZSBub3JtXCJ9IiwyXSxbNCw3XV0=) Then - $H^2 = \ker/\im$, and $\ker = k$ since Frobenius fixes everything, and use that \[ \sum x^{q^i} = x + x^q + x^{q^2} + \cdots = \trace_{\kappa_n {}\slice \kappa}(x) .\] - If $n$ is invertible, so $p\notdivides n$, writing $\Tr(1) = n$ we can take $\Tr(a/n) = a$. - It suffices to show this polynomial isn't identically zero, but it's a polynomial of degree $q^{n-1}$ but $\# \kappa_n = q^n$. - Now use that $a = \trace(x)$ for some $a\nonzero$, then take $b = \trace(bx/a)$. :::