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\title{Lecture 5-3: The cup product and Brauer group}
\date{May 3, 2024}
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We wrap up the material on group cohomology with a brief account of its ring structure, given by the cup product, and give another setting in which factor sets arise.
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As many of you know, groups are not the only mathematical objects with cohomology groups attached to them.  Given any topological space $X$ and a commutative ring $R$, one has \alert {homology groups} $H_n(X,R)$ and \alert{cohomology groups} $H^n(X,R)$ of $X$ \alert{with coefficients in $R$}.  These have the structure of $R$-modules, not just abelian groups; in addition, the latter groups have a ring structure under something called the {\sl cup product}.  The sum $H^*(X,R)=\bigoplus_nH^n(X,R)$ thereby acquires the structure of a graded $R$-algebra, since the cup product of $H^n(X,R)$ and $H^m(X,R)$ lies in $H^{n+m}(X,R)$.  In the setting of group cohomology, assume for simplicity that $G$ acts trivially on a field $k$; then the \alert{group cohomology ring} $H^*(G,k)=\bigoplus_n H^n(G,k)$ is a graded $k$-algebra.  It is \alert{graded commutative} in the sense that $xy = (-1)^{ij}yx$ if $x\in H^i(G,k),y\in H^j(G,k)$.
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Take first $k=F_2$, the field of order 2; here graded commutativity is the same as commutativity.  Also let $G=\mathbb Z_2$, the cyclic group of order 2.  You have seen that $H^n(G,F_2)\cong\mathbb Z_2$ for all $n$; now I can say more precisely that the ring $H^*(G,k)$ is isomorphic to the polynomial ring $k[x]$ in one variable, where the degree of $x$ is 1 (as usual).  For odd primes $p$, the story is a bit more complicated; taking $G\cong \mathbb Z_p$ to be cyclic of order $p$ and $F_p$ to be the field of this order, we still have $H^n(G,k)\cong\mathbb Z_p$ for all $n$, but now it is the sum $\bigoplus_n H^{2n}(G,k)$ of the cohomology groups of even degree that is isomorphic to the polynomial ring $k[x]$, taking the degree of the variable $x$ to be 2, while the sum $\bigoplus_n H^{2n+1}(G,k)$ can be realized as $k[x]y$, where the degree of $y$ is 1 and $y^2=0$.
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Returning to the case $k=F_2$, suppose now that $G$ is replaced by the product $\mathbb Z_2^n$ of $n$ copies of $\mathbb Z_2$.  Then the {\sl K\" unneth formula} in algebraic topology or group cohomology asserts that $H^*(G,k)$ is isomorphic to the tensor product $\otimes^n k[x]$ of $n$ copies of $k[x]$, which may be identified with the polynomial ring $k[x_1,\ldots,x_n]$; here (again as usual) we take the degrees of all variables $x_i$ to be 1.  Thus if $n=2$, we find that $H^2(G,k)$ has basis $x_1^2,x_1x_2,x_2^2$ over $k$, so its cardinality is 8, as previously computed.  A similar $n$th tensor power formula holds for $H^*(\mathbb Z_p,\times\mathbb Z_p,F_p)$.
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Besides group extensions, central simple algebras provide another context in which factor sets arise.  These also provide a nice connection between group cohomology and the Galois theory you studied earlier.  Let $K$ be a finite Galois extension of a field $F$ with Galois group $G$.  Let $f=\{a_{\sigma,\tau}\}_{\sigma,\tau\in G}$ be a normalized factor set of $G$ with values in $K^*$.  Let $B_f$ be the vector space over $K$ with basis $u_\sigma$ for $\sigma\in G$.  Define a ring structure on $B_f$ via $u_\sigma \alpha=\sigma(\alpha)u_\sigma,u_\sigma u_\tau = a_{\sigma,\tau} u_{\sigma\tau}$ for $\sigma,\tau\in G,\alpha\in K$.  We call $B_f$ a \alert{crossed product algebra} for the factor set $f$ (and $G$ and $K^*$); see DF, p. 833.  Note that if $f$ is the constant function 1, then $B_f$ is a kind of twisted analogue of the group algebra $KG$, where $K$ no longer commutes with $G$ but instead elements of $G$ move past elements of $K$ via the Galois group action.  In general, $u_1$ is the multiplicative identity for $B_f$, since $f$ is normalized.
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If the factor set $f$ is replaced by another one $f'$ lying in the same cohomology class, then it is easy to check that the resulting algebra $B_{f'}$ is isomorphic to $B_f$ by a map which is the identity on $K^*$.  Thus there is a bijection between elements of $H^2(G,K^*)$ and $K$-isomorphism classes of crossed product algebras over $F$ containing $K$.  It is easy to see that the center of any algebra $B_f$ is exactly $Fu_1\cong F$:  any combination $\sum\limits_{\sigma\in G}\alpha_\sigma u_\sigma$ commutes with $K$ if and only if $a_\sigma=0$ for all $\sigma\ne1$ (since only $\sigma=1\in G$ fixes all of $K$) and then $ku_1$ commutes with all $u_\sigma$ if and only if $k\in K$ lies in $F$.  Similarly, it is easy to see that $B_f$ has no nonzero proper two-sided ideals:  given a sum $s=\sum \alpha_\sigma u_\sigma$ with as few nonzero terms as possible lying in a proper nonzero ideal $I$, by replacing $s$ with $s\beta-\beta s$ for suitable $\beta\in K$ we get a sum with fewer nonzero terms in $I$, unless $s$ has only one term; but any single term $a\sigma u_\sigma$ is a unit in $B_f$.  Thus $B_f$ is indeed what is called a central simple algebra over $F$ (\lq\lq simple" meaning that it has no nonzero proper two-sided ideals).
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Given any finite-dimensional central simple algebra $A$ over a field $F, A$ has a maximal proper left ideal $L$ (that is, one of maximal dimension), so that $A/L$ is an irreducible left $A$-module. $M$.  Letting $D$ be the ring of endomorphisms of $M$ commuting with the $A$ action, we see from Schur's Lemma (the version you proved in HW some weeks back) that $D$ is a division ring containing $F$ as a subfield.  By the same reasoning that we used to determine the structure of the group algebra $KG$ for $G$ a finite group and $K$ an algebraically closed field of characteristic not dividing the order of $G$, we deduce that $A\cong M_n(D)$, the ring of all $n\times n$ matrices over $D$ (just one such ring rather than the direct sum of two or more since $A$ is simple).  It follows that the dimensions of both $A$ and $D$ over $F$ are squares, at least in the special case $A=B_f$ for some factor set $f$ (and it turns out in general).
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You have already seen an example of such an algebra $A$, namely the ring $H$ of quaternions (which is central simple over the real field $\mathbb R$).  Note that $H$ is a crossed product algebra:  it contains a copy of the unique proper finite extension of $\mathbb R$, namely $\mathbb C$, and it contains an element $j$ acting on $\mathbb C$ by complex conjugation, the unique nontrivial element of the Galois group of $\mathbb C$ over $\mathbb R$ (so that $jzj^{-1}=\bar z$ for $z\in\mathbb C$).  The corresponding factor set $f$ has $f(1,1) = f(1,j) = f(j,1)=1, f(j,j)=-1$.  There is no need here to take $\mathbb R$ as the base field; we could just as easily have started with $\mathbb Q$ and its Galois extension $\mathbb Q[i]$.  In fact, given any $a,b\in\mathbb Q^*$ with $a$ not a square in $\mathbb Q^*$, we could adjoin elements $x,y$ to $\mathbb Q$ to make a central simple $\mathbb Q$-algebra $\mathbb Q_{a,b}$ with the defining relations $x^2=a,y^2=b,xy=-yx$.  Every such algebra is either a division ring or isomorphic to the ring $M_2(\mathbb Q)$ of $2\times 2$ matrices over $\mathbb Q$.
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In general, for any field $F$, we introduce an equivalence relation on central simple algebras $A$ over $F$, which enables us to put a group structure on the set of such algebras.  We have seen that $A\cong M_n((D)$ for some division ring $D$ with center $F$ (and $D$ turns out to be unique); we decree that two such algebras $A,A'$ are equivalent if both are isomorphic to a matrix ring over the same division ring $D$.  The set of equivalence classes $[A]$ of central simple algebras $A$ over $F$ is then called the {\sl Brauer group} Br$(F)$ of $F$.
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Multiplication in this group is defined as follows.  Given two central simple algebras $A,B$ over the same $F$, the tensor product $C=A\otimes_F B$ acquires a multiplication via the recipe $(a\otimes b)(a'\otimes b') = aa'\otimes bb'$ for $a,a',\in A,b,b'\in B$.  One checks immediately that the recipe is compatible with the defining relations of the tensor product; it is also not difficult to see that $C$ is central simple over $F$.  This multiplication is well defined on equivalence classes.  It turns out that the product $[B_f][B_g]$ of the classes of two crossed product algebras relative to the same $F$ and $K$ is just $[B_{fg}]$, the class corresponding to the product of the factor sets $f,$ and $g$.
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Of course $[F]$ is the identity element of Br$(F)$.  If $[A]$ lies in the Brauer group its inverse is the class $[A']$ of $A'=A^{\text{opp}}$, where $A'$ is defined to be $A$ as an additive group but with reversed multiplication, so that $ab$ in $A'$ equals $ba$ in $A$.  Then $A\otimes_F A'\cong M_n(F)$, where $n=\dim_F A$, so that $[A']$ is indeed the inverse of $[A]$.   The crossed product $B_1$ corresponding to the trivial factor set 1 is then isomorphic to a matrix ring over $F$, so that addition in $H^2(G,K^*)$ is compatible with multiplication in the Brauer group. 
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We conclude with a couple of examples.  The Brauer group of $\mathbb C$, or any algebraically closed field, is trivial, since $\mathbb C$ does not admit any proper finite extension that is a division ring.  The Brauer group of $\mathbb R$ has just two elements, namely the classes $[\mathbb R]$ and $[\mathbb H]$.  The Brauer group of a finite field $F$ is trivial, since there was a homework problem last quarter showing that a finite division ring is commutative, so that the only central simple algebras over $F$ are matrix rings over $F$.  The Brauer group of $\mathbb Q$ is huge, incorporating as it does the cohomology groups $H^2(G,K^*)$ for every finite Galois group $G$ of an extension $K$ of $\mathbb Q$ (which is conjecturally every finite group). 
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