The main reference for this series will be the Class Field Theory notes of Milne and the book "Algebra II" by Falko Lorenz. Brauer groups over fields are important objects in number theory, and they have applications in class field theory (e.g. the local reciprocity law by Artin), though they were originally defined in the context of semi-simple algebras over fields. The definition of Brauer groups is deserved to be discussed more than just a brief recall, but I don't have time to type down everything. Readers may refer to the second reference for proofs of results in the first section of this note. And in the second section, we will give a cohomological description for Brauer groups. It leads naturally to discuss about group and Galois cohomology later on.
Notation. In this series, we always fix a field $K$, and an algebra $A$ over $K$ is not assumed to be commutative. We also denote $A^o$ the $K$-algebra with the same group structure as $(A,+)$, but the multiplication in $A^o$ is defined to be $a*b:=b.a$, where $.$ is the multiplication in $A$. $D$ is always denoted a division $K$-algebra.
1. Brief recall about Brauer groups.
We recall that $A$ is simple $K$-algebra if $A$ has no proper two-sided ideal. $A$ is said to be a semi-simple $K$-algebra if $A$ is semi-simple left $A$-module. $A$ is said to be an artinian $K$-algebra if $A$ is an artinian left $A$-module. $A$ is said to be central over $K$ if $Z(A)=K$. $A$ is said to be central-simple $K$-algebra if $A$ is central, $A:K<+\infty$ and $A$ is simple as $K$-algebra. By the theorem of Wedderburn, any semi-simple $K$-algebra is a product of simple artinian $K$-algebras, and each simple artinian $K$-algebra is isomorphic to a matrix ring over a division $K$-algebra $M_n(D)$.
On the set of all simple artinian $K$-algebras, we define a relation $\sim$, $A\sim B$ (say: A is similar to $B$) if there exists a division $K$-algebra $D$ such that $A\cong M_r(D), B\cong M_s(D)$ for some non-negative integers $r,s$. If we restrict this definition to the set of all central-simple $K$-algebras, we obtain the definition of the Brauer group over $K$, i.e. $Br K$ is the group of similarity classes of central-simple $K$-algebras, with the group operation is the tensor product over $K$. The most easiest for Brauer group is that when $K$ is an algebraically closed field, we have $Br K$ is trivial, since any finite dimensional division $K$-algebra is $K$ itself.
There are two important theorems that we always have keep in mind about Brauer groups.
Theorem 1.1 (Centralizer Theorem). Let $A$ be a simple, central, artinian $K$-algebra, $B$ a simple subalgebra of finite dimension over $K$, $C:=Z_A(B)$ where $Z_A(B)$ is the centralizer of $A$ in $B$. Then
(i) $C$ is a simple artinian $K$-algebra.
(ii) $C\sim B^0\otimes A$, and $Z(B)=Z(C)$.
(iii) $C:K<+\infty$ iff $A:K<+\infty$, and in this case $A:K=(B:K)(C:K)$.
(iv) If $Z_A(C)=Z_A(Z_A(B))=B$.
(v) If $L$ is a center of $B$, then $A\otimes_K L\sim B\otimes_L C$.
(vi) If $Z(B)=K$, then $A\cong B\otimes_K C$.
Theorem 1.2 (Skolem-Noether). Let $A$ be a simple, central, artinian $K$-algebra, and $B$ a simple algebra of finite dimension over $K$. Let $f, g$ be two $K$-algebra homomorphism from $B\to A$, then there exists a unit $u\in A^\times$, such that $f(b)=u^{-1}g(b)u$, for all $b\in B$.
We also recall that if $L/K$ is a field extension, then there exists a restriction map $res_{L/K}: Br K\to Br L$ which sends $A\mapsto A\otimes_K L$. This map is a homomorphism of groups, and the kernel of this map is denoted $Br(L/K)$, i.e.
$$Br(L/K):=\{[A]\in Br K| A\otimes_K L\cong M_n(L)\text{ for some integer n}\}$$
For $A\in Br(L/K)$, we say that $A$ splits over $L$. The easiest example for Brauer group above (over algebraically closed field) implies that a central-simple $K$-algebra $A$ always splits in an algebraic closure $\overline{K}$ of $K$. And hence $A_{\overline{K}}:=A\otimes_K \overline{K}\cong M_n(\overline{K})$, and hence the dimension $A:K=A_{\overline{K}}:\overline{K}$ is always a square.
And using Theorem 1.1, one can deduce that
Corollary 1.3. Let $A$ be simple, central, artinian $K$-algebra, and $L$ a subalgebra of $A$ of finite dimension over $K$, then the following are equivalent:
(i) $L$ is a maximal commutative subalgebra of $A$, i.e. $Z_A(L)=L$.
(ii) $A:K=(L:K)^2$
If the one (and hence, two) condition(s) are satisfied, then $A$ splits over $L$. When $D:=A$ is a division $K$-algebra, satisfying the both conditions, then maximal commutative subalgebras of $D$ are exactly maximal subfields of $D$ containing $K$.
Another important statement is that for any $A\in Br K$, there always exists $L/K$: finite, Galois such that $A$ splits over $L$. And hence, we obtain
$$Br K=\bigcup_{L/K:\text{fin. Gal.}}Br(L/K)$$
And our goal in the next section is to give a cohomological description for $Br(L/K)$.
2. Cohomological description of Br(L/K).
In this section, we always fix $L/K$ a finite Galois extension, and $A$ a central-simple $K$-algebra, which splits over $L$, $G:=Gal(L/K)$. By the definition, there exists an $L$-algebra isomorphism
$$h: A\otimes_K L\to End_L(V)$$
where $V$ is a $n$-dimensional $L$-vector space. Let $\sigma\in G$, we can extend the action of $\sigma$ to $A\otimes_K L$ as follows
$$(a\otimes \lambda)^\sigma:=a\otimes \lambda^\sigma$$
We also denote $\rho_\sigma: End_L(V)\to End_L(V)$, where $x^{\rho_\sigma}:=x^{h^{-1}\sigma h}$. It can be seen that $\rho_\sigma\rho_\tau=\rho_{\sigma\tau}$, for all $\sigma,\tau\in G$. Since $h$ is an $L$-algebra isomorphism, we have for all $\lambda\in L$, $\lambda^{\rho_\sigma}=\lambda^\sigma$.
One can also consider $End_L(V)$ as a $K$-subalgebra of $End_K(V)$, and since $\sigma$ fixes $K$ for any $\sigma\in G$, we have $\rho_\sigma$ is a $K$-algebra homomorphism. This yields by the Skolem-Noether's theorem that there exists $u_\sigma\in End_K(V)^\times$ such that
$$ x^{\rho_\sigma}=u_\sigma^{-1}xu_\sigma, \forall x\in End_L(V) \text{ (1)}$$
And for all $\lambda\in L$, we have $\lambda^{\rho_\sigma}=\lambda_\sigma=u_\sigma^{-1}\lambda u_\sigma$, which is equivalent to say
$$\lambda u_\sigma=u_\sigma \lambda^\sigma \text{ (2)}$$
We have from (1) that for all $\sigma,\tau\in G$, and $x\in End_L(V)$
$$x^{\rho_{\sigma\tau}}=u_{\sigma\tau}^{-1}xu_{\sigma\tau}$$
And
$$x^{\rho_\sigma\rho_\tau}=(u_\sigma u_\tau)^{-1} x u_\sigma u_\tau$$
If one writes $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$, for some $c_{\sigma,\tau}\in End_K(V)^\times$, then $c_{\sigma,\tau}\in Z_{End_K(V)}(End_L(V))$. But then, $Z_{End_K(V)}(End_L(V))\supseteq Z(End_L(V))=L$. Assume that $L:K=m$, then $\dim_K End_K(V)=m^2n^2$, $\dim_K (End_L(V))=mn^2$, and hence, the by the centralizer theorem above, $\dim_K Z_{End_K(V)}(End_L(V))=\dim_K L= m$. And hence, $c_{\sigma,\tau}\in L^\times$.
From (1), we can see that $u_1\in L^\times$, by the similar argument as above. Our first goal is to prove that the similarity class of $A$ depends only on $c_{\sigma,\tau}$, but $c_{\sigma,\tau}$ is not uniquely determined by $A$.
Theorem 2.1. Let $L, K, G$ be as above, and $\Gamma$ a $K$-algebra, containing $L$ as a subalgebra. Assume further that $\Gamma$ is generated by $\lambda\in L$, and $u_\sigma$, for $\sigma\in G$ such that $u_1\in L^\times$, and
(i) $\lambda u_\sigma=u_\sigma\lambda^\sigma$.
(ii) $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$
for $c_{\sigma,\tau}\in L^\times$, then $(u_\sigma)_{\sigma\in G}$ forms a basis of $\Gamma$ as $L$-vector space, and $\Gamma:K=(L:K)^2$. Furthermore, $\Gamma$ is a central-simple $K$-algebra, with $L$ is its maximal commutative subalgebra, and hence, $\Gamma\in Br(L/K)$. And the $c_{\sigma,\tau}$ satisfy
$$c_{\sigma,\tau}^\rho c_{\sigma\tau,\rho}=c_{\sigma,\tau\rho}c_{\tau,\rho} \text{ (3)}$$
Proof. First, we have $u_\sigma u_{\sigma^{-1}}=u_1c_{\sigma,\sigma^{-1}}\in L^\times$, and this yields $u_\sigma$ is invertible, for all $\sigma\in G$. For any $\lambda\in L$, consider the left multiplication map by $\lambda$, $\lambda_l: \Gamma\to \Gamma$ defined by $\lambda_l(x)=\lambda x$. Then for any $\gamma\in L$, we have
$$\lambda_l(x\lambda)=\lambda x \gamma=\lambda_l(x)\gamma$$
And so, we can consider $\lambda_l$ as an endomorphism of $L$-right vector space. And the fact that $\lambda u_\sigma=u_\sigma\lambda^{\sigma}$ implies that $u_\sigma$ is an eigenvector and $\lambda^\sigma$ is an eigenvalue of $\lambda_l$. By the primitive element theorem, there exists $\lambda\in L$, such that $(\lambda^\sigma)_{\sigma\in L}$ forms a basis of $L/K$, and this yields $u_\sigma$ are linearly independent over $L$, since they are eigenvectors of distinct eigenvalues. On the other hand, since $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$, which means that $\Gamma:L\le \# G=L:K$. And this implies $\Gamma:K=(L:K)^2$.
To prove that $\Gamma$ is simple, it is sufficient to prove that any surjective ring homomorphism $\phi$ from $\Gamma$ to $\Gamma'$ where $\Gamma'\ne 0$ has the zero kernel. First, we can equip $\Gamma'$ the structure of $L$-vector space via the map $\phi$ as follows $\phi(x)\lambda=\phi(x)\phi(\lambda)$, for all $x\in \Gamma,\lambda\in L$. And the map $\phi$ can be now considered as a map between $L$-algebras. And due to the map $\phi$, we also have $(\phi(u_\sigma))_\sigma$ satisfying the same conditions (i) and (ii). And this yields by our previous argument that $\Gamma':K=\Gamma:K=(L:K)^2$. Hence $\phi$ is an isomorphism, and $\ker \phi=0$.
We will now prove that $Z(\Gamma)=K$. In the case $L\equiv K$ then there is nothing to prove. Otherwise, for any $z\in Z(\Gamma)$, we can represent $z=\sum_{\sigma\in G}u_\sigma\lambda_\sigma$ with $\lambda_\sigma\in L$. Then for any $\lambda\in L$, we have $z\lambda = \lambda z$, where
$$z\lambda = \sum_{\sigma\in G}u_\sigma\lambda_\sigma \lambda \text{ and } \lambda z=\sum_{\sigma\in G}\lambda u_\sigma \lambda_\sigma=\sum_{\sigma\in G}u_\sigma \lambda^{\sigma} \lambda_\sigma$$
And by comparing coefficients, we have $\lambda_\sigma\lambda=\lambda^{\sigma}\lambda_\sigma$ for all $\sigma\in G, \lambda\in L$. Taking $\sigma\ne 1$, if $\lambda_\sigma\ne 0$, we obtain $\lambda=\lambda^\sigma$, for all $\lambda\in L$, which yields $L=K$, a contradiction. Hence for all $\sigma\ne 1$, $\lambda_\sigma= 0$, and this yields $z=u_1\lambda_1\in L$. But then,
$$u_\sigma z=zu_\sigma = u_\sigma z_\sigma$$
Since $u_\sigma$ is invertible, we obtain $z=z^{\sigma}$, and this yields $z\in K$. Hence, $Z(\Gamma)=K$. And this yields $\Gamma$ is central-simple over $K$, and $\Gamma:K=(L:K)^2$, hence, $L$ is a maximal commutative subalgebra of $\Gamma$. Due to Corollary 1.3, we have $\Gamma$ splits over $L$, and hence $\Gamma\in Br(L/K)$.
The final statement follows easily from the comparison between $(u_\sigma u_\tau)u_{\rho}$ and $u_\sigma (u_\tau u_\rho)$. (Q.E.D)
Example. Let $L,K,G$ be as above. Consider the $K$-algebra $End_K(L)$, we note that $G\subset End_K(L)$, and for all $\lambda,\gamma\in L$, we have
$$\sigma(\lambda_l(\gamma))=(\lambda\gamma)^\sigma=\lambda^\sigma \gamma^\sigma=(\lambda^\sigma)_l(\sigma(\gamma))$$
So, in the algebra $\Gamma:=End_K(L)^o$, we have $\lambda*\sigma =\sigma*\lambda^\sigma$. If we denote $u_\sigma:=\sigma$ in $\Gamma$, then $u_1=1$, and $u_\sigma u_\tau=u_{\sigma\tau}$, and this yields $c_{\sigma,\tau}=1$, for all $\sigma,\tau\in G$.
We can now come back to our computation before the statement of Theorem 1. Let $\Gamma$ be a $K$-subalgebra of $End_L(V)$ generated by $L$ and $u_\sigma (\sigma\in G)$, then by Theorem 1, $\Gamma$ is central-simple. We will next prove that $\Gamma^o\sim A$.
Lemma 2.2. $\Gamma\subseteq Z_{End_K(V)}(A^h)$.
Proof. Taking any $x\in A$, then $x^\sigma=x$ for all $\sigma\in G$, hence $h(x)^{\rho_\sigma}=h(x)$. And this implies that $h(x)$ is commutative with $u_\sigma$ for all $\sigma\in G$ due to (1). The commutativity between $h(x)$ and $\lambda\in L$ is clear. So, $\Gamma\subseteq Z_{End_K(V)}(A^h)$. (Q.E.D)
To prove that the equality happen in Lemma 2, we need the following theorem
Theorem 2.3. Let $A$ be a central-simple $K$-algebra, and $L/K$ is a finite extension of fields such that $A$ splits over $L$, then there exists a central-simple $K$-algebra $A'$, such that $A'\sim A$, and $L$ is a maximal commutative subalgebra of $A'$.
Proof. Let $n^2:=A:K$, then there exists an isomorphism of $L$-algebras
$$h: A\otimes_K L\to End_L(V)$$
where $V\cong L^n$. We denote $B:=A^h$ the image of $A$, and $C:=Z_{End_K(V)}(B)$. By the centralizer theorem, we have $Z(C)=Z(B)=K$, and hence, $C$ is a central-simple $K$-algebra, and $C$ contains $L$ since it is obvious that $L\subseteq Z_{End_K(V)}(B)$. Using the theorem again yields
$$End_K(V):K=(C:K)(B:K)$$
And this implies that $C:K=(L:K)^2$. Hence, $L$ is a maximal commutative subalgebra of $C$. And due to the centralizer theorem again, we have $C\sim A^o\otimes_K End_K(V)\sim A^o$. Let $A'=C^o$, then $A'\sim A$. (Q.E.D)
From the theorem above, it follows that $\Gamma=Z_{End_K(V)}(A^h)$, and hence $\Gamma^0\sim A$, as in the proof of Theorem 2.3. This follows that the similarity class of $A$ in $Br(L/K)$ is completely determined by $c_{\sigma,\tau}$.
We will now reason why $[A]$ does not determine $c_{\sigma,\tau}$ exactly. If we change $u_\sigma$ by $u'_\sigma:=u_\sigma a_\sigma$, for some $a_\sigma\in L^\times$, then the shape of $\Gamma$ does not change, since we will obtain another basis $(u'_\sigma)$ for $\Gamma$. But now, by simple computation, we have
$$c'_{\sigma,\tau}=c_{\sigma,\tau}a^\tau_\sigma a_\tau a_{\sigma\tau}^{-1}$$
And this leads us to the following
Definition. Let $L, K, G$ be as above, a tuple $(c_{\sigma,\tau})_{\sigma,\tau\in G}$ where $c_{\sigma,\tau}\in L^\times$ is said to be a cocycle if $c_{\sigma,\tau}$ satisfy the condition of (3) in Theorem 2.1. $(c_{\sigma,\tau})$ is said to be a coboundary if $d_{\sigma,\tau}=a^\tau_\sigma a_\tau a_{\sigma\tau}^{-1}$ for some $a_\sigma\in L^\times,\sigma\in G$.
It can be seen that the set of cocycles forms an abelian group, and the set of coboundaries forms a subgroup of the former group. We denote $H^2(G, L^\times)$ the quotient of the two groups, and it is called the second cohomology group of $G$ with coefficients in $L^\times$.
And the remaining of this note is devoted for the proof of the fact that there exists an isomorphism $Br(L/K) \cong H^2(G,L^\times)$. We first begin with the map
$$f: \{\text {central-simple K algebras splitting over L}\}\to H^2(G,L^\times)$$
as in the beginning of the second section. We note that it is well-defined, and does not depend on the choice of an isomorphism $h$, due to the Stokem-Noether's theorem. And at the first step, we will prove that this depend only on the choice of the similarity class of $A$.
Lemma 2.4. The map $f$ is multiplicative, i.e. $f(A\otimes_K A')=f(A)f(A')$.
Proof. Let us denote $h: A\otimes_K L\to End_L(V)$, and $h': A\otimes_K L\to End_L(V')$ two $L$-algebra isomorphisms. We can introduce the map
$$(h,h'): (A\otimes_K L)\otimes (A'\otimes_K L)\to End_L(V)\otimes_K End_L(V')$$
Let us denote $(u_\sigma, c_{\sigma,\tau}),(u'_\sigma, c'_{\sigma,\tau})(U_\sigma, C_{\sigma,\tau})$ the pairs for $(A\otimes_K A')\otimes_K L$ as in the beginning of the section, then it can be seen that $U_\sigma=u_\sigma\otimes u'_\sigma$. And this follows easily that $C_{\sigma,\tau}=c_{\sigma,\tau}c'_{\sigma,\tau}$. (Q.E.D)
Lemma 2.5.
(i) If $A\sim K$, then $f(A)=1$.
(ii) $f(A^o)=f(A)^{-1}$.
(iii) If $[A]=[B]$ in $Br(L/K)$, then $f(A)=f(B)$.
Proof.
(i) When $A\sim K$ we have $A\cong M_n(K)$, and $A\otimes_K L\cong M_n(K)\otimes_K L\cong M_n(L)$, and in this case, the map $h$ can be seen as the identification maps $1\otimes e_i\mapsto e_i$, where $(e_i)$ is a basis of $L$ over $K$. And one obtains $h\sigma=\sigma h$, for all $\sigma\in G$. And this yields $x^{\rho_\sigma}=x^\sigma$, for all $\sigma\in G$.
And for $x=(x_{ij})\in M_n(L)$, the action of $\sigma\in G$ is defined to be $x^\sigma:=(x^\sigma_{ij})$. And due to the construction of $u_\sigma$, we need to find $u_\sigma\in End_K(L^n)^\times$, such that $xu_\sigma=u_\sigma x^\sigma$. Define
$$u_\sigma((y_i)):=(y_i^{\sigma^{-1}})_i$$
Then it can be seen that $u_\sigma\in End_K(L^n)^\times$, and it is easy to check that $xu_\sigma=u_\sigma x^\sigma$, for all $x\in M_n(L)$. Also, it can be checked easily that $u_\sigma u_\tau=u_{\sigma\tau}$. Hence $c_{\sigma,\tau}=1$, for all $\sigma,\tau \in G$.
(ii) This follows easily from Lemma 2.4 and (i).
(iii) If $A\sim B$, then $A\otimes_K A^o \sim B\otimes_K A^o\sim K$, and $f(B\otimes_K A^o)=1=f(B)f(A)^{-1}$. Hence $f(A)=f(B)$. (Q.E.D)
Hence, one can see that the map $f$ depends only on the similarity class of $A$, and one obtains, by abusing notations, a map
$$f: Br(L/K)\to H^2(G,L^\times)$$
By Lemma 2.4, it is a homomorphism of groups. And it is surjective, due to Theorem 2.1. To prove that $f$ is injective, we can assume that $[A]$ is mapped to $[c]$, where $c_{\sigma,\tau}=1$, for all $\sigma,\tau\in G$. This yields, by our example below Theorem 2.1 that in fact, $A\sim End_K(L)^0\sim K$. And hence, the map $f$ is an isomorphism of groups. In the next part, we will discuss on some number theoretic applications of this observation, for the simplest, but fundamental importance case, which is cyclic extension.
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