4 Brauer Group
4.1 Construction of Brauer Group
Let \(K\) be a field. We denote the class of finite dimensional central simple \(K\)-algebras as \(\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\). When \(K\) is clear, we drop the subscript.
By lemma 1.1.9 and theorem 1.1.10, \(\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\) is closed under tensor product, that is if \(A, B\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\), we have \(A\otimes _{K} B\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\) as well.
For any two \(A, B\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\), we say \(A\) and \(B\) are Brauer equivalent, when there exists \(m, n \in \mathbb {N}_{\ge 0}\) such that \(\operatorname{Mat}_{m}(A)\cong \operatorname{Mat}_{n}B\) as \(K\)-algebras. We denote this relation as \(A\sim _{\operatorname{Br}_{K}} B\), when \(K\) is clear, we drop the subscript.
Isomorphic \(K\)-algebras are Brauer equivalent.
\(\sim _{\operatorname{Br}}\) is reflexive.
Indeed, \(A \cong \operatorname{Mat}_{1}(A)\) as \(K\)-algerbas.
\(\sim _{\operatorname{Br}}\) is symmetric.
Indeed, just exchange \(m\) and \(n\).
\(\sim _{\operatorname{Br}}\) is transitive.
Let \(A\sim _{\operatorname{Br}}B\) and \(B\sim _{\operatorname{Br}}C\); that is for some \(m,n,p, q\in \mathbb {N}_{\ge 0}\) we have \(\operatorname{Mat}_{n}(A)\cong \operatorname{Mat}_{m}(B)\) and \(\operatorname{Mat}_{p}(B)\cong \operatorname{Mat}_{q}(C)\) as \(K\)-algebras. Hence, from construction 3.1.3, we have the following:
In another word, \(A\sim _{\operatorname{Br}}C\).
Hence \(\sim _{\operatorname{Br}}\) is really an equivalence relation, we denote the quotient \({}^{\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}}/_{\sim _{\operatorname{Br}}}\) as \(\operatorname{Br}(K)\).
\((\bullet \otimes _{K}\bullet ):\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\times \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\to \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\) descends to a function on \(\operatorname{Br}(K)\).
We need to prove that for all \(A, B, C, D \in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\) such that \(A\sim _{\operatorname{Br}} B\) and \(C\sim _{\operatorname{Br}}D\), \(A\otimes _{R}C \sim _{\operatorname{Br}} B\otimes _{R}D\) as well. Suppose \(\operatorname{Mat}_{m}(A)\cong \operatorname{Mat}_{n}(B)\) as \(K\)-algebras and \(\operatorname{Mat}_{p}(C)\cong \operatorname{Mat}_{q}(D)\), by construction 3.1.4, we have
\(\operatorname{Br}(K)\) forms a group under \([A]_{\sim _{\operatorname{Br}}}\cdot [B]_{\sim _{\operatorname{Br}}}=[A\otimes _{K}B]_{\sim _{\operatorname{Br}}}\) with neutral element \([K]_{\sim _{\operatorname{Br}}}\) where \(A, B\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\) and \([A]_{\sim _{Br}}^{-1}=[A^{\mathsf{opp}}]_{\sim _{\operatorname{Br}}}\). We need to prove the following properties:
associativity: for all \(A, B, C\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\), \([A]_{\sim _{\operatorname{Br}}}\cdot \left([B]_{\sim _{\operatorname{Br}}}\cdot [C]_{\sim _{\operatorname{Br}}}\right)=\left([A]_{\sim _{\operatorname{Br}}}\cdot [B]_{\sim _{\operatorname{Br}}}\right)\cdot [C]_{\sim _{\operatorname{Br}}}\) because \(A\otimes _{R}\left(B\otimes _{R}C\right)\cong \left(A\otimes _{R}B\right)\otimes _{R}C\) as \(K\)-algebras.
neutral element: for all \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\), \([K]_{\sim _{\operatorname{Br}}}\cdot [A]_{\sim _{\operatorname{Br}}}=[A]_{\sim _{\operatorname{Br}}}=[A]_{\sim _{\operatorname{Br}}}\cdot [K]_{\sim _{\operatorname{Br}}}\). Since \([K]_{\sim _{\operatorname{Br}}}\cdot [A]_{\sim _{\operatorname{Br}}}=[K\otimes _{K} A]_{\sim _{\operatorname{Br}}}\), in construction 3.1.4, we see that \(\operatorname{Mat}_{n}(A)\cong A\otimes _{K}\operatorname{Mat}_{n}(K)\), by lemma 4.1.6, \(A\otimes _{K}\operatorname{Mat}_{n}(K)\) is Brauer equivalent to \(A\otimes _{K} K\) since \(K\sim _{\operatorname{Br}}\operatorname{Mat}_{n}(K)\).
cancellation: for all \(A \in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\), we need \([A]_{\sim _{\operatorname{Br}}}\cdot [A^{\mathsf{opp}}]_{\sim _{\operatorname{Br}}}\), that is we want \(A\otimes _{K}A^{\mathsf{opp}}\sim _{\operatorname{Br}} K\). By construction 3.1.1, we have \(A\otimes _{K}A^{\mathsf{opp}}\cong \operatorname{End}_{K}A\) which is isomorphic to \(\operatorname{Mat}_{\dim _{K}A}(K)\) as \(K\)-algebras.
If \(K\) is algebraically closed, \(\operatorname{Br}(K)\) is trivial; in particular \(\operatorname{Br}_{n}(\mathbb {C})\) is trivial.
We need to show that every \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\) is isomorphic to \(\operatorname{Mat}_{n}(K)\) for some \(K\) when \(K\) is algebraically closed. Indeed, by theorem 3.2.6, \(A \cong \operatorname{Mat}_{n}(D)\) for some division algebra \(D\) and \(n\in \mathbb {N}_{\ge 0}\). Since \(K\) is algebraically closed and \(D\) is an integral domain and finite dimensional, the structure morphism \(\rho : K\to D\) is a isomorphism; therefore \(A\cong \operatorname{Mat}_{n}(K)\).
Let \(A, B \in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\). There exists a division \(K\)-algebra \(D\) and non-zero \(m,n\in \mathbb {N}\) such that \(A\cong \operatorname{Mat}_{m}(D)\) and \(B\cong \operatorname{Mat}_{n}(D)\) as \(K\)-algebras.
By theorem 3.2.6, we can find division algebras \(S_{A}, S_{B}\) and non-zero \(m, n\in \mathbb {N}\) such that \(A\cong \operatorname{Mat}_{n}\left(S_{A}\right)\) and \(B\cong \operatorname{Mat}_{m}\left(S_{B}\right)\) as \(K\)-algebras. Hence \(B\sim _{\operatorname{Br}}A\sim _{\operatorname{Br}}\operatorname{Mat}_{n}\left(S_{A}\right)\sim _{\operatorname{Br}}S_{A}\), in another word, for some non-zero \(a, a'\in \mathbb {N}\), we have \(\operatorname{Mat}_{a}(B)\cong \operatorname{Mat}_{a'}\left(S_{A}\right)\) as \(K\)-algebras. Hence, by theorem 3.2.7, we have that \(S_{A}\cong S_{B}\) as \(K\)-algebras and the lemma is proved.
4.2 Base Change
In this section let \(E/K\) be a field extension. We have seen in corollary 1.1.11 that if \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\) then \(E\otimes _{K}A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{E}\); therefore we have a set-theoretic function \(\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\to \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{E}\). In this section we prove that this descends to a group homomorphism \(\operatorname{Br}(K)\to \operatorname{Br}(E)\). For brevity, if \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\), we denote \(E\otimes _{K}A\) as \(A_{E}\) when this causes no confusion.
We will construct a series of isomorphisms (either over \(K\) or \(E\)) to arrive at the conclusion that \(A\sim _{\operatorname{Br}_{K}}B\) implies \(A_{E}\sim _{\operatorname{Br}_{E}}B_{E}\). Assume \(m,n\in \mathbb {N}_{\ge 0}\) are such that \(\operatorname{Mat}_{m}(A)\cong \operatorname{Mat}_{n}(B)\) are \(K\)-algebras. Then we do the following calculation: as \(E\)-algebras
\(\dagger \): Wee need to check \(\operatorname{Mat}_{m}(E)\cong E\otimes _{K}\operatorname{Mat}_{m}K\) as \(E\)-algebras since construction 3.1.4 only gives a \(K\)-algebra isomorphism. If \(e \in E\), then its image in \(E\otimes _{K}\operatorname{Mat}_{m}(K)\) is \(e\otimes 1\) and its image in \(\operatorname{Mat}_{m}(E)\) is \(\operatorname{diag}(e)\) which under the \(K\)-algebra isomorphism is mapped to \(\sum _{ij}\operatorname{diag}(e)_{ij}\cdot \delta _{ij}=e\otimes 1\).
\(\ddagger \): This is defined by combining two \(E\)-algebra homomorphisms
and
Since \(\left(E\otimes _{K}A\right)\otimes _{E}\left(E\otimes _{K}\operatorname{Mat}_{m}(K)\right)\) is a simple ring, this morphism is automatically injective. It is surjective as well: let \(x\in E\otimes _{K}\left(A\otimes _{K}\operatorname{Mat}_{m}(K)\right)\), without loss of generality, assume \(x=e\otimes (a\otimes \delta _{ij})\) for some \(e\in E\), \(a\in A\). Then precisely \(\left(e\otimes a\right)\otimes \left(1\otimes \delta _{ij}\right)\) is mapped to \(x\).
\(\dagger \! \dagger \): a \(K\)-algebra isomorphism \(A\cong B\) gives an \(E\)-algebra isomorphism \(E\otimes _{K}A \cong E\otimes _{K} B\).
Thus we have a well defined function \(\operatorname{Br}(K)\to \operatorname{Br}(E)\). We now check that this is a group homomorphism. \([K]_{\sim _{\operatorname{Br}_{K}}}\) is mapped to \([E\otimes _{K}K]_{\sim _{\operatorname{Br}_{E}}}\) but \(E\otimes _{K}K\cong E\) as \(E\)-algebra. For \(A, B\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\), we have that \([AB]_{\sim _{\operatorname{Br}_{K}}}\) is mapped to \(\left(A\otimes _{K}B\right)_{E}\cong A_{E}\otimes _{E} B_{E}\) as \(E\)-algebras; hence \([AB]_{\sim _{\operatorname{Br}_{K}}}\) and \([A]_{\sim _{\operatorname{Br}_{K}}}\cdot [B]_{\sim _{\operatorname{Br}_{K}}}\) have the same image under base change.
Denote the base change morphism in construction 4.2.1 as \(\operatorname{Br}_{K}^{E}\).
\(\operatorname{Br}_{K}^{K}\) is identity.
If \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\), then \(A\sim _{\operatorname{Br}}K\otimes _{K}A\).
Consider the tower of field extension \(E/F/K\),
If \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\), then \(E\otimes _{F}\left(F\otimes _{K}A\right)\) is isomorphic to \(E \otimes _{F} A\) as \(E\)-algebras.
\(\operatorname{Br}\) forms a functor from category of field to category of abelian groups.
This is the categorical version of lemma 4.2.1 and lemma 4.2.2.
Let \(E/K\) be a field extension, we define the relative Brauer group \(\operatorname{Br}(E/K)\) to be the kernel of the base change morphism \(\operatorname{Br}_{K}^{E}\).
Unpacking the definition of the relative Brauer group, we see that for any \(A \in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\), if \(E\otimes _{K}A\cong \operatorname{Mat}_{n}(E)\) as \(E\)-algebras, then \(\operatorname{Br}^{E}_{K}\left([A]_{\sim _{\operatorname{Br}}}\right)=1\).
For any field extension \(E/K\) and any \(K\)-algebra \(A\), we say \(E\) is a splitting field of \(A\) if and only if \(E\otimes _{K}A \cong \operatorname{Mat}_{n}(E)\) as \(E\)-algebras for some non-zero \(n\). We also say \(E\) splits \(A\) or \(A\) is splited by \(E\)
Let \(E/K\) be a field extension and \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\), \(E\) splits \(A\) if and only if \([A]_{\sim _{\operatorname{Br}}}\in \operatorname{Br}(E/K)\).
The “only if” part is by definition. For the other direction, we know by definition that \(\operatorname{Mat}_{n}(E\otimes _{K}A)\cong \operatorname{Mat}_{m}(E)\) as \(E\)-algebras for some non-zero \(m, n\). By theorem 3.2.6, we find some division algebra \(D\) and non-zero natural number \(p\) such that \(E\otimes _{K}A\cong \operatorname{Mat}_{p}(D)\) as \(E\)-algebras. Thus \(\operatorname{Mat}_{pm}(E)\cong \operatorname{Mat}_{pn}\left(E\otimes _{K}A\right)\cong \operatorname{Mat}_{p^{2}n}(D)\) as \(E\)-algebras. By theorem 3.2.7, we conclude that \(E\cong D\) as \(E\)-algebras. Hence \(E\otimes _{K}A\cong \operatorname{Mat}_{p}(E)\), in another word, \(E\) splits \(A\).
In light of lemma 4.2.2, if \(K\) is algebraic closed then \(K\) splits any \(K\)-algebra \(A\). Indeed, \(K\) splits \(A\) if and only if \([A]_{\sim \operatorname{Br}}\) but \([A]_{\sim \operatorname{Br}}\) is equal to \(1\).
If two \(\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\) are Brauer equivalent, in another word, \(A \sim _{\operatorname{Br}_{K}} B\), then \(E\) splits \(A\) if and only if \(E\) splits \(B\). Indeed, if \(A\) and \(B\) are equivalent, then \([A]_{\sim \operatorname{Br}} \in \operatorname{Br}(E/K)\) if and only if \([B]_{\sim \operatorname{Br}}\in \operatorname{Br}(E/K)\).
4.3 Good Representative Lemma
In this section, let \(K/F\) be a finite dimensional field extension.
Let \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\) splitted by \(K\). There exists a \(B\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\) such that
\([A]_{\sim _{\operatorname{Br}}}[B]_{\sim _{\operatorname{Br}}}=1\)
there exists \(F\)-algebra map \(K\hookrightarrow B\)
\({\left(\dim _{F}K\right)}^{2}=\dim _{F}B\).
Since \(K\) splits \(A\), we find a non-zero natural number \(n\) such that \(K\otimes _{F}A\cong \operatorname{Mat}_{n}K\cong \operatorname{End}_{K}\left(K^{n}\right)\) as \(K\)-algebras. We define an \(F\)-algebra map \(\iota : A\to \operatorname{End}_{F}\left(K^{n}\right)\) by
where \(|_{F}\) is restriction of scalars. Since \(A\) is simple, \(\iota \) is injective, therefore \(A \cong \iota (A)\) as \(F\)-algebras. Define \(B\) as \(C_{\operatorname{End}_{F}\left(K^{n}\right)}(\iota (A))\), the centralizer of the range of \(\iota \) in \(\operatorname{End}_{F}\left(K^{n}\right)\). We construct an embedding \(K \hookrightarrow B\) by \(r \mapsto (r \cdot \bullet )\)
\(B\) is a central \(F\)-algebra: if \(x \in Z(B)\), then \(x \in \iota (A)\) because by theorem 3.4.13, it is sufficient to prove that \(x\) is in \(C_{\operatorname{End}_{F}\left(K^{n}\right)}\left(B\right)\) which follows from the fact that \(x \in Z(B)\). In fact, \(x \in Z(\iota (A))\): suppose \(a \in A\), we need to check \(x \cdot \iota (a) = \iota (a)\cdot x\), this is the case because \(B\) is defined as the centralizer of \(\iota (A)\). Since \(\iota (A) \cong A\) as \(F\)-algebras, \(\iota (A)\) is \(F\)-central, hence \(x \in F\).
\(B\) is a simple ring: by lemma 3.4.10, it is sufficient to prove that \(\iota (A)\) is a simple ring which comes from \(A \cong \iota (A)\) as \(F\)-algebras.
By corollary 3.4.12, we have \(F\)-algebra isomorphism \(\operatorname{End}_{F}\left(K^{n}\right)\cong \iota (A)\otimes _{F}B \cong A \otimes _{F}B\). Since \(\operatorname{End}_{F}\left(K^{n}\right)\cong \operatorname{Mat}_{\dim _{F}\left(K^{n}\right)}\left(F\right)\) as \(F\)-algebras, we see that \([A]_{\sim _{\operatorname{Br}}}\) and \([B]_{\sim _{\operatorname{Br}}}\) are inverses.
By lemma 3.4.11, \(\dim _{F}B\cdot \dim _{F}\iota (A) = \dim _{F}B\cdot \dim _{F}A = \dim _{F}\operatorname{End}_{F}\left(K^{n}\right) = {\left(\dim _{F}\left(K^{n}\right)\right)}^{2}={\left(\dim _{F}K\cdot \dim _{K}\left(K^{n}\right)\right)}^{2}=n^{2}\cdot {\left(\dim _{F}K\right)}^{2}\). On the other hand, since \(K\otimes _{F}A\cong \operatorname{Mat}_{n}K\), we have \(\dim _{F}K\otimes _{F}A=\dim _{F}K\cdot \dim _{F}A=\dim _{F}\operatorname{Mat}_{n}K=\dim _{F}K\dim _{K}\operatorname{Mat}_{n}K=n^{2}\dim _{F}K\). Since \(\dim _{F}K\ne 0\) , we conclude \(\dim _{F}A=n^{2}\). Since \(n\ne 0\) and \(\dim _{F}(B)\cdot \dim _{F}(A)=n^{2}\dim _{F}(B)=n^{2}{\left(\dim _{F}K\right)}^{2}\), we get the desired result.
Let \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\) splitted by \(K\). There exists a \(B \in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\) such that
\([B]_{\sim _{\operatorname{Br}}}=[A]_{\sim _{\operatorname{Br}}}\)
there exists an \(F\)-algebra map \(K\hookrightarrow B\)
\({\left(\dim _{F}K\right)}^{2}=\dim _{F}B\).
Let \(B\) and \(\iota : K \hookrightarrow B\) be as in lemma 4.3.1. Consider \(B^{\mathsf{opp}}\) and \(K \hookrightarrow B \to B^{\mathsf{opp}}\). This works.
Let \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\). \(K\) splits \(A\) if and only if there exists a \(B\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\) such that
\([B]_{\sim _{\operatorname{Br}}}=[A]_{\sim _{\operatorname{Br}}}\)
there exists an \(F\)-algebra map \(K \hookrightarrow B\)
\({\left(\dim _{F}K\right)}^{2}=\dim _{F}B\).
The “if” direction is corollary 4.3.2. For the “only if” direction, let \(B \in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\) and \(\iota : K \hookrightarrow B\) be given. We give \(B\) a \(K\)-module structure by right multiplication, that is for any \(a \in K\) and \(b \in B\), we define \(a \cdot b := b \cdot \iota (a)\). Since \(B\) is a finite dimensional \(F\)-vector space and \(K/F\) is a finite dimensional field extension, \(B\) is a finite dimensional \(K\)-vector space as well. Since \([B]_{\sim _{\operatorname{Br}}}=[A]_{\sim _{\operatorname{Br}}}\), it is sufficient to show that \(K\) splits \(B\). We define an \(F\)-bilinear map \(\mu : K \to B \to \operatorname{End}_{K}B\) by \((c, a) \mapsto (c \cdot a \cdot \bullet )\) which induce an \(F\)-linear map \(\mu ' : K \otimes _{F} B \to \operatorname{End}_{K}B\). Since for any \(r, c\in K\) and \(a \in B\), we have \(\mu '\left(r \cdot c\otimes a\right)(a') = a a' \iota (rc)=aa'\iota (c)\iota (r)=r\cdot \mu '(c\otimes a)\), that is \(\mu '\) is \(K\)-linear as well. Note that
and that
that is, \(\mu '\) is an \(K\)-algebra map.
If we can show that \(\mu '\) is a bijection, we will prove the result for \(K\otimes _{F} B \cong \operatorname{End}_{K} B \cong \operatorname{Mat}_{\dim _{K} B}K\) as \(K\)-algebras. By corollary 1.1.8, it is sufficient to show \(\dim _{K}K\otimes _{F}B = \dim _{K}\operatorname{End}_{K}B\). Let \(n\) denote \(\dim _{F}K\). Since, \(\dim _{F}K\dim _{K}K\otimes _{F}B =\dim _{F}K\otimes _{F}B=\dim _{F}K \dim _{F}B\). we have \(\dim _{K}K\otimes _{F}B = \dim _{F}B = {\left(\dim _{F}K\right)}^{2}\). On the other hand, since \({\left(\dim _{F}K\right)}^{2}=\dim _{F} B = \dim _{F}K\dim _{K}B\), we have \(\dim _{K} B = \dim _{F}K\); thus \(\dim _{K}\operatorname{End}_{K} B = {\left(\dim _{K}B\right)}^{2}={\left(\dim _{F}K\right)}^{2}\) and the result is proved.
In light of theorem 4.3.3, we isolate the following useful definition:
For any \(X \in \operatorname{Br}(F)\), a \(K\)-good representation of \(X\) is an \(A \in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\) and an \(F\)-algebra map \(K \hookrightarrow A\) such that \([A]_{\sim _{\operatorname{Br}}} = X\) and \(\dim _{F}A={\left(\dim _{F}K\right)}^{2}\). We often denote the \(F\)-algebra map \(K \hookrightarrow A\) as \(\iota \) or \(\iota _{A}\).
When \(K\) is clear from context, we will simply say good representation instead of \(K\)-good representation
For any \(X\in \operatorname{Br}(F)\), \(X\in \operatorname{Br}(K/F)\) if and only if \(X\) admits a good representation.
Rephrase of theorem 4.3.3 and theorem 4.2.5.
4.3.1 Basic Properties
We observe the following easy result about good representations. Let \(X \in \operatorname{Br}(F)\) and \(A\) be a good representation of \(X\).
The range \(\iota _{A}(A)\) is a simple ring.
Because \(K\) is a simple ring, \(\iota _{A}\) is injective therefore \(\iota _{A}(A)\cong K\).
\(C_{A}\left(\iota _{A}(A)\right) = \iota _{A}(A)\).
In the language of section 1.2, \(\iota _{A}(A)\) is a subfield of \(A\), hence by lemma 1.2.3, we only need to show \(\dim _{F}A = {\left(\dim _{F}\iota _{A}(A)\right)}^{2}\). But \(\dim _{F}A={\left(\dim _{F}K\right)}^{2}\) and \(\iota (A)\cong K\).
We give \(A\) a \(K\)-module structure by left multiplication, that is for any \(c \in K\) and \(a \in A\), we define \(c\cdot a\) to be \(\iota _{A}(c)a\). Note that if \(c \in F\) then \(\iota _{A}(c)a = c \cdot a\), in another word, the \(K\)-action and the \(F\)-action on \(A\) are compatible. Then \(A\) is a finite dimensional \(K\)-vector space and \(\dim _{K}A=\dim _{F}K\): indeed \(\dim _{F}K\cdot \dim _{K}A = \dim _{F}K\cdot \dim _{F}K = \dim _{F}A\).
If \(A\) and \(B\) are two good representations of \(X\), then \(A \cong B\) as \(F\)-algebras.
By lemma 4.1.8, we find a division \(F\)-algebra \(D\) and non-zero natural numbers \(m, n\) such that \(A \cong \operatorname{Mat}_{m}(D)\) and \(B \cong \operatorname{Mat}_{n}(D)\) as \(F\)-algebras. Therefore
Therefore \(m = n\) and \(A \cong \operatorname{Mat}_{m}D =\operatorname{Mat}_{n}D\cong B\).
4.3.2 Conjugation Factors and Conjugation Sequences
In this section, let \(K/F\) be a field extension, \(X \in \operatorname{Br}(F)\) and \(A\) be a \(K\)-good representation of \(X\).
Since \(\operatorname{Gal}(K/F)\) acts on \(K^{\star }\), for \(x\in K^{\star }\),we feel free to write \(\sigma \cdot x\) when it feels more readable than \(\sigma (x)\), for example when there are nested brackets.
With respect to \(A\), a conjugation factor of \(\sigma \) is a unit \(x_{\sigma } \in A^{\star }\) such that for all \(c \in K\),
A conjugation sequence is a sequence \(x:\operatorname{Gal}(K/F) \to A^{\star }\) such that for all \(\sigma \in \operatorname{Gal}(K/F)\), \(x_{\sigma }\) is a conjugation factor of \(\sigma \). When we want to stress \(A\), we say \(A\)-conjugation factor and \(A\)-conjugation sequence.
When \(x_{\sigma }\) is a conjugation factor of \(\sigma \), the equalities \(x_{\sigma }\iota _{A}(c) = x_{\sigma }\iota _{A}(\sigma (c))\) and \(\iota _{A}(c)x_{\sigma }^{-1}=x_{\sigma }^{-1}\iota _{A}(\sigma (c))\) are also useful.
\(A\) has a conjugation sequence: let \(\sigma \in \operatorname{Gal}(K/F)\), we have two \(F\)-algebra homomorphisms \(K \to A\) given by \(\iota _{A}\) and \(\iota _{A}\circ \sigma \). Applying theorem 3.3.4 to \(\iota _{A}\) and \(\iota _{A}\circ \sigma \) gives us the desired conjugation factor.
If \(x\) is a conjugation factor of \(\sigma \) and \(y\) of \(\tau \), then \(xy\) is a conjugation factor of \(\sigma \tau \). For any \(c \in K\)
If \(x\) is an \(A\)-conjugation sequence, then \(\{ x_{\sigma }|\sigma \in \operatorname{Gal}(K/F)\} \) is an \(K\)-linearly independent set. When \(K/F\) is finite dimensional and Galois, \(\{ x_{\sigma }|\sigma \in \operatorname{Gal}(K/F)\} \) is a \(K\)-basis for \(A\).
Suppose \(\{ x_{\sigma }\} \) is linearly dependent. Let \(J \subseteq \operatorname{Gal}(K/F)\) be such that \(\{ x_{\sigma }|\sigma \in J\} \) is a maximally linearly independent subset. Then \(J \ne \operatorname{Gal}(K/F)\), let \(\sigma \in \operatorname{Gal}(K/F)\) be an arbitrary automorphism that is not in \(J\). Since \(\{ x_{\tau }|\tau \in J\} \) is maximally linearly independent, \(x_\sigma \in \left\langle x_{\tau }|\tau \in J\right\rangle \). Hence, by construction 4.3.2 we have
for some non-zero \(\lambda _{\tau }\in K\) and \(J' \subseteq J\). For each \(c \in K\), we have the following equality
Since \(\{ x_{\tau }|\tau \in J'\} \) is linearly independent, we have that for each \(\tau \in J'\), \(\lambda _{\tau }\tau (c) = \sigma (c)\lambda _{\tau } = \lambda _{\tau }\sigma (c)\). Note that \(J'\) is not empty, for otherwise \(x_{\sigma }=\sum _{\tau \in \emptyset }\lambda _{\tau }\cdot x_{\tau }=0\) but \(x_{\sigma }\) is invertible. Since for any \(\tau \in J'\), \(\lambda _{\tau }\) is not zero, we have that for all \(c \in K\), \(\sigma (c) = \tau (c)\), i.e. \(\sigma = \tau \). Hence \(\sigma \) is in \(J' \subseteq J\) after all; contradiction.
If \(K/F\) is finite dimensional and Galois, then \(\dim _{F}K\) is equal to the cardinality of \(\operatorname{Gal}(K/F)\), then by the linear independence of \(\{ x_{\sigma }|\sigma \in \operatorname{Gal}(K/F)\} \), we conclude that it is indeed a \(K\)-basis for \(A\).
4.4 The Second Galois Cohomology
In this section, we construct a group isomorphism between \(\operatorname{Br}(K/F) \cong \operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) where \(K/F\) is a finite dimensional Galois extension. To keep alignment of the Brauer group, let us use the multiplicative notation for group cohomology. Recall:
Let \(G\) be a group and \(M\) an abelian group (written multiplicatively) with a \(G\)-action.
A function \(f : G \times G \to M\) is a 2-cocycle if for all \(g,h,j\in G\),
We denote the subgroup of 2-cocycles as \(\mathcal{Z}^{2}(G, M)\).
A function \(f : G\times G \to M\) is a 2-coboundary if there exists an \(x : G \to M\) such that for all \(g, h \in G\)
We denote the subgroup of 2-coboundaries as \(\mathcal{B}^{2}(G, M)\).
The second group cohomology \(\operatorname{H}^{2}\left(G, M\right)\) is defined to be the quotient group of 2-cocycles modulo 2-coboundaries \(^{\mathcal{Z}^{2}(G, M)}/_{\mathcal{B}^{2}(G, M)}\). If \(s, t \in \mathcal{Z}^{2}(G, M)\), we say \(s\) and \(t\) are cohomologous if their equivalence class \([s], [t] \in \operatorname{H}^{2}(G, M)\) are the same; in another word \(st^{-1}\in \mathcal{B}^{2}(G, M)\).
If \(f \in \mathcal{B}^{2}(G, M)\) is a 2-cocycle and \(x\in G\), we have
Indeed:
and
In the following sections of this chapter, we assume that \(X \in \operatorname{Br}(F)\) and \(A\) is a good representation of \(X\). We use \(\rho , \sigma , \tau \) to denote elements of \(\operatorname{Gal}(K/F)\). To improve typographic aesthetics of our proofs, we sometimes use subscript to mean function application.
4.4.1 From \(\operatorname{Br}(K/F)\) to \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F),K^{\star }\right)\)
If \(x\) and \(y\) are two conjugation factors of \(\sigma \), then there exists a unique \(c \in K\) such that \(x = y\iota _{A}(c)\).
The uniqueness is clear: suppose \(x = y\iota _{A}(c)=y\iota _{A}(c')\), then \(c = c'\) because \(x, y\) are units and \(\iota _{A}\) is injective. We first observe that \(y^{-1}x\in C_{A}(\iota (A))\): for any \(z \in K\), \(y^{-1}x\iota _{A}(z)=y^{-1}\iota _{A}(\sigma (z))x=\iota _{A}(z)y^{-1}x\) (by remark 4.3.9). By lemma 4.3.6, \(y^{-1}x\in \iota (A)\), that is for some \(z \in K\), we have that \(y^{-1}x = \iota _{A}(z)\) and the claim is proved.
We denote such \(c\) by \(\operatorname{twist}^{\sigma }(x, y)\) or \(\operatorname{twist}^{\sigma }_{x, y}\), when \(\sigma \) is clear from context, we often omit the superscript. With this notation, \(x = y\iota _{A}(\operatorname{twist}_{x,y})\).
\(\operatorname{twist}(x, x)\) is equal to \(1\) by uniqueness.
In fact, \(\operatorname{twist}(x, y)\) is in \(K^{\star }\) and \(\operatorname{twist}(x, y)^{-1}=\operatorname{twist}(y, x)\).
If \(x\) and \(y\) are conjugation factors for \(\sigma \), \(x=\iota _{A}(\sigma (\operatorname{twist}_{x, y}))y\).
Let \(x\) be a conjugation factor for \(\sigma \), \(y\) for \(\tau \) and \(z\) for \(\sigma \tau \). Since \(xy\) is a also a conjugation factor, we define the comparison coefficient to be \(\operatorname{comp}^{\sigma , \tau }_{x,y,z}:=\sigma \left(\tau \left(\operatorname{twist}_{xy, z}\right)\right)\). We often omit superscript when the context is clear. Note that \(\operatorname{comp}_{x,y,z}\) is a unit in \(K\) with inverse \(\sigma \left(\tau \left(\operatorname{twist}_{z, xy}\right)\right)\). By lemma 4.4.2 and lemma 4.4.5, we have the following useful equalities
Let \(x : \operatorname{Gal}(K/F) \to A^{\star }\) be a conjugation sequence. We have
It is sufficient to make the following calculations:
Then since \(x_{\rho \sigma \tau }\) is invertible and \(\iota _{A}\) is injective, we proved the desired result.
Equation 1 is because: by the first equality in construction 4.4.2 (twice)
2 is because: by definition 4.3.3, we have
therefore by construction 4.2.1
Let \(x\) be an \(A\)-conjugation sequence. We associate with \(x\) a function \(\mathcal{B}^{2}(x) : \operatorname{Gal}(K/F)\times \operatorname{Gal}(K/F)\to K^{\star }\) defined by
We will write \(\mathcal{B}^{2}(x)\) as \(\mathcal{B}^{2}_{A, x}\), \(\mathcal{B}^{2}_{A}(x)\) or \(\mathcal{B}^{2}_{x}\) as well.
For any \(A\)-conjugation sequence \(x\), \(\mathcal{B}^{2}_{x} \in \mathcal{B}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\), that is \(\mathcal{B}_{x}\) is indeed a 2-cocycle.
We need to prove
But this is exactly lemma 4.4.6.
For any good representation \(A\) of \(X \in \operatorname{Br}(K/F)\) and any \(A\)-conjugation sequence \(x\), we have constructed a 2-cocycle \(\mathcal{B}^{2}_{A}(x)\). But to obtain a well-defined function from \(\operatorname{Br}(K/F)\) to \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\), we need to verify that for any other good representation \(B\) of \(X\) and \(B\)-conjugation sequence \(y\), \(\mathcal{B}^{2}_{A}(x)\) and \(\mathcal{B}^{2}_{B}(y)\) are cohomologous. Let us fix another good representation \(B\) of \(X \in \operatorname{Br}(K/F)\) and a \(B\)-conjugation sequence \(y\).
By lemma 4.3.7, \(A\) and \(B\) are isomorphic as \(F\)-algebras, we use \(e_{A,B}\) to denote an arbitrary \(F\)-algebra isomorphism between \(A\) and \(B\). When there is no confusion, we write \(e\) instead of \(e_{A, B}\) Since \(e \circ \iota _{A}\) and \(\iota _{B}\) are two \(F\)-algebra homomorphism from \(K\) to \(B\), by theorem 3.3.4, there exists some \(u \in B^{\star }\) such that for all \(r \in K\), we have \(\iota _{B}(r) = ue\left(\iota _{A}(r)\right)u^{-1}\) (or equivalently, \(u^{-1}\iota _{B}(r)u = e\left(\iota _{A}(r)\right)\)). When there is confusion, we write \(u_{A, B}\) instead of \(u\).
For any \(c \in K\), \(\sigma \in \operatorname{Gal}(K/F)\) and \(A\)-conjugation factor \(x\) of \(\sigma \), we have
From definition 4.3.3, we have \(e\left(\iota _{A}(\sigma \cdot c)\right) = e\left(x\iota _{A}(c)x^{-1}\right)\). Substituting it in construction 4.4.4, we get
If \(x\) is an \(A\)-conjugation factor for \(\sigma \), we can obtain a \(B\)-conjugation factor for \(\sigma \) by defining \(B_{\star }x := ue(x)u^{-1}\) with inverse \(ue\left(x^{-1}\right)u^{-1}\). We use lemma 4.4.8 to check that \(B_{\star }x\) is indeed a conjugation factor for \(\sigma \). If \(y\) is a \(B\)-conjugation factor for \(\sigma \), another useful constant is \(v := \sigma \left(\operatorname{twist}_{y, B_{\star }x}\right)\). We have
We also write \(v_{x,y}\) or even \(v^{A,B}_{x, y}\) when we stress the importance of good representation \(A\) and \(B\) and their conjugation factor \(x\) and \(y\).
Let \(x\) be an \(A\)-conjugation sequence and \(y\) a \(B\)-conjugation sequence. We have
By construction 4.4.2, we have \(y_{\sigma }y_{\tau } = \iota _{B}\left(\operatorname{comp}_{y_{\sigma },y_{\tau },y_{\sigma \tau }}\right)y_{\sigma \tau }\). By repeated application of construction 4.4.5 and construction 4.4.4, we have
Hence
We also have by definition 4.3.3
Hence
Cancelling \(x_{\sigma \tau }\) and by injectivity of \(\iota _{A}\), the result is proved.
Let \(x\) be an \(A\)-conjugation sequence and \(y\) a \(B\)-conjugation sequence. We have
If we unfold construction 4.4.3, we discover the lemma is saying exactly lemma 4.4.9.
We finally arrive at our main conclusion for this section.
Let \(x\) be an \(A\)-conjugation sequence and \(y\) a \(B\)-conjugation sequence. \(\mathcal{B}^{2}_{A,x}\) and \(\mathcal{B}^{2}_{B,y}\) are 2-cohomologous.
By definition 4.4.1, we need to find a function \(f : \operatorname{Gal}(K/F)\to K^{\star }\) such that for all \(\sigma , \tau \in \operatorname{Gal}(K/F)\),
Let \(f(\rho ) := v_{x_{\rho },y_{\rho }}\), by lemma 4.4.10 we see the equality holds.
Let \(X \in \operatorname{Br}(K/F)\), by corollary 4.3.4, \(X\) admits a good representation \(A\); by construction 4.3.4, \(A\) admits a conjugation sequence \(x\). We associate with \(X\) an element \(\operatorname{H}^{2}(X) := \left[\mathcal{B}^{2}_{A,x}\right]\)in \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F),K^{\star }\right)\). By corollary 4.4.11, for any other good representation \(B\) and \(B\)-conjugation sequence \(y\), we have \(\left[\mathcal{B}^{2}_{A,x}\right]=\left[\mathcal{B}^{2}_{B,y}\right]\), hence we have a well-defined function \(\operatorname{H}^{2} : \operatorname{Br}(K/F) \to \operatorname{H}^{2}\left(\operatorname{Gal}(K/F),K^{\star }\right)\).
4.4.2 Cross Product as a Central Simple Algebra
Let \(\mathfrak {a} \in \mathcal{B}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) be any 2-cocycle. In this section, we construct the cross product associated with \(\mathfrak {a}\) which we prove to be \(F\)-central simple. Finally, we show that if \(\mathfrak {a}, \mathfrak {b}\in \mathcal{B}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) are cohomologous, the cross products associated with \(a\) and \(b\) are Brauer equivalent.
Denote \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) to be \(\operatorname{Gal}(K/F) \to K\), i.e. functions from \(\operatorname{Gal}(K/F)\) to \(K\). Notationally, elements of \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) are sequences in \(K\) indexed by \(\operatorname{Gal}(K/F)\); we denote \(\Delta ^{\mathfrak {a}}_{\sigma , c}\) to be the sequence with value \(c\) at \(\sigma \)-th index and zero elsewhere. When \(\mathfrak {a}\) is clear from context, we will omit the superscript. We give \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) the usual zero, addition, negation, that is, we give \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) the normal additive abelian group structure. Since for each \(c \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\),
it is often, if not always, sufficient to consider the special cases of \(\Delta _{\sigma , c}\) and extend the result linearly. For multiplications, we define the result of multiplying \(\Delta _{\sigma , c},\Delta _{\tau , d} \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) to be \(\Delta _{\sigma \tau , c\sigma (d)\, \mathfrak {a}(\sigma ,\tau )}\). Immediately, if either \(c\) or \(d\) is \(0\), the result of multiplication is also zero. That is, for all \(c \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), we have \(c\cdot 0=0\cdot c =0\). For any \(r \in F\) and \(\Delta _{\sigma , c} \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), we define \(r \cdot \Delta _{\sigma , c}\) to be \(\Delta _{\sigma , r\cdot c}\).
When \(K/F\) is infinite dimensional, the correct definition of \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is perhaps \(\bigoplus _{\sigma \in \operatorname{Gal}(K/F)}K\). But in Lean4, function type is easier to manipulate than direct sums. Since our scope is finite dimensional Galois extension, our definition is still accurate.
The cross product \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is a ring with the multiplicative unit \(\Delta _{\mathsf{id}, \mathfrak {a}(\mathsf{id}, \mathsf{id})^{-1}}\). The \(F\)-action on \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) defined by \(r \cdot \Delta _{\sigma , c} := \Delta _{\sigma , r\cdot c}\) makes it an \(F\)-algebra.
We verify the axioms of rings on elements of the form \(\Delta _{\sigma , c}\). Let \(\sigma ,\tau ,\rho \in \operatorname{Gal}(K/F)\) and \(a,b,c\in K\).
associativity of multiplication. We need to check that \(\Delta _{\sigma ,a}\left(\Delta _{\tau ,b}\Delta _{\rho ,c}\right) = \left(\Delta _{\sigma ,a}\Delta _{\tau ,b}\right)\Delta _{\rho ,c}\):
\[ \begin{aligned} \Delta _{\sigma , a}\left(\Delta _{\tau ,b}\Delta _{\rho ,c}\right) & = \Delta _{\sigma ,a}\Delta _{\tau \rho ,b\tau (c) \mathfrak {a}(\sigma ,\tau )}\\ & = \Delta _{\sigma \tau \rho ,a\sigma (b)\sigma (\tau (c)) \sigma (\mathfrak {a}(\sigma , \tau ))};\\ \left(\Delta _{\sigma ,a}\Delta _{\tau ,b}\right)\Delta _{\rho ,c} & = \Delta _{\sigma \tau ,a\sigma (b)\mathfrak {a}(\sigma ,\tau )}\Delta _{\rho ,c}\\ & = \Delta _{\sigma \tau \rho ,a\sigma (b)\mathfrak {a}(\sigma , \tau )\sigma (\tau (c))\mathfrak {a}(\sigma \tau , \rho )}. \end{aligned} \]Hence it is sufficient to check
\[ \sigma (\tau (c))\sigma (\mathfrak {a}(\sigma ,\tau )) = \mathfrak {a}(\sigma ,\tau )\sigma (\tau (c))\mathfrak {a}(\sigma \tau , \rho ). \]This is the 2-cocycle condition in definition 4.4.1 (modulo commutativity of \(K\)).
multiplicative unit: we need to check \(\Delta _{\sigma , a}\Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id}, \mathsf{id})} = \Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})}\Delta _{\sigma , a} = \Delta _{\sigma , a}\). By multiple applications of lemma 4.4.1
\[ \begin{aligned} \Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id}, \mathsf{id})^{-1}}\Delta _{\sigma , a} & = \Delta _{\sigma , \mathfrak {a}(\mathsf{id}, \mathsf{id})^{-1}a\mathfrak {a}(\mathsf{id},\sigma )}\\ & = \Delta _{\sigma , \mathfrak {a}(\mathsf{id}, \mathsf{id}) a\mathfrak {a}(\mathsf{id},\mathsf{id})}\\ & = \Delta _{\sigma , a} \\ \Delta _{\sigma , a}\Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}} & = \Delta _{\sigma , a\sigma (\mathfrak {a}(\mathsf{id},\mathsf{id}))^{-1}\mathfrak {a}(\sigma ,\mathsf{id})}\\ & =\Delta _{\sigma , a\sigma (\mathfrak {a}(\mathsf{id},\mathsf{id}))^{-1}\sigma (\mathfrak {a}(\mathsf{id}, id))}\\ & =\Delta _{\sigma , a} \end{aligned} \]distributivity: We need to check left-distributivity \(\Delta _{\sigma , a}\left(\Delta _{\tau , b} + \Delta _{\rho , c}\right) = \Delta _{\sigma ,a}\Delta _{\tau ,b} + \Delta _{\sigma , a}\Delta _{\rho , c}\) and right distributivity \(\left(\Delta _{\tau ,b} + \Delta _{\rho ,c}\right)\Delta _{\sigma ,a} = \Delta _{\tau ,b}\Delta _{\sigma ,a} + \Delta _{\rho ,c}\Delta _{\sigma ,a}\). This is precisely what “extend linearly” means.
\(F\)-algebra: We need to check for all \(r \in F\), \(\left(r\cdot \Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\right)\Delta _{\sigma ,c} = \Delta _{\sigma ,c}\left(r\cdot \Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\right)\). By lemma 4.4.1
\[ \begin{aligned} \left(r\cdot \Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\right)\Delta _{\sigma ,c} & = \Delta _{\mathsf{id},r\cdot \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\Delta _{\sigma ,c}\\ & = \Delta _{\sigma ,\left(r\cdot \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\right)c\mathfrak {a}(\mathsf{id},\sigma )} \\ & =\Delta _{\sigma ,\left(r\cdot \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\right)c\mathfrak {a}(\mathsf{id},\mathsf{id})} \\ & = \Delta _{\sigma , r\cdot c}\\ \Delta _{\sigma ,c}\left(r\cdot \Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\right) & = \Delta _{\sigma ,c}\Delta _{\mathsf{id},r\cdot \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}} \\ & = \Delta _{\sigma ,c\sigma \left(r\cdot \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\right)\mathfrak {a}(\sigma ,\mathsf{id})}\\ & =\Delta _{\sigma ,c\left(r\cdot \sigma \left(\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\right)\right)\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\\ & = \Delta _{\sigma , c\left(r\cdot 1\right)}\\ & =\Delta _{\sigma ,r\cdot c}. \end{aligned} \]
From now on, we feel free to write \(1 \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) instead of \(\Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\). Then the algebra map \(F \hookrightarrow \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is the map \(r \mapsto r \cdot 1\).
The map \(\iota _{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}: K \to \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) defined by
is an \(F\)-algebra map. Checking that \(\iota _{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}\) preserves \(1\), multiplication and addition uses nothing but axioms of ring. For any \(r \in F\), we need to check \(\iota _{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}(r) = r \cdot 1\). Indeed \(\iota _{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}(r) = \Delta _{\mathsf{id}, r\cdot \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\) and \(r\cdot 1 = r \cdot \Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}} =\Delta _{\mathsf{id},r\cdot \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\). When the context is clear, we also write \(\iota _{\mathfrak {a}}\) instead of \(\iota _{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}\). We give \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) a \(K\)-module structure by left-multiplication, that is for any \(b \in K\) and \(c\in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), we define \(b\cdot c := \iota _{\mathfrak {a}}(b)c\).
We note the following useful equality: for any \(b \in K\)
indeed: \(\iota _{\mathfrak {a}}(b)\Delta _{\sigma , c} = \Delta _{\mathsf{id},b \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\Delta _{\sigma , c} = \Delta _{\sigma , b\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}c\mathfrak {a}(\mathsf{id},\sigma )} = \Delta _{\sigma ,b\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}c\mathfrak {a}(\mathsf{id},\mathsf{id})} = \Delta _{\sigma , bc}\) by lemma 4.4.1. In another word, for any \(b \in F\) and \(c \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), the \(K\)-action of \(b\) on \(c\) and the \(F\)-action of \(b\) on \(c\) agree.
For every \(\sigma \in \operatorname{Gal}(K/F)\), \(\Delta _{\sigma ,1}\) is invertible.
It is sufficient to prove that \(\Delta _{\sigma , 1}\) has a left inverse and right inverse. The left inverse of \(\Delta _{\sigma , 1}\) is
Indeed, for any \(a \in K\), we have
hence substitute \(a = \mathfrak {a}\left(\sigma ^{-1},\sigma \right)^{-1}\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\), we see the right hand side is \(\Delta _{ \mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\) which is precisely \(1 \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\). The right inverse is
Indeed, for any \(a \in K\), we have
hence substitude \(a = \sigma ^{-1}\left(\mathfrak {a}\left(\sigma ,\sigma ^{-1}\right)^{-1}\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\right)\), the right hand side is again \(\Delta _{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\) which is precisely \(1 \in \operatorname{\mathfrak {C}}_{a}\).
For any \(c \in K\), we have
and consequently,
We calculate
We have \(\Delta _{\sigma ,1}\, \Delta _{\tau ,1} = \iota _{\mathfrak {a}}(\mathfrak {a}(\sigma ,\tau ))\Delta _{\sigma \tau ,1} = \mathfrak {a}(\sigma ,\tau )\cdot \Delta _{\sigma \tau ,1}\) Consequently we have for any \(c, d \in K\),
The first equality is in construction 4.4.8. For the second equality, by lemma 4.4.15, we have
The set \(\{ \Delta _{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) forms a \(K\)-basis for \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\).
Suppose some linear combination \(\sum _{\sigma }\lambda _{\sigma }\cdot \Delta _{\sigma , 1}\) is \(0\) for some \(\lambda _{\sigma }\)’s in \(K\). We have, by the equality in construction 4.4.8
Thus, for any \(\tau \in \operatorname{Gal}(K/F)\), we have
which proves linear independence. The fact that \(\{ \Delta _{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) spans \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is easy to see because every \(\Delta _{\tau , a} = a \cdot \Delta _{\tau , 1}\) is certainly in the span.
When \(K/F\) is a finite dimensional Galois extension, the \(K\)-dimension of \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is \(\dim _{F}K\) and the \(F\)-dimension of \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is \({\left(\dim _{F}K\right)}^{2}\).
Now we see that cross product, like a good representation, is a \(K\)-module and \(F\)-algebra with a \(K\)-embedding and correct \(F\)-dimension. In the next sections, we prove that \(\operatorname{\mathfrak {C}}_{a}\) is in fact a central simple \(F\)-algebra.
Central Algebra
We will assume \(K/F\) is a finite dimensional Galois extension.
\(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is a central \(F\)-algebra.
Let \(z \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) that is in the centre. We want to prove that \(z\) is in \(F\). We write \(z\) as \(\sum _{\sigma }\lambda _{\sigma }\cdot \Delta _{\sigma ,1}\). We see that, for any \(\tau \in \operatorname{Gal}(K/F)\), we have
Therefore for any \(d \in K\) and \(\tau \in \operatorname{Gal}(K/F)\), by lemma 4.4.16 and lemma 4.4.15, we have
By lemma 4.4.17, for any \(\sigma ,\tau \in \operatorname{Gal}(K/F)\) and \(d\in K\), we have that
In particular, with \(d = 1\), we have
we substitute back into eq. 3 and get
With \(\tau = \mathsf{id}\), we have
Hence for all \(d \in K\) with \(\lambda _{\sigma } \ne 0\), we have \(\sigma (d) = d\). We immediately deduce that for all \(\sigma \ne \mathsf{id}\), \(\lambda _{\sigma } = 0\) by contraposition. Thus \(z = \lambda _{\mathsf{id}}\Delta _{\mathsf{id}, 1} = \Delta _{\mathsf{id},\lambda _{\mathsf{id}}} = \iota _{\mathfrak {a}}\left(\lambda _{\mathsf{id}}\mathfrak {a}(\mathsf{id},\mathsf{id})\right) = \left(\lambda _{id}\mathfrak {a}(\mathsf{id},\mathsf{id})\right) \cdot 1\). Consequently, to prove \(z\) is in \(F\), it is sufficient to prove that \(\lambda _{\mathsf{id}}\mathfrak {a}(\mathsf{id},\mathsf{id})\) is in \(F\). Since \(K/F\) is finite dimensional and Galois, we only need to prove that \(\lambda _{\mathsf{id}}\mathfrak {a}(\mathsf{id},\mathsf{id})\) is fixed by every \(\tau \in \operatorname{Gal}(K/F)\).Indeed, with \(d = 1\) and \(\sigma = \mathsf{id}\) in eq. 3, we have
Simple Ring
In this section we assume \(K/F\) is a finite dimensional field extension. Let \(I \subseteq \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) be a two sided ideal, we aim to show that either \(I = \{ 0\} \) or \(I = \operatorname{\mathfrak {C}}_{\mathfrak {a}}\). In this section, we use \(\pi \) to denote the canonical ring homomorphism \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\to {}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\). We restrict \(\pi \) to \(\pi |_{\operatorname{im}\left(\iota _{\mathfrak {a}}\right)} : \operatorname{im}\left(\iota _{\mathfrak {a}}\right) \to {}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) and denote the range of \(\pi |_{\operatorname{im}\left(\iota _{\mathfrak {a}}\right)}\) to be \(\Pi \).
The quotient ring \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) is a \(\Pi \)-module defined by \(\pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot \pi (y) := \pi (a \cdot y)\). We first check that the \(\Pi \)-action is well-defined:
Independence of \(a\): Let \(a, b \in K\) be such that \(\pi \left(\iota _{\mathfrak {a}}(a)\right) = \pi \left(\iota _{\mathfrak {a}}(b)\right)\), that is, \(\iota _{\mathfrak {a}}(a-b)\in I\). Since \(I\) is a two sided ideal, \(a\cdot y - b \cdot y = (a - b)\cdot y = \iota _{\mathfrak {a}}(a - b) y\) is also in \(I\). This proves \(\pi (a\cdot y) = \pi (b\cdot y)\).
Independence of \(y\): Let \(y_{1},y_{2}\in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) be such that \(y_{1}-y_{2}\in I\), then for any \(a \in K\), \(a \cdot y_{1} - a \cdot y_{2} = \iota _{\mathfrak {a}}(a)\left(y_{1}-y_{2}\right)\) is in \(I\) because \(I\) is a two sided ideal. This proves \(\pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot \pi (y_{1}) = \pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot \pi (y_{2})\).
Then we check the axioms of module:
Let \(y \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), we check that \(1 \cdot \pi (y) = \pi (y)\) and \(0 \cdot \pi (y) = 0\). This is because \(\Pi \ni 1 = \pi \left(\iota _{\mathfrak {a}}(1)\right)\) and \(\Pi \ni 0=\pi \left(\iota _{\mathfrak {a}}(0)\right)\). Let \(a \in K\), \(\pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot 0 = 0\) because \(0 \in {}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) is equal to \(\pi (0)\).
Let \(a, b \in K\) and \(x, y \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), we check \(\left(\pi \left(\iota _{\mathfrak {a}}(a)\right) + \pi \left(\iota _{\mathfrak {a}}(b)\right)\right)\cdot \pi (x) = \pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot \pi (x) + \pi \left(\iota _{\mathfrak {a}}(b)\right)\cdot \pi (x)\) and \(\pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot (\pi (x)+\pi (y)) = \pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot \pi (x) + \pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot \pi (y)\). These are true because \(\pi \) preserves addition. Similarly \(\pi \left(\iota _{\mathfrak {a}}(a)\right)\cdot \pi \left(\iota _{\mathfrak {a}}(b)\right) \cdot \pi (x) = \pi \left(\iota _{\mathfrak {a}}(ab)\right)\cdot \pi (x)\) because \(\pi \) preserves multiplication as well.
Hence \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) is also a \(K\)-module by pulling back the \(\Pi \)-module structure along \(K \to \Pi \) given by \(a \mapsto \pi \left(\iota _{\mathfrak {a}}(a)\right)\). Note that \(\pi \) is a \(K\)-linear map between \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) by this construction.
If \(I \ne \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), the set \(\{ \pi \left(\Delta _{\sigma , 1}\right)|\sigma \in \operatorname{Gal}(K/F)\} \) forms a \(K\)-basis for \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\).
It is easy to see that the set spans \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) because \(\{ \Delta _{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) spans \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) (lemma 4.4.17). For linear-independence, the idea is the same as in the proof of theorem 4.3.10. We repeat the argument here.
Suppose that \(\{ \pi \left(\Delta _{sigma, 1}\right)|\sigma \in \operatorname{Gal}(K/F)\} \) is linearly dependent. Let \(J \subseteq \operatorname{Gal}(K/F)\) be such that \(\{ \pi \left(\Delta _{\sigma , 1}\right)|\sigma \in J\} \) is the maximally linearly independent set. Let \(\sigma \) be an arbitrary automorphism that is not in \(J\). Therefore, we have \(\pi \left(\Delta _{\sigma , 1}\right) \in \left\langle \pi \left(\Delta _{\tau , 1}\right) | \tau \in J \right\rangle \). Hence we have, by construction 4.4.9 and construction 4.4.8
for some non-zero \(\lambda _{\tau } \in K\) and some \(J' \subseteq J\). Hence, for any \(c \in K\), we have
Since, \(\{ \pi \left(\Delta _{\tau , 1}\right) | \tau \in J\} \) is linearly independent, for all \(c \in K\) and \(\tau \in J\), we have that \(\lambda _{\tau }\tau (c) = \sigma (c)\lambda _{\tau }\). Note that \(J' \ne \emptyset \), otherwise, \(\pi \left(\Delta _{\sigma , 1}\right) = 0\) implying that \(\Delta _{\sigma , 1} \in I\) which by lemma 4.4.14 is invertible but \(I\) does not equal to \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\). Hence for each \(\tau \in J'\), we have that for all \(c \in K\), since \(\lambda _{\tau }\) is not zero, \(\sigma (c) = \tau (c)\), i.e. \(\sigma = \tau \). Therefore, \(\sigma \) is in \(J' \subseteq J\) after all.
If \(I \ne \operatorname{\mathfrak {C}}_{\mathfrak {a}}\), the quotient ring \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) is isomorphic to \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) as \(K\)-modules. In particular \(\pi \) is a \(K\)-linear isomorphism between \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and the quotient ring \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\).
Indeed, by lemma 4.4.17, \(\{ \Delta _{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) is a \(K\)-basis for \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\); and by lemma 4.4.20, \(\{ \pi \left(\Delta _{\sigma , 1}\right) | \sigma \in \operatorname{Gal}(K/F)\} \) is a \(K\)-basis for \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\). The two sets obviously biject. Hence we can define a \(K\)-linear isomorphism by \(\Delta _{\sigma , 1} \mapsto \pi \left(\Delta _{\sigma , 1}\right)\). This isomorphism is equal to \(\pi \) everywhere.
\(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is a simple ring.
For any two-sided-ideal \(I\) that is not equal to \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\), by corollary 4.4.21, \(\pi : cross_{\mathfrak {a}} \to {}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\) is a \(K\)-linear isomorphism, therefore \(I\) is equal to \(0\).
Let \(K/F\) be a finite dimensional and Galois field extension and \(\mathfrak {a}\) be a 2-cocycle in \(\mathcal{B}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\), \(\operatorname{\mathfrak {C}}_{a}\) is a finite dimensional central simple \(F\)-algebra.
4.4.3 From \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) to \(\operatorname{Br}(K/F)\)
For every 2-cocycle \(\mathfrak {a}\), we have defined the cross product \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and proved that it is indeed a finite dimensional central simple \(F\)-algebra in theorem 4.4.23; that is we have a function from \(\mathcal{B}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) to \(\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{F}\). If we want a function from \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) to \(\operatorname{Br}(K/F)\), we need to show that if \(\mathfrak {a}\) and \(\mathfrak {b}\) are cohomologous, \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and \(\operatorname{\mathfrak {C}}_{\mathfrak {b}}\) are Brauer equivalent. We state it as a theorem:
If \(K/F\) is a finite dimensional and Galois field extension, the function \(\operatorname{\mathfrak {C}}: \operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right) \to \operatorname{Br}(K/F)\) defined by
is well-defined.
Let \(\mathfrak {a}\) and \(\mathfrak {b}\) be two cohomologous 2-cocycles. By definition 4.4.1, for some \(\mathfrak {c}: \operatorname{Gal}(K/F) \to K^{\star }\), for all \(\sigma , \tau \in \operatorname{Gal}(K/F)\), we have
Let us denote \(\mathsf{A}\) to be the \(K\)-basis \(\{ \Delta ^{\mathfrak {a}}_{\sigma , 1} | \sigma \in \operatorname{Gal}(K/F)\} \) for \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and \(\mathsf{B}\) to be the \(K\)-basis \(\{ \mathfrak {c}(\sigma )\cdot \Delta ^{\mathfrak {b}}_{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) for \(\operatorname{\mathfrak {C}}_{\mathfrak {b}}\). We immediately have a \(K\)-linear isomorphism \(\phi : \operatorname{\mathfrak {C}}_{\mathfrak {a}}\cong \operatorname{\mathfrak {C}}_{\mathfrak {b}}\) by mapping \(\mathsf{A}\) to \(\mathsf{B}\). Since the \(K\)-action on \(\operatorname{\mathfrak {C}}_{a}\) and \(\operatorname{\mathfrak {C}}_{b}\) agrees with the \(F\)-action on them (construction 4.4.8), \(\phi \) is also an \(F\)-linear isomorphism. We check that \(\phi (1) = 1\) and \(\phi (xy) = \phi (x)\phi (y)\) for all \(x, y \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\):
preservation of one: with \(\sigma = \tau =\mathsf{id}\) in eq. 4, we have \(\mathfrak {c}(\mathsf{id}) = \mathfrak {a}(\mathsf{id},\mathsf{id})\mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}\), thus
\[ \begin{aligned} \phi (1) & = \phi \left(\Delta ^{\mathfrak {a}}_{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}\right)\\ & = \phi \left( \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\cdot \Delta ^{\mathfrak {a}}_{\mathsf{id}, 1} \right) \\ & = \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\cdot \mathfrak {c}(\mathsf{id}) \cdot \Delta ^{\mathfrak {b}}_{\mathsf{id}, 1} \\ & = \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\cdot \mathfrak {c}(\mathsf{id})\cdot \mathfrak {b}(\mathsf{id},\mathsf{id}) \cdot \mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}\cdot \Delta ^{\mathfrak {b}}_{\mathsf{id}, 1} \\ & = \left( \mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\mathfrak {c}(\mathsf{id})\mathfrak {b}(\mathsf{id},\mathsf{id}) \right)\cdot \left( \mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}\cdot \Delta _{\mathsf{id}, 1} \right) \\ & = \left(\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\mathfrak {a}(\mathsf{id},\mathsf{id})\right)\cdot \Delta _{\mathsf{id},\mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}} \\ & = \Delta _{\mathsf{id},\mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}}. \end{aligned} \]preservation of multiplication: let \(\sigma ,\tau \in \operatorname{Gal}(K/F)\) and \(a, b \in K\), we need to prove that \(\phi \left(\Delta ^{\mathfrak {a}}_{\sigma , a}\Delta ^{\mathfrak {a}}_{\tau , b}\right) = \phi \left(\Delta ^{\mathfrak {a}}_{\sigma , a}\right)\phi \left(\Delta ^{\mathfrak {b}}_{\tau , b}\right)\). From eq. 4, we see that
\[ \sigma (\mathfrak {c}(\tau ))\mathfrak {c}(\sigma )\mathfrak {b}(\sigma ,\tau ) = \mathfrak {c}(\sigma \tau )\mathfrak {a}(\sigma ,\tau ). \]Hence, by lemma 4.4.16 and lemma 4.4.15, we have
\[ \begin{aligned} \phi \left(\Delta ^{\mathfrak {a}}_{\sigma , a}\Delta ^{\mathfrak {a}}_{\tau , b}\right) & = \phi \left( \Delta ^{\mathfrak {a}}_{\sigma \tau , a \sigma (b) \mathfrak {a}(\sigma ,\tau )} \right)\\ & = \phi \left( a\sigma (b) \mathfrak {a}(\sigma ,\tau ) \, \cdot \, \Delta ^{\mathfrak {a}}_{\sigma \tau , 1} \right)\\ & = a \sigma (b) \mathfrak {a}(\sigma ,\tau ) \, \cdot \, \phi \left(\Delta ^{\mathfrak {a}}_{\sigma \tau , 1}\right) \\ & = a \sigma (b) \mathfrak {a}(\sigma ,\tau ) \mathfrak {c}(\sigma \tau ) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma \tau , 1}.\\ & = a\sigma (b)\sigma (\mathfrak {c}(\tau ))\mathfrak {c}(\sigma )\mathfrak {b}(\sigma ,\tau ) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma \tau , 1}\\ & = \mathfrak {c}(\sigma ) \sigma (\mathfrak {c}(\tau )) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma \tau ,a\sigma (b)\mathfrak {b}(\sigma \tau )}\\ & = \mathfrak {c}(\sigma )\sigma (\mathfrak {c}(\tau )) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma , a}\Delta ^{\mathfrak {b}}_{\tau , b} \\ \phi \left(\Delta ^{\mathfrak {a}}_{\sigma , a}\right)\phi \left(\Delta ^{\mathfrak {a}}_{\tau , b}\right) & = \phi \left(a \cdot \Delta ^{\mathfrak {a}}_{\sigma , 1}\right) \phi \left(b\cdot \Delta ^{\mathfrak {a}}_{\tau , 1}\right) \\ & = \left(a \cdot \phi \left(\Delta ^{\mathfrak {a}}_{\sigma , 1}\right)\right) \left(b \cdot \phi \left(\Delta ^{\mathfrak {a}}_{\tau , 1}\right)\right) \\ & = \left( a\mathfrak {c}(\sigma ) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma , 1} \right) \left( b \mathfrak {c}(\tau ) \, \cdot \, \Delta ^{\mathfrak {b}}_{\tau , 1} \right)\\ & = a\mathfrak {c}(\sigma ) \, \cdot \, \left(\Delta ^{\mathfrak {b}}_{\sigma , 1} \iota _{\mathfrak {b}}(b\mathfrak {c}(\tau ))\right) \, \Delta ^{\mathfrak {b}}_{\tau , 1} \\ & = a\mathfrak {c}(\sigma ) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma , \sigma (b\mathfrak {c}(\tau ))}\, \Delta ^{\mathfrak {b}}_{\tau , 1} \\ & = a\mathfrak {c}(\sigma )\sigma (b)\sigma (\mathfrak {c}(\tau )) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma , 1}\Delta _{\tau , 1}^{\mathfrak {b}}\\ & = a\mathfrak {c}(\sigma )\sigma (b)\sigma (\mathfrak {c}(\tau ))\mathfrak {b}(\sigma ,\tau ) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma \tau , 1}\\ & = \mathfrak {c}(\sigma )\sigma (\mathfrak {c}(\tau )) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma \tau , a\sigma (b)\mathfrak {b}(\sigma ,\tau )}\\ & = \mathfrak {c}(\sigma )\sigma (\mathfrak {c}(\tau )) \, \cdot \, \Delta ^{\mathfrak {b}}_{\sigma , a}\Delta ^{\mathfrak {b}}_{\tau , b}. \end{aligned} \]
Hence \(\phi \) is actually an \(F\)-algebra isomorphism between \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and \(\operatorname{\mathfrak {C}}_{\mathfrak {b}}\) and isomorphic central simple \(F\)-algebras are certainly Brauer equivalent.
4.4.4 \(\operatorname{H}^{2} \circ \operatorname{\mathfrak {C}}\) and \(\operatorname{\mathfrak {C}}\circ \operatorname{H}^{2}\)
For a finite dimensional Galois extension of field \(K/F\), we have constructed two functions \(\operatorname{H}^{2}\) and \(\operatorname{\mathfrak {C}}\) between the second cohomology group \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) and the relative Brauer group \(\operatorname{Br}(K/F)\). In this section, we prove that they are mutual inverse to one another,
The composition of \(\operatorname{\mathfrak {C}}\) and \(\operatorname{H}^{2}\) is the identity:
![\begin{tikzcd}
\displaystyle \HH^2\left(\gal(K/F), K^\star\right) \arrow{r}{\cross} \arrow[bend right = 15, swap]{rr}{\id} &
\br(K/F) \arrow{r}{\HH^2} & \displaystyle \HH^2\left(\gal(K/F), K^\star\right).
\end{tikzcd}](images/img-0005.svg)
Let \(\mathfrak {a}\) be any 2-cocycle, by lemma 4.4.15, we notice that \(x : \sigma \mapsto \Delta _{\sigma , 1}\) is a conjugation sequence for \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\). Hence by construction 4.4.3, ?? and theorem 4.4.24, we evaluate the composition at \(\mathfrak {a}\) as:
![\begin{tikzcd}
{[\mfa]} \arrow[mapsto]{r} & {\left[\cross_\mfa\right]_{\sim_{\br}}} \arrow[mapsto]{r}& {\left[(\sigma,\tau)\mapsto\comp^{x}_{\Delta_{\sigma, 1}, \Delta_{\tau, 1}, \Delta_{\sigma\tau, 1}}\right]}
\end{tikzcd}](images/img-0006.svg)
That is, we need to show that \(\mathfrak {a}\) and \((\sigma , \tau ) \mapsto \operatorname{comp}_{\Delta _{\sigma , 1}, \Delta _{\tau , 1}, \Delta _{\sigma \tau , 1}}\) are 2-cohomologous. In fact, they are equal. By construction 4.4.2, we have that \(\iota _{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}\left(\operatorname{comp}^{x}_{\Delta _{\sigma , 1}, \Delta _{\tau , 1}, \Delta _{\sigma \tau , 1}}\right) = \Delta _{\sigma , 1}\Delta _{\tau , 1}\Delta _{\sigma \tau , 1}^{-1} = \mathfrak {a}(\sigma ,\tau )\cdot \Delta _{\sigma \tau , 1}\Delta _{\sigma \tau , 1}^{-1} = \mathfrak {a}(\sigma ,\tau )\cdot 1 = \Delta _{\mathsf{id},\mathfrak {a}(\sigma ,\tau )}\) which is precisely \(\iota _{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}(\mathfrak {a}(\sigma ,\tau ))\).
The composition of \(\operatorname{H}^{2}\) and \(\operatorname{\mathfrak {C}}\) is the identity:
![\begin{tikzcd}
\br(K/F) \arrow{r}{\HH^2} \arrow[bend right = 15, swap]{rr}{\id} &
\HH^2\left(\gal(K/F), K^\star\right) \arrow{r}{\cross} &
\br(K/F)
\end{tikzcd}](images/img-0007.svg)
Let \(X \in \operatorname{Br}(K/F)\), \(A\) be an arbitrary good representation of \(X\) and \(x\) be an arbitrary \(A\)-conjugation sequence which exists by corollary 4.3.4 and construction 4.3.4. By definition 4.3.1, \(X = \left[A\right]_{\sim _{\operatorname{Br}}}\). Hence by ?? and theorem 4.4.24, we evaluate the composition at \(X\) as:
![\begin{tikzcd}
{[A]_{\sim_{\br}}} \arrow[mapsto]{r} &
{\left[\mathcal{B}^2_x\right]} \arrow[mapsto]{r} &
{\left[\cross_{\mathcal{B}^2_x}\right]}
\end{tikzcd}](images/img-0008.svg)
Hence we need to prove that \(A\) and \(\operatorname{\mathfrak {C}}_{\mathcal{B}^{2}_{x}}\) are Brauer equivalent. We will show that they are isomorphic as \(F\)-algebras. since \(\{ x_{\sigma }|\sigma \in \operatorname{Gal}(K/F)\} \) is a \(K\)-basis for \(A\) and \(\{ \Delta _{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) is a \(K\)-basis for \(\operatorname{\mathfrak {C}}_{\mathcal{B}^{2}_{x}}\), they are certainly isomorphic as \(K\)-modules. Let \(\phi : \operatorname{\mathfrak {C}}_{\mathcal{B}^{2}_{x}} \cong A\) be the \(K\)-linear isomorphism defined by \(\Delta _{\sigma , 1} \mapsto x_{\sigma }\), since the \(K\)-action on \(A\) and the \(F\)-action on \(A\) are compatible (construction 4.3.2), \(\phi \) is also an \(F\)-linear isomorphism. Like in theorem 4.4.24, we check that \(\phi (1) = 1\) and \(\phi (xy)=\phi (x)\phi (y)\) for all \(x,y \in A\):
preservation of one: by construction 4.4.2, we have
\[ \begin{aligned} \phi (1) & = \phi \left(\Delta _{\mathsf{id},\mathcal{B}^{2}_{x}(\mathsf{id},\mathsf{id})^{-1}}\right) \\ & = \mathcal{B}^{2}_{x}(\mathsf{id},\mathsf{id})^{-1} \phi \left(\Delta _{\mathsf{id}, 1}\right) \\ & = \mathcal{B}^{2}_{x}(\mathsf{id},\mathsf{id})^{-1} x_{\mathsf{id}} \\ & = \operatorname{comp}_{x_{\mathsf{id}},x_{\mathsf{id}},x_{\mathsf{id}}}^{-1} x_{\mathsf{id}} \\ & = \operatorname{comp}_{x_{\mathsf{id}},x_{\mathsf{id}},x_{\mathsf{id}}} x_{\mathsf{id}} x_{\mathsf{id}} x_{\mathsf{id}}^{-1} \\ & = x_{\mathsf{id}} x_{\mathsf{id}}^{-1} \\ & = 1. \end{aligned} \]preservation of multiplication: let \(\sigma ,\tau \in \operatorname{Gal}(K/F)\) and \(c,d \in K\), by construction 4.4.2 and definition 4.3.3, we have
\[ \begin{aligned} \phi \left(\Delta _{\sigma ,c}\Delta _{\tau ,d}\right) & = \phi \left(\Delta _{\sigma \tau , c\sigma (d)\mathcal{B}^{2}_{x}(\sigma ,\tau )}\right) \\ & = c\sigma (d)\mathcal{B}^{2}_{x}(\sigma ,\tau ) \, \cdot \, \phi \left(\Delta _{\sigma \tau , 1}\right) \\ & = c\sigma (d)\mathcal{B}^{2}_{x}(\sigma ,\tau ) \, \cdot \, x_{\sigma \tau } \\ & = c\sigma (d)\operatorname{comp}_{x_{\sigma },x_{\tau },x_{\sigma \tau }} \, \cdot \, x_{\sigma \tau } \\ & = c\sigma (d)\, \cdot \, \iota _{A}\left(\operatorname{comp}_{x_{\sigma },x_{\tau },x_{\sigma \tau }}\right)x_{\sigma \tau } \\ & = c\sigma (d) \, \cdot \, x_{\sigma }x_{\tau }\\ \phi \left(\Delta _{\sigma ,c}\right)\phi \left(\Delta _{\tau ,d}\right) & =\left(c \cdot \phi \left(\Delta _{\sigma , 1}\right)\right) \left(d \cdot \phi \left(\Delta _{\tau , 1}\right)\right) \\ & = \left(c \cdot x_{\sigma }\right) \left(d \cdot x_{\tau }\right)\\ & = c \, \cdot \, x_{\sigma }\iota _{A}(d) x_{\tau } \\ & = c\sigma _{d} \, \cdot \, x_{\sigma }x_{\tau }. \end{aligned} \]
For a finite dimensional and Galois extension of field \(K/F\), the relative Brauer group \(K/F\) bijects to the second cohomology group \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) by the following commutative diagram
![\begin{tikzcd}
{\br(K/F)} \arrow{r}{\HH^2} \arrow[equal]{d} & {\HH^2\left(\gal(K/F), K^\star\right)} \arrow[equal]{d} \\
{\br(K/ F)} & {\HH^2\left(\gal(K/F), K^{\star}\right)} \arrow{l}{\cross}
\end{tikzcd}](images/img-0009.svg)
Exactly lemma 4.4.25 and lemma 4.4.26.
4.4.5 Group Homomorphism
In previous sections, when \(K/F\) is a finite dimensional Galois extension, we have set up a bijection between the relative Brauer group \(\operatorname{Br}(K/F)\) and the second cohomology group \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\). But both functions \(\operatorname{H}^{2}\) and \(\operatorname{\mathfrak {C}}\) are only set-theoretical function. In this section, we aim to upgrade them to group homomorphisms. Technically, we only need to prove either one of them preserves multiplication; we provide a proof that \(\operatorname{H}^{2}\) preserves one anyway because we found the proof to be entertaining.
\(\operatorname{\mathfrak {C}}_{1} = 1\) and \(\operatorname{H}^{2}(1) = 1\)
The function \(\operatorname{\mathfrak {C}}:\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right) \to \operatorname{Br}(K/F)\) preserves one, that is \(\operatorname{\mathfrak {C}}_{}\)
Since \(\{ \Delta _{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) is a \(K\)-basis for \(\operatorname{\mathfrak {C}}_{1}\) where \(1 \in \mathcal{B}^{2}\left(\operatorname{Gal}(K/F),K^{\star }\right)\) is the constant function \(1\) (lemma 4.4.17), we construct a \(K\)-linear map \(\phi : \operatorname{\mathfrak {C}}_{1} \to \operatorname{End}_{F} K\) by \(\Delta _{\sigma , 1} \mapsto \sigma \); note that \(\phi \) is \(F\)-linear as well. In fact, \(\phi \) is also an \(F\)-algebra homomorphism:
\(\phi (1) = 1\): indeed \(\phi \left(\Delta _{\mathsf{id},1}\right) = \mathsf{id}\).
\(\phi (xy) = \phi (x)\phi (y)\): indeed, let \(\sigma ,\tau \in \operatorname{Gal}(K/F)\) and \(c,d\in K\), we need to check that \(\phi \left(\Delta _{\sigma ,c}\Delta _{\tau ,d}\right) = \phi \left(\Delta _{\sigma ,c}\right)\phi \left(\Delta _{\tau ,d}\right)\). The left hand side is equal to
\[ \phi \left(\Delta _{\sigma \tau ,c\sigma (d)}\right) = \phi \left(c\sigma (d)\, \cdot \, \Delta _{\sigma \tau ,1}\right) = c\sigma (d)\, \cdot \, \sigma \tau ; \]and the right hand side is equal to
\[ \phi \left(c\cdot \Delta _{\sigma ,1}\right)\phi \left(d\cdot \Delta _{\tau , 1}\right) = (c \cdot \sigma ) (d \cdot \tau ). \]For any \(x \in K\), applying left hand side to \(x\) will result in \(c\sigma (d)\sigma (\tau (x))\) while right hand side will result in \(c\sigma (d\tau (x))\), hence both sides are equal.
Hence, \(\phi \) is an \(F\)-algebra isomorphism by corollary 1.1.8; that is we have \(\operatorname{\mathfrak {C}}_{1} \cong \operatorname{End}_{F}K \cong \operatorname{Mat}_{\dim _{F}K}(F)\). We conclude that \(\operatorname{\mathfrak {C}}_{1}\) is Brauer equivalent to \(F\) and consequently \(\operatorname{H}^{2}(1) = 1\).
The function \(\operatorname{H}^{2}:\operatorname{Br}(K/F) \to \operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) preserves one, that is \(\operatorname{H}^{2}(1) = 1\).
Apply \(\operatorname{\mathfrak {C}}\) then use lemma 4.4.26 and theorem 4.4.28.
\(\operatorname{\mathfrak {C}}_{\mathfrak {a}\mathfrak {b}}\sim _{\operatorname{Br}}\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\)
The argument in this section is more complicated, because, unlike before, the left hand side and the right hand side are not isomorphic as \(F\)-algebras — left hand side has \(F\)-dimension \(\left(\dim _{F}K\right)\) while the right hand side has \(F\)-dimension \(\left(\dim _{F}K\right)^{4}\). Let \(\mathfrak {a}\) and \(\mathfrak {b}\) be two 2-cocycles in \(\mathcal{B}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\), we denote \(\mathfrak {c}\) to be the 2-cocycle \(\mathfrak {a}\mathfrak {b}\). Intuitively, \(\operatorname{\mathfrak {C}}_{\mathfrak {a}} \otimes _{F} \operatorname{\mathfrak {C}}_{\mathfrak {b}}\) is too “big”, to address this issue we introduce a quotient module.
Consider the quotient module
For any \(a' \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and \(b' \in \operatorname{\mathfrak {C}}_{\mathfrak {b}}\), we can define an \(F\)-linear map \(M \to M\) by descending the \(F\)-linear map \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}} \to \operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes \operatorname{\mathfrak {C}}_{\mathfrak {b}}\)
we need to check that for all \(k \in K, a \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}, b \in \operatorname{\mathfrak {C}}_{\mathfrak {b}}\), the image of \((k\cdot a)\otimes b - a\otimes (k\cdot b)\) is in \(\left\langle (k\cdot a)\otimes b - a \otimes (k\cdot b) | k \in K, a \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}, b \in \operatorname{\mathfrak {C}}_{\mathfrak {b}} \right\rangle \): the image is \((k\, \cdot \, aa')\otimes b - a\otimes (k\, \cdot \, bb')\) which is in the generating set with \(k \in K, aa' \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}, \) and \(bb'\in \operatorname{\mathfrak {C}}_{\mathfrak {b}}\). This map is in fact \(F\)-linear in both \(a'\) and \(b'\), hence we have an \(F\)-bilinear map \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes \operatorname{\mathfrak {C}}_{\mathfrak {b}} \to M \to M\). This gives \(M\) a \(\left(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\right)^{\mathsf{opp}}\)-module structure given by
for any \(a,a' \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and \(b, b' \in \operatorname{\mathfrak {C}}_{\mathfrak {b}}\). All of the module axioms in this case follows from \(F\)-bilinearity.
For any \(c \in \operatorname{\mathfrak {C}}_{\mathfrak {c}}\), we can define another \(F\)-linear map \(M \to M\) by descending the \(F\)-linear map \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}} \to \operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes \operatorname{\mathfrak {C}}_{\mathfrak {b}}\)
we need check that for all \(k \in K, a \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}, b \in \operatorname{\mathfrak {C}}_{\mathfrak {b}}\), the image of \((k\cdot a)\otimes b - a \otimes (k\cdot b)\) is in \(\left\langle (k\cdot a)\otimes b - a \otimes (k\cdot b) | k \in K, a \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}, b \in \operatorname{\mathfrak {C}}_{\mathfrak {b}} \right\rangle \): by lemma 4.4.15 the image is
which is in \(\left\langle (k\cdot a)\otimes b - a \otimes (k\cdot b) | k \in K, a \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}, b \in \operatorname{\mathfrak {C}}_{\mathfrak {b}} \right\rangle \) because for each \(\sigma \in \operatorname{Gal}(K/F)\), the summand is in the generating set with \(\sigma (k) \in K, \Delta ^{\mathfrak {a}}_{\sigma , c(\sigma )}a \in \operatorname{\mathfrak {C}}_{\mathfrak {a}}\) and \(\Delta ^{\mathfrak {b}}_{\sigma , 1}b \in \operatorname{\mathfrak {C}}_{\mathfrak {b}}\). This map is in fact \(F\)-linear in \(c\), therefore we have an \(F\)-bilinear map \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\to M \to M\). (In the above calculation “\(\otimes \)” symbol has low precedence.) This gives \(M\) a \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-module structure given by
In particular, if \(c\) is of the form \(k \cdot \Delta ^{\mathfrak {c}}_{\tau , 1}\), then \(\left(k\cdot \Delta ^{\mathfrak {c}}_{\tau , 1}\right)\cdot [a \otimes b]\) is equal to \(\left[\left(k\cdot \Delta ^{\mathfrak {a}}_{\tau ,1}\right)a \otimes \Delta ^{\mathfrak {b}}_{\tau , 1} b\right]\) because \(\Delta _{\tau , 1}^{\mathfrak {c}}(\sigma ) = 0\) for all \(\sigma \ne \tau \). Two of the module axioms need more than \(F\)-bilinearity:
\(c = 1\): note that \(c = 1 = \mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\, \cdot \, \Delta ^{\mathfrak {c}}_{\mathsf{id}, 1}\), hence
\[ \begin{aligned} 1 \cdot [a\otimes b] & = [\mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}\Delta ^{\mathfrak {a}}_{\mathsf{id},1}a \otimes \Delta ^{\mathfrak {b}}_{\mathsf{id}, 1}b] \\ & = [\Delta ^{\mathfrak {a}}_{\mathsf{id},\mathfrak {a}(\mathsf{id},\mathsf{id})^{-1}}a \otimes \Delta ^{\mathfrak {b}}_{\mathsf{id},\mathfrak {b}(\mathsf{id},\mathsf{id})^{-1}}b] \\ & = [a\otimes b]. \end{aligned} \]\(c_{1}c_{2}\cdot [a \otimes b] = c_{1}\cdot c_{2}\cdot [a\otimes b]\): assume \(c_{1} = k_{1}\cdot \Delta ^{\mathfrak {c}}_{\tau _{1}, 1}\) and \(c_{2}=k_{2}\cdot \Delta ^{\mathfrak {c}}_{\tau _{2},1}\). Then \(c_{1}c_{2} = k_{1}\tau _{1}\left(k_{2}\right)\cdot \Delta ^{\mathfrak {c}}_{\tau _{1}\tau _{2},\mathfrak {c}(\tau _{1},\tau _{2})} = k_{1}\tau _{1}\left(k_{2}\right)\mathfrak {a}(\tau _{1},\tau _{2})\mathfrak {b}(\tau _{1},\tau _{2})\cdot \Delta ^{\mathfrak {c}}_{\tau _{1}\tau _{2},1}\). Therefore, the left hand side is equal to
\[ \begin{aligned} & \left[\left(k_{1}\tau _{1}(k_{2})\mathfrak {a}(\tau _{1},\tau _{2})\mathfrak {b}(\tau _{1},\tau _{2})\cdot \Delta ^{\mathfrak {a}}_{\tau _{1}\tau _{2}, 1}\right)a \otimes \Delta ^{\mathfrak {b}}_{\tau _{1}\tau _{2},1}b\right] \\ =& \left[ k_{1}\tau _{1}(k_{2})\mathfrak {a}(\tau _{1},\tau _{2})\cdot \Delta ^{\mathfrak {a}}_{\tau _{1}\tau _{2},1} a \otimes \mathfrak {b}(\tau _{1},\tau _{2})\Delta ^{\mathfrak {b}}_{\tau _{1}\tau _{2}, 1}b \right]\\ =& \left[ \Delta ^{\mathfrak {a}}_{\tau _{1},k_{1}}\Delta ^{\mathfrak {a}}_{\tau _{2},k_{2}} a \otimes \Delta ^{\mathfrak {b}}_{\tau _{1},1}\Delta ^{\mathfrak {b}}_{\tau _{2},1}b \right]; \end{aligned} \]and the right hand side is also equal to
\[ \begin{aligned} & \left(k_{1}\cdot \Delta ^{\mathfrak {c}}_{\tau _{1}, 1}\right) \left[ k_{2}\cdot \Delta ^{\mathfrak {a}}_{\tau _{2},1}a \otimes \Delta ^{\mathfrak {b}}_{\tau _{2}, 1} b \right] \\ =& \left[ \left(k_{1}\cdot \Delta ^{\mathfrak {a}}_{\tau _{1}, 1}\right)\left(k_{2}\cdot \Delta ^{\mathfrak {a}}_{\tau _{2}, 1}\right)a \otimes \Delta ^{\mathfrak {b}}_{\tau _{1},1}\Delta ^{\mathfrak {b}}_{\tau _{2},1}b \right]\\ =& \left[ k_{1}\tau _{1}(k_{2})\cdot \Delta ^{\mathfrak {a}}_{\tau _{1}\tau _{2},\mathfrak {a}(\tau _{1},\tau _{2})}a \otimes \Delta ^{\mathfrak {b}}_{\tau _{1}, 1}\Delta ^{\mathfrak {b}}_{\tau _{2},1}b \right]\\ =& \left[ \left(k_{1}\cdot \Delta ^{\mathfrak {a}}_{\tau _{1}, 1}\right)\left(k_{2}\cdot \Delta ^{\mathfrak {a}}_{\tau _{2}, 1}\right)a \otimes \Delta ^{\mathfrak {b}}_{\tau _{1},1}\Delta ^{\mathfrak {b}}_{\tau _{2},1}b \right]. \end{aligned} \]
Expanding everything out and checking on the basic elements, we see that for any \(x \in \left(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\right)^{\mathsf{opp}}\), \(y \in \operatorname{\mathfrak {C}}_{\mathfrak {c}}\) and \(z \in M\), \(x \cdot y \cdot z = y \cdot x \cdot z\). In another word, we gave \(M\) a \(\left(\operatorname{\mathfrak {C}}_{\mathfrak {c}}, \operatorname{\mathfrak {C}}_{\mathfrak {a}} \otimes _{F} \operatorname{\mathfrak {C}}_{\mathfrak {b}}\right)\)-bimodule structure.
\(M\) is isomorphic to \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\) as \(F\)-modules.
The map \(M \to \operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\) is obtained by descending the obvious \(F\)-linear map \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\to \operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\). By universal property of tensor product, there is an additive group homomorphism \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}} \to M\) given by \(a \otimes b \mapsto [a\otimes b]\), this map is in fact \(F\)-linear. The two maps are inverse to each other.
The \(F\)-dimension of \(M\) is equal to \({\left(\dim _{F}K\right)}^{3}\), consequently \(M\) is a finitely generated \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-module.
By lemma 4.4.30, the dimension of \(M\) is equal to \(\dim _{F}\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}} = \dim _{K}\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\dim _{F}K\). By lemma 4.4.17, \(\dim _{K}\operatorname{\mathfrak {C}}_{\mathfrak {a}} = \dim _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}} = \dim _{F}K\).
By lemma 2.2.2, there exists some simple \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-module \(S\) such that \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\) is isomorphic to \(\bigoplus _{i\in J}S\) as \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-module for some indexing set \(J\). If we give \(S\) the \(F\)-module structure by pulling back the \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-module structure, by restricting scalars \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\) is isomorphic to \(\bigoplus _{i\in J} S\) as \(F\)-module as well. Since \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\) is a finite dimensional \(F\)-vector space, \(J\) must be finite as well. Note that \(S\) must be a finite dimensional \(F\)-vector space, because \(S\) is finitely generated as \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-module and \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\) has finite \(F\)-dimension. The indexing set \(J\) must be nonempty, otherwise \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\) being isomorphic to \(\bigoplus _{\emptyset } S\) is a trivial ring; but simple rings are non-trivial. Since \(J\) is finite, direct sum over \(J\) and direct product over \(J\) agree. Recall construction 3.1.1 and construction 3.1.2, for all non-zero \(m \in \mathbb {N}\), we have
as \(F\)-algebras, hence
as \(F\)-algebras. Finally
as \(F\)-algebras.
They are both equal to \(\dim _{F}\operatorname{\mathfrak {C}}_{\mathfrak {c}}\) by construction 4.4.11.
They are all equal to \(\dim _{F}\operatorname{\mathfrak {C}}_{\mathfrak {c}} = \dim _{F}S^{|J|}\) by construction 4.4.11.
There exists a \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-linear isomorphism between \(M\) and \(S^{|J|\dim _{F}K}\).
By lemma 2.2.4, we only need to show that \(\dim _{F}M = \dim _{F}S^{|J|\dim _{F}K}\). We already have \(\dim _{F}M = \left(\dim _{F}K\right)^{3}\) by corollary 4.4.31. We also have \(\dim _{F}S^{|J|\dim _{F}K} = |J|\dim _{F}K\dim _{F}S = \dim _{F}K \left(|J|\dim _{F}S\right) = \dim _{F}K\left(\dim _{F}K\right)^{2}\) by corollary 4.4.33.
As \(F\)-vector spaces, \(M \cong S^{|J|\dim _{F}K}\).
Restricting scalars on the \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-linear isomorphism in lemma 4.4.34
As \(F\)-algebras, \(\operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}M \cong \operatorname{Mat}_{|J|\dim _{F}K}(\operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}S)\).
From corollary 4.4.35, we have \(\operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}M\cong \operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}\left(S^{|J|\dim _{F}K}\right)\). By construction 3.1.2, they are isomorphic to \(\operatorname{Mat}_{|J|\dim _{F}K}(\operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}S)\).
where the last equality is by corollary 4.4.32.
The cross product \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\) and the tensor product \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\) are Brauer equivalent.
We define an \(F\)-algebra homomorphism \(\phi : \left(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\right)^{\mathsf{opp}} \to \operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}M\) by \(x \mapsto (x \cdot \bullet )\). By corollary 4.4.37, both sides has \(F\)-dimension \(\left(\dim _{F}K\right)^{4}\), therefore, \(\phi \) is an \(F\)-algebra isomorphism by corollary 1.1.8. Hence we have another \(F\)-algebra isomorphism by composing the isomorphism in corollary 4.4.36:
In the meantime, by construction 4.4.11, we have \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\cong \operatorname{Mat}_{|J|}\left(\left(\operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}S\right)\mathsf{opp}\right)\); hence \(\operatorname{Mat}_{\dim _{F}K}\left(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\right)\) is isomorphic to \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{F}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\).
For a finite dimensional Galois field extension \(K/F\), the relative Brauer group \(\operatorname{Br}(K/F)\) is isomorphic to the second group cohomology \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F),K^{\star }\right)\).
In corollary 4.4.27, we have seen that \(\operatorname{H}^{2}\) and \(\operatorname{\mathfrak {C}}\) form a bijection, thus it is sufficient to check either one of them preserves multiplication. The function \(\operatorname{\mathfrak {C}}: \operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\) preserves multiplication: let \([\mathfrak {a}], [\mathfrak {b}]\) be two elements in \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right)\), by theorem 4.4.38\(, \operatorname{\mathfrak {C}}(\mathfrak {a}\mathfrak {b})\) is indeed Brauer equivalent to \(\operatorname{\mathfrak {C}}(\mathfrak {a})\otimes _{F}\operatorname{\mathfrak {C}}(\mathfrak {b})\) that is