Dependencies

A ring \(R\) is simple if the only two-sided-ideals of \(R\) are \({0}\) and \(R\). An algebra is simple if it is simple as a ring.

Division rings are simple.

Let \(R\) be a ring and \(A\) an \(R\)-algebra, we say \(A\) is central if and only if the centre of \(A\) is \(R\)

Every commutative ring is a central algebra over itself.

Simpleness is invariant under ring isomorphism and centrality is invariant under algebra isomorphism.

If \(A\) is a simple \(R\)-algebra, “ring homomorphism” in lemma 1.1.6 can be replaced with \(R\)-algebra homomorphism.

For any field \(K\) and \(K\)-algebra \(A\), a subfield \(B \subseteq A\) is a commutative \(K\)-subalgebra of \(A\) that is closed under inverse for any non-zero member.

Subfields inherit a natural ordering from subalgebras.

Note that all modules are assumed to be left modules; when we need to consider right \(R\)-modules, we will consider left \(R^{\mathsf{opp}}\)-modules instead. We use \(\delta _{ij}\) to denote the matrix whose \((i,j)\)-th entry is \(1\) and \(0\) elsewhere. \(\delta _{ij}\) forms a basis for matrices.

Note that by lemma 2.2.1, any two simple \(A\)-module are isomorphic, hence for any \(A\)-module \(M\) and any simple \(A\)-module \(S\), we can write \(M\) as a direct sum of copies of \(S\).

Note that any \(A\)-linear endomorphism of \(D^{n}\) is \(\operatorname{Mat}_{n}(D)\)-linear, and vice versa. Thus we have \(\operatorname{End}_{A}\left(D^{n}\right)\cong \operatorname{End}_{\operatorname{Mat}_{n}(D)}\left(D^{n}\right)\) as \(k\)-algebras.

In particular, if \(M\) is a simple \(A\)-module, then \(\operatorname{End}_{A}M\) is a simple \(k\)-algbera.

Let \(M\) be a simple \(A\)-module, then \(\operatorname{End}_{A}M\) has finite \(k\)-dimension.

Note that for all \(A\)-module \(M\), \(\operatorname{End}_{\operatorname{End}_{A}M}M\) is a \(k\)-algebra as well, with \(k \hookrightarrow \operatorname{End}_{\operatorname{End}_{A}M}M\) given by \(a \mapsto (x \mapsto a\cdot x)\). Thus, we always have a \(k\)-algebra homomorphism \(A \to \operatorname{End}_{\operatorname{End}_{A}M}M\) given by the \(A\)-action on \(M\). When \(A\) is a simple ring, this map is injective.

Balancedness is invariant under linear isomorphism.

Any simple \(A\)-module is balanced.

A useful special case is when \(S=A\), then since \(C_{A}(A)=Z(A)\), we have \(C_{A\otimes _{R}B}\left(\operatorname{im}\left(A\to A\otimes _{R} B\right)\right)\) is equal to \(Z(A)\otimes _{R} B\). Since \(\operatorname{im}\left(R\otimes _{R}B\to A\otimes _{R}B\right)=\operatorname{im}\left(A\to A\otimes _{R}B\right)\), we conclude its centralizer in \(A\otimes _{R}B\) is \(Z(A)\otimes _{R}B\).

Let \(A\) be a simple ring and \(I\) a non-trivial left ideal. One can write \(1 \in A\) as \(\sum _{i=0}^{n}x_{i}y_{i}\) for some \(x_{i}\in I\) and \(y_{i} \in A\).

The \(n\), \(x_{i}\) and \(y_{i}\) are all non-zero.

For any \(K\)-algebra homomorphism \(f : B \to A\), we give \(M\) a \(B \otimes _{K}\operatorname{End}_{A}M\)-module structure by defining \((b\otimes l)\cdot m\) to be \(f(b)\cdot l(m)\). To emphasis \(f\), we denote \(M\) with the \(B\otimes _{K}\operatorname{End}_{A}M\)-module structure by \(M^{f}\).

Let \(f : B \to A\) be a \(K\)-algebra homomorphism, \(M^{f}\) is finitely generated as a \(B\otimes _{K}\operatorname{End}_{A}M\)-module.

Given that \(B\) is simple, any \(k\)-algebra homomorphism \(f : B\to A\) injective; therefore by finite \(K\)-dimensionality of \(A\), \(B\) is finite \(K\)-dimensional as well.

\(A\) is a \(Z(A)\)-algebra whose algebra structure is given by \(Z(A)\hookrightarrow A\). By lemma 1.1.2, \(Z(A)\) is a field. \(A\) is finite dimensional as a \(Z(A)\)-module because of the tower \(A/Z(A)/F\).

Let \(B\) be any \(F\)-algebra and \(S\subseteq B\) an \(F\)-subalgebra. For any \(x\in B^{\times }\), we have that \(x S x^{-1}:=\{ xsx^{-1}|s\in S\} \) is an \(F\)-subalgebra of \(B\) as well. We have the obvious \(F\)-algebra isomorphism \(S \cong xSx^{-1}\) given by \(s\mapsto xsx^{-1}\) and \(x^{-1}tx\leftarrow \@undefined t\). Therefore \(\dim _{F}S=\dim _{F}xSx^{-1}\) and \(S\) is a simple ring if and only if \(xSx^{-1}\) is a simple ring.

For any finite dimensional \(F\)-module \(B\), we have isomorphism \(\operatorname{End}_{F}B \cong \operatorname{Mat}_{\dim _{F}B}F\) as \(F\)-algebras. Hence \(\operatorname{End}_{F} B\) is a finite-dimensional central simple algebra over \(F\).

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.

Isomorphic \(K\)-algebras are Brauer equivalent.

\(\sim _{\operatorname{Br}}\) is reflexive.

\(\sim _{\operatorname{Br}}\) is symmetric.

\(\sim _{\operatorname{Br}}\) is transitive.

\(\operatorname{Br}\) forms a functor from category of field to category of abelian groups.

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 \(E\)

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 \([B]_{\sim \operatorname{Br}}\in \operatorname{Br}(E/K)\).

The range \(\iota _{A}(A)\) is a simple ring.

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.

If \(x\) is a conjugation factor of \(\sigma \) and \(y\) of \(\tau \), then \(xy\) is a conjugation factor of \(\sigma \tau \). For any \(c \in K\)

\(\operatorname{twist}(x, x)\) is equal to \(1\) by uniqueness.

In fact, \(\operatorname{twist}(x, y)\) is in \(K^{\star }\) and \(\operatorname{twist}(x, y)^{-1}=\operatorname{twist}(y, x)\).

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.

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.

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 \(\operatorname{H}^{2}(1) = 1\).

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.

For a finite dimensional Galois field extension \(K/F\), the relative Brauer group \(\operatorname{Br}(K/F)\) is isomorphic to the second group cohomology \(\operatorname{H}^{2}\left(\operatorname{Gal}(K/F),K^{\star }\right)\).

\(\operatorname{Br}(K)\) 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:

1. 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.

2. 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 \(K\sim _{\operatorname{Br}}\operatorname{Mat}_{n}(K)\).

3. 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.

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.

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 \(\operatorname{H}^{2} : \operatorname{Br}(K/F) \to \operatorname{H}^{2}\left(\operatorname{Gal}(K/F),K^{\star }\right)\).

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

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 \(\Delta _{\sigma , r\cdot c}\).

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.

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.

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.

Let \(A\) be an \(R\)-algebra and \(M\) an \(A\)-module. We have isomorphism \(\operatorname{End}_{A}\left(M^{n}\right)\cong \operatorname{Mat}_{n}\left(\operatorname{End}_{A}M\right)\) as \(R\)-algebras. For any \(f \in \operatorname{End}_{A}\left(M^{n}\right)\), we define a matrix \(M\) whose \((i,j)\)-th entry is

where \(x\) is at the \(j\)-th position. On the other hand, if \(M \in \operatorname{Mat}_{n}\left(\operatorname{End}_{A}M\right)\), we define an \(A\)-linear map \(f : M^{n}\to M^{n}\) by

\(A\) 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.

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 \(u\).

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 \(\dim _{F}K\cdot \dim _{K}A = \dim _{F}K\cdot \dim _{F}K = \dim _{F}A\).

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.

Let \(A\) be an \(R\)-algebra. Then \(\operatorname{Mat}_{m}\left(\operatorname{Mat}_{n}(A)\right) \cong \operatorname{Mat}_{mn}(A)\). The trick is to think \(\operatorname{Mat}_{m}A\) as \(\{ 0,\dots ,m-1\} \times \{ 0,\dots ,m-1\} \to A\). Since the indexing set \(\{ 0,\dots ,mn-1\} \) bijects with \(\left(\{ 0,\dots ,m-1\} \times \{ 0,\dots ,n-1\} \right)\), the isomorphism is just function currying, function uncurrying, precomposing and postcomposing bijections.

Let \(A, B\) be \(R\)-algebras. Then \(\operatorname{Mat}_{mn}\left(A\otimes _{R}B\right)\cong \operatorname{Mat}_{m}(A)\otimes _{R}\operatorname{Mat}_{n}B\) as \(K\)-algebras. We first construct \(R\)-algebra isomorphism \(A \otimes _{R}\operatorname{Mat}_{n}(R) \cong \operatorname{Mat}_{n}(A)\):

where \(\operatorname{diag}\) is the diagonal matrix and \(\delta _{ij}\) the matrix whose only non-zero entry is at \((i,j)\)-th and is equal to \(1\). Thus \(\operatorname{Mat}_{m}(A)\otimes _{R}\operatorname{Mat}_{n}(B)\cong \left(A\otimes _{R}B\right)\otimes _{R} \left(\operatorname{Mat}_{m}(R)\otimes _{R} \operatorname{Mat}_{n}(R)\right)\) as \(R\)-algebra. The Kronecker product gives us an \(R\)-algebra map \(\operatorname{Mat}_{m}(R)\otimes _{R}\operatorname{Mat}_{n}(R) \to \operatorname{Mat}_{mn}(R)\). We want this map to be an isomorphism. By lemma 1.1.6, we only need to prove it to be surjective: for all \(\delta _{ij} \in \operatorname{Mat}_{mn}(R)\), we interpret \(\operatorname{Mat}_{mn}(R)\) as a function \(\{ 0,\dots ,m-1\} \times \{ 0,\dots ,n-1\} \to R\), then \(\delta _{ij}\) is the image of \(\delta _{ab}\otimes \delta _{cd} \in \operatorname{Mat}_{m}(R)\otimes _{R}\operatorname{Mat}_{n}(R)\) where \(i = (a, c)\) and \(j = (b,d)\). Combine everything together, we see \(\operatorname{Mat}_{mn}\left(A\otimes _{R}B\right)\) is isomorphic to \(\operatorname{Mat}_{mn}\left(A\otimes _{R} B\right)\) as \(R\)-algebras.

If \(M\) is an \(R\)-module, we have a natural \(\operatorname{Mat}_{n}(R)\)-module structure on \(\widehat{M}:=M^{n}\) given by \((m_{ij})\cdot (v_{k})=\sum _{j}m_{ij}\cdot v_{j}\). If \(f : M \to N\) is an \(R\)-linear map, then \(\widehat{f} : M^{n}\to N^{n}\) given by \((v_{i}) \mapsto (f(v_{i}))\) is a \(\operatorname{Mat}_{n}(R)\)-linear map. Thus we have a well-defined functor \(\operatorname{\mathfrak {Mod}}_{R} \Longrightarrow \operatorname{\mathfrak {Mod}}_{\operatorname{Mat}_{n}(R)}\).

If \(M\) is a \(\operatorname{Mat}_{n}(R)\)-module, then \(\widetilde{M} := \{ \delta _{ij}\cdot m | m \in M\} \subseteq M\) is an \(R\)-module whose \(R\)-action is given by \(r \cdot (\delta _{ij}\cdot m) := (r\cdot \delta _{ij})\cdot m\). More over if \(f : M \to N\) is a \(\operatorname{Mat}_{n}(R)\)-linear map, \(\widetilde{f} : \widetilde{M} \to \widetilde{N}\) given by the restriction of \(f\) is \(R\)-linear. Hence, we have a functor \(\operatorname{\mathfrak {Mod}}_{\operatorname{Mat}_{n}(R)}\Longrightarrow \operatorname{\mathfrak {Mod}}_{R}\).

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 \(R\) be a commutative ring and \(A\) an \(R\)-algebra. Then we have an \(R\)-algebra homomorphism \(A\otimes _{R}A^{\mathsf{opp}} \cong \operatorname{End}_{R}A\) given by \(a\otimes 1 \mapsto (a\cdot \bullet )\) and \(1\otimes a \mapsto (\bullet \cdot a)\). When \(R\) is a field and \(A\) is a finite dimensional central simple algebra, this morphism is an isomorphism by corollary 1.1.8.

Assume \(R\) is a field. Let \(A, B\) be finite dimensional \(R\)-algebras where \(A\) is simple as well. Then any \(R\)-algebra homomorphism \(f:A\to B\) is bijective if \(\dim _{R}A=\dim _{R}B\).

Assume \(R\) is a field. The centre of \(A\otimes _{R} B\) is \(Z\left(A\right)\otimes _{R}Z\left(B\right)\).

Assume \(R\) is a field. Let \(S\) and \(T\) be \(R\)-subalgebras of \(A\) and \(B\) respectively. Then \(C_{A\otimes _{R}B}\left(S\otimes _{R}T\right)\) is equal to \(C_{A}(S)\otimes _{R}C_{B}(T)\)

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 \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\).

\(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is a simple ring.

Central simple algebras are stable under base change. That is, if \(L/K\) is a field extension and \(D\) is a central simple algebra over \(K\), then \(L\otimes _{K} D\) is central simple over \(L\).

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 \({\left(\dim _{F}K\right)}^{2}\).

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.

As \(F\)-algebras, \(\operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}M \cong \operatorname{Mat}_{|J|\dim _{F}K}(\operatorname{End}_{\operatorname{\mathfrak {C}}_{\mathfrak {c}}}S)\).

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 \(K\hookrightarrow B\)

• \({\left(\dim _{F}K\right)}^{2}=\dim _{F}B\).

As \(F\)-vector spaces, \(M \cong S^{|J|\dim _{F}K}\).

For any \(X\in \operatorname{Br}(F)\), \(X\in \operatorname{Br}(K/F)\) if and only if \(X\) admits a good representation.

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

.

Let \(S \subseteq A\) be a central simple \(F\)-subalgebra,

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.

For any ring \(A\) and \(A\)-module \(M\), we say \(M\) is a balanced \(A\)-module, if the \(A\)-linear map \(A \to \operatorname{End}_{\operatorname{End}_{A}M}M\) is surjective.

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.

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.

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 \(\iota _{A}\).

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

For any ring \(A\), \(A\) is balanced as \(A\)-module.

\(\operatorname{Br}_{K}^{K}\) is identity.

Consider the tower of field extension \(E/F/K\),

\((\bullet \otimes _{K}\bullet ):\operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\times \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\to \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}\) descends to a function on \(\operatorname{Br}(K)\).

Let \(A\) be a simple ring, then centre of \(A\) is a field.

Let \(B\) be any \(F\)-algebra, \(x \in B^{\times }\) and \(S\subseteq B\) be an \(F\)-subalgebra of \(B\), then \(C_{B}(xSx^{-1})=x\left(C_{B}(S)\right)x^{-1}\).

If we assume \(B\) is free as \(R\)-module, then for any \(R\)-subalgebra \(S\), we have that \(C_{A\otimes _{R}B}\left(\operatorname{im}\left(S\to A\otimes _{R}B\right)\right)\) is \(C_{A}(S) \otimes _{R} B\)

Let \(B\) be any simple \(F\)-algebra (not  necessarily central). The centralizer of \(\mathcal{L}_{B}\) in \(\operatorname{End}_{F}B\) is equal to \(\mathcal{R}_{B}\).

The centralizer of \(\mathcal{L}_{A}\) in \(\operatorname{End}_{F}A\) is smaller than or equal to \(\mathcal{R}_{A}\):

\(C_{A}\left(\iota _{A}(A)\right) = \iota _{A}(A)\).

Let \(S\subseteq A\) be a simple \(F\)-subalgebra, then there exists an \(x\in \left(A\otimes _{F}\operatorname{End}_{F}S\right)^{\times }\) such that \(C_{A}(S)\otimes _{F}\operatorname{End}_{F}S\) is isomorphic to \(x\left(A\otimes _{F}\mathcal{R}_{S}\right)x^{-1}\) as \(F\)-algebras.

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.

Let \(x : \operatorname{Gal}(K/F) \to A^{\star }\) be a conjugation sequence. We have

Let \(x\) be an \(A\)-conjugation sequence and \(y\) a \(B\)-conjugation sequence. We have

Let \(x\) be an \(A\)-conjugation sequence and \(y\) a \(B\)-conjugation sequence. We have

The set \(\{ \Delta _{\sigma , 1}|\sigma \in \operatorname{Gal}(K/F)\} \) forms a \(K\)-basis for \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\).

For any \(c \in K\), we have

and consequently,

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 \(c, d \in K\),

For every \(\sigma \in \operatorname{Gal}(K/F)\), \(\Delta _{\sigma ,1}\) is invertible.

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 \({}^{\operatorname{\mathfrak {C}}_{\mathfrak {a}}}/_{I}\).

Let \(S\subseteq A\) be a simple \(F\)-subalgebra. Then

Let \(k\) be a maximal subfield of \(D\),

Let \(M\) be an \(A\)-module, there exists a simple \(A\)-module \(S\) such that \(M\) is a direct sum of copies of \(S\), i.e. \(M \cong \bigoplus _{i \in \iota } S\) for some indexing set \(\iota \).

Let \(M\) be a simple \(A\)-module, then \(\operatorname{End}_{A} M\cong D^{\mathsf{opp}}\) as \(k\)-algebras.

\(\operatorname{End}_{A}\left(D^{n}\right)\) is isomorphic to \(D^{\mathsf{opp}}\) as \(k\)-algebras.

\(D^{n}\) is a simple \(A\)-module where the module structure is given by pulling back the \(\operatorname{Mat}_{n}(D)\)-module structure of \(D^{n}\).

Let \(M\) and \(N\) be \(R\)-modules such that \(\mathcal{C}_{i\in \iota }\) is a basis for \(N\), then every elements of \(x \in M \otimes _{R} N\) can be uniquely written as \(\sum _{i\in \iota }m_{i}\otimes \mathcal{C}_{i}\) where only finitely many \(m_{i}\)’s are non-zero

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 \(K\hookrightarrow B\)

• \({\left(\dim _{F}K\right)}^{2}=\dim _{F}B\).

If \(A\) and \(B\) are two good representations of \(X\), then \(A \cong B\) as \(F\)-algebras.

For any \(c \in K\), \(\sigma \in \operatorname{Gal}(K/F)\) and \(A\)-conjugation factor \(x\) of \(\sigma \), we have

For any simple \(A\)-module \(M\), we have \(A \cong \operatorname{End}_{\operatorname{End}_{A}M}M\) as \(k\)-algebras.

Let \(f,g : B \to A\) be two \(K\)-algebra homomorphisms. Then \(M^{f}\) and \(M^{g}\) are isomorphic as \(B\otimes _{K}\operatorname{End}_{A}M\)-module.

Let \(M\) and \(N\) be two finite \(A\)-module with compatible \(k\)-action. Then \(M\) and \(N\) are isomorphic as \(A\)-module if and only if \(\dim _{k} M\) and \(\dim _{k} N\) are equal.

Let \(M\) and \(N\) be simple \(A\)-modules, then \(M\) and \(N\) are isomorphic as \(A\)-modules.

There exists a \(\operatorname{\mathfrak {C}}_{\mathfrak {c}}\)-linear isomorphism between \(M\) and \(S^{|J|\dim _{F}K}\).

\(M\) is isomorphic to \(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\otimes _{K}\operatorname{\mathfrak {C}}_{\mathfrak {b}}\) as \(F\)-modules.

As \(F\)-algebras, we have \(\mathcal{R}_{A}\cong A^{\mathsf{opp}}\).

If \(A\) is a central \(R\)-algerba, \(A^{\mathsf{opp}}\) is also central. .

The composition of \(\operatorname{\mathfrak {C}}\) and \(\operatorname{H}^{2}\) is the identity:

The composition of \(\operatorname{H}^{2}\) and \(\operatorname{\mathfrak {C}}\) is the identity:

.

Let \(S\subseteq A\) be a simple \(F\)-subalgebra, then \(C_{A}(S)\) is simple as well.

\(R\) is a simple ring if and only if any ring homomorphism \(f : R \to S\) either injective or \(S\) is the trivial ring.

Let \(K\) be a field, \(M\) and \(N\) be flat \(K\)-modules. Suppose \(p \subseteq M\) and \(q \subseteq N\) are \(K\)-submodules, then \((p \otimes _{K} N) \sqcap (M \otimes _{K} q) = p \otimes _{K} q\) as \(K\)-submodules.

Let \(S, T\) be two subalgebras of \(A\), then \(C_{A}(S\sqcup T)=C_{A}(S)\sqcap C_{A}(T)\).

If \(A\) and \(B\) are central \(K\)-algebras, \(A\otimes _{K}B\) is a central \(K\)-algebra as well.

Let \(S\subseteq A\) be a simple \(F\)-subalgebra, then \(A \otimes _{F} \mathcal{R}_{S}\) is a simple ring.

Suppose \(L\) is a subfield of \(A\), the following are equivalent:

1. \(L\) = \(C_{A}(L)\)

2. \(\dim _{K}A = {\left(\dim _{K}L\right)}^{2}\)

3. for any commutative \(K\)-subalgebra \(L' \subseteq A\), \(L \subseteq L'\) implies \(L = L'\)

If \(x\) and \(y\) are two conjugation factors of \(\sigma \), then there exists a unique \(c \in K\) such that \(x = y\iota _{A}(c)\).

If \(x\) and \(y\) are conjugation factors for \(\sigma \), \(x=\iota _{A}(\sigma (\operatorname{twist}_{x, y}))y\).

Let \(A\) be a ring and \(I\) a non-trivial minimal left ideal of \(A\), then \(I\) is a simple \(A\)-module.

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.

For any \(F\)-algebra \(B\), every element in \(C_{\operatorname{End}_{F}B}\left(\mathcal{L}_{B}\right)\) is in fact \(Z(B)\)-linear. Let \(x\in C_{\operatorname{End}_{F}B}\left(\mathcal{L}_{B}\right)\), \(z\in Z(B)\) and \(b \in B\), we have \(x(z\cdot b) = z\cdot x(b)\) because \(x\) commutes with the linear map \(\left(z\cdot \bullet \right)\).

If \(K\) is algebraically closed, \(\operatorname{Br}(K)\) is trivial; in particular \(\operatorname{Br}_{n}(\mathbb {C})\) is trivial.

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 \(A\).

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.

\(\operatorname{\mathfrak {C}}_{\mathfrak {a}}\) is a central \(F\)-algebra.

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.

Let \(S\subseteq A\) be a simple \(F\)-subalgebra, we have

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 function \(\operatorname{\mathfrak {C}}:\operatorname{H}^{2}\left(\operatorname{Gal}(K/F), K^{\star }\right) \to \operatorname{Br}(K/F)\) preserves one, that is \(\operatorname{\mathfrak {C}}_{}\)

The functors constructed in construction 2.1.1 and construction 2.1.2 form an equivalence of category.

Let \(f, g : B \to A\) be two \(K\)-algebra homomorphism. Then \(f\) and \(g\) differ only by a conjugation. That is there exists a unit \(x \in A^{\times }\) such that \(g = x f x^{-1}\).

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 \(E/K\) be a field extension and \(A\in \operatorname{\mathsf{C}\! \mathsf{S}\! \mathsf{A}}_{K}\), \(E\) splits \(A\) if and only if \([A]_{\sim _{\operatorname{Br}}}\in \operatorname{Br}(E/K)\).

If \(A\) is a simple \(K\)-algebra and \(B\) is a central simple \(K\)-algebra, \(A\otimes _{K}B\) is a central simple \(K\)-algebra as well.

If \(A\otimes _{K} B\) is a simple ring, then \(A\) and \(B\) are both simple.

Let \(A\) be a simple ring and \(I\) a non-trivial minimal left ideal. Then there exists a non-zero \(n \in \mathbb {N}\) such that \(A \cong I^{n}\) as \(A\)-modules.

Let \(K\) be a field and \(B\) an finite dimensional simple algebra over \(K\). There exists a non-zero \(n\in \mathbb {N}\) and a division \(K\)-algebra \(S\) such that \(B \cong \operatorname{Mat}_{n}(S)\).

Let \(A\) be an Artinian simple ring. There exists a non-zero \(n\) and an ideal \(I \subseteq A\) such that \(I\) is simple as an \(A\)-module and \(A \cong I^{n}\) as \(A\)-module.

Let \(B\) be a finite-dimensional simple \(K\)-algebra. Suppose \(B\) is isomorphic as \(k\)-algebras to both \(\operatorname{Mat}_{n}(D)\) and \(\operatorname{Mat}_{n'}(D')\) where \(n, n'\) are non-zero natural numbers and \(D, D'\) are \(k\)-division algebra, then \(n = n'\) and \(D \cong D'\) as \(k\)-algebras.