3 Results in Noncommutative Algebra
3.1 A Collection of Useful Lemmas
In section, we collect some lemmas that does not really belong to anywhere.
3.1.1 Tensor Product
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
Given the basis \(\mathcal{C}\), we have \(R\)-linear isomorphism \(N \cong \bigoplus _{i\in \iota }R\), hence \(M\otimes _{R}N \cong \bigoplus _{i\in \iota }(M\otimes _{R}R)\cong \bigoplus _{i\in \iota }M\) as \(R\)-modules.
By switching \(M\) and \(N\), the symmetric statement goes without saying.
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.
The hard direction is to show \((p \otimes _{R} N) \sqcap (M \otimes _{R} q) \le p \otimes _{R} q\). Consider the following diagram:
![\begin{tikzcd}
p \otimes_{K} q \ar[r, "u"] \ar[d, "\alpha"] & M \otimes_{K} q \ar[r, "v"] \ar[d, "\beta"] & {}^{M}/_{p} \otimes_{K} q \ar[d, "\gamma"] \\
p \otimes_{K} N \ar[r, "u'"] & M \otimes_{K} N \ar[r, "v'"] & ^{M}/_{p}
\otimes_{K} N
\end{tikzcd}](images/img-0003.svg)
Since \(^{M}/_{p}\) is flat, \(\gamma \) is injective. Let \(z \in (p \otimes _{R} N) \sqcap (M \otimes _{R} q) = \operatorname{im}\beta \sqcap \operatorname{im}u'\). By abusing notation, replace \(z\) with some elements of \(M \otimes _{K} q\) and continue with \(\beta (z)\in \operatorname{im}\beta \sqcap \operatorname{im}u'\). Since \(v'(\beta (z))=\gamma (v(z))\) and that \(\beta (z)\in \operatorname{im}u'\), we conclude that \(\gamma (v(z))=0\), that is \(z\in \ker v=\operatorname{im}u\). We abuse notation again, let \(z \in p \otimes _{K} q\), we need to show \(\beta (u(z))\in \operatorname{im}\beta \sqcap \operatorname{im}u'\), but \(\beta \circ u=u'\circ \alpha \), we finish the proof.
3.1.2 Centralizer and Center
Let \(R\) be a commutative ring and \(A\), \(B\) be two \(R\)-algebras. We denote centralizer of \(S\subseteq A\) by \(C_{A}S\) and centre of \(A\) by \(Z(A)\).
Let \(S, T\) be two subalgebras of \(A\), then \(C_{A}(S\sqcup T)=C_{A}(S)\sqcap C_{A}(T)\).
This lemma can be generalized to centralizers of arbitrary supremum of subalgebras.
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 \(C_{A}(S) \otimes _{R} B\)
A symmetric statement goes without saying.
Let \(w\in C_{A\otimes _{R}B}\left(\operatorname{im}\left(S\to A\otimes _{R}B\right)\right)\). Since \(B\) is free, we choose an arbitrary basis \(\mathcal{B}\); by lemma 3.1.1, we write \(w = \sum _{i}m_{i}\otimes _{K}\mathcal{B}_{i}\). It is sufficient to show that \(m_{i}\in C_{A}(S)\) for all \(i\). Let \(a \in S\), we need to show that \(m_{i}\cdot a = a \cdot m_{i}\). Since \(w\) is in the centralizer, \(w \cdot (a\otimes 1) = (a\otimes 1)\cdot w\). Hence we have \(\sum _{i}(a\cdot m_{i})\otimes \mathcal{B}_{i}=\sum _{i}(m_{i}\cdot a)\otimes \mathcal{B}_{i}\). By the uniqueness of lemma 3.1.1, we conclude \(a\cdot m_{i}=m_{i}\cdot a\).
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 \(Z(A)\otimes _{R}B\).
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 \(C_{A}(S)\otimes _{R}C_{B}(T)\)
From lemma 3.1.2, \(C_{A}(S)\otimes _{R}C_{B}(T)\) is equal to \(\left(C_{A}(S)\otimes _{R} B\right)\sqcap \left(A\otimes _{R}C_{B}(T)\right)\). The left hand side \(C_{A}(S)\otimes _{R} B\) is equal to \(C_{A\otimes _{R}B}\left(\operatorname{im}\left(S\to A\otimes _{R}B\right)\right)\) and the right hand side is equal to \(C_{A\otimes _{R}B}\left(\operatorname{im}\left(T\to A\otimes _{R}B\right)\right)\). Hence by lemma 3.1.3, their intersection is equal to
This is precisely \(C_{A\otimes _{R}B}\left(S\otimes _{R}T\right)\).
Assume \(R\) is a field. The centre of \(A\otimes _{R} B\) is \(Z\left(A\right)\otimes _{R}Z\left(B\right)\).
Special case of corollary 3.1.6.
3.1.3 Some Isomorphisms
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.
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
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.
3.2 Wedderburn-Artin Theorem for Simple Rings
3.2.1 Classification of Simple Rings
Let \(A\) be a ring and \(I\) a non-trivial minimal left ideal of \(A\), then \(I\) is a simple \(A\)-module.
Let \(J\le I\) be an \(A\)-submodule of \(I\), suppose \(J\) is non-trivial, we prove that \(J=I\). Then the image \(J'\) of \(J\) under \(I \hookrightarrow A\) is a non-trivial left ideal of \(A\). Since \(I \hookrightarrow A\) is injective, it is sufficient to prove that \(J' = I\). This is because \(J'\le I\) and \(J' \not{\lt} J\).
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 \(y_{i} \in A\).
Let \(I'\) be the two-sided ideal spanned by \(I\). Then since \(A\) is a simple ring, \(I' = A\). Thus \(1 \in I'\). One can write \(1 \in A\) as \(\sum _{i}a_{i}x_{i}b_{i}\) for some \(x_{i}\in I\) and \(a_{i},b_{i}\in A\), since \(I\) is a left ideal \(a_{i}x_{i}\in I\) as well.
Now, we can find the smallest \(n\) such that \(1\in A\) can be written as \(\sum _{i=0}^{n}x_{i}y_{i}\) for some \(x_{i}\in I\) and \(y_{i}\in A\). Let us fix the notations \(n\), \(x_{i}\) and \(y_{i}\)
The \(n\), \(x_{i}\) and \(y_{i}\) are all non-zero.
If \(n\) is \(0\), then \(1 = 0\) in \(A\), but all simple rings are non-trivial. We argue by contradiction to prove that all \(x_{i}\) and \(y_{i}\) are non-zero. Assume there exists a \(j\) such that \(y_{j}\ne 0\) implies \(x_{j} = 0\). Without loss of generality, we assume \(j = 0\). Then \(1 = \sum _{i=0}^{n}x_{i}y_{i}=\sum _{i=1}^{n}x_{i}y_{i}\). This contradicts the minimality of \(n\).
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.
We continue to write \(1 = \sum _{i=0}^{n}x_{i}y_{i}\) in the shortest possible manner. Then we can define an \(A\)-linear map \(g : I^{n}\to A\) by \((v_{i})\mapsto \sum v_{i}y_{i}\). Then \(g\) is surjective: if \(a \in A\), then \((ax_{i})\) is mapped to \(a\) under \(g\). \(g\) is injective as well: support \(g(v_{i})=0=\sum _{i}v_{i}y_{i}\) with \((v_{i})\) not all zero. Without loss of generality, we assume \(v_{0} \ne 0\), then the ideal \(\langle v_{0}\rangle \) is equal to \(I\) (since \(I\) is simplelemma 3.2.1). Thus \(x_{0}\in I = \langle v_{0}\rangle \); implying that \(x_{0}=r\cdot v_{0}\) for some \(r\in A\). Thus \(1=1 - r\cdot 0 = \sum _{i=0}^{n}x_{i}y_{i}-\sum _{i=0}^{n}r\cdot v_{i}y_{i}\). In this way, we cancelled the term at \(i=0\), contradicting the minimality of \(n\). Hence \(g\) is an isomorphism.
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.
By theorem 3.2.4, we only need a minimal left ideal. Since \(A\) is Artinian, such ideal exists.
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 \(B \cong \operatorname{Mat}_{n}(S)\).
By theorem 3.2.5, we can find a \(n\) and a minimal left ideal \(I\) such \(A \cong I^{n}\) as \(A\)-modules. Note that \({\left(\operatorname{End}_{B}I\right)}^{\mathsf{opp}}\) is a division ring. Then since \(B^{\mathsf{opp}}\cong \operatorname{End}_{B}B\cong {\operatorname{End}_{B}(I^{n})}\cong \operatorname{Mat}_{n}\left(\operatorname{End}_{B}I\right)\) as rings where the final isomorphism is from construction 3.1.2, we have \(e : B\cong \operatorname{Mat}_{n}\left(\operatorname{End}_{B}I\right)^{\mathsf{opp}}\) as rings. We also have a \(K\)-algebra structure on \({\left(\operatorname{End}_{B} I\right)}^{\mathsf{opp}}\) given by \((a \cdot f)(x)=f(a\cdot x)\), and this algebra structure promotes the ring isomorphism \(e\) to a \(k\)-algebra isomorphism.
3.2.2 Uniqueness of the Classification
In the previous section, we know that finite dimensional simple \(K\)-algebra \(B\) over is in fact a matrix algebras of a division \(K\)-algebra \(S\). In this section, we prove that the division algebra \(S\) is essentially unique.
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.
Since \(D^{n}\) is a simple \(B\)-module, by lemma 2.2.8, we see that \(\operatorname{End}_{A}D^{n}\cong D^{\mathsf{opp}}\) and \(\operatorname{End}_{A}D^{n}\cong D'^{\mathsf{opp}}\) as \(k\)-algebras. Thus \(D^{\mathsf{opp}}\cong D'^{\mathsf{opp}}\) as \(k\)-algebras, consequently \(D\cong D'\) as \(k\)-algebras as well. Since \(A \cong \operatorname{Mat}_{n}(D)\cong \operatorname{Mat}_{n'}(D')\cong \operatorname{Mat}_{n'}(D)\) as \(k\)-algebras and \(A\) is finite \(k\)-dimensional, a dimension argument shows that \(n=n'\).
3.3 Skolem-Noether Theorem
Let \(K\) be a field, \(A\), \(B\) be \(K\)-algebras where \(A\) is central simple and finite \(K\)-dimensional and \(B\) is simple. Let \(M\) be a simple \(A\)-module.
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.
Since \(M\) is a finite \(A\)-module and \(A\) a finite dimensional \(K\)-vector space, \(M\) is a finite dimensional \(K\)-vector space as well. Suppose \(S \subseteq M\) generates \(M\) as \(K\)-module, the claim is that \(S\) generates \(M^{f}\) as well. Let \(x \in M^{f}\), we write \(x = \sum _{i}\lambda _{i}\cdot s_{i}\) with \(\lambda _{i}\in K\) and \(s_{i}\in S\). Note that \(\lambda _{i}\cdot s_{i} = (\rho (\lambda _{i})\otimes \mathbf{1}_{M})\) in \(M^{f}\) where \(\rho : K \to B\) is the map giving \(B\) its \(K\)-algebra structure. Hence \(x\) is in the span of \(S\) in \(M^{f}\) as well.
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.
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.
By lemma 2.2.4, it is sufficient to prove \(\dim _{K}M^{f}=\dim _{K}M^{g}\). But as \(K\)-vector space, \(M^{f}\) and \(M^{g}\) are literally \(M\).
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 \(g = x f x^{-1}\).
Let \(M\) be any simple \(A\)-module (which exists by lemma 2.2.5). By lemma 3.3.3, we have some isomorphism \(\phi : M^{f}\cong M^{g}\) as \(B\otimes _{K}\operatorname{End}_{A}M\)-module. Since \(M\) is simple, we have that \(M\) is a balanced \(A\)-module by lemma 2.2.15. Let \(e\) denote the \(k\)-algerba isomorphism \(A \cong \operatorname{End}_{\operatorname{End}_{A}M}M\) given by the \(A\)-action on \(M\). Since both \(\phi \) and \(\phi ^{-1}\) defines an element of \(\operatorname{End}_{\operatorname{End}_{A}M}M\), we define \(a := e^{-1}(\phi )\) and \(b := e^{-1}(\phi ^{-1})\). Then \(ab = 1\) since \(e(ab)=e(a)\cdot e(b)=\phi \phi ^{-1}=1\). We prove that the image of \(f\) and \(afb\) under \(e\) are the same; that is for all \(x\in B\) and \(m\in M\), \(e(g(x))(m) = e(a f(x) b)(m)\). The right hand side is equal to
Similarly, the left hand side is equal to \(g(x)\cdot m\). Note that \(\phi \) is \(B\otimes \operatorname{End}_{A}M\)-linear. Therefore \(\phi \left(\left(x\otimes \mathbf{1}\right)\cdot \phi ^{-1}(m)\right) = \left(x\otimes \mathbf{1}\right)\cdot m\). Unfolding the definition of \(M^{f}\) and \(M^{g}\), we see this is saying precisely \(\phi \left(f(x)\cdot \phi ^{-1}(m)\right)=g(x)\cdot m\).
3.4 Double Centralizer Theorem
In this section let \(F\) be a field and \(A\) an \(F\)-algebra. Define \(\mathcal{L}_{A} \subseteq \operatorname{End}_{F}A\) to be
i.e. \(F\)-linear maps defined by left multiplication; similarly define \(\mathcal{R}_{A}\). Note that \(\mathcal{L}_{A}\) and \(\mathcal{R}_A\) are \(F\)-subalgebras of \(\operatorname{End}_{F}A\). When we need to stree the underlying field is \(F\), we also write \(\mathcal{L}_{A}^{F}\) and \(\mathcal{R}_{A}^{F}\). We assume \(A\) to be a finite dimensional central simple \(F\)-algebra.
The centralizer of \(\mathcal{L}_{A}\) in \(\operatorname{End}_{F}A\) is smaller than or equal to \(\mathcal{R}_{A}\):
Indeed, let \(x \in C_{\operatorname{End}_{F}A}\left(\mathcal{L}_{A}\right)\). Recall from construction 3.1.1 that \(e : A\otimes _{F}A^{\mathsf{opp}}\cong \operatorname{End}_{F}A\) as \(F\)-algebras. Then \(e^{-1}(x)\) is in \(C_{A\otimes _{F}A^{\mathsf{opp}}}\left(\operatorname{im}\left(A\to A\otimes _{F}A^{\mathsf{opp}}\right)\right)\) (for \(e\) sends \(a\otimes 1\) to the \(F\)-linear map \((a\cdot \bullet )\)). Since \(C_{A\otimes _{F}A^{\mathsf{opp}}}\left(\operatorname{im}\left(A\to A\otimes _{F}A^{\mathsf{opp}}\right)\right) = Z(A)\otimes _{F}A^{\mathsf{opp}}=F\otimes _{F}A^{\mathsf{opp}}=\operatorname{im}\left(A^{\mathsf{opp}}\to A\otimes _{F}A^{\mathsf{opp}}\right)\), we find some \(y\in A^{\mathsf{opp}}\) such that \(1 \otimes y = e^{-1}(x)\). Therefore \(e\left(1\otimes y\right) = x\); but \(e\left(1\otimes y\right)\) is in \(\mathcal{R}_{A}\) for it is the linear map \((\bullet \cdot y)\).
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 \(\left(z\cdot \bullet \right)\).
\(A\) 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 \(A/Z(A)/F\).
As \(F\)-algebras, we have \(\mathcal{R}_{A}\cong A^{\mathsf{opp}}\).
We prove the map \(A^{\mathsf{opp}} \to \mathcal{R}_{A}\) is bijective. It is injective because if \(\left(\bullet \cdot a\right) = \left(\bullet \cdot b\right)\), then \(a = 1 \cdot a = 1 \cdot b = b\). The map is surjective by the definition of \(\mathcal{R}_{A}\).
Let \(B\) be any simple \(F\)-algebra (not necessarily central). The centralizer of \(\mathcal{L}_{B}\) in \(\operatorname{End}_{F}B\) is equal to \(\mathcal{R}_{B}\).
It is straightforward to show \(\mathcal{R}_{B}^{F}\le C_{\operatorname{End}_{F}A}\left(\mathcal{L}_{B}^{F}\right)\). So we only need to prove \(C_{\operatorname{End}_{F}A}\left(\mathcal{L}_{B}^{F}\right)\le \mathcal{R}_{B}^{F}\). By lemma 3.4.1, since \(B\) is a central simple finite dimensional \(Z(B)\)-algebra, we have that
Suppose \(f\in \operatorname{End}_{F}B\) is in \(C_{\operatorname{End}_{F}B}B\), by remark 3.4.2, \(f\) is \(Z(B)\)-linear as well. Then \(f\) is in \(\mathcal{R}_{B}^{Z(B)}\); that is \(f\) is equal to \(\left(\bullet \cdot b\right)\) for some \(b \in B\) as \(Z(B)\)-linear maps. Then \(f\) is also equal to \(\left(\bullet \cdot b\right)\) as \(F\)-linear maps.
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.
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 \(a \in C_{B}\left(xSx^{-1}\right)\), then \(x^{-1}ax\) is in \(C_{B}(S)\). Conversely if \(a\) is equal to \(xbx^{-1}\) with \(b\in C_{B}(S)\), then it is in \(C_{B}\left(xSx^{-1}\right)\) as well.
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 \(F\).
Let \(S\subseteq A\) be a simple \(F\)-subalgebra, then \(A \otimes _{F} \mathcal{R}_{S}\) is a simple ring.
By lemma 3.4.4, we have \(A\otimes \mathcal{R}_{S}\cong A\otimes S^{\mathsf{opp}}\) as \(F\)-algebras. The claim follows from theorem 1.1.10.
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.
By lemma 1.1.9 and theorem 1.1.10, \(A \otimes _{F}C_{A}(S)\) is a central simple \(F\)-algebra. Let \(f : S \to A\otimes _{F}\operatorname{End}_{F}S\) be an \(F\)-algebra homomorphism defined by \(s \mapsto s \otimes \mathbf{1}_{S}\) and \(g : S \to A \otimes _{F}\operatorname{End}_{F}S\) be an \(F\)-algebra homomorphism defined by \(\mathbf{1}_{A} \otimes \left(s\cdot \bullet \right)\). Then by theorem 3.3.4, we that there exists some \(x\in \left(A\otimes _{F}\operatorname{End}_{F}S\right)^{\times }\) such that \(f = xgx^{-1}\). Then we have \(S\otimes _{F}\operatorname{End}_{F}S\) is equal to \(x\left(A\otimes _{F}\mathcal{L}_{S}\right)x^{-1}\): indeed the left hand side is \(\operatorname{im}f\) while the right handside is \(x\left(\operatorname{im}g\right)x^{-1}\). Therefore \(C_{A\otimes _{F}\operatorname{End}_{F}S}\left(S\otimes _{F}\operatorname{End}_{F}S\right) = C_{A\otimes _{F}\operatorname{End}_{F}S}\left(x\left(A\otimes _{F}\mathcal{R}_{S}\right)x^{-1}\right)\). By lemma 3.4.6, the right hand side is equal to \(xC_{A\otimes _{F}\operatorname{End}_{F}S}\left(A\otimes _{F}\mathcal{L}_{S}\right)x^{-1}\) which is \(x\left(A\otimes _{F}C_{\operatorname{End}_{F}S}\left(\mathcal{L}_{S}\right)\right)x^{-1}\) by lemma 3.1.4 which is \(x\left(A\otimes _{F}\mathcal{R}_{S}\right)x^{-1}\) by lemma 3.4.5.
Let \(S\subseteq A\) be a simple \(F\)-subalgebra, then \(C_{A}(S)\) is simple as well.
By lemma 3.4.9, \(C_{A}(S) \otimes _{F}\operatorname{End}_{F}S\) is isomorphic to \(x\left(A\otimes _{F}\mathcal{R}_{S}\right)x^{-1}\) as \(F\)-algebras. Then \(C(S)\otimes _{F}\operatorname{End}_{F}S\) is simple since \(A\otimes _{F}\mathcal{R}_{S}\) is simple by lemma 3.4.8. By theorem 1.1.12, \(C_{A}(S)\) is simple.
Let \(S\subseteq A\) be a simple \(F\)-subalgebra. Then
By lemma 3.4.9, \(C_{A}(S) \otimes _{F}\operatorname{End}_{F}S\) is isomorphic to \(x\left(A\otimes _{F}\mathcal{R}_{S}\right)x^{-1}\) as \(F\)-algebras. Hence \(\dim _{F}\left(C_{A}(S)\otimes _{F}\operatorname{End}_{F}S\right)=\dim _{F}\left(A\otimes _{F}\mathcal{R}_{S}\right)\) where the left hand side is \(\dim _{F}C_{A}(S)\cdot \dim _{F}\operatorname{End}_{F}S\) and the right hand side is \(\dim _{F}A\cdot \dim _{F}\mathcal{R}_{S}\). Since \(\dim _{F}\operatorname{End}_{F}S=\dim _{F}S^{2}\) and \(\dim _{F}\mathcal{R}_{S}=\dim S\) (by lemma 3.4.4), we proved this lemma.
Let \(S \subseteq A\) be a central simple \(F\)-subalgebra,
By lemma 3.4.10, \(C_{A}(B)\) is simple and by theorem 1.1.10, \(B \otimes _{F} C_{A}(B)\) is simple. Hence the map \(B \otimes _{F} C_{A}(B) \to A\) induced by \(B \hookrightarrow A\) and \(C_{A}(B) \hookrightarrow A\) is injective. By corollary 1.1.8, we only need to show \(\dim _{F}B\otimes _{F}C_{A}(B) =\dim _{F}A\) which is precisely lemma 3.4.11.
Let \(S\subseteq A\) be a simple \(F\)-subalgebra, we have
It is straightforward that \(S \le C_{A}\left(C_{A}(S)\right)\). By lemma 3.4.10, \(C_{A}(S)\) is simple, hence \(\dim _{F}C_{A}\left(C_{A}(S)\right)\cdot \dim _{F}C_{A}(S)=\dim _{F}A=\dim _{F}C_{A}(S)\cdot \dim _{F}S\) (by applying lemma 3.4.11 twice), i.e. \(\dim _{F}C_{A}\left(C_{A}(S)\right) = \dim _{F}S\). This equality of dimension gives us the desired result.