2 Morita Equivalence
This chapter intertwine with section 3.2: section 2.2 depends on section 3.2.1; while section 3.2.2 depends on section 2.2.
2.1 Construction of the equivalence
Let \(R\) be a ring and \(0 \ne n\in \mathbb {N}\). In this chapter, we prove that the category \(R\)-modules and the category of \(\operatorname{Mat}_{n}(R)\)-modules are equivalent. Then we use the equivalence to prove several useful lemmas.
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 \(\operatorname{\mathfrak {Mod}}_{R} \Longrightarrow \operatorname{\mathfrak {Mod}}_{\operatorname{Mat}_{n}(R)}\).
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.
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 \(\operatorname{\mathfrak {Mod}}_{\operatorname{Mat}_{n}(R)}\Longrightarrow \operatorname{\mathfrak {Mod}}_{R}\).
The functors constructed in construction 2.1.1 and construction 2.1.2 form an equivalence of category.
Let \(M\) be an \(R\)-module, then the unit \(\widetilde{\widehat{M}} \cong M\) is given by
Let \(M\) be an \(\operatorname{Mat}_{n}(R)\)-module, then the counit \(\widehat{\widetilde{M}}\cong M\) is given by \(m \mapsto (\delta _{i0}\cdot m)\). This map is both injective and surjective.
2.2 Stacks 074E
Let \(A\) be a finite dimensional simple \(k\)-algebra.
Let \(M\) and \(N\) be simple \(A\)-modules, then \(M\) and \(N\) are isomorphic as \(A\)-modules.
By theorem 3.2.6, there exists non-zero \(n\in \mathbb N\), \(k\)-division algebra \(D\) such that \(A\cong \operatorname{Mat}_{n}(D)\) as \(k\)-algebras. Then by theorem 2.1.2, we have equivalence of category \(e : \operatorname{\mathfrak {Mod}}_{A}\cong \operatorname{\mathfrak {Mod}}_{D}\). Since simple module is a categorical notion (it can be defined in terms monomorphisms), \(e(M)\) and \(e(N)\) are simple \(D\)-modules. Since \(D\) is a division ring, \(e(M)\) and \(e(N)\) are isomorphic as \(D\)-modules, therefore \(M\) and \(N\) are isomorphic as \(A\)-modules.
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 \(\iota \).
By theorem 3.2.6, there exists non-zero \(n\in \mathbb N\), \(k\)-division algebra \(D\) such that \(A\cong \operatorname{Mat}_{n}(D)\) as \(k\)-algebras. Then by theorem 2.1.2, we have equivalence of category \(e : \operatorname{\mathfrak {Mod}}_{A}\cong \operatorname{\mathfrak {Mod}}_{D}\). Since simple module is a categorical notion (it can be defined in terms monomorphisms), \(e^{-1}(D)\) is a simple module over \(A\). Since \(e(M)\) is a free module over \(D\), we can write \(e(M)\) as \(\bigoplus _{i\in \iota } D\) for some indexing set \(\iota \). By precomposing the unit of \(e\), we get an isomorphism \(M \cong e^{-1}\left(\bigoplus _{i\in \iota } D\right)\). We only need to prove \(e^{-1}\left(\bigoplus _{i\in \iota } D\right)\cong \bigoplus _{i\in \iota }e^{-1}\left(D\right)\). This is because direct sum is the categorical coproduct.
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 \(S\).
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.
The forward direction is trivial as an \(A\)-linear isomorphism is a \(k\)-linear isomorphism as well. Conversely, suppose \(\dim _{k}M = \dim _{k}N\). By lemma 2.2.2, there exists a simple \(A\)-module \(S\) such that \(M \cong \bigoplus _{i\in \iota } S\) and \(N \cong \bigoplus _{i\in \iota '} S\) as \(A\)-modules. Without loss of generality \(S \ne 0\), for otherwise we have \(M \cong N\) anyway. If either of \(\iota \) or \(\iota '\) is empty, then \(\dim _{k}M = \dim _{k} N = 0\) implying that \(M = N = 0\), we again have \(M \cong N\). Thus, we assume both \(\iota \) and \(\iota '\) are non-empty. By pulling back the \(A\)-module structure on \(S\) to a \(k\)-module structure along \(k \hookrightarrow A\), \(M, N, S, \bigoplus _{i\in \iota } S, \bigoplus _{i\in \iota '} S\) are all finite dimensional \(k\)-vector spaces. Hence \(\iota \) and \(\iota '\) are finite. The equality \(\dim _{k}M=\dim _{k}N\) tells us that \(\iota \cong \iota '\) as set, hence \(M \cong \bigoplus _{i\in \iota }S\cong \bigoplus _{i\in \iota '}S\cong N\) as required.
Let \(A \cong \operatorname{Mat}_{n}(D)\) as \(k\)-algebras for some \(k\)-division algebra and \(n\ne 0\).
\(D^{n}\) is a simple \(A\)-module where the module structure is given by pulling back the \(\operatorname{Mat}_{n}(D)\)-module structure of \(D^{n}\).
By theorem 2.1.2, we have \(\operatorname{\mathfrak {Mod}}_{A}\cong \operatorname{\mathfrak {Mod}}_{D}\cong \operatorname{\mathfrak {Mod}}_{\operatorname{Mat}_{n}(D)}\). Since \(D\) is a simple \(D\)-module, \(D^{n}\) is a simple \(\operatorname{Mat}_{n}(D)\) module and consequently, a simple \(A\)-module.
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.
\(\operatorname{End}_{A}\left(D^{n}\right)\) is isomorphic to \(D^{\mathsf{opp}}\) as \(k\)-algebras.
Indeed, we calculate:
Let \(M\) be a simple \(A\)-module, then \(\operatorname{End}_{A} M\cong D^{\mathsf{opp}}\) as \(k\)-algebras.
By theorem 2.1.2, \(D^{n}\) is simple as \(A\)-module; hence by lemma 2.2.1, \(D^{n}\) and \(M\) are isomorphic as \(A\)-module. Lemma 2.2.7 gives the desired result.
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.
By theorem 3.2.4, such \(D\) and \(n\) always exists. Hence we only need to show \(D^{\mathsf{opp}}\) has finite \(k\)-dimension. Since \(\dim _{k}A=\dim _{k}\operatorname{Mat}_{n}(D)\) are both finite, we conclude \(D^{\mathsf{opp}}\) is finite as a \(k\)-vector space by pulling back the finiteness along \(D \hookrightarrow \operatorname{Mat}_{n}(D)\).
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.
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.
Balancedness is invariant under linear isomorphism.
For any ring \(A\), \(A\) is balanced as \(A\)-module.
If \(f \in \operatorname{End}_{\operatorname{End}_{A}M}A\), then the image of \(f(1)\) under \(A \to \operatorname{End}_{\operatorname{End}_{A}}A\) is \(f\) again.
We assume again that \(A\) is a finite dimensional simple \(k\)-algebra.
Any simple \(A\)-module is balanced.
Indeed, if \(M\) is a simple \(A\)-module, then \(A \cong \bigoplus _{i\in \iota } M\) for some indexing set \(\iota \) by lemma 2.2.2. Since \(A\) is balanced, \(\bigoplus _{i\in \iota }M\) is balanced. Let \(g \in \operatorname{End}_{\operatorname{End}_{A}M}M\), we can define a corresponding \(G \in \operatorname{End}_{\operatorname{End}_{\bigoplus _{i}M}}\left(\bigoplus _{i}M\right)\) by sending \((v_{i})\) to \((g(v_{i}))\). Since \(\bigoplus _{i}M\) is balanced, we know that for some \(a\in A\), \(G\) is the image of \(a\) under \(A \to \operatorname{End}_{\operatorname{End}_{\bigoplus _{i}M}}\left(\bigoplus _{i}M\right)\). Then the image of \(a\) under \(A \to \operatorname{End}_{\operatorname{End}_{A}M}M\) is \(g\).
For any simple \(A\)-module \(M\), we have \(A \cong \operatorname{End}_{\operatorname{End}_{A}M}M\) as \(k\)-algebras.
The canonical map \(A \to \operatorname{End}_{\operatorname{End}_{A}M}M\) is both injective and surjective, as \(M\) is a balanced \(A\)-module and \(A\) is a simple ring.