Bourbaki

Example 2

Tag 007R, permanent. Cite it as [Bourbaki, Tag 007R].

Algebra, VIII, p. 113, § 6. Morita Equivalence of Modules and Algebras

Let A be a kk-algebra and n⩾1n\geqslant 1 an integer. We identify the matrix algebra Mn(A)\mathbf{M}_n(A) with the endomorphism algebra of the right A-module AdnA^n_d. We have seen that A and Mn(A)\mathbf{M}_n(A) are Morita equivalent. For every left A-module M, identify Adn⊗AMA^n_d\otimes_AM with MnM^n. The algebra Mn(A)\mathbf{M}_n(A) then has a left action on MnM^n, and we have

(a⋅m)i=∑j=1naijmj (a\cdot m)_i=\sum_{j=1}^na_{ij}m_j

for a=(aij)a= (a_{ij}) in Mn(A)\mathbf{M}_n(A) and m=(mi)m= (m_i) in MnM^n. Morita’s theorem implies the following results:

a) Every left module over the algebra Mn(A)\mathbf{M}_n(A) is isomorphic to a module of the form MnM^n, where M is a left A-module.

b) Let M be a left A-module. The mapping N↦NnN\mapsto N^n is a bijection from the set of A-submodules of M to the set of Mn\mathbf{M}_n(A)-submodules of MnM^n.

c) Let M and N be left A-modules. For an A-linear mapping g:M→Ng: M\rightarrow N, let gng_n be the mapping (mi)↦(g(mi))(m_i)\mapsto (g(m_i)) from MnM^n to NnN^n. Then the mapping g↦gng\mapsto g_n is a bijection from HomA(M,N)_A(M,N) to HomMn(A)(Mn,Nn)_{\mathbf{M}_n(A)}(M^n,N^n).

d) Let M be a left A-module. The module MnM^n over the ring Mn(A)\mathbf{M}_n(A) is indecomposable (resp. semisimple, simple, Artinian, Noetherian, finitely generated) if and only if the A-module M is.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.