By Rabib Islam

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Additional resources for Morita equivalence

Example text

R1 sn ) + · · · + (0, . . , 0, 1) ⊗ (rn s1 , . . , rn sn ) = 0 =⇒ (r1 . . , rn ) ⊗ (s1 , . . , sn ) = 0. Since g˜ is Mn (R)-balanced, it is a bimodule homomorphism, and hence a bimodule isomorphism. 3. 16 from a bicategorical perspective. That is, we explore what it means for a bimodule to be faithfully balanced in Bim. 20. The map f : R → EndR (RR ) defined by f (r) = φr , where φr is left multiplication by r, is a ring isomorphism. Proof. We first show that all the endomorphisms of RR are left multiplications by elements of R.

Since c ∈ ker β2 , we know that c ∈ im α2 , so we can pick d ∈ T L such that α2 (d) = c. Since ψ L is surjective, we can pick e ∈ Q ⊗R L such that ψ L (e) = d. Since ψ F is injective, b is the unique preimage of c through ψ F . Since the diagram commutes, we know that α1 (e) = b. Finally, by exactness, a = β1 (b) = β1 (α1 (e)) = 0. Thus, ψ M is injective, and hence an isomorphism. We can now use our knowledge of Bim to prove the following corollary. 1]). The rings R and S are Morita equivalent if and only if there is an (R, S)-bimodule P and an (S, R)-bimodule Q such that R PS ⊗S S QR R RR and S QR ⊗R R PS S SS as bimodules.

Sn ) = (1, 0, . . , 0) ⊗ (r1 s1 , . . , r1 sn ) + · · · + (0, . . , 0, 1) ⊗ (rn s1 , . . , rn sn ). We thus also see that g˜ is injective: ˜ 1 , . . , rn ) ⊗ (s1 , . . , sn )) = 0 =⇒ ∀i, ∀j, ri sj = 0 g((r =⇒ (1, 0, . . , 0) ⊗ (r1 s1 , . . , r1 sn ) + · · · + (0, . . , 0, 1) ⊗ (rn s1 , . . , rn sn ) = 0 =⇒ (r1 . . , rn ) ⊗ (s1 , . . , sn ) = 0. Since g˜ is Mn (R)-balanced, it is a bimodule homomorphism, and hence a bimodule isomorphism. 3. 16 from a bicategorical perspective. That is, we explore what it means for a bimodule to be faithfully balanced in Bim.