By Jenca G.
We turn out that if E1 and E2 are a-complete impression algebras such that E1 is an element of E2 and E2 is an element of E1, then E1 and E2 are isomorphic.
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Extra info for A Cantor-Bernstein type theorem for effect algebras
RIGHT-EXACTNESS 39 sequence to obtain the top-most exact row in the following diagram I ⊗R M f 0 / IM / R ⊗R M g /M / (R/I) ⊗R M /0 h / M/IM / 0. The vertical maps are given by f (i⊗m) = im and g(r⊗m) = rm and h(r⊗m) = rm. Show that these maps are well-defined and make the diagram commute. The map f is an epimorphism, and g is an isomorphism. 15. Let ϕ : R → S be a homomorphism of commutative rings. Let M be an R-module, and let N be an S-module. (a) Prove that if S is flat over R and N is flat over S, then N is flat over R.
The other parts of bilinearity are verified similarly. Since h is R-bilinear, the universal property for tensor products yields a welldefined R-module homomorphism H : M ⊗R N → (M ⊗R N )/K satisfying the following: for m ⊗ n ∈ M ⊗R N , fix m ∈ M and n ∈ N such that f (m) = m and g(n) = n ; then H(m ⊗ n ) = m ⊗ n). It follows readily that the composition Hφ : (M ⊗R N )/K → (M ⊗R N )/K is the identity on (M ⊗R N )/K, so φ is injective as desired. 2. The maps µZ2 : Z → Z and ✶Z/2Z : Z/2Z → Z/2Z are both injective.
4. Let R be a commutative ring, and let M and N be R-modules. Let I, J ⊆ R be ideals such that IM = 0 and JN = 0. Then (I + J)(M ⊗R N ) = 0. To show this, it suffices to show that I(M ⊗R N ) = 0 and J(M ⊗R N ) = 0. Let a ∈ I, and let µM a : M → M denote the homothety m → am. Our assumption implies that M µM a = 0 = µ0 . 3 implies that the induced map M ⊗R N → M ⊗R N is given by multiplication by a, and by multiplication by 0. That is, multiplication by a on M ⊗R N is 0. This implies that a(M ⊗R N ) = 0, and hence I(M ⊗R N ) = 0.