Álgebra Extraordinaria by I. M. Yaglom

By I. M. Yaglom

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Symbolic Reasoning allied to boring A lgebra. the manager examples of such structures are Hamilton sQ uaternions, Grassmann sC alculus of Extension and Boole sS ymbolic good judgment. Such algebras have an intrinsic price for separate distinct examine; they are also priceless of a comparative learn, for the sake of the sunshine thereby thrown at the normal idea of symbolic reasoning, and on algebraic symbolism particularly.

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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.

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