Math 515 homework #9 Due date: 19980216 D. J. Bernstein 19980209 On this sheet “R-module homomorphism” is abbreviated “R-map.”

Base change. Let R be a ring. Let A be an R-algebra. Let M be an R-module. The tensor product A ⊗_{R} M is an A-module with an R-map m : M → A ⊗_{R} M. For any A-module U and any R-map h : M → U, there is an A-map φ : A ⊗_{R} M → U such that h = φ ∘ m. Furthermore, φ is unique: if φ′ is another A-map, and if φ′ ∘ m = φ ∘ m, then φ′ = φ.

Right exactness of base change. Let R be a ring. Let L, M be R-modules. Let f be an R-map from L to M. Define N = M/ Im f; N is the cokernel of f. Let g be the natural R-map from M to N. Then Im f = Ker g; in particular, g ∘ f = 0. Let A be an R-algebra. Then A ⊗_{R} L, A ⊗_{R} M, A ⊗_{R} N are A-modules, with R-maps l : L → A ⊗_{R} L, m : M → A ⊗_{R} M, n : N → A ⊗_{R} N. Now m ∘ f is an R-map from L to A ⊗_{R} M, so there is an A-map f′ : A ⊗_{R} L → A ⊗_{R} M such that f′ ∘ l = m ∘ f. Similarly, there is an A-map g′ : A ⊗_{R} M → A ⊗_{R} N such that g′ ∘ m = n ∘ g. The following exercises show that Im f′ = Ker g′. Note that the same is true for tensor products in general, not just base change; a proof is outlined in Eisenbud page 574. Exercise 20. Prove that g′ ∘ f′ = 0. (Hint: Prove that g′ ∘ f′ ∘ l = 0 ∘ l.) Define T = (A ⊗_{R} M)/ Im f′. Let π be the natural A-map from A ⊗_{R} M to T; then π ∘ f′ = 0. Exercise 21. Prove that there is an R-map u : N → T such that u ∘ g = π ∘ m. Exercise 22. Prove that there is an A-map v : A ⊗_{R} N → T such that v ∘ g′ = π. By Exercise 20, Im f′ ⊆ Ker g′. By Exercise 22, Ker g′ ⊆ Ker π = Im f′. Therefore Im f′ = Ker g′. In other words, the sequence A ⊗_{R} L → A ⊗_{R} M → A ⊗_{R} N is exact. Note that g′ is surjective. (Replace L, M, f with M, N, g to see that Im g′ is the kernel of A ⊗_{R} N → A ⊗_{R} (N/N); but A ⊗_{R} (N/N) = 0.) Thus (A ⊗_{R} M)/f′, the cokernel of f′, is isomorphic to A ⊗_{R} N. In short, tensor products preserve cokernels. Tensor products can fail to preserve kernels. For example, the inclusion 2Z → Z is injective, but 2Z/4Z → Z/2Z is the zero map.