By Thomas Markwig Keilen
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Additional resources for Commutative Algebra [Lecture notes]
22 (Snake lemma). Let the following commutative diagram of R-linear maps be given: 0 / M′ 0 / N′ α /M α′ /N ϕ′ β /0 ϕ′′ ϕ 31 / M ′′ β ′ / N ′′ /0 2. Modules and linear maps Then consider the following diagram: 0 0 0 (⋄) 0 / ker(ϕ′ ) α| / ker(ϕ) β| / ker(ϕ′′ ) (∗) 0 / M′ α /M β / M ′′ (∗) 0 / N′ α′ /N β′ / N ′′ /0 (⋄) ... Coker(ϕ′ ) α′ / Coker(ϕ) β′ / Coker(ϕ′′ ) /0 ϕ′ δ/ / ... /0 ϕ′′ ϕ 0 δ 0 0 If the two (*) -rows are exact, then the (⋄) - sequence is exact for a suitable “connecting homomorphism” δ.
So there is a unique α′ : λ∈Λ M ⊗ Mλ → M ⊗ (m ⊗ mλ )λ∈Λ → λ∈Λ Mµ µ∈Λ aλ (m ⊗ mλ ) Obviously: (α′ ◦ α)(m ⊗ (mλ )λ ) = ... = m ⊗ (mλ )λ =⇒ α′ ◦ α = id =⇒ α is injective. 41 Mµ , such that: 2. Modules and linear maps (d) Clearly γ : ϕ → ϕ˜ is an R-linear map. Our claim is now, that γ is bijective: If ψ : M → HomR (N, P ) is R-linear, then ψ ′ :M × N → P (m, n) → ψ(m)(n) is bilinear. Thus there exists a unique homomorphism ϕ :M ⊗ N → P m ⊗ n → ψ(m)(n) = ϕ(m ⊗ n) = ϕ(m)(n) ˜ = γ(ϕ)(m)(n) Thus ψ = γ(ϕ) ∈ Im(γ) and γ is surjective.
Ri ∈ R, s ∈ S : srii = r˜si k r˜i ⊗ mi s =⇒ x = i=1 k 1 ⊗ r˜i mi s = i=1 k = Thus 0 = α(x) = 1 1 r˜i mi ), x ∈ ker α ⊗( s i=1 k ˜i mi i=1 r s k =⇒ ∃u ∈ S : u · = =⇒ x = 1 su ⊗ k i=1 u˜ ri mi = 0 52 r˜i mi = 0 i=1 (u˜ ri )mi 3. Localisation (b) clear (c) We show two inclusion: “⊇”: n s “⊆”: Let = n′ s′ with n ∈ N, n′ ∈ N ′ , s, s′ ∈ S. =⇒ ∃u ∈ S : us′ n = usn′ ∈ N ∩ N ′ ∈N ′ ∈N =⇒ n s = ′ us n us′ s ∈ S −1 (N ∩ N ′ ) (d) We know that /M /M /N 0 /0 N is exact. 6 we know that / S −1 N 0 / S −1 (M / S −1 M N) /0 is exact.