By Thomas Markwig Keilen
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M1 ) ... , Mi ≤ M ′′ . Then β −1 (M0 ) ⊆ β −1 (M1 ) ⊆ β −1 (M2 ) ⊆ ... , Mi ≤ M . Then by assumption there exists a k, such that ∀i ≥ k we have α−1 (Mi ) = α−1 (Mk ) and β(Mi ) = β(Mk ). 6. 61 4. , pn − 1}, n ≥ 0 pn Zp∞ := = is artinian, but not noetherian (the so-called Pr¨ ufer group). To prove this, we claim that: N 1 pn Zp∞ a Z- submodule ⇐⇒ ∃ n ∈ N : N = =: Nn Z Proof. • “⇐=”: • “=⇒”: Let a pn ∈ N , such that p ∤ a. =⇒ gcd(a, pn ) = 1 =⇒ ∃ b, q ∈ Z : 1 = ba + qpn =⇒ a 1 a pn =b n +q n =b n ∈N n p p p p =0 =⇒ 1 pn ⊆N We now have to consider two cases: a pn (1) ∃ n maximal, such that there exists 1 pn N= (2) 1 pn ∈ N with p ∤ a.
Next we show that δ is R-linear: Let m′′ , m ˜ ′′ ∈ ker(ϕ′′ ); r, r˜ ∈ R and let m, m ˜ ∈ M and n′ , n ˜ ′ ∈ N ′ as in the definition of δ. =⇒ β(rm + r˜m) ˜ = rm′′ + r˜m ˜ ′′ , since β is linear ′ ′ ′ =⇒ α (rn + r˜n ˜ ) = ϕ(rm + r˜m), ˜ since α′ , ϕ are linear ˜ ′ = rδ(m′′ ) + r˜δ(m ˜ ′′ ) =⇒ δ(rm′′ + r˜m ˜ ′′ ) = rn′ + r˜n It remains to show, that the sequence is exact - we only prove this for the interesting part ker(δ) = Im(β| ): • “⊇”: Let m′′ ∈ Im(β| ) =⇒ ∃ m ∈ ker ϕ : β(m) = m′′ and thus (α′ )−1 (ϕ(m)) = δ(m′′ ) = 0 • “⊆”: Let m′′ ∈ ker(δ) and let m ∈ M, n′ ∈ N ′ as in the definition of δ.
32. Let M, N, P ; Mλ , λ ∈ Λ be R-modules. Then: (a) M ⊗R N ∼ = N ⊗R M via: m⊗n→n⊗m (b) (M ⊗R N ) ⊗R P ∼ = M ⊗R N ⊗R P via: = M ⊗R (N ⊗R P ) ∼ (m ⊗ n) ⊗ p → m ⊗ (n ⊗ p) → m ⊗ n ⊗ p (c) M ⊗ ( λ∈Λ Mλ ) ∼ = λ∈Λ (M ⊗ Mλ ) via: m ⊗ (mλ )λ∈Λ → (m ⊗ mλ )λ∈Λ In particular: M ⊗R Rn ∼ = Mn 40 2. Modules and linear maps (d) HomR (M ⊗ N, P ) ∼ = HomR (M, HomR (N, P )) via: ϕ → (ϕ˜ : M → HomR (N, P ) : m → (N → P : n → ϕ(m ⊗ n))) Proof. (a) clear, since N ⊗R M satisfies the universal property. (b) Exercise (c) M × bilin.