11.1 Wedderburn's Theorem
1. If RM is simple, then M = Rx for any 0 ≠ x 
M. Hence α : R → M given by α(r) = rx is an onto R-linear map, so M ≅ R/L where L = ker(α), and L is maximal because M is simple. Conversely, R/L is simple for every maximal left ideal L. Conversely, R/L is simple for every maximal left ideal L again by Theorem 6 §8.1.
M. Hence α : R → M given by α(r) = rx is an onto R-linear map, so M ≅ R/L where L = ker(α), and L is maximal because M is simple. Conversely, R/L is simple for every maximal left ideal L. Conversely, R/L is simple for every maximal left ideal L again by Theorem 6 §8.1.
3. Define σ : Ra → Rb by σ(ra) = rb. This is well defined because ra = 0 implies rbR = r(aR) = 0, so rb = 0 . So σ is an onto homomorphism of left R-modules with σ(a) = b. Finally if σ(ra) = 0 then rb = 0 so (ra)R = r(aR) = r(bR) = 0, so ra = 0 . Hence σ is one-to-one.
5. If L1⊆ L2 ⊆ 
are left ideals of eRe, then RL1⊆ RL2 ⊆ are left ideals of R so RLn = RLn+1 = for some n by hypothesis. If i ≥ n, the fact that Li ⊆ eRe for each i, gives Li = eLi ⊆ eRLi = eRLn = eReLn ⊆ Ln. Hence Ln = Ln+1 = , as required.
are left ideals of eRe, then RL1⊆ RL2 ⊆ are left ideals of R so RLn = RLn+1 = for some n by hypothesis. If i ≥ n, the fact that Li ⊆ eRe for each i, gives Li = eLi ⊆ eRLi = eRLn = eReLn ⊆ Ln. Hence Ln = Ln+1 = , as required.
7. If RM is finitely generated, then 143 M is an image of Rn for some n ≥ 1 by Theorem 5 §7.1 and its Corollaries. Since Rn is noetherian as a ...
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