Commutative algebra - Department of Mathematical Sciences - old ...

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84 6. FINITE MODULES (1) E is an injective R-module. (2) HomR(S, E) is an injective S-module. Proof. (1) ⇒ (2): This is 3.6.10. (2) ⇒ (1): Let E ⊂ E ′ be an injective envelope, 3.6.15. HomR(S, E) → HomR(S, E ′ ) is injective and identifies HomR(S, E) as an essential submodule, since S is a finite R-module. E ′ is injective, so also HomR(S, E ′ ). It follows that HomR(S, E) HomR(R, E) E HomR(S, E ′ ) HomR(R, E ′ ) E ′ with the right down map being surjective. So E = E ′ . 6.6.8. Exercise. (1) Show that Z → Z[ √ −5] is finite. (2) Let p be a prime number. Show that Z → Z (p) is not finite.
7 Modules of finite length 7.1. Simple Modules 7.1.1. Definition. A nonzero module M is a simple module if 0 and M are the only submodules. 7.1.2. Proposition. Let f : M → M ′ be a homomorphism (1) If M is simple, then f is either zero or injective. (2) If M ′ is simple, then f is either zero or surjective. (3) If M, M ′ are both simple, then f is either zero or an isomorphism. Proof. This follows from 2.3.3. 7.1.3. Proposition. A simple module is isomorphic to a factor module R/P for some uniquely determined maximal ideal P . A simple module is finite, in fact generated by one element. Proof. A nonzero f : R → M must be surjective and R/P is simple exactly when P is maximal. Clearly P = Ann(M). 7.1.4. Corollary. Any nonzero finite module has a simple factor module. Proof. If M = 0 then MP = 0 for some maximal ideal. By 6.4.1 M ⊗R k(P ) = M/P M = 0. Choose a nonzero linear projection M/P M → k(P ), giving M → k(P ) → 0 exact. 7.1.5. Example. (1) If K is a field, a one dimensional vector space is simple. (2) If R is a principal ideal domain, then R/(p) is a simple module for all irreducible (p). (3) Z/(p) are simple for all prime numbers p. (4) Let K be a field. K[X]/(X − a) is a simple K[X]-module. 7.1.6. Lemma. Let I ⊂ R be an ideal. A simple R/I-module is a simple Rmodule. 7.1.7. Exercise. (1) Show that Z does not contain any simple modules. (2) Show that Q does not have any simple factor Z-module. (3) Let L, L ′ ⊂ M be simple submodules. Show that either L ∩ L ′ = 0 of L = L ′ . (4) Let L = L ′ ⊂ M be simple submodules. Show that L ⊕ L ′ = L + L ′ . 7.2. The Length 7.2.1. Definition. Let R be a ring. A module M is of finite length if it admits a composition series by submodules 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn−1 ⊂ Mn = M such that each factor Mi/Mi−1 is a simple module. 85
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ELEMENTARY COMMUTATIVE ALGEBRA LECT
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Contents Prerequisites 7 1. A dicti
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Prerequisites The basic notions fro
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10 1. A DICTIONARY ON RINGS AND IDE
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22 2. MODULES (5) If R is a field,
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24 2. MODULES (1) IM = {a1y1 + ·
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26 2. MODULES 2.3.6. Corollary. Let
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28 2. MODULES (2) Give an example o
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30 2. MODULES 2.4.11. Proposition.
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32 2. MODULES Proof. By 2.5.3 the c
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Page 36 and 37: 36 2. MODULES 2.6.12. Example. Let
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