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114 10. DEDEKIND RINGS (2) If R has only finitely many maximal ideals then it is a principal ideal domain. Proof. (1) Any nonzero prime ideal is principal. Conclusion by 10.3.3. (2) Let P, P2 . . . , Pn be the finitely many maximal ideals. Choose a ∈ P \P 2 ∪P2 · · ·∪Pn, 5.1.3. Then (a) is P primary. By 10.3.3 (a) = P k , so (a) = P . As all maximal ideals are principal, conclusion by 10.3.3. 10.3.5. Proposition. Let R be Dedekind domain. An ideal I is generated by at most two elements. Proof. Let Ass(R/I) = {P1, . . . , Pn} and U = R\P1 ∪ · · · ∪ Pn, then by 10.3.4 U −1 R is a principal ideal domain. By 10.1.6 U −1 R U −1 I, so choose by 8.2.9 a homomorphism f : R → I such that U −1 f is an isomorphism. Then f is injective 9.2.8 and the ideal f(R) = (a) ⊂ R satisfies: Pi /∈ Ass(R/(a)) for any i. Let Q1, . . . , Qm ∈ Ass(I/(a)) and choose b ∈ I\Q1 ∪ · · · ∪ Qm. b is a nonzero divisor on I/(a) and therefore b I/(a) is an isomorphism as I/(a) has finite length. It follows that I = (a, b). 10.3.6. Theorem. Let R be Dedekind domain. (1) A torsion free module is flat. (2) A finite torsion free module is projective. (3) Any ideal is projective. (4) Let F be a finite torsion free module. Then there is a number n and an ideal I such that F R n ⊕ I Proof. (1), (2), (3) These follow from 10.1.6, 6.5.13. (4) Let R have fraction field K and assume rankK F ⊗RK = n+1. Choose a nonzero homomorphism F → R and get by induction on n, F I1⊕· · ·⊕In+1 for nonzero ideals Ij in R. It suffices to treat the case n = 1. Let Ass(R/I1) = {P1, . . . , Pm} and U = R\P1∪· · ·∪Pm, then by 10.3.4 U −1 R is a principal ideal domain. By 10.1.6 U −1 I2 U −1 R, so choose by 8.2.9 a homomorphism f : I2 → R such that U −1 f is an isomorphism. Then f is injective 9.2.8 and the ideal f(I2) ⊂ R satisfies: Pi /∈ Ass(R/f(I2)) for any i. It follows that I1 + f(I2) = R. Conclusion by a surjection I1 ⊕ I2 → R. 10.3.7. Proposition. Let R be a Dedekind domain. A finite module M decomposes M = T (M) ⊕ F as a direct sum of the torsion submodule and a finite torsion free submodule F . Proof. By 10.3.6 the projection M → M/T (M) splits. 10.3.8. Proposition. Let R be a Dedekind domain. A finite torsion module M has decomposition M = R/P n1 nk 1 ⊕ · · · ⊕ R/Pk Where P1, . . . , Pk are not necessarily distinct maximal ideals. Proof. This follows from 9.5.1 and 10.2.9. 10.3.9. Corollary. Let R be a Dedekind domain and M a finite module. Then M has decomposition M = R/P n1 1 ⊕ · · · ⊕ R/P nk k ⊕ Rn ⊕ Q1 · · · Ql Where P1, . . . , Pk, Q1, . . . , Ql are not necessarily distinct maximal ideals.
10.3. DEDEKIND DOMAINS 115 10.3.10. Exercise. (1) Show that the ring Z[ √ −5] is a Dedekind domain. (2) Show that the ring Z[ √ 5] is not a Dedekind domain.
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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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34 2. MODULES (3) The formation of
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36 2. MODULES 2.6.12. Example. Let
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38 2. MODULES (1) The change of rin
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40 3. EXACT SEQUENCES OF MODULES 3.
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42 3. EXACT SEQUENCES OF MODULES Fo
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44 3. EXACT SEQUENCES OF MODULES an
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46 3. EXACT SEQUENCES OF MODULES Pr
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58 4. FRACTION CONSTRUCTIONS 4.1.4.
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60 4. FRACTION CONSTRUCTIONS 4.2.7.
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62 4. FRACTION CONSTRUCTIONS and ge
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Commutative algebra - Department of Mathematical Sciences - old ...

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