Commutative Algebra/Fractions, annihilator

Fractions within rings 0 100% #FFFFFF Definition 9.1: Let be a commutative ring, and let be an arbitrary subset. is called multiplicatively closed iff the following two conditions hold: 0 100% #FFFFFF Definition 9.2: Let be a ring and a multiplicatively closed subset. Define , where the equivalence relation is defined as . Equip this with addition and multiplication . The following two lemmata ensure that everything is correctly defined. Lemma 9.3: is an equivalence relation. Proof: For reflexivity and symmetry, nothing interesting happens. For transitivity, there is a little twist. Assume and . Then there are such that and . But in this case, we have ; note because is multiplicatively closed. Lemma 9.4: The addition and multiplication given above turn into a ring. Proof: We only prove well-definedness; the other rules follow from the definition and direct computation. Let thus and . Thus, we have and for suitable . We want and . These translate to and for suitable . We get the desired result by picking and observing and . Note that we were heavily using commutativity here. 0 100% #FFFFFF Theorem 9.5 (properties of augmentation): Let a ring and multiplicatively closed. Set , the projection morphism. Then: is a unit. for some . Every element of has the form for suitable , . Let be ideals. Then , where . Let an ideal. If , then . We will see further properties like 4. when we go to modules, but we can't phrase it in full generality because in modules, we may not have a product of two module elements. Proof: 1.: If , then the rules for multiplication for indicate that is an inverse for . 2.: Assume . Then there exists such that . 3.: Let be an arbitrary element of . Then . 4. 5. Let , that is, . Then , where is a unit in . Further, is an ideal within since is a morphism. Thus, . 0 100% #FFFFFF Theorem 9.6 (universal property): Let be a ring, multiplicatively closed, let be another ring and let be a morphism, such that for all , . Then there exists a unique morphism such that . Proof: We first prove uniqueness. Assume there exists another such morphism . Then we would have . Then we prove existence; we claim that defines the desired morphism. First, we show well-definedness. Firstly, exists for . Secondly, let , that is, . Then The multiplicativity of this morphism is visually obvious (use that is a morphism and commutativity); additivity is proven as follows: It is obvious that the unit is mapped to the unit. Theorem 9.7: Category theory context Fractions within modules 0 100% #FFFFFF Definition 9.8: Let be a ring, a multiplicative subset of and an -module. Set to be the ring augmented by inverses of . We define the -module as follows: (the formal fractions), where again , with addition and module operation . Note that applying this construction to a ring that is canonically an -module over itself, we obtain nothing else but canonically seen as an -module over itself, since multiplication and addition coincide. Thus, we have a generalisation here! That everything is well-defined is seen exactly as in the last section; the proofs carry over verbatim. 0 100% #FFFFFF Theorem 9.9 (properties of the augmented module): Let be an -module, let be a multiplicatively closed subset of , and let be submodules. Then , , and ; in the first two equations, all modules are seen as submodules of (as above with ), and in the third isomorphy relation, the modules are seen as independent -modules. Proof: 1. note that to get from the third row back to the second, we used that submodules are closed under multiplication by an element of to equalize denominators and thus get a suitable ( is closed under multiplication). 2. to get from the second to the first row, we note for a suitable , and in particular for example , where . 3. We set and prove that this is an isomorphism. First we prove well-definedness. Indeed, if , then , hence and thus . Then we prove surjectivity. Let be given. Then obviously is mapped to that element. Then we prove injectivity. Assume . Then , where and , that is for a suitable . Then and therefore . 0 100% #FFFFFF Theorem 9.10: functor relating tensor product and fractions 0 100% #FFFFFF Theorem 9.11: Let be -modules and multiplicatively closed. Then . Proof: Exercises Exercise 9.2.1: Let be -modules and an ideal. Prove that is a submodule of and that (this exercise serves the purpose of practising the proof technique employed for theorem 9.11). The annihilator, faithfulness 0 100% #FFFFFF Definition 9.12: Let be a ring, a module over and an arbitrary subset. Then the annihilator of with respect to is defined to be the set . 0 100% #FFFFFF Theorem 9.13: Let be a ring, a module over and an arbitrary subset. Then is an ideal of . Proof: Let and . Then for all , . Hence the theorem by lemma 5.3. 0 100% #FFFFFF Definition 9.14: An -module is called faithful iff . 0 100% #FFFFFF Theorem 9.15: Let be a ring. Then regarded as an module over itself is faithful. Proof: Let such that . Then in particular . 0 100% #FFFFFF Theorem 9.16: Let be an -module and an arbitrary subset. Let be the submodule of generated by . Then . Proof: From the definition it is clear that , since annihilating all elements of is a stronger condition than only those of . Let now and , where and . Then . Local properties Definition 9.17: Let be an -module (where is a ring) and let be a prime ideal. Then the localisation of with respect to , denoted by , is defined to be with ; note that is multiplicatively closed because is a prime ideal. 0 100% #FFFFFF Definition 9.18: A property which modules can have (such as being equal to zero) is called a local-global property iff the following are equivalent: has property (*). has property (*) for all multiplicatively closed . has property (*) for all prime ideals . has property (*) for all maximal ideals . Theorem 9.19: Being equal to zero is a local-global property. Proof: We check the equivalence of 1. - 4. from definition 9.12. Clearly, 4. 1. suffices. Assume that is a nonzero module, that is, we have such that . By theorem 9.11, is an ideal of . Therefore, it is contained within some maximal ideal of , call (unfortunately, we have to refer to a later chapter, since we wanted to separate treatments of different algebraic objects. The required theorem is theorem 12.2). Then for we have and therefore in . The following theorems do not really describe local-global properties, but are certainly similar and perhaps related to those. Theorem 9.20: If is a morphism, then the following are equivalent: surjective. surjective for all multiplicatively closed. surjective for all prime. surjective for all maximal. Proof: