Commutative Algebra/Basics on prime and maximal ideals and local rings

Prime ideals 0 100% #FFFFFF Definition 12.1: Let be a ring. A prime ideal is an ideal of such that whenever , either or . Lemma 12.2: Let be a ring and an ideal. is prime if and only if is an integral domain. Proof: prime is equivalent to . This is equivalent to . 0 100% #FFFFFF Theorem 12.3: Let be multiplicatively closed. Then there exists a prime ideal not intersecting . Proof: Order all ideals of not intersecting by set inclusion, and let a chain be given. The ideal (this is an ideal, since , hence , ) is an upper bound of the chain, since cannot intersect for else one of the would intersect . Since the given chain was arbitrary, Zorn's lemma implies the existence of a maximal ideal among all ideals not intersecting . This ideal shall be called ; we prove that it is prime. Let , and assume for contradiction that and . Then , are strict superideals of and hence intersect , that is, , , , , . Then , contradiction. Projection to the quotient ring In this section, we want to fix a notiation. Let be a ring and an ideal. Then we may form the quotient ring consisting of the elements of the form , . Throughout the book, we shall use the following notation for the canonical projection : 0 100% #FFFFFF Definition 12.4: Let an ideal. The map is the canonical projection of to . Maximal ideals 0 100% #FFFFFF Definition 12.5: Let be a ring. A maximal ideal of is an ideal that is not the whole ring, and there is no proper ideal such that . Lemma 12.6: An ideal is maximal iff is a field. Proof: A ring is a field if and only if its only proper ideal is the zero ideal. For, in a field, every nonzero ideal contains , and if is not a field, it contains a non-unit , and then does not contain . By the correspondence given by the correspondence theorem, corresponds to , the zero ideal of corresponds to , and any ideal strictly in between corresponds to an ideal such that . Hence, is a field if and only if there are no proper ideals strictly containing . Lemma 12.7: Any maximal ideal is prime. Proof 1: If is a ring, maximal, then is a field. Hence is an integral domain, hence is prime. Proof 2: Let be maximal. Let . Assume . Then for suitable , . But then . 0 100% #FFFFFF Theorem 12.8: Let be a ring and an ideal not equal to all of . Then there exists a maximal with . Proof: We order the set of all ideals such that and by inclusion. Let be a chain of those ideals. Then set . Clearly, all are contained within . Since , . Further, assume . Then for some , contradiction. Hence, is a proper ideal such that , and hence an upper bound for the given chain. Since the given chain was arbitrary, we may apply Zorn's lemma to obtain the existence of a maximal element with respect to inclusion. This ideal must then be maximal, for any proper superideal also contains . 0 100% #FFFFFF Lemma 12.9: Let be a ring, . Then via , maximal ideals of containing correspond to maximal ideals of . Proof: From the correspondence theorem. Local rings 0 100% #FFFFFF Definition 12.10: A local ring is a ring that has exactly one maximal ideal. 0 100% #FFFFFF Theorem 12.11 (characterisation of local rings): Let be a ring. The following are equivalent: is a local ring. If is a unit, then either or is a unit, where arbitrary. The set of all non-units forms a maximal ideal. If where is a unit, then one of the is a unit. If is arbitrary, either or is a unit. Proof: 1. 2.: Assume and are both non-units. Then and are proper ideals of and hence they are contained in some maximal ideal of by theorem 12.7. But there is only one maximal ideal of , and hence , thus . Maximal ideals can not contain units. 2. 3.: The sum of two non-units is a non-unit, and if is a non-unit and , is a non-unit (for if , is an inverse of ). Hence, all non-units form an ideal. Any proper ideal of contains only non-units, hence this ideal is maximal. 3. 4.: Assume the are all non-units. Since the non-units form an ideal, is contained in that ideal of non-units, contradiction. 4. 5.: Assume , are non-units. Then is a non-unit, contradiction. 5. 1.: Let two distinct maximal ideals. Then , hence , , , that is, . is not a unit, so is, contradiction. Localisation at prime ideals In chapter 9, we had seen how to localise a ring at a multiplicatively closed subset . An important special case is , where is a prime ideal. Lemma 12.12: Let be a prime ideal of a ring. Then is multiplicatively closed. Proof: Let . Then can't be in , hence . 0 100% #FFFFFF Definition 12.13: Let be a prime ideal of a ring. Set . Then is called the localisation of at . Theorem 12.14: Let be a ring, be prime. is a local ring. Proof: Set , then . Set . All elements of are non-units, and all elements of are of the form , , and thus are units. Further, is an ideal since is and by definition of addition and multiplication in and since is multiplicatively closed. Hence is a local ring. This finally explains why we speak of localisation.