Commutative Algebra/Radicals, strong Nakayama

Radicals of ideals 0 100% #FFFFFF Definition 13.1: Let be a ring, an ideal. The radical of , denoted , is . A radical ideal is an ideal such that . 0 100% #FFFFFF Theorem 13.2: Let be a ring, an ideal. Then . Proof: For , note that . For , assume for no . Form the quotient ring . By theorem 12.3, pick a prime ideal disjoint from the multiplicatively closed set . Form the ideal . is a prime ideal which contains and does not intersect . Hence is not in the right hand side. 0 100% #FFFFFF Corollary 13.3: The radical of an ideal is an ideal. Proof: Intersection of ideals is an ideal. 0 100% #FFFFFF Definition 13.4: Let be a ring, . The Jacobson radical of is defined as thus: . 0 100% #FFFFFF Theorem 13.5: Let be a ring, . is a radical ideal. Proof: Clearly, . Further from theorem 13.2; the last equality from . The radicals of the zero ideal 0 100% #FFFFFF Definition 13.6: Let be a ring. is an ideal. The nilradical of , written , is defined as . Note that by definition , the set of nilpotent elements. 0 100% #FFFFFF Theorem 13.7: . Proof: Theorem 13.2. 0 100% #FFFFFF Definition 13.8: Let be a ring. is an ideal. The Jacobson radical of , written , is defined as . We have . If is a ring, is the set of units of . 0 100% #FFFFFF Theorem 13.9: Let be a ring, its Jacobson radical. Then . Proof: : Let , . Assume . Form the ideal ; by theorem 12.8 there exists maximal with , hence . If , then , contradiction. : Assume for all and . Then there is a maximal ideal not containing . Hence and for a and an . Hence is not a unit. Radicals and localisation 0 100% #FFFFFF Definition 13.10: If is a ring, is a multiplicative subset, an ideal, set , the localisation of the ideal with respect to . 0 100% #FFFFFF Theorem 13.11: Let be a ring, an ideal, a multiplicatively closed subset. Then . Proof: Let , that is, . Then , , . There exists such that . Thus , whence and . Thus, . Let . We may assume . Choose such that . Then , whence . Strong Nakayama lemma Exercises Prove that whenever is a reduced ring, then the canonical homomorphism is injective.