2 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 1. Ideals In this course, all rings A will be commutative with unity 1. An ideal a A is allowed to be all of A. So A/a = 0 is a ring. (Zero is the only ring⊆ in which 1 = 0.) The ideals we are interested in are the proper ideals a A. All homomorphisms of rings are required to take 1 to 1. All subrings are required to contain 1. We want to discuss the following subsets of ring and how they are expressed in terms of maximal and prime ideals. (1) D := the set of zero divisors D := a A b =0, ab =0 { ∈ |∃ } So, 0 D iff A =0. (2) U :=∈ the set (group) of units in A. U = U(A) := a A b A, ab =1 { ∈ |∃ ∈ } (3) n := the set (ideal) of nilpotent elements of A. n := x A n, xn =0 { ∈ |∃ } Exercise: Show that 1 + n U. ⊆ 1.1. maximal ideals. Definition 1.1. m is a maximal ideal if it is maximal among all proper ideals. Proposition 1.2. m A is maximal iff A/m is a field. ⊂ Note: In a field, 1 = 0. So, every field F has exactly two ideals: 0 and F . Also, any element of a ring which is not a unit generates a proper ideal. The proposition follows. Theorem 1.3 (A-M 1.3,1.4). Every nonzero ring has maximal ideals and every proper ideal is contained in a maximal ideal. (by Zorn’s lemma) Lemma 1.4 (A-M 1.5). a A is a nonunit iff a is contained in some maximal ideal. ∈ Proposition 1.5. The set of nonunits of A is equal to the union of all maximal ideals of A: A U = m − MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 3 1.1.1. Jacobson and nil radicals. Definition 1.6. The Jacobson radical is defined to be the intersec- tion of all maximal ideals of A. rad A := m Lemma 1.7. If x rad A then u + xis a unit for all u U. ∈ ∈ Proof. Otherwise, u + x m u m , a contradiction. ∈ i ⇒ ∈ i Proposition 1.8 (A-M 1.9, variant). Let a be any ideal. Then a rad A 1+a U ⊆ ⇐⇒ ⊆ Proof. By the lemma, a rad A 1+a U. Conversely, suppose that a m for some maximal⊆ m ⇒. Then x⊆ a,x / m which implies ⊆ 1 1 ∃ ∈ ∈ that x is a unit modulo m. So, y A s.t. xy 1+m1. But then 1+x( y) m is not invertible.∃ So,∈ 1 + a U. ∈ − ∈ 1 ⊆ If x A is nilpotent then 1 + x is a unit. Threrefore: ∈ Corollary 1.9. The set of nilpotent elements of A forms an ideal nilrad A which is contained in the Jacobson radical: nilrad A rad A ⊆ Definition 1.10. A local ring is a ring with a unique maximal ideal. A semilocal ring is a nonzero ring with only finitely many maximal ideals. Corollary 1.11 (A-M 1.6.ii). Let m be a maximal ideal. Then 1+m U iff A is a local ring (iffthere are no other maximal ideals.) ⊆ Proof. By the proposition, 1+m U m m no other maximal ideals ⊆ ⇐⇒ ⊆ i ⇐⇒ 1.1.2. problems. (1) Let A[x] be the polynomial ring in one generator x. Show that f(x)=a + a x + a xn is a unit in A[x] iff a is a unit in A 0 1 ··· n 0 and a1, ,an are nilpotent. (2) Show that··· rad A[x] = nilrad A[x]. (3) Show that the only idempotent (e2 = e) contained in rad A is 0..