22 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA

3.4. review. x 1 = 0 in S− M xu = 0 for some u S s ⇔ ∈ and (Corollary 3.9): 1 1 S− M = S− A M ∼ ⊗ 1 1 1 Lemma 3.12 (A-M 3.7). S− (M N) = S− M 1 S− N.In A ∼ S− A particular: ⊗ ⊗ (M N) = M N ⊗A p ∼ p ⊗Ap p 3.5. local properties. A property P describing a class of rings or modules is called local if property P holds for a ring A (or module) iff it holds for every localization Ap at a prime has property P .For example, the property of being equal to 0 is local: Proposition 3.13 (A-M 3.8). If M is an A-module, the following are equivalent. (1) M =0 (2) Mp =0for every prime ideal p in A (3) Mm =0for every maximal ideal m in A. Proof. (1) (2) (3) is clear. So, we just need to show (1) (3). If x =0 ⇒M consider⇒ Ann(x) m. ¬ ⇒¬ ￿ ∈ ⊆ ￿ Using the exactness of localization, we can easily get: Corollary 3.14 (A-M 3.9). Suppose that ϕ : M N is a homomor- phism of A-modules. Then the following are equivalent.→ (1) ϕ is a monomorphism. (2) ϕ : M N is a monomorphism for every prime ideal p in A p p → p (3) ϕm : Mm Nm is a monomorphism for every maximal ideal m in A. → And the same holds with “monomorphism” replace with “epimorphism.” This, and Lemma 3.12 imply that flatness is a local property: Proposition 3.15 (A-M 3.10). If M is an A-module, the following are equivalent. (1) M is flat. (2) Mp is a flat Ap-module for every prime ideal p in A (3) Mm is a flat Am-module for every maximal ideal m in A. MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 23

3.6. extended and contracted ideals in localizations. Recall: for 1 the ring homomorphism f : A S− A the extension of an ideal a A e 1 → ⊆ is given by a = S− a and the contraction of an ideal b B is, by c 1 ⊆ definition, b = f − (b). It is clear that the contraction of a prime ideal is prime. Also, there is always a between the of contracted ideals C = bc = a A a = aec { } { ⊆ | } and the set of extended ideals: E = ae = b B b = bce { } { ⊆ | } 1 Proposition 3.16 (A-M 3.11). (i) Every ideal in S− A is an ex- tended ideal. (ii) If a is an ideal in A then aec = (a : s) s S ￿∈ e 1 So, a = S− A iff S a = . (iv) p pe gives a bijection∩ ￿ between∅ the primes in A disjoint from ↔ 1 S and the primes in S− A. 1 c Proof. (i), (ii) are easy. For (iv): If q is a prime ideal in S− A then q is prime in A and qce = q. So, qc is disjoint from S by (ii). Going the other way, suppose that p is a prime ideal in A disjoint from S. In other words, S T = A p. Then, ⊆ − 1 t S− T = t T, s S s | ∈ ∈ ￿ ￿ 1 is a multiplicative of S− A and 1 1 1 S− A = S− T S− p

e 1 1 Therefore p = S− p is a prime ideal in￿S− A. ￿ Corollary 3.17 (A-M 3.13). If p A is prime then the prime ideals ⊆ of Ap are qp where q is a prime contained in p. Exercise 3.18. (1) Show that the ring homomorphism f : A 1 → S− A induces an injective mapping: 1 f ∗ : Spec(S− A) Spec(A) → 2 3 1 (2) If S = 1,h,h ,h , then S− A is denoted A and the { ···} h of Spec(Ah) in Spec(A) is the set X = p h/p h { | ∈ } 24 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA

(3) Show that the mapping in (1) is a homeomorphism onto its im- age. Recall that the basic open sets in Spec(A) are the X above. (The closed sets are V (E)= p E p .) h { | ⊆ } 1 1 1 Proposition 3.19 (A-M 3.11(v)). (1) S− (a b)=S− a S− b 1 1 1 ∩ ∩ (2) S− (ab)=(S− a)(S− b) 1 1 1 (3) S− (a + b)=S− a + S− b 1 1 (4) S− (r(a)) = r S− a Putting a = 0 in (4) we get: 1 1 Corollary 3.20 (A-M 3.12). S− nilrad A = nilrad S− A 1 1 Lemma 3.21. Ann(S− (A/a)) = S− a Proposition 3.22 (A-M 3.14). If M is a f.g. A-module then 1 1 Ann(S− M)=S− Ann(M) N+P Since (N : P ) = Ann N we get: Corollary 3.23 (A-M￿ 3.15).￿ If N,P are submodules of M and P is f.g. Then 1 1 1 S− (N : P )=(S− N : S− P )