MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 43 8.3. characterization of Artin rings. Recall 7.14 (7.7): In a Noe- therian ring, every ideal a contains a power of its radical r(a). Taking a = 0 and r(0) = nilrad A we get: Corollary 8.13 (A-M 7.15). A Noetherian nilrad A is nilpotent. ⇒ Lemma 8.14. Suppose m A is a maximal ideal (A/m field) and A is Artinian or Noetherian. ⊂Then m/m2 is a fin dim vector space over A/m. Proof. A/m2 is Noetherian or Artinian and its proper ideals are exactly the vector subspaces of m/m2. ! Lemma 8.15 (A-M 6.11). Suppose that 0 A is a finite product of ⊆ (not necessarily distinct) maximal ideals 0 = m1m2 mn. Then the following are equivalent: · · · (1) A is Noetherian. (2) A is Artinian. (3) Each quotient m m /m m is finite dimensional as a 1 · · · i 1 · · · i+1 vector space over the field A/mi+1. The proof is by induction on n using the exact sequence 0 m1 mn 1 A A/m1 mn 1 0 → · · · − → → · · · − → Theorem 8.16 (A-M 8.5). A Artinian A Noetherian and dim A = 0. ⇐⇒ Proof. Both sides imply dim A = 0. So, assuming dim A = 0, we need to show A Artinian iff A Noetherian. By the lemma, it suffices to show that 0 is a finite product of maximal ideals. k ( ) A Artinian n = mi nilpotent (n = 0). So, mi mi = 0. So,⇒A is Noetherian.⇒ ∩ ⊆ ∩ ( ) A Noetherian 0 has primary decomposition.! So A has only finitely⇐ many minimal⇒primes. But dim A = 0 implies all primes are maximal. So, A has only finitely many maximal ideals mi. By 7.15 (8.13) this implies n = m is nilpotent. By 6.11 (8.15), A is Artinian. ∩ i ! Proposition 8.17. Suppose A is Noetherian local with max ideal m. Then either n (1) m = 0 for some n ( 6.11 A is Artin local) or (2) mn = mn+1 = ⇒ ' ' · · · Proof. By Nakayama (ma = a a = 0), ⇐⇒ mn = mn+1 mn = 0 ⇐⇒ ! 44 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Theorem 8.18 (A-M 8.7: Structure thm for Artin rings). n A Artin A = A A : Artin local ⇐⇒ ∼ i i i=1 " k n Proof. ( ) Suppose n = 0, n = i=1 mi. ⇒ ϕ Claim: A n A/mk is an isomorphism. i=1 i # Proof: (Using−→ properties of coprime ideals from 1.3.3) ! ϕ onto and m coprime mk coprime i ⇒ i ⇒ mk = mk $ i ∩ i ker ϕ = mk = mk = 0 !(ϕ is 1-1) ⇒ ∩ i i ( ) This follows from " Claim⇐ : A, B Artin A B Artin. ⇒ × Proof of claim: A 0 " A B # 0 B is a short exact sequence of A B modules where× the first× and third× terms satisfy the DCC for submo×dules since the submodules of A 0 are the ideals in A and the submodules of 0 B are the ideals of B×. So, the middle satisfies the × DCC for submodules = ideals. ! Proposition 8.19 (A-M 8.8). If A is Artin local then TFAE: (1) Every ideal is principal. (2) m is principal. (3) dim m/m2 1 (k = A/m). k ≤ Proof. (1) (2) (3) is obvious. So, suppose (3). ⇒ ⇒ 2 Case 1. dimk m/m = 0 m = m2 m = 0 A = A/m = k is a field ⇒ ⇒NAK ⇒ and 0 = (0) is the only ideal. 2 Case 2. dimk m/m = 1 m = (x) mr = (xr) ⇒ ⇒ m = rad A is nilpotent (mn = 0) Claim: These are all the ideals of A. Proof: Let a A be a nontrivial proper ideal. ⊂ a = A a m a = 0 a mn = 0 ' ⇒ ⊂ ' ⇒ '⊂ So, a mr, a mr+1 for some r ⊂ '⊂ r+1 r+1 r r+1 r r m ! a + m m a + m = m NAK a = m ⇒ ⊂ ⇒ ⇒ which shows (1). The book proof of the last line is: y = axr a, y / mr+1 a / m a : unit xr a a = (xr) = mr ∃ ∈ ∈ ⇒ ∈ ⇒ ⇒ ∈ ⇒ MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 45 ! 8.4. examples, problems. Example 8.20. A = k[x , x , ] is generated, a = (x , x2, x3, ). 1 2 · · · ∞ 1 2 3 · · · Then B = A/a is local with unique maximal ideal m = (x1, x2, ). Also m is the unique prime ideal. So, dim B = 0. But m is· ·not· nilpotent. So, A is not Artinian and not Noetherian. In fact, m is not finitely generated. Example 8.21. A = k[x2, x3]/(x4) k[x]/(x4). Then A is Artin local, m = (x2, x3), m2 = 0 and dim m = dim⊂ m/m2 = 2. Problem. Show that a local ring A with maximal ideal m is Artinian iff m is nilpotent and m/m2 is a finite dimension vector space over A/m. Use the fact that there is an epimorphism from the n-fold tensor power of m/m2 onto mn/mn+1. Problem #6, p.92 A Noetherian, q a p-primary ideal. Consider chains of primary ideals from q to p. Show that all such chains are bounded and all maximal chains have the same length. Step 1. We may assume q = 0. Proof: For simplicity suppose the maximal length is 3: q q q p in A ⊂ 1 ⊂ 2 ⊂ 0 q /q q /q p/q in A/q ⊂ 1 ⊂ 2 ⊂ p/q-primary prime p q q q q p q / is prime in A/ . i%/ are&' / (-primary%&'( and distinct. So, there is a 1-1 correspondence between p-primary ideals between q and p and p/q-primary ideals between 0 and p/q. So, we may replace A by A/q and assume that q = 0. Step 2. 0 is p-primary A p has no zero divisors. Proof: If s A p is a⇒z.d. −then st = 0 for some nonzero t. But 0 is p-primary. So,∈ either− t = 0 or sn = 0. But both are impossible. ı 1 Step 3. S = A p has no z.d. a " S− A = Ap and p-primary q: − ⇒ → ∀ 1 1 (1) S− q is S− p-primary (2) qec = q There is a 1-1 correspondence between p-primary ideals in A and ⇒ 1 S− p-primary ideals in the local ring Ap. 1 A Ap = S− A 1 e 1 p-primary q S− q = q S− p-primary c 0→ 1 p-primary a a S− p-primary ← 46 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Step 4. Ap is Artin local. Proof: dim Ap = 0. p is a minimal prime (It is the only prime associated to 0.) Step 5 In an Artin local ring all ideals are m-primary. 2 dimk m/m = d1 i i+1 dimk m /m = di m m+1 dimk m /m = dr r+1 m = 0 dr+1 = 0 Claim. All maximal chains of proper ideals have length d1 + d2 + + dk · · ·Proof: For each ideal a define a mi + mi+1 a = dim ∩ i k mi+1 (1) If a ! b then 0 ai < bi di ≤ ≤ (2) If bi ai 2 then c s.t. a ! c ! b [Let i be minimal so − ≥ ) ∃)i ) i i+1 that bi > ai. Take b b m , b / a m + m . Let c = a + (b). Then) c =)a + 1 and∈c ∩= a for∈ j ∩= i.] i i j j ' Therefore, the numbers ai take all values from 0 to di for any maximal chain of ideals. ) ).