REGULAR SEQUENCES AND REGULAR RINGS HYEONGKWAN KIM Abstract. Regular sequence is defined and its homological properties are explored in relation to Koszul complex and Ext. The notion of depth is intro- duced and related to properties of regular rings. Regular local rings are shown to belong to a larger class of rings called Cohen-Macaulay rings. 1. Reference Contents of this paper are freely drawn from Matsumura, Commutative ring theory, Cambridge University Press, 1980. 2. Background The (Krull) dimension of a ring is the supremum of the lengths of chains of prime ideals. The height of a prime ideal p is the dimension of Ap, and is denoted by ht p. Let (A, m) be a Noetherian local ring. It is a well-known result from dimension theory that the dimension of A is equal to the least number of generators of an m-primary ideal of A. If dim A = d and a1,...,ad generate an m-primary ideal, then a1,...,ad is called a system of parameters. Hence the number of minimal generators of m is at least the dimension of A, and if it is equal to the dimension the ring is called a regular local ring, and the generators are called a regular system of parameters. A regular local ring is a UFD (Auslander and Buchsbaum), and localization of a regular local ring at every prime ideal is again a regular local ring(Serre). A regular ring is a Noetherian ring whose localization at every prime ideal is a regular local ring. A ring is normal if its localization at every prime is an integrally closed domain. It is easy to show that a local normal ring is a domain and that a regular ring is normal. A ring is said to be catenary if for every pair of prime ideals, there is a maximal chain of prime ideals between them and all such chains have the same (finite) length. A ring is said to be universally catenary if it is Noetherian and every finitely generated A-algebra is catenary. If M is an A-module and I ⊂ A an ideal, then the I-adic topology on M is defined by taking neighborhoods of 0 to be those subsets containing InM for some n. The set of prime ideals of a ring A is called the spectrum of A, and is denoted Spec(A). If I ⊂ A is an ideal, denote V (I) := {p ∈ Spec(A) | I ⊂ p}. Let A be a ring and a1,...,an ∈ A. The Koszul complex is defined as follows: K0 = A, Ki = 0 for i< 0 and i>n. For0 ≤ i ≤ n, Ki is the free A-module of rank n! − (n−i)!i! with basis {ej1...ji |1 ≤ j1 < · · · ji ≤ n}. The differential d : Ki −→ Ki 1 is 1 2 HYEONGKWAN KIM given by i k−1 d(ej1...ji )= X(−1) ajk ej1...jˆk ...ji . k=1 The complex is denoted by K•(a1,...,an) or K•(a). If M is an A-module, define K•(a,M) := K•(a) ⊗A M and denote its homology group by Hi(a,M). 3. Regular sequences Definition 3.1. Let A be a ring and M an A-module. An element a ∈ A is said to be M-regular if it is not a zero divisor in M, that is, if m ∈ M, am = 0 implies m = 0. Definition 3.2. a1,...,an ∈ A is said to be an M-sequence if the following condi- tions hold: (1) a1 is M-regular, a2 is M/a1M-regular, a3 is M/(a1,a2)M-regular,...,an is M/(a1,...,an−1)M-regular. (2) M/(a1,...an)M 6= 0. Note that if a1,...,an is an M-sequence then so is a1,...,ai for any i ≤ n. Lemma 3.3. If a1,...,an is an M-sequence, and if a1m1+. .+anmn =0, mi ∈ M, then mi ∈ a1M + ...anM for all i. Proof. If n = 1 then a1m1 = 0 implies m1 = 0 since a1 is an M-sequence and hence M-regular. In particular, m1 ∈ a1M. Now assume it is true for n − 1. Since an is non-zero divisor modulo a1,...,an−1, the condition a1m1 + . + anmn = 0 implies (3.1) mn = a1q1 + . + an−1qn−1, for some qi ∈ M. Therefore n−1 a1m1 + . + anmn = X ai(mi + anqi)=0. i=1 By induction, mi + anqi ∈ a1M + ...an−1M for 1 ≤ i ≤ n − 1. Hence mi ∈ a1M + . + anM, 1 ≤ i ≤ n − 1. This is also true for i = n by (3.1). p1 pn Theorem 3.4. If a1,...,an ∈ A is an M-sequence then so is a1 ,...,an for any positive integers p1,...,pn. p Proof. It suffices to show that if a1,a2,...,an is an M-sequence, so is a1,a2,...,an. p1 For, if it is the case, then a1 ,a2,...,an is an M-sequence, and a2,a3,...,an is p1 p2 p1 an (M/a1 M)-sequence and hence a2 ,a3,...,an is an (M/a1 M)-sequence, and pi p1 pn so on. So ai is (M/(a1,...,ai−1)M)-regular, and M/(a1 ,...,an )M 6= 0 since M/(a1,...,an)M 6= 0. Prove by induction on p. It is clearly true for p = 1. Assume it is true for p − 1. p Since a1 is M-regular, so is a1. For i> 1, suppose p aim = a1m1 + a2m2 + . + ai−1mi−1, REGULAR SEQUENCES AND REGULAR RINGS 3 p−1 for some mj ∈ M. Since a1 ,a2,...,ai is an M-sequence by assumption, this implies that p−1 (3.2) m = a1 q1 + a2q2 + . + ai−1qi−1, for some qj ∈ M. Hence we get p−1 0= a1 (a1m1 − aiq1)+ a2(m2 − aiq2)+ . + ai−1(mi−1 − aiqi−1). By the above lemma, we have p−1 a1m1 − aiq1 ∈ a1 M + a2M + . + ai−1M. Hence aiq1 ∈ a1M +. .+ai−1M. So by regularity q1 ∈ a1M +. .+ai−1M and (3.2) p p gives m ∈ a1M +a2M +. .+ai−1M. This shows that ai is (M/(a1,a2,...,ai−1)M)- p−1 p regular. Also, since M/(a1 ,a2,...,ai−1)M) 6= 0, we have M/(a1,a2,...,ai)M 6= 0. p So a1,a2,...,ai is an M-sequence, as required. Definition 3.5. Let M be an A-module. We write M[x1,...,xn] for M⊗AA[x1,...,xn] and call its elements polynomials with coefficients in M. If a1,...,an ∈ A, xi 7→ ai induces an A-module homomorphism M[x1,...,xn] → M. Denote the image of p ∈ M[x1,...,xn] under this map by p (a1,...,an). Note that M[x1,...,xn] is a direct sum of copies of M indexed by monomials in A[x1,...,xn]. Hence it makes sense to speak of coefficients and degree of elements of M[x1,...,xn]. Definition 3.6. Let a1,...,an ∈ A, I = (a1,...,an). a1,...,an is said to be an M-quasi-regular sequence if IM 6= M and for every positive integer d and every d+1 homogeneous polynomial F ∈ M[x1,...,xn] of degree d, F (a) ∈ I M implies that all the coefficients of F are in IM. Note that unlike M-regular sequence, this notion is independent of the order d+1 of a1,...,an. We get an equivalent definition if the condition F (a) ∈ I M is replaced by F (a) = 0. For, if F (a) ∈ Id+1M then there is a homogeneous polynomial G of degree d + 1 such that F (a) = G(a). If we lower its degree by 1 by replacing one variable xi from each term by ai and subtracting from F, we get a new homogeneous polynomial F ′ of degree d with F ′(a) = 0. Furthermore, if all coefficients of F ′ are in IM then the same is true for F . Lemma 3.7. Let a1,...,an be an M-quasi-regular sequence and I = (a1,...,an). If (IM : b)= IM for some b ∈ A then (IdM : b)= IdM for every positive integer d. Proof. Assume it is true for d − 1. Let m ∈ M and bm ∈ IdM. Then bm ∈ Id−1M so by the assumption m ∈ Id−1M. Hence m = F (a) where F is a homogeneous d polynomial of degree d − 1. So we have bF (a) = bm ∈ I M. Since a1,...,an is M-quasi-regular, all coefficients of bF are in IM, and therefore m = F (a) ∈ IdM. This shows (IdM : b)= IdM. Theorem 3.8. If a1,...,an is an M-sequence then it is M-quasi-regular. d Proof. If n = 1, a1 is M-regular so M 6= a1M. Also, if m ∈ M and m ⊗ x is a d d+1 d d+1 ′ homogeneous polynomial of degree d and a1m ∈ a1 M then a1m = a1 m for ′ d ′ ′ some m ∈ M, a1(m − a1m ) = 0, and again by M-regularity of a1, m = a1m , i.e., m ∈ a1M. 4 HYEONGKWAN KIM Now assume it is true for n − 1. Then in particular a1,...,an−1 is M − quasi − regular. Let F be a homogeneous polynomial of degree d such that F (a) = 0. Consider first the case d = 1. Let F = m1 ⊗ x1 + . + mn ⊗ xn, where mi ∈ M. Then F (a)= a1m1 +. .+anmn = 0. Since a1,...,an is an M-regular sequence, we ′ ′ ′ have mn = a1m1 + . + an−1mn−1 for some mi ∈ M and in particular mn ∈ IM where I = (a1,...,an). If we let ′ ′ h = m1 ⊗ x1 + . + mn−1 ⊗ xn−1 + an(m1 ⊗ x1 + . + mn−1 ⊗ xn−1), we have h(a1,...,an−1) = 0 and since a1,...,an−1 is M-quasi-regular, all coef- ficients of h are in (a1,...,an−1)M, and this implies m1,...,mn−1 ∈ IM.