Lectures on Algebraic Theory of D-Modules
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1 Lectures on Algebraic Theory of D-Modules Dragan Miličić
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3 Contents Chapter I. Modules over rings of differential operators with polynomial coefficients 1 1. Hilbert polynomials 1 2. Dimension of modules over local rings 6 3. Dimension of modules over filtered rings 9 4. Dimension of modules over polynomial rings Rings of differential operators with polynomial coefficients Modules over rings of differential operators with polynomial coefficients Characteristic variety Holonomic modules Exterior tensor products Inverse images Direct images Kashiwara s theorem Preservation of holonomicity 51 Chapter II. Sheaves of differential operators on smooth algebraic varieties Differential operators on algebraic varieties Smooth points of algebraic varieties Sheaves of differential operators on smooth varieties 73 Chapter III. Modules over sheaves of differential operators on smooth algebraic varieties Quasicoherent D X -modules Coherent D X -modules Characteristic varieties Coherentor D-modules on projective spaces 90 Chapter IV. Direct and inverse images of D-modules The bimodule D X Y Inverse and direct images for affine varieties Inverse image functor Projection formula Direct image functor Direct images for immersions Bernstein inequality Closed immersions and Kashiwara s theorem Local cohomology of D-modules Base change 128 iii
4 iv CONTENTS Chapter V. Holonomic D-modules Holonomic D-modules Connections Preservation of holonomicity under direct images A classification of irreducible holonomic modules Local cohomology of holonomic modules Preservation of holonomicity under inverse images 140 Bibliography 143
5 CONTENTS v These notes represent a brief introduction into algebraic theory of D-modules. The original version was written in 1986 when I was teaching a year long course on the subject. Minor revisions were done later when I was teaching similar courses. A major reorganization was done in I would like to thank Dan Barbasch and Pavle Pandžić for pointing out several errors in previous versions of the manuscript.
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7 CHAPTER I Modules over rings of differential operators with polynomial coefficients 1. Hilbert polynomials Let A = n Z An be a graded noetherian commutative ring with identity 1 contained in A 0. Then A 0 is a commutative ring with identity 1. Assume that A n = 0 for n < Lemma. (i) A 0 is a noetherian ring. (ii) A is a finitely generated A 0 -algebra. Proof. (i) Put A + = n=1 An. Then A + is an ideal in A and A 0 = A/A +. (ii) A + is finitely generated. Let x 1, x 2,..., x s be a set of homogeneous generators of A + and denote d i = deg x i, 1 i s. Let B be the A 0 -subalgebra generated by x 1, x 2,..., x s. We claim that A n B, n Z +. Clearly, A 0 B. Assume that n > 0 and y A n. Then y A + and therefore y = s i=1 y ix i where y i A n di. It follows that the induction assumption applies to y i, 1 i s. This implies that y B. The converse of 1.1 follows from Hilbert s theorem which states that the polynomial ring A 0 [X 1, X 2,..., X n ] is noetherian if the ring A 0 is noetherian. Let M = n Z M n be a finitely generated graded A-module. Then each M n, n Z, is an A 0 -module. Also, M n = 0 for sufficiently negative n Z Lemma. The A 0 -modules M n, n Z, are finitely generated. Proof. Let m i, 1 i k, be homogeneous generators of M and deg m i = r i, 1 i k. For j Z + denote by z i (j), 1 i l(j), all homogeneous monomials in x 1, x 2,..., x s of degree j. Let m M n. Then m = k i=1 y im i where y i A n ri, i i k. By 1.1, y i = j a ijz j (n r i ), with a ij A 0. This implies that m = i,j a ijz j (n r i ) m i ; hence M n is generated by (z j (n r i ) m i ; 1 j l(n r i ), 1 i k). Let M fg (A 0 ) be the category of finitely generated A 0 -modules. Let λ be a function on M fg (A 0 ) with values in Z. The function λ is called additive if for any short exact sequence: we have Clearly, additivity implies that λ(0) = 0. 0 M M M 0 λ(m) = λ(m ) + λ(m ). 1
8 2 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS 1.3. Lemma. Let 0 M 0 M 1 M 2 M n 0 be an exact sequence in M fg (A 0 ). Then n ( 1) i λ(m i ) = 0. Proof. Evident. i=0 Let Z[[t]] be the ring of formal power series in t with coefficients in Z. Denote by Z((t)) the localization of Z[[t]] with respect to the multiplicative system {t n n Z + }. Let M be a finitely generated graded A-module. Then the Poincaré series P (M, t) of M (with respect to λ) is P (M, t) = n Z λ(m n ) t n Z((t)). For example, let A = k[x 1, X 2,..., X s ] be the algebra of polynomials in s variables with coefficients in a field k graded by the total degree. Then, A 0 = k and for every finitely generated graded A-module M, we have dim k M n <. Hence, we can define the Poincaré series for λ = dim k. In particular, for the A-module A itself, we have P (A, t) = n Z dim k A n t n = ( s + n 1 s 1 n=0 ) t n = 1 (1 t) s. The next result shows that Poincaré series in general have an analogous form Theorem (Hilbert, Serre). For any finitely generated graded A-module M we have f(t) P (M, t) = s i=1 (1 ) tdi where f(t) Z[t, t 1 ]. Proof. We prove the theorem by induction in s. If s = 0, A = A 0 and M is a finitely generated A 0 -module. This implies that M n = 0 for sufficiently large n. Therefore, λ(m n ) = 0 except for finitely many n Z and P (M, t) is in Z[t, t 1 ]. Assume now that s > 0. The multiplication by x s defines an A-module endomorphism f of M. Let K = ker f, I = im f and L = M/I. Then K, I and L are graded A-modules and we have an exact sequence This implies that 0 K M f M L 0. 0 K n M n x s M n+d s L n+ds 0 is an exact sequence of A 0 -modules for all n Z. In particular, by 1.3, λ(k n ) λ(m n ) + λ(m n+ds ) λ(l n+ds ) = 0,
9 1. HILBERT POLYNOMIALS 3 for all n Z. This implies that (1 t ds ) P (M, t) = n Z λ(m n ) t n n Z λ(m n ) t n+ds = n Z(λ(M n+ds ) λ(m n )) t n+ds i.e., = n Z(λ(L n+ds ) λ(k n )) t n+ds = P (L, t) P (K, t) t ds, (1 t ds ) P (M, t) = P (L, t) t ds P (K, t). From the construction it follows that x s act as multiplication by 0 on L and K, i.e., we can view them as A/(x s )-modules. Hence, the induction assumption applies to them. This immediately implies the assertion. Since the Poincaré series P (M, t) a rational function, we can talk about the order of its pole at a point. Let d λ (M) be the order of the pole of P (M, t) at 1. By the theorem, f(t) = k Z a kt k with a k Z and a k = 0 for all k Z except finitely many. Let p be the order of zero of f at 1. Assume that p > 0. Then f(t) = (1 t)g(t) where g(t) = k Z b kt k, with b k Q and b k = 0 for all k Z except finitely many. Moreover, we have a k = b k b k 1 for all k Z. By induction in k this implies that b k Z. By repeating this procedure if necessary, we see that f(t) = (1 t) p g(t) where g(t) = k Z b kt k, with b k Z and b k = 0 for all k Z except finitely many. Moreover, g(1) Corollary. If d i = 1 for 1 i s, the function n λ(m n ) is equal to a polynomial with rational coefficients of degree d λ (M) 1 for sufficiently large n Z. Proof. Let p be the order of zero of f at 1. Then we can write f(t) = (1 t) p g(t) with g(1) 0. In addition, we put d = d λ (M) = s p, hence Now, (1 t) d = k=0 P (M, t) = g(t) (1 t) d. d(d + 1)... (d + k 1) k! and if we put g(t) = N k= N a kt k we get λ(m n ) = for all n N. This is equal to N k= N a k (d + n k 1)! (d 1)!(n k)! = N k= N N k= N t k = ( ) d + k 1 t k, d 1 k=0 ( ) d + n k 1 a k d 1 (n k + 1)(n k + 2)... (n k + d 1) a k, (d 1)!
10 4 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS hence λ(m n ) is a polynomial in n with the leading term ( N ) n d 1 nd 1 a k = g(1) (d 1)! (d 1)! 0. k= N We call the polynomial which gives λ(m n ) for large n Z the Hilbert polynomial of M (with respect to λ). From the proof we see that the leading coefficient of the Hilbert polynomial of M is equal to g(1) (d 1)!. Returning to our example of A = k[x 1, X 2,..., X s ], we see that ( ) s + n 1 dim k A n = = ns 1 s 1 (s 1)! Hence, the degree of the Hilbert polynomial for A = k[x 1, X 2,..., X s ] is equal to s 1. Now we are going to prove a characterization of polynomials (with coefficients in a field of characteristic 0) having integral values for large positive integers. First, we remark that, for any s Z + and q s, we have ( ) q q s = s! + Q(q) s where Q is a polynomial of degree s 1. Therefore any polynomial P of degree d, for large q, can be uniquely written as ( ) ( ) ( ) q q q P (q) = c 0 + c c d 1 + c d, d d 1 1 with suitable coefficients c i, 0 i d. Since binomial coefficients are integers, if c i, 0 i d, are integers, the polynomial P has integral values for integers n d. The next result is a converse of this observation Lemma. If the polynomial ( ) ( ) ( ) q q q q P (q) = c 0 + c c d 1 + c d d d 1 1 takes integral values P (n) for large n Z, all its coefficients c i, 0 i d, are integers. Proof. We prove the statement by induction in d. If d = 0 the assertion is obvious. Also d ( ) q + 1 d ( ) q P (q + 1) P (q) = c i c i d i d i i=0 i=0 d (( ) ( )) q + 1 q d 1 ( ) q = c i = c i, d i d i d i 1 i=0 using the identity ( ) ( ) ( ) q + 1 q q = + s s s 1 for q s 1. Therefore, q P (q + 1) P (q) is a polynomial with coefficients c 0, c 1,..., c d 1, and P (n) Z for large n Z. By the induction assumption all c i, 0 i d 1, are integers. This immediately implies that c d is an integer too. i=0
11 1. HILBERT POLYNOMIALS 5 We shall need another related remark. If F is a polynomial of degree d with the leading coefficient a 0, G(n) = F (n) F (n 1) = (a 0 n d + a 1 n d ) (a 0 (n 1) d + a 1 (n 1) d ) = a 0 dn d is polynomial in n of degree d 1 with the leading coefficient da 0. The next result is a converse of this fact Lemma. Let F be a function on Z such that G(n) = F (n) F (n 1), is equal to a polynomial in n of degree d 1 for large n Z. Then F is equal to a polynomial in n of degree d for large n Z. Proof. Assume that G(n) = P (n 1) for n N d, where P is a polynomial in n of degree d 1. Then by 1.6 we have d 1 ( ) n P (n) = c i d i 1 i=0 Hence, for n N + 1, n F (n) = (F (k) F (k 1)) + F (N) = k=n+1 n k=n+1 G(k) + F (N) = n P (k 1) + C where C is a constant. Also, by the identity used in the previous proof, ( ) q q (( ) ( )) j j 1 q ( ) j 1 q ( ) j 1 = +1 = + 1 = s s s s 1 s 1 j=s+1 for q > s 1. This implies that n P (k 1) = k=d n d 1 k=d i=0 ( d 1 = c i i=0 for some constant C. n k=d i j=s+1 ( ) k 1 d 1 c i = d i 1 ( k 1 d i 1 i=0 ) ) d 1 i=1 c i ( n k=d ( d 1 c i k=d j=s ( ) ) k 1 d i 1 k=d i ( ) ) k 1 d i 1 d 1 ( ) n = c i + C d i In particular, it follows that the sum n N λ(m n ) is equal to a polynomial of degree d λ (M) for large N Z. In addition, if we put λ(m n ) = a 0 N d + a 1 N d a d 1 N + a d n N for large N Z, then d! a 0 is an integer. i=0
12 6 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS For example, if A = k[x 1, X 2,..., X s ], the dimension of the space of all polynomials of degree N is equal to N N ( ) ( ) s + n 1 s + N dim k (A n ) = = = N s s 1 s s! n=0 n=0 2. Dimension of modules over local rings 2.1. Lemma (Nakayama). Let A be a local ring with the maximal ideal m. Let V be a finitely generated A-module such that mv = V. Then V = 0. Proof. Assume that V 0. Then we can find a minimal system of generators v 1,,..., v s of V as an A-module. By the assumption, v s = s i=1 m iv i for some m i m. Therefore, (1 m s )v s = s 1 i=1 m iv i. Since 1 m s is invertible, this implies that v 1,..., v s 1 generate V, contrary to our assumption. In the following we assume that A is a noetherian local ring, m its maximal ideal and k = A/m the residue field of A Lemma. dim k (m/m 2 ) < +. Proof. By the noetherian assumption m is finitely generated. If a 1,..., a p are generators of m, their images ā 1,..., ā p in m/m 2 span it as a vector space over k. Let s = dim k (m/m 2 ). Then we can find a 1,..., a s m such that ā 1,..., ā s form a basis of m/m 2. We claim that they generate m. Let I be the ideal generated by a 1,..., a s. Then I + m 2 = m and m(m/i) = m/i. Hence, by 2.1, we have m/i = 0. Therefore, we proved: 2.3. Lemma. The positive integer dim k (m/m 2 ) is equal to the minimal number of generators of m. Any s-tuple (a 1,..., a s ) of elements from m such that (ā 1,..., ā s ) form a basis of m/m 2 is called a coordinate system in A. Clearly, m p, p Z +, is a decreasing filtration of A. Therefore, we can form Gr A = p=0 mp /m p+1. We claim that Gr A is a finitely generated algebra over k and therefore a noetherian graded ring. Actually, the map X i ā i m/m 2 Gr A extends to a surjective morphism of k[x 1,..., X s ] onto Gr A. Let M be a finitely generated A-module. Then we can define a decreasing filtration of M by m p M, p Z +, and consider the graded Gr A-module Gr M = p=0 mp M/m p+1 M Lemma. If M is a finitely generated A-module, Gr M is a finitely generated Gr A-module. Proof. From the definition of the graded module Gr M we see that m Gr p M = Gr p+1 M for all p Z +. Hence Gr 0 M = M/mM generates Gr M. On the other hand, M/mM is a finite dimensional linear space over k. This implies, by 1.2, that dim k (m p M/m p+1 M) < +, in particular, the A- modules m p M/m p+1 M are of finite length. Since length is clearly an additive function, by 1.5 we see that p length A (m p M/m p+1 M) = dim k (m p M/m p+1 M)
13 2. DIMENSION OF MODULES OVER LOCAL RINGS 7 is equal to a polynomial in p with rational coefficients for large p Z +. Moreover, the function p 1 p length A (M/m p M) = length A (m q M/m q+1 M) is equal to a polynomial with rational coefficients for large p Z +, and its leading coefficient is of the form e pd d!, where e, d Z +. We put d(m) = d and e(m) = e, and call these numbers the dimension and multiplicity of M. Now we want to discuss some properties of the function M d(m). The critical result in controlling the filtrations of A-modules is the Artin-Rees lemma Theorem (Artin, Rees). Let M be a finitely generated A-module and N its submodule. Then there exists m 0 Z + such that for all p Z +. q=0 m p+m0 M N = m p (m m0 M N) Proof. Put A = n=0 mn. Then A has a natural structure of a graded ring. Let (a 1,..., a s ) be a coordinate system in A. Then we have a natural surjective morphism A[a 1,..., a s ] A, and A is a graded noetherian ring. Let M = n=0 mn M. Then M is a graded A -module. It is clearly generated by M 0 = M as an A -module. Since M is a finitely generated A-module, we conclude that M is a finitely generated A -module. In addition, put N = n=0 (N mn M) M. Then m p (N m n M) m p N m n+p M N m n+p M implies that N is an A -submodule of M. Since A is a noetherian ring, N is finitely generated. There exists m 0 Z + such that m 0 n=0 (N mn M) generates N. Then for any p Z +, m 0 N m p+m0 M = m p+m0 s (N m s M) m p (N m m0 M) N m p+m0 M. s=0 This result has the following consequence the Krull intersection theorem Theorem (Krull). Let M be a finitely generated A-module. Then m p M = {0}. p=0 Proof. Put E = p=0 mp M. Then, by 2.5, E = m p+m0 M E = m p (m m0 M E) = m p E, in particular, me = E, and E = 0 by Nakayama lemma Lemma. Let 0 M M M 0 be an exact sequence of finitely generated A-modules. Then (i) d(m) = max(d(m ), d(m )); (ii) if d(m) = d(m ) = d(m ), we have e(m) = e(m ) + e(m ).
14 8 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS Proof. We can view M as a submodule of M. If we equip M with the filtration m p M, p Z +, and M and M with the induced filtrations M m p M, p Z +, and m p M, p Z +, we get the exact sequence This implies that for any p Z + length A (m p M/m p+1 M) 0 Gr M Gr M Gr M 0. = length A ((M m p M)/(M m p+1 M)) + length A (m p M /m p+1 M ) and, by summation, length A (M/m p M) = length A (M /(M m p M)) + length A (M /m p M ). Therefore the function p length A (M /(M m p M)) is equal to a polynomial in p for large p Z +. On the other hand, by 2.5, m p+m0 M m p+m0 M M m p M ; hence, for large p Z +, the functions p length A (M /(M m p M)) and p length A (M /m p M ) are given by polynomials in p with equal leading terms Corollary. Let A be a noetherian local ring with s = dim k (m/m 2 ). Then, for any finitely generated A-module M we have d(m) s. Proof. By 2.7 it is enough to show that d(a) s. This follows immediately from the existence of a surjective homomorphism of k[x 1,..., X s ] onto Gr A, and the fact that the dimension of the space of polynomials of degree n in s variables is a polynomial in n of degree s. A noetherian local ring is called regular if d(a) = dim k (m/m 2 ) Theorem. Let A be a noetherian local ring and (a 1, a 2,..., a s ) a coordinate system in A. Then the following conditions are equivalent: (i) A is a regular local ring; (ii) the canonical morphism of k[x 1, X 2,..., X s ] into Gr A defined by X i ā i, 1 i s, is an isomorphism. Proof. By definition, the canonical morphism of k[x 1,..., X s ] into Gr A is surjective. Let I be the graded ideal which is the kernel of the natural surjection of k[x 1,..., X s ] onto Gr A. If I 0, it contains a homogeneous polynomial P of degree d > 0. Let J be the ideal in k[x 1, X 2,..., X s ] generated by P. Then its Poincaré series is P (J, t) = td (1 t). Clearly, s P (k[x 1, X 2,..., X s ]/J, t) = P (k[x 1, X 2,..., X s ], t) P (J, t) = 1 td (1 t) s = 1 + t + + td 1 (1 t) s 1. The order of the pole of the Poincaré series P (k[x 1, X 2,..., X s ]/J, t) at 1 is s 1, and by 1.5 the function dim k (k[x 1, X 2,..., X s ]/J) n is given by a polynomial in n of degree s 2 for large n Z +. It follows that the function dim k (k[x 1,..., X s ]/I) n = dim k Gr n A is given by a polynomial in n of degree s 2 for large n Z +. This implies that d(a) s 1. Therefore, I = 0 if and only if d(a) = s Theorem. Let A be a regular local ring. Then A is integral.
15 3. DIMENSION OF MODULES OVER FILTERED RINGS 9 Proof. Let a, b A and a 0, b 0. Then, by 2.6, we can find p, q Z + such that a m p, a / m p+1, and b m q, b / m q+1. Then their images ā Gr p A and b Gr q A are different form zero, and since Gr A is integral by 2.9, we see that ā b 0. Therefore, ab 0. Finally we want to discuss an example which will play an important role later. Let k be a field, A = k[x 1, X 2,..., X n ] be the ring of polynomials in n-variables with coefficients in k and  = k[[x 1, X 2,..., X n ]] the ring of formal power series in n-variables with coefficients in k. It is easy to check that  is a local ring with maximal ideal ˆm generated by X 1, X 2,... X n. Also, the canonical morphism from k[x 1, X 2,..., X n ] into Gr  is clearly an isomorphism. For any x k n we denote by m x the maximal ideal in A generated by X i x i, 1 i n. Then its complement in A is a multiplicative system in A, and we denote by A x the corresponding localization of A. It is isomorphic to the ring of all rational functions on k n regular at x. This is clearly a noetherian local ring. The localization of m x is the maximal ideal n x = (m x ) x of all rational functions vanishing at x. The automorphism of A defined by X i X i x i, 1 i n, gives an isomorphism of A 0 with A x for any x k n. On the other hand, the natural homomorphism of A into  extends to an injective homomorphism of A 0 into Â. This homomorphism preserves the filtrations on these local rings and induces a canonical isomorphism of Gr A 0 onto Gr Â. Therefore we have the following result Proposition. The rings A x, x k n, are n-dimensional regular local rings. 3. Dimension of modules over filtered rings Let D be a ring with identity and (D n ; n Z) an increasing filtration of D by additive subgroups such that (i) D n = {0} for n < 0; (ii) n Z D n = D ; (iii) 1 D 0 ; (iv) D n D m D n+m, for any n, m Z; (v) [D n, D m ] D n+m 1, for any n, m Z. Then Gr D = n Z Grn D = n Z D n/d n 1 is a graded ring with identity. The property (v) implies that it is commutative. In particular, D 0 = Gr 0 D is a commutative ring with identity. Therefore, we can view Gr D as an algebra over D 0. Let s assume in addition that D satisfies (vi) Gr D is a noetherian ring; (vii) Gr 1 D generates Gr D as a D 0 -algebra. Then, by 1.1, D 0 is a noetherian ring. Moreover, by (vi), (vii) and 1.2 we know that we can choose finitely many elements x 1, x 2,..., x s Gr 1 D such that Gr D is generated by them as a D 0 -algebra. Clearly, by (vii), we also have Gr n+1 D = Gr 1 D Gr n D for n Z + and therefore D n+1 = D n D 1 for n Z +.
16 10 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS Let D be the opposite ring of D. Then the filtration (D n ; n Z) has the same properties with respect to the multiplication of D. Moreover, the identity map D D induces an isomorphism of graded rings Gr D and Gr D. Let M be a D-module. An increasing filtration F M = (F n M ; n Z) of M by additive subgroups is a D-module filtration if D n F m M F m+n M, for n, m Z. In particular, F n M are D 0 -modules. A D-module filtration F M is hausdorff if n Z F nm = {0}. It is exhaustive if n Z F n M = M. It is called stable if there exists m 0 Z such that D n F m M = F m+n M for all n Z + and m m 0. A D-module filtration is called good if (i) F n M = {0} for sufficiently negative n Z; (ii) the filtration F M is exhaustive; (iii) F n M, n Z, are finitely generated D 0 -modules; (iv) the filtration F M is stable. In particular, a good filtration is hausdorff Lemma. Let F M be an exhaustive hausdorff D-module filtration of M. Then the following statements are equivalent: (i) F M is a good filtration; (ii) Gr D-module Gr M is finitely generated. Proof. First we prove (i) (ii). There exists m 0 Z such that D n F m0 M = F n+m0 M for all n Z +. Therefore Gr n D Gr m0 M = Gr n+m0 M for all n Z +. It follows that n m 0 Gr n M generates Gr M as a Gr D-module. Since F n M are finitely generated D 0 -modules, Gr n M are finitely generated D 0 -modules too. This implies, since F n M = {0} for sufficiently negative n Z, that n m 0 Gr n M is a finitely generated D 0 -module. (ii) (i). Clearly, Gr n M = {0} for sufficiently negative n Z. Also, by 1.2, all Gr n M are finitely generated D 0 -modules. The exact sequence 0 F n 1 M F n M Gr n M 0 implies that F n M = F n 1 M for sufficiently negative n, hence there exists n 0 Z such that n Z F n M = F n0 M. Since the filtration F M is hausdorff, F n0 M = {0}. This implies, by induction in n, that all F n M are finitely generated D 0 -modules. Let m 0 Z be such that n m 0 Gr n M generates Gr M as Gr D-module. Let m m 0. Then Gr m+1 M = Gr m+1 k D Gr k M k m 0 = Gr 1 D Gr m k D Gr k M Gr 1 D Gr m M Gr m+1 M, k m 0 i.e., Gr 1 D Gr m M = Gr m+1 M. This implies that and by induction in n, F m+1 M = D 1 F m M + F m M = D 1 F m M F m+n M = D 1 D 1... D 1 F m M D n F m M F m+n M. Therefore, F m+n M = D n F m M for all n Z +. Hence, F M is a good filtration.
17 3. DIMENSION OF MODULES OVER FILTERED RINGS 11 In particular, (D n ; n Z) is a good filtration of D considered as a D-module for left multiplication Remark. From the proof it follows that the stability condition in the definition of a good filtration can be replaced by an apparently weaker condition: (iv) There exists m 0 Z such that D n F m0 M = F m0+nm for all n Z Lemma. Let M be a D-module with a good filtration F M. Then M is finitely generated. Proof. By definition, n Z F n M = M and F n+m0 M = D n F m0 M for n Z + and some sufficiently large m 0 Z. Therefore, F m0 M generates M as a D- module. Since F m0 M is a finitely generated D 0 -module, the assertion follows Lemma. Let M be a finitely generated D-module. Then M admits a good filtration. Proof. Let U be a finitely generated D 0 -module which generates M as a D- module. Put F n M = 0 for n < 0 and F n M = D n U for n 0. Then U = Gr 0 M, and Gr n M = F n M/ F n 1 M = (D n U)/(D n 1 U) Gr n D Gr 0 M Gr n M, i.e., Gr n M = Gr n D Gr 0 M for all n Z +. Hence, Gr M is finitely generated as a Gr D-module. The statement follows from 3.1. The lemmas 3.1 and 3.3 imply that the D-modules admitting good filtrations are precisely the finitely generated D-modules Proposition. The ring D is a left and right noetherian. Proof. Let L be a left ideal in D. The natural filtration of D induces a filtration (L n = L D n ; n Z), on L. This is evidently a D-module filtration. The graded module Gr L is naturally an ideal in Gr D, and since Gr D is a noetherian ring, it is finitely generated as Gr D-module. Therefore, the filtration (L n ; n Z) is good by 3.1, and L is finitely generated by 3.3. This proves that D is left noetherian. To get the right noetherian property one has to replace D with its opposite ring D. If we have two filtrations F M and F M of a D-module M, we say that F M is finer than F M if there exists a number k Z + such that F n M F n+k M for all n Z. If F M is finer than F M and F M finer than F M, we say that they are equivalent Lemma. Let F M be a good filtration on a finitely generated D-module M. Then F M is finer than any other exhaustive D-module filtration on M. Proof. Fix m 0 Z + such that D n F m0 M = F n+m0 M for all n Z +. Let F M be another exhaustive D-module filtration on M. Then F m0 M is finitely generated as a D 0 -module. Since F M is exhaustive, it follows that there exists p Z such that F m0 M F p M. Since F M is a good filtration, there exists n 0 such that F n0 M = {0}. Put k = p + n 0. Clearly, for m n 0, we have
18 12 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS F m M = 0 F m+k M. For n 0 < m m 0, we have n 0 n 0 < m and p = n 0 + k < m + k. This yields F m M F m0 M F p M F m+k M. Finally, for m > m 0, we have m m 0 m since m 0 is positive, and p k. It follows that F m M = D m m0 F m0 M D m F p M F m+p M F m+k M Corollary. Any two good filtrations on a finitely generated D-module are equivalent. Let M be a finitely generated D-module and F M a good filtration on M. Then Gr M is a finitely generated Gr D-module, hence we can apply the results on Hilbert polynomials from 1. Let λ be an additive function on finitely generated D 0 -modules. Assume also that λ takes only nonnegative values on objects of the category M fg (D 0 ) of finitely generated D 0 -modules. Then, by 1.5, λ(f n M) λ(f n 1 M) = λ(gr n M) is equal to a polynomial in n for large n Z +. By 1.7 this implies that λ(f n M) is equal to a polynomial in n for large n Z +. If F M is another good filtration on M, by 3.7 we know that F M an F M are equivalent,i.e., there is a number k Z + such that F n M F n+k M F n+2k M for all n Z. Since λ is additive and takes nonnegative values only, we conclude that λ(f n M) λ(f n+k M) λ(f n+2k M) for all n Z. This implies that the polynomials representing λ(f n M) and λ(f n M) for large n have equal leading terms. We denote the common degree of these polynomials by d λ (M) and call it the dimension of the D-module M (with respect to λ). By 1.6 the leading coefficient of these polynomials has the form e λ (M)/d λ (M)! where e λ (M) N. We call e λ (M) the multiplicity of the D-module M (with respect to λ). Let 0 M f g M M 0 be an exact sequence of D-modules. If M is equipped by a D-module filtration F M, it induces filtrations F M = (f 1 (f(m ) F n M) ; n Z) on M and F M = ( g(f n M) ; n Z) on M. Clearly, these filtrations are D-module filtrations. Moreover, the sequence 0 Gr M Gr f Gr M Gr g Gr M 0 is exact. If the filtration F M is good, Gr M is a finitely generated Gr D-module, hence both Gr M and Gr M are finitely generated Gr D-modules. By 3.1, F M and F M are good filtrations. Therefore, we proved the following result Lemma. Let 0 M M M 0 be an exact sequence of D-modules. If F M is a good filtration on M, the induced filtrations F M and F M are good.
19 3. DIMENSION OF MODULES OVER FILTERED RINGS 13 By the preceding discussion λ(gr n M) = λ(gr n M ) + λ(gr n M ) for all n Z. This implies, by induction in n, that λ(f n M) = λ(f n M ) + λ(f n M ) for all n Z. This leads to the following result Proposition. Let 0 M M M 0 be an exact sequence of finitely generated D-modules. Then (i) d λ (M) = max(d λ (M ), d λ (M )); (ii) if d λ (M) = d λ (M ) = d λ (M ), then e λ (M) = e λ (M ) + e λ (M ). Finally, let φ be an automorphism of the ring D such that φ(d 0 ) = D 0. We can define a functor φ from the category M(D) of D-modules into itself which attaches to a D-module M a D-module φ(m) with the same underlying additive group structure and with the action of D given by (T, m) φ(t )m for T D and m M. Clearly, φ is an automorphism of the category M(D), and it preserves finitely generated D-modules Proposition. Let M be a finitely generated D-module. Then d λ ( φ(m)) = d λ (M). Proof. Let T 1, T 2,..., T s be the representatives in D 1 of classes in Gr 1 D generating Gr D as a D 0 -algebra. Then there exists d N such that φ(t i ) D d for 1 i s. Since T 1, T 2,..., T s and 1 generate D 1 as a D 0 -module, we conclude that φ(d 1 ) D d. Let F M be a good filtration of M. Define a filtration F φ(m) by F p φ(m) = Fdp M for p Z. Clearly, F φ(m) is an increasing filtration of φ(m) by finitely generated D 0 -submodules. Also, D 1 F m φ(m) = φ(d1 ) F dm M D d F dm M F d(m+1) M = F m+1 φ(m) for m Z. Hence, by induction, we have D n F m φ(m) = D1 D n 1 F m φ(m) D1 F m+n 1 φ(m) Fm+n φ(m) for all n, m Z, i.e., F φ(m) is a D-module filtration. By 3.6, there exists a good filtration F φ(m) which is finer than this filtration, i.e, there exists k Z+ such that F n φ(m) F n+k φ(m) = Fd(n+k) M for all n Z. Therefore, λ(f n φ(m)) λ(f d(n+k) M) for n Z. For large n Z, λ(f d(n+k) M) is equal to a polynomial in n with the leading term equal to e λ (M)d d λ(m) n dλ(m). d λ (M)!
20 14 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS Since λ(f n φ(m)) is also given by a polynomial of degree d λ ( φ(m)) for large n Z, we conclude that d λ ( φ(m)) d λ (M). By applying the same reasoning to φ 1 we also conclude that d λ (M) = d λ ( φ 1 ( φ(m))) d λ ( φ(m)). 4. Dimension of modules over polynomial rings Let A = k[x 1,..., X n ] where k is an algebraically closed field. We can filter A by degree of polynomials, i.e., we can put A m = { c I x I c I k, I m}. Then Gr A = k[x 1,..., X n ], hence A satisfies properties (i)-(vii) from the preceding section. Since A 0 = k we can take for the additive function λ the function dim k. This leads to notions of dimension d(m) and multiplicity e(m) of a finitely generated A-module M.We know that for any p Z +, we have ( ) n + p dim k A p = = pn + lower order terms in p, n n! i.e., d(a) = n and e(a) = 1. In addition, for any finitely generated A-module M we have an exact sequence 0 K A p M 0, hence, by 3.9, d(m) n. We shall give later a geometric interpretation of d(m). Let x k n and denote by m x be the maximal ideal in k[x 1,..., X n ] of all polynomials vanishing at x. We denote by A x the localization of A at x, i.e., the ring of all rational k-valued functions on k n regular at x. As we have seen in 2.11, A x is an n-dimensional regular local ring with the maximal ideal n x = (m x ) x consisting of all rational k-valued functions on k n vanishing at x. Let M be an A-module. Its localization M x at x is an A x -module. We define the support of M by supp(m) = {x k n M x 0} Lemma. Let 0 M M M 0 be an exact sequence of A-modules. Then supp(m) = supp(m ) supp(m ). Proof. By exactness of localization we see that 0 M x M x M x 0 is an exact sequence of A x -modules. This immediately implies our statement. For an ideal I k[x 1,..., X n ] we denote V (I) = {x k n f(x) = 0 for f I} Proposition. Let M be a finitely generated A-module and I its annihilator in A. Then supp(m) = V (I). Proof. We prove the statement by induction in the number of generators of M. Assume first that M has one generator, i.e., M = A/I. Then M x = (A/I) x = A x /I x. Let x V (I). Then I m x and I x n x. Hence I x A x. It follows that (A/I) x 0 and x supp(m). Conversely, if x / V (I), there exists f I such
21 4. DIMENSION OF MODULES OVER POLYNOMIAL RINGS 15 that f(x) 0, i.e., f / m x. Therefore, f is invertible in the local ring A x and f I x implies that I x = A x. Hence (A/I) x = 0 and x / supp(a/i). Therefore, supp(a/i) = V (I). Now we consider the general situation. Let m 1,..., m p be a set of generators of M. Denote by M the submodule generated by m 1,..., m p 1. Then we have the exact sequence 0 M M M 0 and M is cyclic. Moreover, by 4.1, supp(m) = supp(m ) supp(m ). Hence, by the induction assumption, supp(m) = V (I ) V (I ) where I and I are the annihilators of M and M respectively. Clearly, I I is in the annihilator I of M. On the other hand, I annihilates M and M, hence I I I. It follows that This implies that I I I I I. V (I ) V (I ) V (I I ) V (I) V (I I ). Let x / V (I ) V (I ). Then there exist f I and g I such that f(x) 0 and g(x) 0. It follows that (f g)(x) = f(x) g(x) 0 and x / V (I I ). Hence, V (I I ) V (I ) V (I ) and all inclusions above are equalities. Hence, we have V (I) = V (I ) V (I ) and supp(m) = V (I). This immediately implies the following consequence Corollary. Let M be a finitely generated A-module. Then its support supp(m) is a Zariski closed subset in k n. The next lemma is useful in some reduction arguments Lemma. Let B be a noetherian commutative ring and M 0 be a finitely generated B-module. Then there exist a filtration 0 = M 0 M 1 M n 1 M n = M of M by B-submodules, and prime ideals J i of B such that M i /M i 1 = B/J i, for 1 i n. Proof. For any x M we put Ann(x) = {a B ax = 0}. Let A be the family of all such ideals Ann(x), x M, x 0. Because B is a noetherian ring, A has maximal elements. Let I be a maximal element in A. We claim that I is prime. Let x M be such that I = Ann(x). Then ab I implies abx = 0. Assume tha b / I, i.e., bx 0. Then I Ann(bx) and a Ann(bx). By the maximality of I, a Ann(bx) = I, and I is prime. Therefore, there exists x M such that J 1 = Ann(x) is prime. If we put M 1 = Bx, M 1 = B/J1. Now, denote by F the family of all B-submodules of M having filtrations 0 = N 0 N 1 N k = N such that N i /N i 1 = B/Ji for some prime ideals J i. Since M is a noetherian module, F contains a maximal element L. Assume that L M. Then we would have the exact sequence: 0 L M L 0, and by the first part of the proof, L would have a submodule N of the form B/J for some prime ideal J, contradicting the maximality of L. Hence, L = M. This proves the existence of the filtration with required properties.
22 16 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS 4.5. Theorem. Let M be a finitely generated A-module and supp(m) its support. Then d(m) = dim supp(m). This result has the following companion local version. The localization A x of A at x k n is a noetherian local ring. Moreover, its maximal ideal n x is the ideal generated by the polynomials X i x i, 1 i n, and their images in n x /n 2 x span it as a vector space over k. Therefore, X i x i, 1 i n, form a coordinate system in A x. For any finitely generated A-module M, its localization M x at x is a finitely generated A x -module, hence we can consider its dimension d(m x ). For any algebraic variety V over k and x V we denote by dim x V the local dimension of V at x Theorem. Let M be a finitely generated A-module and x supp(m). Then d(m x ) = dim x (supp(m)). We shall simultaneously prove 4.5 and 4.6. First we observe that if we have an exact sequence of A-modules: 0 M M M 0 and 4.5 and 4.6 hold for M and M, we have, by 3.9 and 4.1, that d(m) = max(d(m ), d(m )) = max(dim supp(m ), dim supp(m )) = dim(supp(m ) supp(m )) = dim supp(m). Also, for any x supp(m), by the exactness of localization we have the exact sequence: 0 M x M x M x 0; hence, by 2.7 and 4.1, d(m x ) = max(d(m x), d(m x )) = max(dim x supp(m ), dim x supp(m )) = dim x (supp(m ) supp(m )) = dim x supp(m). Assume that 4.5 and 4.6 hold for all M = A/J where J is a prime ideal. Then the preceding remark, 4.4 and an induction in the length of the filtration would prove the statements in general. Hence we can assume that M = A/J with J prime. Assume first that J is such that A/J is a finite-dimensional vector space over k. Then A/J is an integral ring and it is integral over k. Hence it is a field which is an algebraic extension of k. Since k is algebraically closed, A/J = k and J is a maximal ideal. In this case, by Hilbert Nulstellenatz, supp(m) = V (J) is a point x in k n, i.e., dim supp(m) = 0. On the other hand, since M x is one-dimensional linear space, d(m x ) = 0, and the assertion is evident. It follows that we can assume that J is not of finite codimension in A, in particular it is not a maximal ideal. Let J 1 J be a prime ideal different form J. Then there exists f J 1 such that f / J. It follows that J (f) + J J 1 and J (f) + J. Therefore, A/J 1 is a quotient of A/((f) + J), and A/((f) + J) is a quotient of A/J. In addition, A/((f) + J) = M/fM. Consider the endomorphism of M given by multiplication by f. Then, if g + J is in the kernel of this map, 0 = f(g + J) = fg + J and fg J. Since J is prime and f / J it follows that
23 4. DIMENSION OF MODULES OVER POLYNOMIAL RINGS 17 g J, g + J = 0 and the map is injective. Therefore, we have an exact sequence of A-modules: 0 M f M M/fM 0. This implies, by 3.9, that d(m/fm) d(m). If d(m/fm) = d(m), we would have in addition that e(m) = e(m) + e(m/fm), hence e(m/fm) = 0. This is possible only if d(m/fm) = 0, and in this case it would also imply that d(m) = 0 and M is finite-dimensional, which is impossible by our assumption. Therefore, d(m/f M) < d(m). Since A/J 1 is a quotient of M/fM, this implies that d(a/j 1 ) < d(a/j). Let x V (J 1 ). Then, by localization, we get the exact sequence: 0 M x f Mx M x /fm x 0 of A x -modules. This implies, by 2.7, that d(m x /fm x ) d(m x ). If d(m x /fm x ) = d(m x ), we would have in addition that e(m x ) = e(m x ) + e(m x /fm x ), hence e(m x /fm x ) = 0. This is possible only if d(m x /fm x ) = 0, and in this case it would imply that m x (M x /fm x ) = M x /fm x and, by Nakayama lemma, M x /fm x = 0. It would follow that the multiplication by f is surjective on M x, and, since f m x, by Nakayama lemma this would imply that M x = 0 contrary to our assumptions. Therefore, d(m x /fm x ) < d(m x ). Since A/J 1 is a quotient of M/fM this implies that d((a/j 1 ) x ) < d((a/j) x ). Let Z 0 = {x} Z 1 Z n 1 Z n = k n be a maximal chain of nonempty irreducible closed subsets of k n. Then I(Z 0 ) = m x I(Z 1 ) I(Z n 1 ) I(Z n ) = {0} is a maximal chain of prime ideals in A. By the preceding arguments we have the following sequences of strict inequalities and 0 d(a/i(z 0 )) < d(a/i(z 1 )) < < d(a/i(z n )) = d(a) = n, 0 d((a/i(z 0 )) x ) < d((a/i(z 1 )) x ) < < d((a/i(z n )) x ) = d(a x ) = n, by It follows that d((a/i(z j )) x ) = d(a/i(z j )) = j = dim Z j for 0 j n. Since every closed irreducible subset Z can be put in a maximal chain, it follows that d((a/i(z)) x ) = d(a/i(z)) = dim Z for any closed irreducible subset Z k n and any x Z. On the other hand, this implies that d((a/j) x ) = d(a/j) = dim V (J) for any prime ideal J in A and x V (J). By 4.2, this ends the proof of 4.5 and 4.6. Next result follows immediately from 4.5 and Corollary. Let M be a finitely generated A-module. Then d(m) = sup d(m x ). x supp(m) Finally, we prove a result we will need later Lemma. Let be I an ideal in A. Then dim V (I) = dim V (Gr I).
24 18 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS Proof. The short exact sequence of A-modules 0 I A A/I 0, where the modules are equipped with the filtrations induced by the natural filtration of A leads to the short exact sequence of graded A-modules. Hence, we have dim k F p (A/I) = = 0 Gr I A Gr(A/I) 0 p (dim k F q (A/I) dim k F q 1 (A/I)) q=0 p dim k Gr p (A/I) = q=0 p (dim k Gr p A dim Gr p I) q=0 = dim k F p A dim k F p Gr I = dim k F p (A/ Gr I). Therefore, d(a/i) = d(a/ Gr I). The assertion follows from 4.4 and Rings of differential operators with polynomial coefficients Let k be a field of characteristic zero. Let A be a commutative algebra over k. Let End k (A) be the algebra of all k-linear endomorphisms of A. It is a Lie algebra with the commutator [S, T ] = ST T S for any S, T End k (A). Clearly, End k (A) contains, as a subalgebra, the set End A (A) of all A-linear endomorphisms of A. To any element a A we can attach the A-linear endomorphism of A defined by b ab for b A. Since this endomorphism takes the value a on 1, this map is clearly an injective morphism of algebras. On the other hand, if T End A (A), we have T (b) = bt (1) = T (1)b for any b A, i.e., T is given by multiplication by T (1). This implies the folowing result Lemma. The algebra homomorphism which attaches to an element a A the A-linear endomorphism b ab, b A, is an isomorphism of A onto End A (A). In the following, we identify A with the subalgebra End A (A) of End k (A). A k-derivation of A is a T End k (A) such that T (ab) = T (a)b + at (b) for any a, b A. In particular, [T, a](b) = T (ab) at (b) = T (a)b, i.e., [T, a] = T (a) A for any a A. This implies that [[T, a 0 ], a 1 ] = 0 for any a 0, a 1 A. This leads to the following definition. Let n Z +. We say that an element T End k (A) is a (k-linear) differential operator on A of order n if [... [[T, a 0 ], a 1 ],..., a n ] = 0 for any a 0, a 1,..., a n A. We denote by Diff k (A) the space of all differential operators on A Lemma. Let T, S be two differential operators of order n, m respectively. Then T S is a differential operator of order n + m.
25 5. RINGS OF DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS 19 Proof. We prove the statement by induction in n + m. If n = m = 0, T, S End A (A), hence T S End a (A) and it is a differential operator of order 0. Assume now that n + m p. Then [T S, a] = T Sa at S = T [S, a] + [T, a]s, and [T, a] and [S, a] are differential operators of order n 1 and m 1 respectively. By the induction assumption, this differential operator is of order n+m 1. Therefore T S is of order n + m. Therefore Diff k (A) is a subalgebra of End k (A). We call it the algebra of all k-linear differential operators on A. Also, we put F n Diff k (A) = {0} for n < 0 and F n Diff k (A) = {T Diff k (A) order(t ) n} for n 0. clearly, this is an increasing exhaustive filtration of Diff k (A) by vector subspaces over k. This filtration is compatible with the ring structure of Diff k (A), i.e., it satisfies F n Diff k (A) F m Diff k (A) F n+m Diff k (A) for any n, m Z Lemma. (i) F 0 Diff k (A) = A. (ii) F 1 Diff k (A) = Der k (A) A. (iii) [F n Diff k (A), F m Diff k (A)] F n+m 1 Diff k (A) for any n, m Z +. Proof. (i) is evident. (ii) As we remarked before, Der k (A) F 1 Diff k (A). Also, for any T Der k (A), we have T (1) = T (1 1) = 2T (1), hence T (1) = 0. This implies that Der k (A) A = 0. Let S F 1 Diff k (A) and T = S S(1). Then T (1) = 0, hence T (a) = [T, a](1), and T (ab) = [T, ab](1) = ([T, a]b)(1) + (a[t, b])(1) = (b[t, a])(1) + (a[t, b])(1) = T (a)b + at (b), i.e., T Der k (A). (iv) Let T, S be of order n, m respectively. We claim that [T, S] is of order n + m 1. We prove it by induction on n + m. If n = m = 0, there is nothing to prove. In general, by Jacobi identity, we have [[T, S], a] = [[T, a], S] + [T, [S, a]] where [T, a] and [S, a] are of order n 1 and m 1 respectively. Hence, by the induction assumption, [[T, S], a] is of order n + m 2 and [T, S] is of order n + m 1. This implies that the graded ring Gr Diff k (A) is a commutative A-algebra. In addition, Diff k (A) satisfies properties (i)-(v) from 3. Let n 1. Let T be a differential operator on A of order n. Then we can define a map from A n into Diff k (A) by σ n (T )(a 1, a 2,..., a n 1, a n ) = [[... [[T, a 1 ], a 2 ],..., a n 1 ], a n ]. Since σ n (T )(a 1, a 2,..., a n 1, a n ) is of order 0, we can consider this map as a map from A n into A Lemma. Let T be a differential operator on A of order n. Then: (i) the map σ n (T ) : A n A is a symmetric k-multilinear map;
26 20 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS (ii) the operator T is of order n 1 if and only if σ n (T ) = 0. Proof. (i) We have to check the symmetry property only. To show this, we observe that, by the Jacobi identity, we have [[S, a], b] = [[S, b], a] for any S Diff k (A) and a, b A. This implies that σ n (T )(a 1, a 2,..., a i, a i+1,..., a n 1, a n ) hence σ(t ) is symmetric. (ii) is obvious. = [[... [[... [[T, a 1 ], a 2 ],..., a i ], a i+1 ]..., a n 1 ], a n ] = [[... [[... [[T, a 1 ], a 2 ],..., a i+1 ], a i ]..., a n 1 ], a n ] = σ n (T )(a 1, a 2,..., a i+1, a i,..., a n 1, a n ), Now we want to discuss a special case. Let A = k[x 1, X 2,..., X n ]. Then we put D(n) = Diff k (A). We call D(n) the algebra of all differential operators on k n. Let 1, 2,..., n be the standard derivations of k[x 1, X 2,..., X n ]. For I, J Z n + we put X I = X i1 1 Xi Xin n and J = j1 1 j jn n. Then X I J D(n), and it is a differential operator of order J = j 1 +j 2 + +j n. Moreover, if T is a differential operator given by T = P I (X 1, X 2,..., X n ) I, I p with polynomials P I k[x 1, X 2,..., X n ], we see that T is of order p Lemma. The derivations ( i ; 1 i n) form a basis of the free k[x 1,..., X n ]- module Der k (k[x 1, X 2,..., X n ]). Proof. Let T Der k (k[x 1, X 2,..., X n ]). Put P i = T (X i ) for 1 i n, and define S = n i=1 P i i. Clearly, n S(X i ) = P j j (X i ) = P i = T (X i ) j=1 for all 1 i n. Since X 1, X 2,..., X n generate k[x 1, X 2,..., X n ] as a k-algebra it follows that T = S. Therefore, ( i ; 1 i n) generate the k[x 1, X 2,..., X n ]- module Der k (k[x 1, X 2,..., X n ]). Assume that n i=1 Q i i = 0 for some Q i k[x 1, X 2,..., X n ]. Then 0 = ( n j=1 Q j j )(X i ) = Q i for all 1 i n. This implies that i, 1 i n, are free generators of Der k (k[x 1, X 2,..., X n ]). Let T be a differential operator of order p on k[x 1, X 2,..., X n ]. If p < 0, T = 0 and we put Symb p (T ) = 0. If p = 0, T A, and we put Symb 0 (T ) = T. For p 1, we define a polynomial Symb p (T ) in k[x 1, X 2,..., X n, ξ 1, ξ 2,..., ξ n ] in the following way. Let (ξ 1, ξ 2,..., ξ n ) k n. Then we can define a linear polynomial l ξ = n i=1 ξ ix i k[x 1, X 2,..., X n ] and the function (ξ 1, ξ 2,..., ξ n ) 1 p! σ p(t )(l ξ, l ξ,..., l ξ )
27 5. RINGS OF DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS 21 on k n with values in k[x 1, X 2,..., X n ]. Clearly, one can view this function as a polynomial in X 1, X 2,..., X n and ξ 1, ξ 2,..., ξ n homogeneous of degree p in ξ 1, ξ 2,..., ξ n, and denote it by Symb p (T ). The polynomial Symb p (T ) is called the p-symbol of the differential operator T. By its definition, Symb p (T ) vanishes for T of order < p. Therefore, for p 0, it induces a k-linear map of Gr p D(n) into k[x 1, X 2,..., X n, ξ 1, ξ 2,..., ξ n ]. We denote by Symb the corresponding k-linear map of Gr D(n) into k[x 1, X 2,..., X n, ξ 1, ξ 2,..., ξ n ] Theorem. The map Symb : Gr D(n) k[x 1, X 2,..., X n, ξ 1, ξ 2,..., ξ n ] is a k-algebra isomorphism. The proof of this result consists of several steps. First we prove the symbol map is an algebra morphism Lemma. Let T, S D(n) of order p and q respectively. Then Symb p+q (T S) = Symb p (T ) Symb q (S). Proof. Let ξ k n, and define the map τ ξ : D(n) D(n) by τ ξ (T ) = [T, l ξ ]. Then τ ξ (T S) = [T S, l ξ ] = T Sl ξ l ξ T S = [T, l ξ ]S + T [S, l ξ ] = τ ξ (T )S + T τ ξ (S). Therefore, for any k Z +, we have This implies that Symb p+q (T S) = τ k ξ (T S) = k i=0 ( ) k τ k i ξ (T ) τ i i ξ(s). 1 (p + q)! σ 1 p+q(t )(l ξ, l ξ,..., l ξ ) = (p + q)! τ p+q ξ (T S) = 1 p!q! τ p ξ (T ) τ q ξ (S) = Symb p(t ) Symb q (S). Since Symb 0 (X i ) = X i and Symb 1 ( i ) = ξ i, 1 i n, we see that for X I J with p = J we have In particular, for T = Symb p (X I J ) = X I ξ J. J p P I (X 1, X 2,..., X n ) I, with polynomials P I k[x 1, X 2,..., X n ], we see that Symb p (T ) = P I (X 1, X 2,..., X n )ξ I. I =p Hence, the symbol morphism is surjective. It remains to show that the symbol map is injective Lemma. Let T F p D(n). Then Symb p (T ) = 0 if and only if T is of order p 1.
28 22 I. DIFFERENTIAL OPERATORS WITH POLYNOMIAL COEFFICIENTS Proof. We prove the statement by induction in p. It is evident if p = 0. Therefore we can assume that p > 0. Let ξ k n, and define the map τ ξ : D(n) D(n) by τ ξ (T ) = [T, l ξ ]. Then, for any λ k and η k n, we have τ ξ+λη (T ) = [T, l ξ+λη ] = [T, l ξ ] + λ[t, l η ] = τ ξ (T ) + λτ η (T ). Since τ ξ and τ η commute we see that, for any k Z +, we have τ k ξ+λη(t ) = k i=0 ( ) k λ i τ k i ξ (τ i i η(t )). By our assumption, τ p ξ+λη (T ) = 0 for arbitrary λ k. Therefore, since the field k is infinite, τ p i ξ (τη(t i )) = 0 for 0 i p. In particular, we see that τ p 1 ξ (τ η (T )) = 0 for any ξ, η k n. This implies that Symb p 1 ([T, l η ]) = 0 for any η k n, in particular Symb p 1 ([T, X i ]) = 0 for 1 i n, and by the induction assumption, [T, X i ], 1 i n, are of order p 2. Let P, Q k[x 1, X 2,..., X n ]. Then [T, P Q] = T P Q P QT = [T, P ]Q + P [T, Q], hence the order of [T, P Q] is less than or equal to the maximum of the orders of [T, P ] and [T, Q]. Since X i, 1 i n, generate k[x 1, X 2,..., X n ] we conclude that the order of [T, P ] is p 2 for any polynomial P. This implies that the order of T is p 1. This also ends the proof of 4.6. In particular, we see that D(n) satisfies properties (i)-(vii) from 3. From 3.5 we immediately deduce the following result Theorem. The ring D(n) is right and left noetherian. k Corollary. (X I J ; I, J Z n +) is a basis of D(n) as a vector space over Proof. If J = p, the p-symbol of X I J is equal to X I ξ J and (X I ξ J ; I, J Z n +) form a basis of k[x 1,..., X n, ξ 1,..., ξ n ] as a vector space over k. The following caracterization of D(n) is frequently useful Theorem. The k-algebra D(n) is the k-algebra generated by X 1, X 2,..., X n and 1, 2,..., n satisfying the defining relations [X i, X j ] = 0, [ i, j ] = 0 and [ i, X j ] = δ ij for all 1 i, j n. Proof. Let B be the k-algebra generated by X 1, X 2,..., X n and 1, 2,..., n satisfying the defining relations [X i, X j ] = 0, [ i, j ] = 0 and [ i, X j ] = δ ij for all 1 i, j n. Since these relations hold in D(n) and it is generated by X 1, X 2,..., X n and 1, 2,..., n we conclude that there is a unique surjective morphism of B onto D(n) which maps generators into the corresponding generators. Clearly, B is spanned by (X I J ; I, J Z n +). Therefore, by 5.10, this morphism is also injective Proposition. The center of D(n) is equal to k 1.
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