Rings of Differential Operators and Étale Homomorphisms G´Isli Másson

Rings of differential operators and étale homomorphisms by G´ısli Másson B.S., University of Iceland (1985) Submitted to the Department of Mathematics in partial fulfillment of the requirements for the degree of Doctor of Philosophy at the MASSACHUSETTS INSTITUTE OF TECHNOLOGY June 1991 c Massachusetts Institute of Technology 1991

Signature of Author ...... Department of Mathematics May 2, 1991

Certified by ...... Michael Artin Professor of Mathematics Thesis Supervisor Accepted by ...... Sigurdur Helgason Chairman, Departmental Graduate Committee Department of Mathematics Rings of differential operators and étale homomorphisms by G´ısli Másson

Submitted to the Department of Mathematics on May 2, 1991, in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

Abstract Rings of differential operators over rings of krull dimension 1 were studied by Musson, Smith and Stafford, and Muhasky. In particular it was proved that if k is a field of characteristic zero, and A is a finitely generated reduced k-algebra, then (i) D(A), the ring of differential operators on A, has a minimal essential twosided ideal J , (ii) A contains a minimal essential left D(A)-submodule, J,and (iii) the algebras C(A):=A/J and H(A):=D(A)/J are finite dimensional vector spaces over k. In this case the algebras C(A)andH(A) were studied by Brown and Smith. We use étale homomorphisms to obtain similar results in a somewhat more general setting. While an example of Bernstein, Gelfand and Gelfand shows that the statements above do not hold for all reduced finitely generated k-algebras of higher krull dimension, we present sufficient conditions on commutative, noetherian and reduced rings so that analogues of the ideal J and module J can be constructed, and for which statements similar to (i) and (ii) hold. Besides being applicable to rings of higher krull dimension, our method also reveals that the earlier results are a reflection of the (rather natural) fact that differential operators preserve supports of regular functions. We then use our results to prove the existence of minimal reflexive modules for rings of differential operators over domains which satisfy the condition S2. Thisisrelatedto results of Chamarie and Stafford. We also compute the dimension of C(A) (in terms of intersection multiplicities) when A is a completion of an affine plane curve over an algebraically closed field of characteristic zero, and we give a description of H(A) for some algebras A of higher krull dimension, an analogue of a theorem of Brown. Finally we point out that we give a generalization of the formula for applying differential operators to inverses, extending a result of Nakai and Hart.

Thesis Supervisor: Michael Artin Title: Professor of Mathematics

2 Acknowledgements

First of all I would like to thank my advisor Michael Artin, who always had the right ideas, and who set the record straight on countless matters. I would also like to thank him for the financial support I received as a research assistant the last two years of my studies. The mathematics department at MIT I thank for its support and virtually unlimited resources, and I would in particular like to thank Phyllis Ruby, our graduate administra- tor, for all the letters and other help which made real life considerably simpler. I would like to thank all the friends I made during my time at MIT, especially all my officemates for providing me with good surroundings and nice time on and off work. Special thanks go to Amnon Yekutieli for all the mathematical discussions and for the good time in Texas, to Wolfram Gerdes for the beer and the intercontinental cultural excursions, to Tom Roby for being my primary reference on the english language, and to Jan Pedersen for the coffee and Danish. You all made my time in the Boston area an unforgettable experience. I thank my teachers at the University of Iceland for providing me with a solid under- graduate education, and I also thank them for their hospitality during my stay in Iceland in the spring and summer of 1990. All my friends in Iceland I thank for making it a place where I want to return to, and Hans Jakob Beck and Snorri Bergmann get special thanks for keeping me in the illusion that every place has the same luxurious computer resources as I got spoiled by at MIT. My parents, Sigrún G´ısladóttir and Jóhann Már Mar´ıusson, deserve very special thanks for always being supportive of my decisions, and father, this is probably what our early games with a pencil and graph paper finally lead to. Finally my very best thanks go to my wife, my companion and my very best friend, Freyja Hreinsdóttir, for taking care of me and our son, Nökkvi, while I was out working. I thank her for the mathematics, and for pointing out all the mistakes and the places where there was room for improvement in my thesis. Last but not least I do thank her for all the patience and willingness to go along even if it meant that she had to sacrifice the most. Freyja, I hope that now I can start giving something in return for all your help.

3 Contents

Introduction 6

1 Definition and first properties of rings of differential operators 9 1.1 Basic properties of differential operators ...... 9 1.1.1 Endomorphisms, derivations and the derivation ring ...... 9 1.1.2 Definition of rings of differential operators ...... 11 1.2 Elementary results ...... 14 1.2.1 A collection of basic lemmas ...... 14 1.2.2 Modules of differentials ...... 17

2 Etale´ homomorphisms and differential operators 20 2.1 Basic properties of étale homomorphisms ...... 21 2.1.1 Formally étale homomorphisms ...... 21 2.1.2 Etale´ homomorphisms, essentially of finite type ...... 24 2.2 Behavior of differential operators under étale homomorphisms ...... 26 2.2.1 Differentials and formally étale homomorphisms ...... 26 2.2.2 Differential operators and formally étale morphisms ...... 30 2.2.3 Finitely presented modules of differentials ...... 36 2.2.4 Etale´ descent for differential operators ...... 43 2.3 Differential operators and algebraic relations ...... 47 2.3.1 Elements satisfying polynomial relations ...... 47

4 2.3.2 Proof of theorem 2.3.2 ...... 52 2.3.3 Right ideals and ´etale homomorphisms ...... 63

3 Minimal twosided ideals in rings of differential operators 68 3.1 Supports in reduced commutative rings ...... 69 3.1.1 Associated prime ideals ...... 69 3.1.2 Geometrical unibranches ...... 76 3.2 Minimal twosided ideals and minimal left submodules ...... 84 3.2.1 Differential operators and supports ...... 84 3.2.2 Essential submodules ...... 88 3.2.3 Minimal nonzero ideals in rings of differential operators ...... 91 3.2.4 Applications of theorem 3.2.11 ...... 96

4 Reflexive modules and rings of differential operators 101 4.1 Some noncommutative ring theory ...... 102 4.1.1 Rings of quotients ...... 102 4.1.2 Reflexive modules ...... 105

4.2 Rings satisfying the condition S2 ...... 111 4.2.1 Minimal reﬂexive modules ...... 111

5 Algebras related to rings of diﬀerential operators 117 5.1 The algebra C(A) ...... 118 5.1.1 A dimension formula related to aﬃne plane curves ...... 118 5.2 The algebra H(A) ...... 123 5.2.1 A description when the irreducible components of A are regular . 123

Bibliography 130

5 Introduction

In [25] it was proved that if k is an algebraically closed field of characteristic zero, and A is a finitely generated k-algebra, and a domain of krull dimension 1, then (i) D(A), the ring of differential operators on A, has a minimal essential twosided ideal J , (ii) A contains a minimal essential left D(A)-submodule, J,and (iii) the algebras C(A):=A/J and H(A):=D(A)/J are finite dimensional vector spaces over k. A similar result was independently obtained in [17], in the slightly more general case that k is (any) field of characteristic zero and A is a finitely generated reduced k-algebra of krull dimension 1. In [25] it was moreover shown that the result above also holds if A is replaced by its localization at a maximal ideal. In both papers, one of the main techniques was to use the fact that if A is as above, and the map Spec Ae → Spec A is set theoretically injective (where Ae denotes the normalization of A), then the rings D(A)andD(Ae) are Morita equivalent (a result motivated by an example of [18]). Since Ae is regular in this case, its ring of differential operators is well known, and using the Morita equivalence it is hence possible to derive many of the properties of D(A) from the same properties of D(Ae). There are two reasons why this method fails for rings of higher krull dimension: First of all it may happen that D(A)andD(Ae) are no longer Morita equivalent if A is not of krull dimension 1; and secondly, rings of differential operators over normal domains may be rather unpleasant. A well known example is the ring of differential operators on the

6 cubic cone, originally given in [1]. The goal of this thesis is to generalize the results stated at the beginning of the introduction to some rings of higher krull dimension, and to show how étale homomorphisms may be used to do that. In particular we show that if A is a commutative, noetherian and reduced k-algebra such that A has an étale covering B with irreducible components that are geometrical unibranches, then an analogue of the module J, as above, may be constructed. We call this analogue χA. If furthermore the modules of differentials of A

are finitely presented, the ideal D(A, χA) corresponds to J . We then prove that if the rings of differential operators on the irreducible components of B are simple, results as (i), (ii) from above are true for A. The key to our method is the fact that if f : A → B is a formally étale homomorphism of k-algebras, then it extends in a natural way to a homomorphism f : D(A) →D(B)of filtered rings. This result is implicit in [26] albeit not stated explicitly there, so we give a complete proof. If in addition the modules of differentials of A are finitely presented, ∼ we show that there is a natural isomorphism B ⊗A D(A) = D(B); b⊗D 7→ bf(D), i.e., differential operators on B are given by simply extending the coefficients of differential operators on A. We prove an étale descent theorem for left modules over rings of differential operators, and in order to apply it to twosided ideals, we prove the following formula: If B is a k- algebra and b ∈ B is a root of a polynomial P ∈ B[T ] such that P 0(b) is an invertible element of B,then X∞ Db = bD + νd(D) D ∈D(B) d=1

where the νdsareB-bilinear endomorphisms on D(B), computed recursively from b and the coefficients of the polynomial P . This is a generalization of the formula for applying differential operators to inverses, proved in [20, 8]. As mentioned earlier, the example in [1] shows that D(A) can be quite complicated, even when A is a finitely generated normal domain over an algebraically closed field of characteristic zero. However, it was proved in [4] that if k is a field of characteristic zero,

7 and A is a geometrical unibranch domain, which is essentially of finite type over k and satisfies the condition S2,thenD(A) has some nice properties. For example it has no nontrivial reflexive twosided ideals. As an application of our results, we prove a related theorem. Namely if A is a domain, essentially of finite type over a field of characteristic zero, and it satisfies the condition S2, then the set

{J|J is a reflexive nonzero ideal of A and a D(A)-submodule of A }, ordered by inclusion, has a unique minimal element. We also apply our results to the algebras C(A)andH(A), mentioned at the beginning of this introduction. They have been studied in [3, 17, 23, 24, 25]. We give a formula for the dimension of C(A), in terms of intersection multiplicities, when A is the completion at a maximal ideal of an affine reduced plane curve over an algebraically closed field of characteristic zero, thus giving a partial answer to a question asked in [24]. The algebra

H(A) embeds naturally in the ring D(A/χA), and our last application is to identify the image of this embedding when A is a reduced algebra, essentially of ﬁnite type over a ﬁeld of characteristic zero, and with irreducible components that are regular. This is an analog of [3, theorems 4.5, 4.6], and the proof is similar.

8 Chapter 1

Definition and first properties of rings of differential operators

This chapter contains most of the deﬁnitions and elementary lemmas needed in this thesis. All results stated here are well known, and most of them may be found e.g. in [7], [15] or [25].

1.1 Basic properties of diﬀerential operators

1.1.1 Endomorphisms, derivations and the derivation ring

Let A be a commutative k-algebra, and let M and N be A-modules. Let Homk(M,N) denote the linear space of all k-linear mappings M → N. This space obviously contains

HomA(M,N) as a subset. We write EndkM for Homk(M,M), and we call the elements

of EndkM (k-linear) endomorphisms on M.

If φ ∈ Homk(M,N)andm ∈ M,weletφ∗m ∈ N denote the image of m under φ.For a ∈ A and φ ∈ Homk(M,N), we let aφ and φa in Homk(M,N) be the elements deﬁned by (aφ)∗m := a(φ∗m)and(φa)∗m := φ∗(am) for all m ∈ M. In this way it is clear that

Homk(M,N) has the structure of an A-bimodule, and that HomA(M,N)isatwosided

A-submodule of Homk(M,N). If N = M, then we can also deﬁne multiplication on

9 EndkM, and it is clear that it then becomes a k-algebra. We deﬁne one more structure:

Deﬁnition Let A, M and N be as above. For φ ∈ Homk(M,N)anda ∈ A,welet

[φ, a] denote the element φa − aφ ∈ Homk(M,N). We call it the commutator of φ and

a. Similarly if φ, ψ ∈ EndkM,welet[φ, ψ]:=φψ − ψφ ∈ EndkM and we call it the commutator of φ and ψ.

With this, the following lemma is obvious:

Lemma 1.1.1 Let A be a commutative k-algebra, and let M and N be A-modules. With the deﬁnitions as above, we have that

HomA(M,N)={ψ ∈ Homk(M,N)|[ψ,a] = 0 for all a ∈ A}

Let A be a commutative k-algebra and let M be an A-module. A k-derivation on

M,orderivation on M for short, is an homomorphism δ ∈ Homk(A, M) such that

δ ∗ (ab)=a(δ ∗ b)+(δ ∗ a)b for all a, b ∈ A.WeletDerk(A, M)denotethek-linear space

of all derivations on M. We write DerkA for Derk(A, A). Since A is commutative, it is

clear that Derk(A, M)isaleft A-submodule of Homk(A, M). On the other hand it is in

general not arightA-submodule of HomA(A, M), even if M = A.

Let A be a commutative k-algebra. Since A is naturally isomorphic to HomA(A, A), it is now clear by lemma 1.1.1 that we may regard A in a natural way as a k-subalgebra of

EndkA. We will do so without further comment.

Deﬁnition Let A be a commutative k-algebra. We let ∆(A)denotethek-subalgebra of

EndkA generated by the left A-submodules A and DerkA, and we call it the derivation ring on A. For all d ≥ 0welet∆d(A) be the space generated by all multiples of at most d derivations on A.

It is not hard to see that if D ∈ ∆d(A), then [D, a] ∈ ∆d−1(A) for all a ∈ A.By

0 1 deﬁnition we have that ∆ (A)=A and it is also easy to see that ∆ (D)=A ⊕ DerkA

(the deﬁnitions readily imply that A ∩ DerkA = 0). The structure of ∆(A) is fairly

10 well known, especially in the case when A is regular and of finite type over a field of characteristic zero. In that case it turns out that ∆(A) is a simple noetherian ring and it is generated by finitely many elements over A. See for example [15, chapter 15] for an account of the derivation ring.

1.1.2 Deﬁnition of rings of diﬀerential operators

We can now deﬁne the central objects of this work. Let A be a commutative k-algebra, and let M and N be A-modules. We deﬁne subsets of Homk(M,N) as follows: Let

0 D (M,N):={D ∈ Homk(M,N)|[D, a] = 0 for all a ∈ A},

and then for all integers d>0 deﬁne subsets inductively by

d d−1 D (M,N):={D ∈ Homk(M,N)|[D, a] ∈D (M,N) for all a ∈ A}.

From these deﬁnitions it is clear that all the sets Dd(M,N)areinfacttwosidedA-

submodules of Homk(M,N) and that we have a chain of inclusions

D0(M,N) ⊂D1(M,N) ⊂ ....

We deﬁne [∞ D(M,N):= Dd(M,N). d=0

0 0 If L is a third A-module, we have by induction that Dd(N,L)Dd (M,N) ⊂Dd+d (M,L) 0 for all integers d, d ≥ 0. Hence it follows that D(M,M)isak-subalgebra of EndkM,and it is ﬁltered (in the sense of [15, 1.6.1]) by the subgroups Dd(M,M), d ≥ 0. Moreover D(M,N)isaD(N)-D(M)-bimodule, and it is also ﬁltered (on both sides) by the subgroups Dd(M,N), d ≥ 0. We write D(A)andDd(A) instead of D(A, A)andDd(A, A) respectively. We make:

11 Definition Let A be a commutative k-algebra. We call the filtered k-algebra D(A), defined as above, the ring of (k-linear) differential operators on A. We call the elements of D(A) (k-linear) differential operators on A,andifD ∈Dd(A) we say that the differential operator D has order ≤ d.

Remarks 1. If M and N are A-modules, we call elements of D(M,N) (k-linear) differential operators from M into N, and we deﬁne order of diﬀerential operators in the same way.

0 2. We note that lemma 1.1.1 implies that we may identify D (M,N)andHomA(M,N) for all A-modules M and N. It follows especially that D0(A)=A for all commutative k-algebras A.

The ﬁrst result on rings of diﬀerential operators is the following well known lemma:

Lemma 1.1.2 Let A be a commutative k-algebra. Then

1 1 D (A)=A ⊕ DerkA =∆(A)

Proof: First let D ∈D1(A) be such that D ∗ 1 = 0. Then for a and b in A we have that

0=[[D, a],b] ∗ 1=D ∗ (ab) − aD ∗ b − bD ∗ a − baD ∗ 1

and since by assumption D ∗ 1 = 0, we thus have that

D ∗ (ab)=aD ∗ b + bD ∗ a.

1 This shows that D ∈ DerkA. More generally if D is any element of D (A), the element 0 1 0 D = D − D ∗ 1isinD (A)andD ∗ 1=0,soD ∈ A ⊕ DerkA. The converse inclusion is obvious from the deﬁnitions. 2

From this lemma it may be tempting to think that ∆d(A)=Dd(A) for all d ≥ 0, and

12 hence that ∆(A)=D(A). If k is a field of characteristic zero and A is a regular finitely generated algebra over k, this is in fact the case (cf. e.g. [15, corollary 15.5.6]), and hence the study of rings of differential operators reduces to the study of the derivation ring. In particular it follows in this case that D(A) is a simple noetherian ring. However, it is not always true that D(A)=∆(A). Regularity of A seems in fact to be essential for this equality to hold, and it has been conjectured by Nakai (cf. [15, section 15.6.5]) that if A is a domain and an algebra of finite type over a field of characteristic zero, then D(A)=∆(A) if and only if A is regular.

We end this section by some remarks on notation. If f : A → B is a homomorphism of commutative k-algebras, and M and N are B-modules, the notation D(M,N) is clearly

ambiguous. Hence we will write DA/k(M,N) if we must clarify all the objects which go into the deﬁnition. Usually, however, we can omit the underlying ring k from this

notation, and we write DA(M,N) instead. The same considerations apply for the spaces d D (M,N). Aside from this, we will reserve the notation D(A) for the object DA(A, A). At this point it is also appropriate to justify the notation D ∗ a for the action of D ∈D(A)ona ∈ A, rather than the more conventional notations D(a)orevenDa.At several points throughout this work, we will be regarding D(A)asarightA-module, while we regard A as a left D(A)-module. Hence there is a danger that the traditional notation D(a) leads to confusion, as is exempliﬁed by the expression D(a+b). For the same reason would juxtaposition be ambiguous. Therefore we will follow the example of [25] and write D ∗ a for the operation of D ∈D(A)ona ∈ A and retain Da for the right operation of a ∈ A on D ∈D(A). We will also use this notation for elements of A-modules whenever there is a danger of confusion. We note of course that Da ∗ b = Dab ∗ 1=D ∗ ab for all D ∈D(A)andalla, b ∈ A.

13 1.2 Elementary results

1.2.1 A collection of basic lemmas

We will now state some basic results on diﬀerential operators, starting with:

Lemma 1.2.1 Let A be a commutative k-algebra, and let M be an A-module. Let N be an A-submodule of M. Then there is a canonical isomorphism

∼ D(M,N) = {D ∈D(M)|D ∗ M ⊂ N}⊂D(M)

Proof: Trivial. 2

This lemma applies especially if we take M = A and N = I, an ideal of A.The lemma shows that we can identify D(A, I) with the subset {D ∈D(A)|D ∗ A ⊂ I},a right ideal of D(A). In the sequel we will do so without further comment. From this description it follows immediately that I ⊂D(A, I) is a left (and hence a twosided) ideal of D(A) if and only if I is a left D(A)-submodule of A. We have in fact a lemma:

Lemma 1.2.2 Let A be a commutative k-algebra and assume that I ⊂ A is a left D(A)- submodule of A (and hence an ideal of A). Let π : A → A/I denote the canonical projection. Then there is a canonical homomorphism π : D(A) →D(A/I) of ﬁltered rings, and for all D ∈D(A), π(D) is the unique element of D(A/I) such that π(D) ∗ π(a)=π(D ∗ a) for all a ∈ A. Moreover we have an exact sequence

π 0 −→ D (A, I) −→ D (A) −→ D(A/I)

14 Proof: We get the canonical ring homomorphism π : D(A) →D(A/I) by chasing in the diagram π 0 - I - A - A/I - 0 p p p p p p p D|I D p π(D) p p ?p ? ?p π 0 - IA- - A/I - 0

The diagram shows that π(D), defined by the formula π(D) ∗ π(a)=π(D ∗ a) for all a ∈ A, is a well defined differential operator on A/I, of order less than or equal to the order of D. Uniqueness follows immediately and subsequently we have that π is a ring homomorphism. The formula implies directly that D(A, I)isthekernelofπ. 2

Important Warning Later we will show that the homomorphism π : D(A) →D(A/I) need not be surjective.

As mentioned above, D(A, I) is a twosided ideal of D(A) whenever I is a left D(A)- submodule of A.Moreover,sinceI = D0(A, I)itisclearthatD(A, I) is nontrivial whenever I is. On the other hand it may happen that D(A) has nontrivial twosided ideals while A is a simple left D(A)-module. The next lemma helps in applying commutative theory to the study of twosided ideals in rings of diﬀerential operators:

Lemma 1.2.3 Let A be a commutative k-algebra and let I ⊂D(A) be a twosided ideal. Then I =06 if and only if I contains a nonzero element of A (i.e., a nonzero diﬀerential operator of order zero).

Proof: The implication ⇐ is clear. For the other direction assume that D ∈ I \ A.For all a ∈ A we have that ord([D, a]) < ord(D), and [D, a]=Da − aD ∈ I since I is a twosided ideal. Since D 6∈ A, it follows that [D, a] =6 0 for some a ∈ A. An induction then implies that I ∩ A =6 ∅. 2

15 Recall that a ring is said to be semiprime if it has no nonzero nilpotent twosided ideals (cf. [15, section 0.2.7]). The lemma above immediately implies another:

Lemma 1.2.4 Let A be a commutative and reduced k-algebra. Then D(A) is a semiprime ring.

Proof: Assume that I is a nonzero nilpotent twosided ideal of D(A). By lemma 1.2.3, there is a nonzero element a ∈ A ∩ I, and by assumption a must be nilpotent. 2

We will on several occasions need to study rings of differential operators on direct products of commutative k-algebras. The structure of rings of differential operators on those is closely related to the rings of differential operators on the factors, as the next lemma shows:

Lemma 1.2.5 Let A := A1 × ...× An where n is an integer ≥ 1 and A1,...,An are commutative k-algebras. Then there is a natural isomorphism:

∼ D(A) = D(A1) × ...×D(An)

given by the formula D 7→ (π1(D),...,πn(D)), where for i =1,...,n, πi(D) ∈D(Ai)

is deﬁned by the formula πi(D) ∗ πi(a)=πi(D ∗ a).(πi denotes the natural projection

A → Ai.)

Proof: [17, proposition 1.12] 2

This implies that we may identify the rings D(A)andD(A1) × ...×D(An). We will do so without further comment. We end this section with a triviality that will be needed later

16 Lemma 1.2.6 Let φ : A → B be a homomorphism of commutative k-algebras, and let

M and N be B-modules. If φ is surjective, then DB(M,N) = DA(M,N)

Proof: Trivial. 2

1.2.2 Modules of diﬀerentials

In [10, p. 214] and [7, IV (16.3)] the modules of n-diﬀerentials, for n ≥ 0, are deﬁned. We recall some basic properties of them.

Let A be a commutative k-algebra. Then A ⊗k A carries a natural k-algebra structure

by componentwise multiplication. We have two natural morphisms A → A ⊗k A of k- algebras, a 7→ a⊗1anda 7→ 1⊗a. We ﬁx these morphisms, and from now on we will

always regard A ⊗k A as a (left) A-module via the former morphism and we denote the

latter morphism by dA. 0 0 Let mA : A ⊗k A → A denote the natural multiplication map a⊗a 7→ aa .Itisa morphism of k-algebras, and we observe that it is A-linear by the A-module structure we

just deﬁned on A ⊗k A. This morphism is of course surjective, so if we let IA denote the

kernel of mA, we have an exact sequence

mA 0 −→ IA −→ A ⊗k A −→ A −→ 0.

of A-modules, and we note that IA is an ideal of A ⊗k A. For all integers n ≥ 0welet n ⊗ n+1 PA denote the k-algebras A k A/IA . These algebras are left A-modules and for all ≥ n ≥ n → n n 0wecallPA the module of n-diﬀerentials on A.Forn 0, we let dA : A PA ⊗ → n denote the composite of dA with the natural projection A k A PA. These maps ≥ n → n are homomorphisms of k-algebras, and for n 0wecalldA : A PA the universal n-derivation on A.

17 Fix an integer n ≥ 0. The morphism mA : A ⊗k A → A factors naturally to an n → n+1 A-linear k-algebra map PA A with kernel IA/IA . By abuse of language we will also denote this map by mA and we denote its kernel also by IA. Having set that notation, we have an exact sequence

−→ −→ n −→mA −→ 0 IA PA A 0.

n n+1 of left A-modules, and IA is an ideal of PA such that IA = 0. The deﬁnitions of n → n n → n dA : A PA and mA : PA A now imply that mA is a left inverse for dA.Thatis,we have a commutative diagram 1 A - A @ 6 @ @ mA dn A @ @R n PA

This leads to

Lemma 1.2.7 Let A be a commutative k-algebra and let n ≥ 0 be an integer. Then n ∈Dn n dA (A, PA), and it has the following universal property: For all A-modules M and ∈Dn ∈ n all D (A, M) there is a unique φ HomA(PA,M) completing the diagram

n dA - n A PA p @ p @ p @ p φ D @ p @R ?p M

Proof: [10, theorem 2.2.6] 2

This lemma implies that if A is a commutative k-algebra and M is an A-module, then

18 ≥ Dn ∼ n for all integers n 0 there are natural isomorphisms (A, M) = HomA(PA,M). ≥ n Againﬁxanintegern 0. We note that by deﬁnition of PA it is in fact isomorphic n+1 n+1 to PA /IA , and hence we have a natural surjective A-linear k-algebra homomorphism n+1 → n π : PA PA making the diagram

n+1 PA @ dn+1 @ m A π @ A @ ? @R - Pn - A n A mA A dA commutative. Therefore for all A-modules M, we get a natural injective A-linear ho- n → n+1 momorphism HomA(PA,M) HomA(PA ,M). It is then an easy task to verify that this map in fact corresponds to the inclusion Dn(A, M) ⊂Dn+1(A, M). That is, for all integers n ≥ 0 we have a commutative diagram

Dn+1 ∼ n+1 (A, M) = HomA(PA ,M) 6 6

Dn ∼ n (A, M) = HomA(PA,M)

19 Chapter 2

Etale´ homomorphisms and diﬀerential operators

In this chapter, the theory of étale homomorphisms is applied to rings of differential operators. Thefirstresultisthatiff : A → B is a formally étale homomorphism of k-algebras, then it extends in a natural way to a homomorphism f : D(A) →D(B) of filtered rings. This result is implicit in [26] and the arguments in the proof are essentially the same, but the proof is carried out here for the sake of completion and for the convenience of the reader. If f : A → B is as above, and in addition the modules of differentials of A are finitely presented, we prove that the morphism f induces in a natural way an isomorphism ∼ D(B) = B ⊗A D(A)ofB-D(A)-bimodules. This is then used to prove a descent theorem for left modules over rings of differential operators, a technique which will be useful in chapter 3. Let B be a commutative k-algebra, and let b ∈ B bearootofapolynomialP ∈ B[T ] such that P 0(b) is an invertible element of B (here P 0 denotes the formal derivative of

20 the polynomial P ). We give a formula

X∞ Db = bD + νd(D) D ∈D(B) d=1

where the νdsareB-bilinear endomorphisms on D(B), computed recursively from b and the coeﬃcients of the polynomial P . A model for this result is the formula

∂a−1 = a−1∂ − a−2∂ ∗ a,

where ∂ is a derivation on an algebra and a is an invertible element in the algebra. We show also how an earlier result of [20, 8] is a special case of our formula. Finally use this last result to show that if f : A → B is an ´etale homomorphism,

and the modules of diﬀerentials of A are ﬁnitely presented, then B ⊗A J is naturally isomorphic to a twosided ideal of D(B) whenever J is a twosided ideal of D(A).

2.1 Basic properties of ´etale homomorphisms

2.1.1 Formally ´etale homomorphisms

As in [11, Chapter II.1] we make the following deﬁnitions:

Deﬁnition Let A denote a ring. A covariant functor

F : A − algebras −→ Sets

is said to be formally ´etale if for all pairs (C, I)whereC is an A-algebra and I is an ideal in C with I2 =0,themap F (C) −→ F (C/I)

21 is bijective.

Deﬁnition A ring homomorphism f : A → B of commutative rings is said to be formally

´etale if HomA(B,−) is a formally ´etale functor from A − algebras to Sets.

Remark We define formally smooth and formally unramified functors and ring homomorphisms in a similar way, by replacing the word “bijective” with the weaker conditions “surjective” or “injective” respectively. It follows that a ring homomorphism is formally étale if and only if it is both formally smooth and formally unramified.

Let f : A → B be a homomorphism of commutative rings. Then the deﬁnition implies that f is formally ´etale if and only the following condition holds: For all commutative diagrams u B - C/I 6 6

- A v C

where C is a commutative ring, I ⊂ C is an ideal such that I2 =0andu, v are ring homomorphisms, there is a unique ring homomorphism u0 : B → C completing the diagram u B - C/I p p 6 p 6 p p p u0 f p p p p Rp - A v C

The ring homomorphism f is formally smooth if and only if there is at least one such u0, and it is formally unramified if and only if there is at most one such u0.Inthiswayare 0-étale, 0-smooth and 0-unramified homomorphisms defined in [14].

Remarks 1. An induction shows that this last condition is equivalent to the same condition with “I2 = 0” replaced by “I nilpotent”.

22 2. If in the diagrams above, A, B and C are algebras over some commutative ring k, and the morphisms f, u and v are k-algebra homomorphisms, it follows immediately that u0 is also a k-algebra homomorphism whenever it exists.

We list some well known properties of formally ´etale homomorphisms:

Lemma 2.1.1 The following properties hold: (i) The composite of two formally étale homomorphisms is formally étale. (ii) If f : A → B is formally étale, and A0 is any commutative A-algebra, then 0 0 f ⊗ 1:A → B ⊗A A is formally étale. (iii) Let f : A → B be a formally unramified morphism and let g : B → C be any homomorphism of commutative rings. If g ◦f is formally étale, then g is formally étale. (iv) For any commutative ring A and any multiplicatively closed subset S ⊂ A,the localization map A → S−1A is formally étale. (v) If f : A → B is formally étale and S and T are multiplicatively closed subsets of A and B such that f(S) ⊂ T , then the corresponding morphism S−1A → T −1B is also formally étale.

Remark Properties (i), (ii), (iv) and (v) of the lemma hold also with “´etale” replaced by either “smooth” or “unramiﬁed”.

Proposition 2.1.2 Formally smooth morphisms are ﬂat.

Proof: By part (v) of lemma 2.1.1 and the remark following it, we have that localizations of formally smooth morphisms are formally smooth, and since ﬂatness is a local property (cf. [14, Theorem 7.1]) it suﬃces to prove the proposition for local homomorphisms.

This case, however, follows directly from [7, (0III, 19.7.1)] once we observe that formal

23 smoothness in the sense of Grothendieck implies formal smoothness in our sense (cf. [14, §28]). 2

2.1.2 Etale´ homomorphisms, essentially of ﬁnite type

Definition We say that a homomorphism f : A → B of noetherian commutative rings is of finite type if B is generated (as an algebra over A) by finitely many elements. We say that f is essentially of finite type if it can be factored as A → C → B where A → C ∼ − is of finite type and B = S 1C for some multiplicatively closed subset S of C.

We follow [11, Chapter III] for the definition of étale homomorphisms (note however that since we are assuming our rings to be noetherian, the notions “essentially of finite type” and “essentially of finite presentation” are the same).

Definition A homomorphism of commutative noetherian rings is said to be étale if it is formally étale and essentially of finite type.

Remark We define smooth and unramified morphisms in a similar manner, and it follows that a morphism is étale if and only if it is both smooth and unramified.

We list some properties of ´etale homomorphisms:

Lemma 2.1.3 Lemma 2.1.1 holds with the word “formally” deleted. Furthermore we have: (vi) Let f : A → B be essentially of ﬁnite type. Then f is ´etale if and only if

Af −1(P ) → BP is étale for all P ∈ Spec B if and only if Af −1(M) → BM is étale for all maximal ideals M of B. (vii) Let k be a field. A ring homomorphism k → A is étale if and only if A is a finite direct product of finite separable field extensions of k.

24 We have a local structure theorem for ´etale morphisms:

Theorem 2.1.4 Let A and B be local noetherian rings, and let f : A → B be an ´etale ∼ local morphism. Then B = (A[T ]/P )M , where: (i) P ∈ A[T ] is a monic polynomial, (ii) M isaprimeidealinA[T ] such that P ∈ M and A ∩ M is the maximal ideal of A, and (iii) P 0 6∈ M, where P 0 denotes the formal derivative of P . Conversely for any P and M satisfying these three condition, the natural map A →

(A[T ]/P )M is an ´etale local morphism.

Corollary 2.1.5 An unramified morphism is étale if and only it is flat.

We will be working with ´etale algebras over reduced rings, so it is useful to note the following:

Lemma 2.1.6 If A is a reduced noetherian ring, and A → B is ´etale, then B is reduced too.

Proof: The ring A is reduced if and only if Ap is reduced for all p ∈ Spec A.Thus by lemma 2.1.3 we may assume that f : A → B is an étale local homomorphism. Since étale homomorphisms are flat, the lemma now follows from [14, Theorem 23.9] and lemma 2.1.3. 2

25 2.2 Behavior of diﬀerential operators under ´etale homomorphisms

2.2.1 Diﬀerentials and formally ´etale homomorphisms

Let f : A → B be a homomorphism of commutative k-algebras and ﬁx an integer n ≥ 0. From section 1.2.2 we recall the exact sequence

−→ −→ n −→mA −→ 0 IA PA A 0

of A-modules. By tensoring it on the left by B, we get an exact sequence of B-modules,

⊗ −→ ˜ −→ ⊗ nB−→AmA −→ 0 IB B A PA B 0

˜ ⊗ → ⊗ n where IB is the image of the map B A IA B A PA. n We recall that the (left) A-module PA is a k-algebra and that mA is a homomorphism of k-algebras. Therefore we can deﬁne a k-algebra structure on the (left) B-module ⊗ n ⊗ B A PA by componentwise multiplication. Then B A mA becomes a homomorphism ˜ ⊗ n ˜ of k-algebras, and its kernel, IB, is an ideal of B A PA.Moreover,IB is generated by ⊗ ∈ ∈ n+1 elements of the form b x where b B and x IA, and since IA = 0 it follows that ˜n+1 IB =0. ⊗ n → ⊗ n 7→ ⊗ n Let 1 dA : A B A PA denote the map a 1 dA(a). It is obviously a morphism of k-algebras, and for all a ∈ A we have the formula

⊗ ◦ ⊗ n ⊗ ⊗ n (B A mA) (1 dA)(a)=(B A mA)(1 dA(a)) = f(a).

26 Therefore we have a commutative diagram

1 BB- 6 6

f B⊗AmA

- B ⊗ Pn A ⊗ n A A 1 dA

˜ ⊗ ˜n+1 of k-algebras and k-algebra homomorphisms, and IB =ker(B A mA) is such that IB = 0. From these considerations we now get:

Proposition 2.2.1 Let f : A → B beaformallyétale morphism of k-algebras, and let ≥ ñ → ⊗ n n 0 be an integer. Then there is a unique k-algebra morphism dB : B B A PA making the diagram 1 B - B p p 6 p 6 p p p dñ p B ⊗ f p B AmA p p Rp - B ⊗ Pn A ⊗ n A A 1 dA commutative. This map is a differential operator of order ≤ n

˜ ⊗ ñ+1 Proof: As remarked earlier, IB =kerB A mA is nilpotent (IB = 0). Thus the ñ → ⊗ n existence and uniqueness of dB : B B A PA follows directly from the definition of étale morphisms. It is a morphism of k-algebras by a remark in section 2.1.1. From ñ − ñ ∈ ⊗ ∈ the diagram we now have clearly that bdB(1) dB(b) ker B A mA for all b B,and ⊗ n+1 ñ ≤ since (ker B A mA) = 0 it then follows that dB is a differential operator of order n ñ 2 (recall that dB is a ring homomorphism).

n → n By this proposition and the universal property of dB : B PB, it follows that if f : A → B is a formally ´etale homomorphism of k-algebras, then for all n ≥ 0, there is

27 n → ⊗ n ˜n ◦ n a unique B-linear map ψB : PB B A PA such that dB = ψB dB. The next theorem shows that this homomorphism is in fact an isomorphism, and thus we have a very close n n ≥ → relation of the k-algebras PA and PB for all integers n 0iff : A B is a formally ´etale morphism of k-algebras.

Theorem 2.2.2 Let f : A → B be a formally ´etale homomorphism of k-algebras, and ∼ ≥ ⊗ n →= n let n 0 be an integer. Then there is a unique B-module isomorphism B A PA PB making the diagram d˜n B - ⊗ n B B A PA p @ p @ p @ p dn p B @ p @R ?p n PB

˜n → ⊗ n n → commutative. Hence dB : B B A PA solves the same universal problem as dB : B n PB.

Proof: Once we have shown the existence of this isomorphism, uniqueness follows from n → n the universal property of dB : B PB. We will show existence by constructing an n → ⊗ n inverse to the map ψB : PB B A PA mentioned above. n → n ≤ The map dB : B PB is a diﬀerential operator of B-modules of order n.Sincef n ◦ → n is A-linear we have that dA f : A PB is a diﬀerential operator of A-modules of order ≤ n → n n. Therefore by the universal property of dA : A PA there is a unique A-linear n → n morphism φA : PA PB such that

◦ n n ◦ φA dA = dB f.

By adjointness of Hom and ⊗ we have a natural isomorphism of groups

n n ∼ ⊗ n n HomA(PA, PB) = HomB(B A PA, PB),

28 ⊗ n → n ◦ ⊗ so there is a unique B-linear homomorphism φB : B A PA PB such that φB (1 id) = n φA. Then by composing this with dA and using the formula above, we get that

◦ ⊗ n n ◦ φB (1 dA)=dB f

n → ⊗ ⊗ n → By the deﬁnitions of mB : PB B and B A mA : B A PA B we now have for all a ∈ A that

◦ ⊗ n n ⊗ ⊗ n (mB φB)(1 dA(a))=mB(dB(f(a))) = f(a)=(B A mA)(1 dA(a)).

n ⊗ n By the construction of PA it follows that the B-module B A PA is generated by the ⊗ n ∈ elements 1 dA(a), a A and therefore we have that

mB ◦ φB = B ⊗A mA

ñ The two formulas we have now proved, along with the definition of dB,nowimply that ◦ ◦ ñ ⊗ ◦ ñ mB (φB dB)=B A mA dB =idB and ◦ ñ ◦ ◦ ⊗ n n ◦ (φB dB) f = φB (1 dA)=dB f

I.e, we have a commutative diagram

1 B - B 6@ 6 @ ◦ ˜n φB dB f @ mB @ @R - Pn A n ◦ B dB f

◦ ˜n → and it is trivial to show that φB dB is a ring homomorphism. Since f : A B is formally ´etaleandthekernelofmB is nilpotent, the diagonal is unique. However, it can

29 n → n n ◦ ˜n be replaced by the map dB : B PB,sowemusthavedB = φB dB.

Now we can ﬁnish the proof. By combining the formulas we have for φB and ψB,we get that n ◦ ˜n ◦ ◦ n dB = φB dB =(φB ψB) dB

n → n ◦ and so the universal property of dB : B PB implies that φB ψB = id. Conversely we ◦ n ˜n recall that ψB dB = dB, and then we can read from our diagram that

◦ ◦ ⊗ n ◦ n ◦ ˜n ◦ ⊗ n ψB φB (1 dA)=ψB dB f = dB f =1 dA.

⊗ n ⊗ n ∈ Since B A PA is generated as a (left) B-module by the elements 1 dA(a), a A,this

implies that ψB ◦ φB =id. 2

2.2.2 Diﬀerential operators and formally ´etale morphisms

In this section we will study the relationship between differential operators on A-modules and differential operators on B-modules if f : A → B is a formally étale homomorphism of k-algebras.

We start with a general observation. Let f : A → B be a homomorphism of k- ≥ n algebras and let n 0 be an integer. If M is a B-module, the groups HomA(PA,M) Dn ∈ ∈Dn and A(A, M) have natural (left) B-module structures: For b B, D A(A, M) and a ∈ A,welet(bD) ∗ a = b(D ∗ a). Moreover it is clear that the isomorphism n ∼ Dn HomA(PA,M) = A(A, M) is an isomorphism of B-modules. This implies that the n − Dn − functors HomA(PA, )and A(A, ) are naturally equivalent. We will use this in the sequel without further comment. We start with a lemma:

30 Lemma 2.2.3 Let f : A → B be a formally ´etale homomorphism of k-algebras and let n ≥ 0 be an integer. Then (i) there is a natural equivalence

n Dn − −→ D n − φ : A(A, ) B(B, )

of functors from B − Mod into B − Mod; ∈Dn n (ii) if M is a B-module and D A(A, M), then φM (D) is the unique element of Dn B(B,M) completing the commutative diagram

f A - B p p p p p p D p p p n p φM (D) ? ©p M

˜n → ⊗ n Proof: By theorem 2.2.2, dB : B B A PA has the same universal property as n → n ⊗ dB : B PB. Hence by using the adjointness of Hom and we have a chain of natural equivalences

Dn − ∼ ⊗ n − ∼ n − ∼ Dn − B(B, ) = HomB(B A PA, ) = HomA(PA, ) = A(A, )

of functors from B − Mod into B − Mod. This proves the ﬁrst part. For the second part we look at the following diagram:

d˜n B - ⊗ n B B A PA p p p p 6 p (3) p 6 p p p p R © (2) f M ι p D Ip p p (1) p - n A n PA dA

31 We are just repeating the argument from the first part of the proof. We fill in the first n → n ⊗ arrow by the universal property of dA : A PA. The adjointness of Hom and gives us the second arrow, and finally the third arrow comes from the universal property of ñ → ⊗ n n dB : B B A PA. It is clear that this last arrow is equal to φM (D). All these arrows 0 ∈Dn are unique, and we note that if we start with any differential operator D B(B,M) such that D = D0 ◦ f, we can successively fill in the diagram in reverse order. Hence we 0 n 2 conclude that D = φM (D).

Proposition 2.2.4 Let f : A → B be a formally ´etale homomorphism of k-algebras. n Dn − →Dn − The natural equivalences φ : A(A, ) B(B, ) commute with the inclusions Dn − ⊂Dn+1 − Dn − ⊂Dn+1 − ≥ A(A, ) A (A, ) and B(B, ) B (B, ) for all n 0, and thus they deﬁne a natural equivalence

φ : DA(A, −) −→ D B(B,−) of functors from B − Mod into B − Mod. Moreover if M is a B-module and D ∈

DA(A, M) then ord(φM (D)) ≤ ord(D) and φM (D) is the unique element of DB(B,M) which completes the diagram f A - B p p p p p p D p p p p ? ©p M

≥ ∈Dn ⊂ Proof: Fix an integer n 0, let M be a B-module and assume that D A(A, M) Dn+1 n n+1 ∈Dn+1 A (A, M). We must show that φM (D)=φM (D) B (B,M). But this is clear since both these element ﬁll in the diagram in the second part of lemma 2.2.3 (with n replaced by n + 1). Hence the natural equivalences commute with inclusions, and

32 we conclude that they deﬁne a natural equivalence DA(A, −) →DB(B,−) of functors B − Mod into B − Mod.

The definition of the map φM implies that it preserves order of differential operators. 0 Finally for the last part let D ∈DA(A, M). If D ∈DB(B,M) fills in the diagram, we ∈Dn 0 ∈Dn pick n large enough so D A(A, M)andD B(B,M). Then uniqueness follows from lemma 2.2.3. 2

Theorem 2.2.5 Let f : A → B be a formally étale homomorphism of k-algebras. It induces a homomorphism D(A) →D(B) of filtered k-algebras, which we also denote by f. The filtered homomorphism f : D(A) →D(B) is characterized by the following property: For D ∈D(A), f(D) is the unique element of D(B) such that

f(D) ∗ f(a)=f(D ∗ a) for all a ∈ A.

Proof: We apply proposition 2.2.4 with M = B, and thus get a morphism φB :

DA(A, B) →D(B) of ﬁltered groups. The map f : A → B is a morphism of A-modules, so it clearly induces a morphism ψ : D(A) →DA(A, B) of ﬁltered groups, given by the

formula ψ(D):=f ◦ D.Weletf := φB ◦ ψ.Withf : D(A) →D(B) so deﬁned, it is now clear that f is k-linear and a morphism of ﬁltered groups.

Let D be an element of D(A). Then f(D)=φB(f ◦D), and hence by proposition 2.2.4 it follows that f(D) is the unique element of D(B) which completes the diagram

f A - B p p p p p p f◦D p p p p ? ©p B

33 This is precisely what we want. It remains to show that f : D(A) →D(B) is a homomorphism of k-algebras. We have already seen that f is k-linear. We must show that f preserves multiplication. Let D, D0 ∈D(A). Then for all a ∈ A we have that

(f(D)f(D0)) ∗ a = f(D ∗ (D0 ∗ a)) = f(DD0 ∗ a)=f(DD0) ∗ a

so by the uniqueness already proved it follows that f(D)f(D0)=f(DD0). The uniqueness also shows that f(1) = 1. 2

We have an immediate corollary to this:

Corollary 2.2.6 Let f : A → B be an injective formally ´etale homomorphism of k- algebras. Then f : D(A) →D(B) is also injective, and its image is equal to the ﬁltered ring {D ∈D(B)|D ∗ f(A) ⊂ f(A)}.

Proof: Let D ∈D(A) be such that f(D) = 0. Then for all a ∈ A,0=f(D) ∗ f(a)= f(D ∗ a), and since f : A → B is injective, this implies that D = 0. For the second claim we ﬁrst observe that f(D) ∗ f(A) ⊂ f(A) for all D ∈D(A). Conversely let D ∈D(B) be such that D ∗ f(A) ⊂ f(A). Since f is injective, D induces a unique 0 0 element D of EndkA with the property that f(D ∗ a)=D ∗ f(a) for all a ∈ A.From this it follows that D0 ∈D(A)isoforder≤ ord(D). Then we have for all a ∈ A that f(D0) ∗ f(a)=f(D0 ∗ a)=D ∗ f(a), so by uniqueness we must have that D = f(D0). 2

Remarks 1. Using this corollary, we will from now on identify the rings D(A)and {D ∈D(B)|D ∗ f(A) ⊂ f(A)} whenever f : A → B is an injective formally ´etale homomorphism of k-algebras, and we ﬁnd it convenient.

34 2. The corollary applies especially to the case where A is a commutative reduced k-algebra and K is its total ring of fractions. We conclude that

D(A)={D ∈D(K)|D ∗ A ⊂ A}

3. Let f : A → B be a faithfully ﬂat and formally ´etale homomorphism of k-algebras. Then there is an exact sequence

f −→f1 0 −→ A −→ B −→ B ⊗A B, f2

and from the corollary it is more or less immediate that this exact sequence induces an exact sequence f −→f1 0 −→ D (A) −→ D(B) −→ D(B ⊗A B) f2

of ﬁltered rings. 4. Note that the proof of the corollary did not use the fact that f : A → B was formally ´etale, it only used the existence and uniqueness of the morphism f : D(A) →D(B)with the given properties, and injectivity of f. Later we will in fact encounter situations where this is useful to know.

For later usage we now return to the map

φB : DA(A, B) −→ D (B)

deﬁned in proposition 2.2.4 when f : A → B is a formally ´etale homomorphism of k-algebras. We had already seen that it is an isomorphism of left B-modules.

It is clear that DA(A, B) has a natural right D(A)-module structure via composite of differential operators. Moreover by using last theorem, we can define a right D(A)-module structure on D(B). We fix these structures and then we have:

35 Proposition 2.2.7 Let f : A → B beaformally´etale k-algebras homomorphism. The morphism φB : DA(A, B) →D(B), deﬁned in proposition 2.2.4, is an isomorphism of B-D(A)-bimodules.

Proof: It only remains to show that φB is right D(A)-linear. For that let D ∈DA(A, B) and D0 ∈D(A). By the last part of theorem 2.2.5 we have that f(D0) ◦ f = f ◦ D0.It follows that

0 0 0 φB(D) ◦ f(D ) ◦ f = φB(D) ◦ f ◦ D = DD 0 = φB(DD ) ◦ f

0 Since φB(D) ◦ f(D ) ∈D(B), last part of proposition 2.2.4 thus implies that φB(D) ◦ 0 0 f(D )=φB(DD ). This is precisely to say that φB is right D(A)-linear. 2

2.2.3 Finitely presented modules of diﬀerentials

Theorem 2.2.5 implies that if f : A → B is a formally ´etale homomorphism of k-algebras,

then we can deﬁne the homomorphism B ⊗A D(A) →D(B)ofB-D(A)-bimodules by the formula b ⊗ D 7→ bf(D). We will now introduce some conditions which ensure that this homomorphism is in fact an isomorphism.

We start by recalling some homological algebra: Let f : A → B be a morphism of commutative rings. We can form the following two bifunctors

B ⊗A HomA(−1, −2)

and

HomA(−1,B⊗A −2)

36 from A − Mod×A − Mod into B − Mod. Both these bifunctors are contravariant in the first variable and covariant in the second variable. We can define a natural transformation τ between these two functors as follows: For A-modules M and N we define a B-linear homomorphism

τN,M : B ⊗A HomA(N,M) −→ HomA(N,B ⊗A M)

by τN,M(b⊗g)=gb where gb(n):=b⊗g(n). This transformation is functorial in both

variables, that is, τN,− : B ⊗A HomA(N,−) → HomA(N,B ⊗A −) is a natural transfor-

mation of functors for all N and so is τ−,M : B ⊗A HomA(−,M) → HomA(−,B⊗A M) for all M.

Deﬁnition An A-module N is said to be ﬁnitely presented if there are integers r and s and an exact sequence of A-modules

Ar −→ As −→ N −→ 0

With this in hand we can state the following:

Lemma 2.2.8 If f : A → B is ﬂat homomorphism of commutative rings, and N is a

ﬁnitely presented A-module, then τN,− : B ⊗A HomA(N,−) → HomA(N,B ⊗A −) is a natural equivalence of functors.

Proof: This follows directly from the observations above and [22, lemma 3.83]. 2

We can apply this result to diﬀerential operators:

Proposition 2.2.9 Let f : A → B be a ﬂat homomorphism of commutative k-algebras, ≥ n and let n 0 be an integer. If PA is a ﬁnitely presented A-module, then there is a

37 canonical isomorphism of B-modules,

n ⊗ Dn −→ D n τB : B A (A) A(A, B),

n ⊗ ∗ ∗ deﬁned by by the formula τB(b D):=Db, where Db a = bf(D a). n ≥ n If the A-modules PA are ﬁnitely presented for all integers n 0, the isomorphisms τB commute with inclusions, and hence they induce an isomorphism of B-D(A)-bimodules

τB : B ⊗A D(A) −→ D A(A, B) deﬁned by the same formula as above.

n Proof: The ﬁrst statement follows directly by applying lemma 2.2.8 to N = PA and the natural equivalence of functors

Dn − ∼ n − A(A, ) = HomA(PA, ).

n It is trivial to show that the maps τB commute with inclusions, and by passing to direct limits we get the B-linear isomorphism

τB : B ⊗A D(A) −→ D A(A, B)

0 0 It remains to show that τB is right D(A)-linear. Let D ∈D(A). Then (b⊗D)D = 0 0 0 b⊗(DD ) by deﬁnition, and hence τB((b⊗D)D )=(DD )b. On the other hand we have 0 0 τB(b⊗D)D = DbD . But for all a ∈ A we have that

0 0 0 0 DbD ∗ a = Db ∗ (D ∗ a)=bf(D ∗ (D ∗ a)) = bf(DD ∗ a)=(DD )b ∗ a

0 0 and therefore is DbD =(DD )b.ThisimpliesthatτB is right D(A)-linear. 2

38 We can now prove:

→ n Theorem 2.2.10 If f : A B is a formally ´etale homomorphism of k-algebras and PA is ﬁnitely presented for all integers n ≥ 0, then f induces an isomorphism of B-D(A)- bimodules

B ⊗A D(A) −→ D (B),

given by (b⊗D) 7→ bf(D).

Proof: From proposition 2.1.2 it follows that f : A → B is flat, and hence this isomorphism is just the composite of the B-D(A)-isomorphisms from propositions 2.2.7 and 2.2.9. It remains only to establish the formula stated in the theorem. For that let b ∈ B and D ∈D(A). The isomorphism from proposition 2.2.9 maps b ⊗ D to the differential operator Db ∈DA(A, B), and then proposition 2.2.4 implies that φB(Db)is the unique differential operator in D(B) such that

φB(Db) ∗ f(a)=Db ∗ a = bf(D ∗ a)

for all a ∈ A. It is clear, however, that bf(D) is a diﬀerential operator in D(B), and theorem 2.2.5 implies that bf(D) ∗ a = bf(D ∗ a) for all a ∈ A. Thereforewehavethat

φB(Db)=bf(D). 2

Remark A similar statement holds for modules. In fact if A and B are as in theorem 2.2.10 and M is an A-module, then we can proceed as in the proof of proposi- ∼ tion 2.2.9 to get a canonical isomorphism B ⊗A DA(A, M) = DA(A, B ⊗A M). Further- ∼ more proposition 2.2.4 implies that there is a canonical isomorphism DA(A, B ⊗A M) =

DB(B,B ⊗A M), so by combining these we have a canonical isomorphism

∼ B ⊗A DA(A, M) = DB(B,B ⊗A M)

39 n Having theorem 2.2.10, it becomes natural to ask when PA is ﬁnitely presented for all integers n ≥ 0. The following two results show that this is often the case:

Proposition 2.2.11 If f : A → B isaformallyétale homomorphism of k-algebras, and n ≥ n PA is finitely presented for some integer n 0, then PB is also finitely presented.

Proof: By assumption we have an exact sequence

r −→ s −→ n −→ A A PA 0

of A-modules, and by applying the right exact functor B ⊗A − to this sequence, we get an exact sequence of B-modules

r −→ s −→ ⊗ n −→ B B B A PA 0

→ ⊗ n Finally, since f : A B is formally ´etale, theorem 2.2.2 tells us that B A PA is n 2 isomorphic to PB as a B-module.

n Proposition 2.2.12 If A is a ﬁnitely generated k-algebra, then PA is ﬁnitely generated for all integers n ≥ 0.

Proof: This follows directly from [7, IV (16.4.22)]. However, for the sake of completion we supply a proof here:

As before, we let IA denote the kernel of the multiplication map mA : A ⊗k A → A. 2 We will ﬁrst prove that IA/IA is a ﬁnitely generated (left) A-module. It is well known

(and an easy exercise to see) that IA is generated by the elements {a ⊗ 1 − 1 ⊗ a|a ∈ A}.

40 Hence we have for all a, b ∈ A that

2 1⊗ab = a ⊗ b + b ⊗ a − ab ⊗ 1(modIA ), and from this it follows that

2 ab ⊗ 1 − 1 ⊗ ab ≡ ab ⊗ 1 − a ⊗ b − b ⊗ a + ab ⊗ 1(modIA ) = a(b ⊗ 1 − 1 ⊗ b)+b(a ⊗ 1 − 1 ⊗ a)

for all a, b ∈ A. Hence if the finitely many elements {ai} generate A as a k-algebra, 2 then the left A-module IA/IA is generated by the canonical images of the elements ai ⊗ 1 − 1 ⊗ ai. r−1 r Now let r>0 be an integer and assume that IA /IA is finitely generated. We want P r r+1 ∈ r to show that IA /IA is finitely generated. Let x IA and write x = j xjyj where r−1 xj ∈ IA and yj ∈ IA for all j. We then use the assumption and the first part of the n 2 proof to write the xisandtheyis in terms of generators (modulo IA and IA )andthe result follows. n ≥ Finally to prove that the modules PA are finitely generated for all n 0weusethe exact sequences −→ n n+1 −→ n −→ n−1 −→ 0 IA /IA PA PA 0.

We have already proved that the terms on the left end are ﬁnitely generated and since

0 ∼ n PA = A as a left A-module, we get immediately by induction that PA is finitely generated for all n (we bear in mind the elementary fact that if the two ends in a short exact sequence are finitely generated modules, then the middle term is a finitely generated module as well.) 2

Remark We recall of course that all ﬁnitely generated modules over noetherian rings are automatically ﬁnitely presented.

41 Earlier it was pointed out that localizations of commutative rings are always formally ´etale. Therefore these two propositions give a large class of algebras to which theorem 2.2.10 applies, e.g. all algebras which are essentially of ﬁnite type over a noetherian ring.

Next we point out a lemma which follows directly from theorem 2.2.10:

Lemma 2.2.13 Let f : A → B beaformally´etale homomorphism of k-algebras, and n ≥ assume that PA is ﬁnitely presented for all integers n 0. Then there is a natural equivalence

B ⊗A −−→D(B) ⊗D(A) −

of functors from D(A) − Mod into B − Mod.

∼ Proof: By theorem 2.2.10, D(B) = B ⊗A D(A)asaB-D(A)-bimodule. Then cancella- tion gives the desired result. 2

This lemma implies that when studying the behavior of the functor D(B) ⊗D(A) − on exact sequences of left D(A)-modules, we may just as well look at the better known

functor B ⊗A −. For example we have immediately that D(B) ⊗D(A) − is exact. Hence

in the sequel we will sometimes use B ⊗A M in place of D(B) ⊗D(A) M if M is a left D(A)-module (and the assumptions of the lemma hold).

We end this section by discussing localizations. Localizations are formally ´etale and

n hence ﬂat. Therefore if A is a commutative k-algebra such that PA is ﬁnitely presented for all integers n ≥ 0, and S ⊂ A is any multiplicatively closed subset of A, by theorem 2.2.10 − ∼ − we have a natural isomorphism D(S 1A) = S 1D(A)(whereweregardD(A)asaleft A-module). We will thus from now on treat these two objects as being equal. If J is a left

42 A-submodule of D(A), by the above remarks and ﬂatness we may regard S−1J as a left S−1A-submodule of D(S−1A). Furthermore, if J is a left ideal of D(A) then lemma 2.2.13

−1 −1 −1 applies,andwehavethatS J = D(S A) ⊗D(A) J is a left ideal of D(S A). With the assumptions and notation as above and for p ∈ Spec A, we will from now on write Jp for Ap ⊗A J, and we will regard it as a left Ap-submodule of D(Ap). It is a left ideal if J is.

2.2.4 Etale´ descent for diﬀerential operators

If f : R → S is a ring homomorphism, we will sometimes write f∗− for the functor S⊗R− from the category of left R-modules into the category of left S-modules. This notation lacks the reference to the rings, but has some obvious advantages, for example if we need to study tensor products with respect to two or more ring homomorphisms R → S. For now we note that if f : A → B is a homomorphism of commutative k-algebras such that the conditions of lemma 2.2.13 hold, the equivalence of the functors D(B) ⊗D(A) − and B ⊗A − implies that there is no danger of confusion when we write f∗M where M is a left D(A)-module.

For f : A → B, a homomorphism of commutative rings, let f1,f2 : B → B ⊗A B be the two natural base extensions of f,deﬁned by f1(b)=1⊗b, f2(b)=b⊗1. For a B-module ∼ M, it follows immediately that we have natural B ⊗A B-isomorphisms, f1∗M = B ⊗A M ∼ and f2∗M = M ⊗A B. As in [12, p. 34], we let θi, i = 1, 2, 3 denote the three B⊗AB⊗AB- homomorphisms induced by a B ⊗A B-homomorphism θ : B ⊗A M → M ⊗A B. We will ﬁx this notation in what follows.

We will now deﬁne descent data. In light of the notational convention we have established and [12, proposition II.3.1], it follows that our deﬁnition is equivalent to the one given in [12]:

Deﬁnition Let f : A → B be a homomorphism of commutative rings and let M be

43 a B-module. An isomorphism θ : B ⊗A M → M ⊗A B of B ⊗A B-modules such that

θ2 = θ3θ1 is called a descent data for M over A.

n Now assume that A is a commutative k-algebra, and that PA is ﬁnitely presented for all integers n ≥ 0. Furthermore assume that f : A → B is a formally ´etale homomorphism

n of k-algebras. Then by proposition 2.2.11, we have that PB is also finitely presented for all n ≥ 0, and by lemma 2.1.1 it follows that the morphisms fi : B → B ⊗A B, i =1, 2 are formally étale. Hence if M is a left D(B)-module, we can form the left D(B ⊗A B)- modules f1∗M and f2∗M. It was pointed out earlier that these modules are naturally isomorphic as B ⊗A B-modules to the modules B ⊗A M and M ⊗A B respectively, and therefore we can define a left D(B ⊗A B)-module structure on those as before. If θ :

B ⊗A M → M ⊗A B is a D(B ⊗A B)-homomorphism, restriction of scalars implies that it is especially a B ⊗A B-homomorphism, and hence we can deﬁne θi, i =1, 2, 3 as before.

n Definition Let A be a commutative k-algebra such that PA is finitely presented for all n ≥ 0, and let f : A → B be a formally étale homomorphism of k-algebras. Let M be a left D(B)-module. An isomorphism θ : B ⊗A M → M ⊗A B of left D(B ⊗A B)-modules such that θ2 = θ3θ1 is called a descent data for M over D(A).

We get immediately:

n Theorem 2.2.14 Let A be a commutative k-algebra such that PA is finitely presented for all n ≥ 0, and let f : A → B be a faithfully flat and formally étale extension of k-algebras. Let M bealeftD(B)-module and let θ : B ⊗A M → M ⊗A B be a descent data for M over D(A). Then (i) there is a left D(A)-module N and an isomorphism

η : B ⊗A N −→ M

44 of left D(B)-modules such that the diagram

1⊗η - B ⊗A B ⊗A N B ⊗A M

∼ = θ

? ? B ⊗ N ⊗ BM- ⊗ B A A η⊗1 A

commutes; (ii) the pair (N,η) is determined up to a unique isomorphism; (iii) we may in fact take N = {m ∈ M|θ(1⊗m)=m⊗1}, and the isomorphism η :

B ⊗A N → M to be the multiplication map b⊗n 7→ bn.

Proof: It is clear from the definition that θ : f1∗M → f2∗M is also a descent data for the B-module M over A. Hence the existence and uniqueness of the pair (N,η) (as a B-module) is a direct consequence of the theorem of faithfully flat descent (cf. [12, Théorème II.3.2]). The only fact remaining to be checked is that N, defined as in part (iii), is actually a D(A)-submodule of M. It then follows directly from the definition of the

left D(B)-module structure on B ⊗A N that η : B ⊗A N → M is D(B)-linear. We must show that θ(1⊗(f(D)∗m)) = (f(D)∗m)⊗1ifD ∈D(A)andm ∈ M is such

that θ(1⊗m)=m⊗1. We recall that B ⊗A M is naturally isomorphic to f1∗M. Under ⊗ ∗ ⊗ ⊗ ∗ that isomorphism, the element 1 (f(D) m) corresponds to (1 1) f1 (f(D) m)= ⊗ ⊗ ◦ ◦ f1(f(D)) f1 m = f2(f(D)) f1 m, since by deﬁnition f1 f = f2 f. By assumption on the element m we have that θ(1⊗m)=m⊗1 which in turn corresponds to the element ⊗ ⊗ ∈ D ⊗ ⊗ ∗ (1 1) f2 m f2∗M.Sinceθ is (B A B)-linear, this implies that θ(1 (f(D) m)) ⊗ ⊗ ⊗ ∗ ∈ corresponds to the element f2(f(D)) f2 m =(1 1) f2 (f(D) m) f2∗M. Finally, as ⊗ ⊗ ∗ ∈ ∗ before, we have that the element (1 1) f2 (f(D) m) f2∗M corresponds to (f(D)

m)⊗1 ∈ M ⊗A B, and thus we have that θ(1⊗(f(D) ∗ m))=(f(D) ∗ m)⊗1. 2

n ≥ Let A be a commutative k-algebra such that PA is ﬁnitely presented for all n 0

45 and let f : A → B be a faithfully ﬂat formally ´etale homomorphism of k-algebras. We can immediately apply theorem 2.2.14 to two cases: Taking M = D(B), we have natural

∼ ∼ 1 isomorphisms f1∗D(B) = D(B ⊗A B) = f2∗D(B). Let θ : B ⊗A D(B) →D(B) ⊗A B

be the D(B ⊗A B)-linear homomorphism that corresponds to the composite of these ∼ natural isomorphisms. Since D(B ⊗A B) = (B ⊗A B)⊗AD(A), it follows that θ is in fact a descent data for D(B)overD(A). Hence the theorem implies that there is a left

D(A)-module N, and an isomorphism η : B ⊗A N →D(B), making the diagram there

commutative. Of course D(A) and the multiplication map B ⊗A D(A) →D(B); b⊗D 7→ ∼ bf(D) is such a pair, and thus we have that D(A) = {D ∈D(B)|f1(D)=f2(D)}. ∼ Another natural case to consider is M = B. Inthatcasewehavethatf1∗B = ∼ B ⊗A B = f2∗B,andthisD(B ⊗A B)-isomorphism simply corresponds to the identity map. Hence it is obviously a descent data for B over D(A), and the theorem implies

that there is a left D(A)-module N,andanD(B)-isomorphism η : B ⊗A N → B.We ∼ get of course that N = A. This leads however to a little less trivial statement that will be useful later, namely:

n Proposition 2.2.15 Let A be a commutative k-algebra such that PA is finitely presented for all n ≥ 0, and let f : A → B be a faithfully flat and formally étale extension

of k-algebras. Let I be a left D(B)-submodule of B such that B ⊗A I = I ⊗A B as

subsets of B ⊗A B. Then there is a unique D(A)-submodule J of A such that B ⊗A J is isomorphic to I via the ordinary multiplication map b⊗x 7→ bf(x). The module J is in

fact isomorphic to {x ∈ I|f1(x)=f2(x)}.

Proof: Uniqueness of J follows from faithful ﬂatness. The identity map on B ⊗A B

restricts to a D(B ⊗A B)-isomorphism θ : B ⊗A I → I ⊗A B, and it follows immediately

that all the maps θi, i =1, 2, 3 are also the identity. Therefore is θ a descent data for I over D(A). Then the existence and the description of J follows from theorem 2.2.14. 2

1This is rather unfortunate notation. We are considering D(B)asaleftB-module here

46 2.3 Diﬀerential operators and algebraic relations

2.3.1 Elements satisfying polynomial relations P ∈ n r ∈ Let B be a commutative k-algebra. Let b B and P = r=0 arT B[T ] be such that P (b)=0andP 0(b) is an invertible element of B (P 0 denotes the formal derivative of the polynomial P ). For such a pair and all integers i ≥ 0weletγ(P,b),i and α(P,b),i be the B-B-linear endomorphisms on D(B) deﬁned on D ∈D(B) by the formulas:

Xn − 0 −1 r r−i γ(P,b),i(D)= P (b) i b ar D r=0 and Xn − 0 −1 r r−i α(P,b),i(D)= P (b) i b [D, ar] r=0 We will drop the subscript (P, b) from this notation when the context makes it clear. P n ∈D Sometimes we will write γi(D)=biD and αi(D)= r=0 cr,i[D, ar] for D (B)ifthe precise formulas for the coeﬃcients bi and cr,i are unimportant. Fix a b ∈ B and P ∈ B[T ] as above. From the deﬁnitions, and the assumption that

B is commutative, it follows that all the endomorphisms γi and αi commute. We let Z D P,b be the commutative k-subalgebra of EndB (B)B generated by them, and we let

JP,b be the ideal of ZP,b generated by the endomorphisms αi, i ≥ 0. We observe that

γ1 = −id = −1, γ0 =0andγi = αi = 0 whenever i>n, so the algebra ZP,b is in fact

finitely generated over k,1∈ZP,b and JP,b is a finitely generated ideal of ZP,b. The usefulness of these definitions lies in the following: By definition every element of ZP,b is a polynomial in the variables γi, αi, and for all D ∈D(B)wemaywrite P n ∈Z γi(D)=biD and αi(D)= r=0 cr,i[D, ar]. Therefore it follows that for every ν P,b

47 0 ∈ 0 0 ∈{ } there are bj B and aj,1,...,aj,mj a0,...,an such that

X 0 0 0 0 ν(D)= bj[[...[[D, aj,1],aj,2],...],aj,mj ] j

for all D ∈D(B)(ifmj = 0 we deﬁne the iterated commutator to be equal to D). With this we can now prove the following useful lemma:

Lemma 2.3.1 Let B be a commutative k-algebra, and let b ∈ B, P ∈ B[T ] be such that 0 P (b)=0and P (b) is an invertible element of B.LetZP,b be deﬁned as above and let

ν ∈ZP,b. Assume that A is a subring of B which contains all the coeﬃcients of the

polynomial P . Then there are ﬁnitely many elements bi ∈ B, ai ∈ A such that

X ν(D)= biDai i

for all D ∈D(B).

Proof: This follows simply by expanding the commutators in the expression preceding the lemma. 2

We can now state the following theorem, which in certain sense tells us how move an element b, as above, to the left of every diﬀerential operator on B.

Theorem 2.3.2 Let B be a commutative k-algebra, and let b ∈ B and P ∈ B[T ] be such that P (b)=0and P 0(b) is an invertible element of B. Then there are endomorphisms ∈Jd ⊂Z νd P,b P,b for d =1, 2,...such that

X∞ Db = bD + νd(D) d=1

for all D ∈D(B).

48 Moreover, a family of such endomorphisms νd may be computed using the following

formulas:   1 σ1 if d =1 νd =   d − d−1 σ1 σ1 if d>1

≥ ≥ d where for integers d i 1, the elements σi are deﬁned by the recursive formulas

1 σ1 = α0 Xd d+1 d d−(i−1) ≥ σ1 = α0 + σi (γi+1σ1 + αi)ifd 1 i=1 d+1 d d+1 ≥ σi+1 = σi σ1 if d 1andi =1,...,d

The proof of this theorem is elementary, but it uses some rather involved inductions, and is highly unilluminating in general. Hence it will be deferred until the next section. Here we only point out the following fact:

d d Remark From the deﬁnition of the γisandαisitisobviousthatγi(D (B)) ⊂D(B) d d−1 −1 and αi(D (B)) ⊂D (B) for all i =0,...,nand all integers d ≥ 0(weletD (B):=0). d d−1 Since JP,b is generated by the elements αi, it thus follows that ν(D (B)) ⊂D (B) for d all ν ∈JP,b and all integers d ≥ 0. This implies that if D ∈D (B) for some d ≥ 0, then P ∈Jd0 0 ∞ ν(D) = 0 whenever ν P,b and d >d. Therefore the sum d=1 νd(D) is well deﬁned for every D ∈D(B).

The above also implies that if we are given any D ∈D(B), say of order ≤ d,then

we may write Db = bD + ν(D)whereν = ν1 + ...+ νd ∈ZP,b. From this we get the following consequence of theorem 2.3.2:

Lemma 2.3.3 Let B be a commutative k-algebra, and let b ∈ B and P ∈ B[T ] be such that P (b)=0and P 0(b) is an invertible element of B.LetA be a subring of B which contains all the coeﬃcients of the polynomial P . Then for all D ∈D(B) and all integers

49 j ≥ 0 there are ﬁnitely many elements bi ∈ B, ai ∈ A such that

X j Db = biDai i

Proof: Fix an element D ∈D(B), say of order ≤ d, and an integer j ≥ 0. By theorem 2.3.2 we have that Db = bD + ν(D)whereν = ν1 + ...+ νd. We recall that ν is B-bilinear and hence by using the formula for Db twice, we get that

Db2 = bDb + ν(Db)=b(bD + ν(D)) + ν(bD + ν(D))

= b2D +2bν(D)+ν2(D)

An induction then shows that

Xj j j j−l l Db = l b ν (D). l=0

and the result follows by applying lemma 2.3.1 to all the endomorphisms νl. 2

We end this section by showing how it is possible to use theorem 2.3.2 to prove an older result on invertible elements (cf. [8, 20]): Let B be a commutative k-algebra and let s be an invertible element of B.Thens−1 is a root of the polynomial P = sT − 1 ∈ B[T ] and P 0(s−1)=s is invertible. Thus theorem 2.3.2 applies.

First we compute αi for i ≥ 0. We have that deg P = 1, so by our deﬁnitions we have for all D ∈D(B)that

− 0 −1 −1 0 −1 0−i − 1 −1 1−i αi(D)= P (s ) i (s ) [D, 1] + i (s ) [D, s]) − −1 1 i−1 = s i s [D, s]

50   −  −s 2[D, s]ifi =0  −1 =  −s [D, s]ifi =1    0ifi ≥ 2

d ≥ Next we compute the σi s. Since γi =0=αi if i 2, the recursive formulas become

1 σ1 = α0 d+1 d ≥ σ1 = α0 + σ1 α1 if d 1 d+1 d d+1 ≥ ≥ σi+1 = σi σ1 if i 1andd 1 and then an easy induction on the ﬁrst two formulas shows that

dX−1 d i σ1 = α0(α1) i=0 for all d ≥ 1. Thus from the deﬁnition of the νds it follows directly that

d−1 νd = α0(α1) for all d ≥ 1. That is, for all d ≥ 1andallD ∈D(B)wehavethat

− d −d−1 νd(D)=( 1) s [D, s](d)

where [D, s](d) := [...[[D, s],s],...,s] is the commutator repeated d times. Having this, theorem 2.3.2 immediately implies the following

Proposition 2.3.4 If B is a commutative k-algebra and s is an invertible element of B, then for all D ∈D(B) we have the formula

X∞ −1 −1 − d −d−1 Ds = s D + ( 1) s [D, s](d) d=1

51 ≥ where for all d 1, [D, s](d) := [...[[D, s],s],...,s] is the commutator repeated d times. Especially, if we apply both sides of this equation to 1 ∈ B, we get that

X∞ ∗ −1 −1 ∗ − d −d−1 ∗ D s = s D 1+ ( 1) s [D, s](d) 1 d=1

As before we note that [D, s](d) =0ifd>ord(D), so the inﬁnite sums in the proposition are well deﬁned. This proposition has also been proved in [8, 20].

2.3.2 Proof of theorem 2.3.2

Let k be a commutative ring and let B be a commutative k-algebra. It is clear by deﬁnition that D(B), the ring of (k-linear) diﬀerential operators on B,isaB-B-bimodule D in a natural way. We let EndB (B)B denote the space of B-B-linear endomorphism on D D (B). It is clear that EndB (B)B is a k-algebra in a natural way. Since B is commutative, the formulas

ρ(b)(D):=Db λ(b)(D):=bD

∈ ∈D → D for b B and D (B) deﬁne two maps ρ, λ : B EndB (B)B. It is trivial to check that ρ and λ are in fact injective morphisms of k-algebras. Therefore the morphisms ρ D and λ are in fact two canonical ways of embedding B into EndB (B)B.Wewantto study the interplay of these two embeddings and hence we make the following deﬁnition:

A D Deﬁnition We let (B) denote the smallest k-subalgebra of EndB (B)B containing A D the images of ρ and λ.Thatis, (B)isthek-subalgebra of EndB (B)B generated by the set {ρ(b),λ(b)|b ∈ B}.

We get immediately:

Lemma 2.3.5 Let B be a commutative k-algebra. Then

52 (i) A(B) is commutative, (ii) D(B) is a (left) A(B)-module via the natural action, and (iii) for every d ≥ 0, Dd(B) is a A(B)-submodule of D(B).

Remark In light of the lemma we will sometimes write σD instead of σ(D) for elements σ ∈A(B)andD ∈D(B). Moreover, if there is no confusion about the underlying k-algebra B, we will from now on write A, D and Dd instead of A(B), D(B)andDd(B).

In general, elements of D do not commute with elements of B and hence are ρ(b)and λ(b) usually diﬀerent elements of A. We want to study the diﬀerence of them and thus we make

Deﬁnition Let B be a commutative k-algebra. We let

ad := ρ − λ

and we let I be the ideal of A generated by the set {ad(b)|b ∈ B}.

We observe that for all b ∈ B,alld ≥ 0andallD ∈Dd,wehavethat

ad(b)(D)=ρ(b)(D) − λ(b)(D)=Db − bD =[D, b] ∈Dd−1

(we recall that D−1 = 0 by deﬁnition). It follows immediately that IDd ⊂Dd−1 for all d ≥ 0, and a trivial induction then implies that

InDn+d ⊂Dd

for all integers n ≥ 0andd ≥−1.

We will now establish some technical lemmas concerned with computations in A, starting with:

53 Lemma 2.3.6 Let B be a commutative k-algebra, and let b ∈ B. (i) For all integers r ≥ 0 we have a formula

rX−1 ρ(br)=λ(br)+ λ(bi)ad(b)ρ(br−1−i) i=0 P n r ∈ (ii) Assume that b is a root of the polynomial P = r=0 arT B[T ]. Then

Xn rX−1 Xn i r−1−i r 0= λ(arb )ad(b)ρ(b )+ ad(ar)ρ(b ) r=0 i=0 r=0

Proof: (i) We prove this by an induction on r.Forr = 0 the formula reads ρ(1) = λ(1) so there is nothing to prove. For higher powers we have

ρ(br+1)=ρ(b)ρ(br) ! rX−1 ind.=hyp. ρ(b) λ(br)+ λ(bi)ad(b)ρ(br−1−i) i=0 rX−1 =(ad(b)+λ(b)) λ(br)+ λ(bi)ad(b)ρ(br−i) i=0 rX−1 = λ(br+1)+ λ(bi)ad(b)ρ(br−i)+λ(br)ad(b) i=0 Xr = λ(br+1)+ λ(bi)ad(b)ρ(br−i) i=0

(ii) By assumption P (b) = 0 and hence is

0=ρ(P (b)) Xn r = ρ(arb ) r=0 Xn r = λ(ar)+ad(ar) ρ(b ) r=0 Xn Xn r r = λ(ar)ρ(b )+ ad(ar)ρ(b ) r=0 r=0

54 By part (i) of the lemma we have that

! Xn Xn rX−1 r r i r−1−i λ(ar)ρ(b )= λ(ar) λ(b )+ λ(b )ad(b)ρ(b ) r=0 r=0 i=0 Xn Xn rX−1 r i r−1−i = λ(arb )+ λ(arb )ad(b)ρ(b ) r=0 r=0 i=0 Xn rX−1 i r−1−i = λ(P (b)) + λ(arb )ad(b)ρ(b ) r=0 i=0 Xn rX−1 i r−1−i = λ(arb )ad(b)ρ(b ) r=0 i=0 and by combining these two formulas the proof is complete. 2

Before we can state our next lemma we need to introduce some additional notation. d For all integers d ≥ 0 we deﬁne equivalence relations ≡ on A as follows:

d σ ≡ σ0 ⇔ σ − σ0 ∈Id

d The careful reader notices of course that ≡ is just the same equivalence relation as “congruence modulo Id” but for our purposes this somewhat diﬀerent notation is more convenient. We observe immediately the following: Let d, d0 ≥ 0 be integers and let 0 d 0 d+d σ, σ0 ∈Abe such that σ ≡ σ0. Then for all τ ∈Id , στ ≡ σ0τ. With this notation in hand we can now state:

Lemma 2.3.7 Let d ≥ 0 be an integer and let σ ∈A, b ∈ B be such that

d ρ(b) ≡ λ(b)+σ

Then for all integers r ≥ 0 we have that

Xr d ρ(br) ≡ λ(br)+ λ(br−i)σρ(bi−1) i=1

55 d Proof: For r = 0 the formula reads ρ(1) ≡ λ(1), a triviality and for r = 1 the implication of the lemma becomes just our assumption. Then we proceed by induction on r:

r+1 r ρ(b )=ρ(b)ρ (b ) ! Xr d ≡ ρ(b) λ(br)+ λ(br−i)σρ(bi−1) i=1 Xr = ρ(b)λ(br)+ λ(br−i)σρ(bi) i=1 Xr d ≡ (λ(b)+σ) λ(br)+ λ(br−i)σρ(bi) i=1 Xr = λ(br+1)+λ(br)σ + λ(br−i)σρ(bi) i=1 rX+1 = λ(br+1)+ λ(br+1−i)σρ(bi−1) i=1 where the last equality we get by a change of variables. 2

P ∈ n r ∈ Let B be a commutative k-algebra, and let b B and P = r=0 arT B[T ]besuch that P (b)=0andP 0(b) is an invertible element of B. As in section 2.3.1 we can then define the endomorphisms γi and αi, the algebra ZP,b and the ideal JP,b of ZP,b.Itis clear from the definitions that ZP,b is a subalgebra of A and that JP,b ⊂I. It follows of course that J n Dn+d ⊂Dd P,b (B) (B) for all integers n ≥ 0andd ≥−1. We now recall the recursive definitions from the statement of theorem 2.3.2:

1 σ1 = α0 Xd d+1 d d−(i−1) ≥ σ1 = α0 + σi (γi+1σ1 + αi)ifd 1 i=1 d+1 d d+1 ≥ σi+1 = σi σ1 if d 1andi =1,...,d

56 d Z It is clear from these deﬁnitions that all the endomorphisms σi are elements of P,b.But we have more, namely:

P ∈ n r ∈ Proposition 2.3.8 Let b B be a root of the polynomial P = r=0 arT B[T ] such P 0 n r−1 d that P (b)= r=0 rarb is an invertible element in B. Let the elements σi be deﬁned as above. Then: ≤ ≤ d ∈Ji (i) For all integers i, d such that 1 i d we have that σi P,b and

d+1 − d ∈Jd+1 σi σi P,b ;

(ii) For all integers s, d ≥ 0 we have a formula

Xd s d≡+1 s s s−i d ρ(b ) λ(b )+ i λ(b )σi ; i=1

d ∈Ji Proof: (i) First we prove that σi P,b. We do this by a double induction, noting that the case i = d = 1 follows right from the deﬁnitions. We assume that the statement holds 0 ≤ ≤ 0 ≤ 0 d−(j−1) ∈J for all d d and all 1 i d . From that it follows directly that γj+1σ1 +αj P,b for all j such that 1 ≤ j ≤ d and then we have by the deﬁnition that

Xd d+1 d d−(j−1) ∈J σ1 = α0 + σj (γj+1σ1 + αj) P,b j=1

Then we use the induction hypothesis again to get that

d+1 d d+1 ∈Ji J J i+1 σi+1 = σi σ1 P,b P,b = P,b for all 1 ≤ i ≤ d. d+1 − d ∈Jd+1 Next we prove that σi σi P,b . We also do that by a double induction.

57 First of all, for d = i = 1 we get from the recursive formulas that

2 − 1 1 1 − σ1 σ1 = α0 + σ1 (γ2σ1 + α1) α0 ∈J2 = α0(γ2α0 + α1) P,b

since α0 and α1 are in JP,b. We now assume that d>1 and that we have proved the statement for all d0

Xd d+1 − d d d−(j−1) σ1 σ1 = α0 + σj (γj+1σ1 + αj) j=1 dX−1 − − d−1 d−1−(j−1) α0 σj (γj+1σ1 + αj) j=1 d 1 = σd (γd+1σ1 + αd) dX−1 d d−j+1 − d−1 d−j d − d−1 + γj+1(σj σ1 σj σ1 )+(σj σj )αj j=1

Here we pause to observe that we may write

d d−j+1 − d−1 d−j d d−j+1 − d d−j d d−j − d−1 d−j σj σ1 σj σ1 = σj σ1 σj σ1 + σj σ1 σj σ1 d d−j+1 − d−j d − d−1 d−j = σj (σ1 σ1 )+(σj σj )σ1

Thus, by putting this into the previous equation we get that

dX−1 d+1 − d d d d−j+1 − d−j σ1 σ1 = σd(γd+1α0 + αd)+ γj+1σj (σ1 σ1 ) j=1 dX−1 dX−1 d − d−1 d−j d − d−1 + γj+1(σj σj )σ1 + (σj σj )αj j=1 j=1

d ∈Jj ≤ ≤ We have already shown that σj P,b for all integers j such that 1 j d.Wealso recall that αj ∈JP,b for all j ≥ 0. We use this and our induction hypothesis to conclude

58 that d ∈Jd J J d+1 σd(γd+1α0 + αd) P,b P,b = P,b d d−j+1 − d−j ∈Jj J d−j+1 J d+1 γj+1σj (σ1 σ1 ) P,b P,b = P,b d − d−1 d−j ∈Jd J J d+1 γj+1(σj σj )σ1 P,b P,b = P,b d − d−1 ∈Jd J J d+1 (σj σj )αj P,b P,b = P,b for all j such that 1 ≤ j ≤ d − 1. Then it follows directly from the equation above that d+1 − d ∈Jd+1 σ1 σ1 P,b . In order to ﬁnish the induction we must show the lemma holds for all i such that 1 1and1

d+1 − d d d+1 − d−1 d σi σi = σi−1σ1 σi−1 σ1 d d+1 − d d d d − d−1 d = σi−1σ1 σi−1σ1 + σi−1σ1 σi−1 σ1 d d+1 − d d − d−1 d = σi−1(σ1 σ1 )+(σi−1 σi−1 )σ1 ∈Ji−1J d+1 J d J P,b P,b + P,b P,b J d+1 = P,b

This proves part (i) of the proposition. (ii) We observe that the formula makes sense for all s, d ≥ 0. This is because σd is i ≤ ≤ s deﬁned for all 1 i d and i = 0 whenever s

for all d ≥ 0, a triviality since ρ(1) = λ(1)=1.Alsoifd = 0, then the statement reads

1 ρ(bs) ≡ λ(bs)

for all s ≥ 0, again a triviality which follows right from the deﬁnitions. Having taken care of all the trivial cases, we will do the main body of the proof by

59 an induction on d. We assume that the formula holds for all integers d0 ≤ d and we prove it for d +1.

First we recall that ad(b)andad(ar)areallinI, so by lemma 2.3.6 and the induction hypothesis (on d0 = d)wehavethat

Xn rX−1 Xn i r−1−i r 0= λ(arb )ad(b)ρ(b )+ ad(ar)ρ(b ) r=0 i=0  r=0  Xn rX−1 Xd d≡+2 i  r−1−i r−1−i r−1−i−j d λ(arb )ad(b) λ(b )+ j λ(b )σj r=0 i=0 j=1 ! Xn Xd r r r−i d + ad(ar) λ(b )+ i λ(b )σi r=0 i=1

We rewrite this last expression, using the elementary fact that

rX−1 rX−1 r−1−i i r s = s = s+1 i=0 i=0 and the deﬁnitions of γi and αi,toobtain ! ! Xn rX−1 Xn Xd rX−1 d≡+2 r−1 r−1−i r−1−j d 0 λ(arb ) ad(b)+ j λ(arb )ad(b)σj r=0 i=0 r=0 j=1 i=0 ! Xn Xd Xn r r r−i d + λ(b )ad(ar)+ i λ(b )ad(ar) σi r=0 i=1 r =0 ! Xn Xd Xn r−1 r r−(j+1) d = λ( rarb )ad(b)+ λ( j+1 arb )σj ad(b) r=0 j=1 r=0 ! Xn Xd Xn r r r−i d + λ(b )ad(ar)+ i λ(b )ad(ar) σi r=0 i=1 r=0 Xd 0 − 0 d = λ(P (b))ad(b)+ λ( P (b))γj+1σj ad(b) j=1 Xd − 0 − 0 d λ(P (b))α0 λ(P (b)) αiσi i=1

If we now multiply trough this equation by λ(P 0(b)−1)andwriteρ(b) − λ(b) for ad(b),

60 we get that Xd Xd d≡+2 − − d − − d 0 ρ(b) λ(b) γj+1σj ad(b) α0 αiσi j=1 i=1 or Xd Xd d≡+2 d d ρ(b) λ(b)+ γj+1ad(b)σj + α0 + αiσi j=1 i=1

d ∈Jj By part (i) of the proposition we have that σj P,b for all j =1,...,d. Then, since ad(b) ∈I, we can apply the induction hypothesis again (on d0 = d +1−j for j =1,...,d and s = 1) and we have

d d − d ad(b)σj = ρ(b)σj λ(b)σj d+1X−j d≡+2 d 1 1−i d+1−j d − d λ(b)σj + i λ(b )σi σj λ(b)σj i=1 d+1−j d = σ1 σj for all j =1,...,d. If we combine this with the formula we had before, we then have that

Xd Xd d≡+2 d+1−j d d ρ(b) λ(b)+ γj+1σ1 σj + α0 + αiσi j=1 i=1 Xd d d−(i−1) = λ(b)+α0 + σi (γi+1σ1 + αi) i=1 d+1 = λ(b)+σ1 where Xd d+1 d d−(i−1) σ1 = α0 + σi (γi+1σ1 + αi) i=1 This proves the statement we want to prove in the special case when s = 1. For higher powers on b we use lemma 2.3.7 to get the formula

Xs s d≡+2 s s−i d+1 i−1 ρ(b ) λ(b )+ λ(b )σ1 ρ(b ) i=1

61 d+1 ∈J Then we recall that σ1 P,b and thus we can use the induction hypothesis still once more (on d0 = d)togettheequations

Xd d+1 i−1 d≡+2 i−1 d+1 i−1 i−1−j d d+1 σ1 ρ(b ) λ(b )σ1 + j λ(b )σj σ1 j=1 for i =1,...,s. If we put this into the formula above, we ﬁnally get that   Xs Xd s d≡+2 s s−i  i−1 d+1 i−1 i−1−j d d+1 ρ(b ) λ(b )+ λ(b ) λ(b )σ1 + j λ(b )σj σ1 i=1 j=1 ! Xs Xd Xs s s−1 d+1 i−1 s−1−j d d+1 = λ(b )+ λ(b )σ1 + j λ(b )σj σ1 i=1 j=1 i=1 Xd s s s−1 d+1 s s−(j+1) d d+1 = λ(b )+ 1 λ(b )σ1 + j+1 λ(b )σj σ1 j=1 dX+1 s s s−j d+1 = λ(b )+ j λ(b )σj j=1

d+1 d d+1 2 where σj+1 = σj σ1 for j =1,...,d

This proposition contains in fact everything we need in order to prove theorem 2.3.2:

{ }∞ Proof of theorem 2.3.2: It suﬃces to demonstrate that the family νd d=1, deﬁned by the formulas   1 σ1 if d =1 νd =   d − d−1 σ1 σ1 if d>1 does satisfy the requirements of the theorem. 1 ∈J By part (i) of proposition 2.3.8 we have that ν1 = σ1 P,b and that

d − d−1 ∈Jd νd = σ1 σ1 P,b for all d>1.

62 0 ≥ ∈Dd0 J d+1 Dd Nowﬁxanintegerd 0, and an element D (B). Since P,b annihilates (B) 0 for all d ≥ 0, it follows that νd(D) = 0 for all d>d. Hence to ﬁnish the proof we must show that Xd0 Db = bD + νd(D). d=1

But now by the deﬁnition of νd we have immediately that

Xd0 d0 νd = σ1 d=1 so by applying part (ii) of proposition 2.3.8 to d = d0 and s = 1 we have that

0 d≡+1 d0 ρ(b) λ(b)+σ1 Xd0 = λ(b)+ νd d=1

0 P d≡+1 − − d0 ∈Jd0+1 By deﬁnition of the symbol this implies that ρ(b) λ(b) d=1 νd P,b .Since 0 P J d +1 Dd0 − − d0 2 P,b annihilates the space (B), it follows that Db bD d=1 νd(D)=0.

2.3.3 Right ideals and ´etale homomorphisms

n ≥ Let A be a commutative k-algebra such that PA is finitely presented for all integers n 0. Let f : A → B be an étale homomorphism of k-algebras. Then theorem 2.2.10 gives an ∼ isomorphism B ⊗A D(A) = D(B)ofB-D(A)-bimodules. This implies immediately that if I is a left ideal of D(A), then B ⊗A I is a left ideal of D(B). A similar argument is obviously not available to us if I is a right ideal of D(A) (it need not even be a left A- submodule). However, if a right ideal I of D(A)isatthesametimealeftA-submodule, we will in this section show that B ⊗A I is indeed a right ideal of D(B). This applies especially to twosided ideals of D(A). We start with a lemma and proposition which proves this claim for localizations and local étale morphisms:

63 Lemma 2.3.9 Let A be a commutative k-algebra and let P ∈ A[T ] be a monic polynomial. Let B be a localization of the k-algebra A[T ]/P such that P 0(T ) is invertible in B. Let f : A → B be the composite A → A[T ] → A[T ]/P → B. Then f is ´etale, and for all

b ∈ B and D ∈D(A) there are bi ∈ B and ai ∈ A such that

X f(D)b = bif(Dai) i

→ 0 Proof: By [16, Example 3.4] we have that A (A[T ]/P )P 0(T ) =: B is étale (and thus formally étale). The algebra B is a localization of B0, and then the results of section 2.1.2 imply that f : A → B is étale. By theorem 2.2.5, f then extends to a homomorphism f : D(A) →D(B) of filtered rings. For convenience of notation, let us write C := A[T ]/P ,lett ∈ C be the image of T ∈ A[T ], and let λ : C → B be the localization map. Let us now fix elements D ∈D(A)andb ∈ B. Having set the notation as above, we may write b = λ(c0)λ(s)−1 with c0,s ∈ C. Then by expanding the commutators in

proposition 2.3.4 it follows that there are bi ∈ B and integers ni ≥ 0 such that

X 0 −1 0 ni f(D)b =(f(D)λ(c ))λ(s) = bi(f(D)λ(c ))λ(s) i X 0 ni = bif(D)λ(c s ) i

This implies that we may assume that b = λ(c) for some c ∈ C.SinceA[T ] → C is surjective, and t is the image of T under this map, it follows that we may write c = Q(t) P 0 i ∈ for some Q = i aiT A[T ]. Then

X 0 i f(D)λ(c)= f(Dai)λ(t ) i

and therefore we may assume that b = λ(c)=λ(ti) for some integer i ≥ 0. Finally we

64 observe that by our assumptions P (λ(t)) = 0 and P 0(λ(t)) is invertible in B,andallthe coeﬃcients of the polynomial P are in f(A). Hence lemma 2.3.3 applies and we conclude that X X i f(D)λ(t )= bjf(D)f(aj)= bjf(Daj) j j

for some bj ∈ B, aj ∈ A. 2

Remark Taking P = T ∈ B[T ] in this lemma shows especially that it holds when B is any localization of A. Furthermore, theorem 2.1.4 implies that the lemma holds also whenever f : A → B is a local ´etale homomorphism of local rings.

n Proposition 2.3.10 Let A be a commutative k-algebra such that PA is ﬁnitely presented for all integers n ≥ 0, and assume that f : A → B is either a localization of k-algebras, or a local ´etale homomorphism of local k-algebras. Let J bealeftA-submodule of D(A).

If J is a right ideal of D(A), then B ⊗A J is a right ideal of D(B).

Proof: By definition, the left B-submodule B ⊗A J of D(B) is generated by the elements ∼ f(D), D ∈ J, and since f is étale, we have by theorem 2.2.10 that B ⊗A D(A) = D(B) via b⊗D 7→ bf(D). Since we are assuming that J is a right ideal of D(A), it therefore suffices to show that f(D)b ∈ B ⊗A J if D ∈D(A)andb ∈ B. However by lemma 2.3.9 and the remark following it, we have that

X f(D)b = bif(Dai) i

for some bi ∈ B and ai ∈ A.EveryDai ∈ J by assumption and we are done. 2

Next we prove a lemma which shows that in order to determine if a left A-submodule of D(A) is a right ideal, it is necessary and suﬃcient to do so locally:

65 n Lemma 2.3.11 Let A be a commutative k-algebra such that PA is ﬁnitely presented for all integers n ≥ 0, and let J be a left A-submodule of D(A). Then J is a right ideal of

D(A) if and only if Jp(= Ap⊗J) is a right ideal of D(Ap) for all p ∈ Spec A.

Proof: The direction “⇒” follows directly from proposition 2.3.10.

For the other direction assume that Jp is a right ideal of D(Ap) for all p ∈ Spec A.

For p ∈ Spec A,letλp : D(A) →D(Ap) denote the localization map. By deﬁnition we have that λp(J) ⊂ Jp,andifD ∈D(A) is such that λp(D) ∈ Jp for all p ∈ Spec A, then [14, theorem 4.6] implies that D ∈ J (the localizations are just localizations as left

A-modules). But all the λps are ring homomorphisms, so if Jp is a right ideal for all p, we have for D ∈ J and D0 ∈D(A)that

0 0 λp(DD )=λp(D)λp(D ) ∈ JpD(Ap) ⊂ Jp, so DD0 ∈ J. 2

We can now prove the theorem we have been aiming at in this section:

n Theorem 2.3.12 Let A be a commutative k-algebra such that PA is ﬁnitely presented for all integers n ≥ 0, and let f : A → B be an ´etale homomorphism of k-algebras. Let

J bealeftA-submodule of D(A).IfJ is a right ideal of D(A), then B ⊗A J isaright ideal of D(B).

66 Proof: Fix a P ∈ Spec B and let p = f −1(P ) ∈ Spec A.Thenf induces a local ´etale morphism Ap → BP , and we get a commutative diagram

D(A) f - D(B)

? ? - D(Ap) D(BP )

⊗ ⊗ It follows that (B A J)P = BP Ap Jp, and by applying proposition 2.3.10 twice, we thus have that (B ⊗A J)P is a right ideal of D(BP ). This holds for all P ∈ Spec B and then lemma 2.3.11 implies that B ⊗A J is a right ideal of D(B). 2

Left ideals of D(A) are automatically left A-submodules, and thus the theorem implies the following result which we state as a corollary:

n Corollary 2.3.13 Let A be a commutative k-algebra such that PA is ﬁnitely presented for all integers n ≥ 0, and let f : A → B be an ´etale homomorphism of k-algebras. If J is a twosided ideal of D(A), then B ⊗A J is a twosided ideal of D(B).

67 Chapter 3

Minimal twosided ideals in rings of diﬀerential operators

Let k be an algebraically closed field of characteristic zero, and let A be a finitely generated k-algebra and a domain of krull dimension 1. In this case it was proved in [25] that A contains a minimal nonzero D(A)-module J,andthatD(A) contains a minimal nonzero twosided ideal J . The same result, in a slightly more general setting but with a similar proof, was independently obtained in [17]. In [25] it was moreover shown that the result above also holds if A is replaced by its localization at a singular maximal ideal. The method employed in [17, 25] was to compare D(A) to the ring of differential operators on the normalization of A and use the fact that in this case the normalization is a regular ring. In this chapter we prove some generalizations of the results described above, notably to certain reduced rings of krull dimension more than 1. If the krull dimension of A is more than 1, the method described above can no longer be used, partly because the normalization need not be regular. Instead, for a commutative, noetherian and reduced k-algebra A, we use supports of regular functions on Spec A to construct a canonical

D(A)-submodule of A, called ΓA. Then if the irreducible components of A have rings of diﬀerential operators that are simple, we prove that statements analogous to those

68 of [17, 25] hold for the module ΓA and the ideal D(A, ΓA). In many cases this method gives no new information, e.g. when A is a domain. However, if A is as before and it has an étale covering B which is a componentwise geometrical unibranch(we define this in the chapter), we can construct the D(B)-module ΓB as above, and we use the descent theory from chapter 2 to construct a canonical D(A)-submodule χA of A such ∼ that ΓB = B ⊗A χA. If in addition A has finitely presented modules of differentials, and the irreducible components of B have rings of differential operators that are simple, we show that χA is the minimal essential left D(A)-submodule of A and that D(A, χA)is the minimal essential twosided ideal of D(A). Hence these are proper analogues of the module J and the ideal J mentioned at the beginning. Our method applies especially to local reduced algebras of krull dimension 1. Hence apart from being generalizations to rings of higher krull dimension, the results of this chapter show how the theorems of [17, 25] reflect the (rather natural) fact that differential operators preserve supports of functions. We should point out that from [1] it follows that the ring of differential operators on the cubic cone has an infinite descending chain of twosided ideals (cf. [4]). Hence one can not hope for a generalization of the main results of this chapter which holds for all reduced finitely generated algebras over algebraically closed fields of characteristic zero. Finally we note that rings of differential operators over commutative rings with zero divisors have been studied in [19], but there the theory is approached in a somewhat different way and results of different type are obtained.

3.1 Supports in reduced commutative rings

3.1.1 Associated prime ideals

The following deﬁnition appears in [14].

69 Deﬁnition Let A be a commutative ring and let p ⊂ A be a prime ideal. (i) p is said to be associated if p = Ann(x) for some nonzero x ∈ A. (ii) An associated prime of A is embedded if it strictly contains some other associated prime of A. Otherwise it is said to be isolated.

We let Ass(A) denote the set of associated prime ideals of A. We list a few useful elementary facts about associated primes in noetherian rings:

Lemma 3.1.1 Let A be a commutative, noetherian ring. Then: (i) the set Ass(A) is ﬁnite; (ii) the ideal \ p p∈Ass(A) is equal to the set of all nilpotent elements of A; (iii) the set [ p p∈Ass(A) is equal to the set of all zero-divisors of A. (iv) If A is also reduced, then it has no embedded primes.

Proof: The ﬁrst three parts are implied by [14, Theorems 1.2, 6.1, 6.5], and the last past is an easy exercise. 2

For a commutative noetherian ring A, we let Min(A) denote the set of minimal prime ideals of A, ordered by inclusion. It is not hard to see that Min(A) ⊂ Ass(A) whenever A is a noetherian commutative ring (cf. [14, Theorem 6.5]). Thus it follows from part 4 of last lemma that Min(A)=Ass(A) for a reduced noetherian commutative ring. In the sequel we will mostly be working with reduced rings and we will use this equality without further comment. We need another deﬁnition from [14]:

70 Deﬁnition If A is a commutative ring and I ⊂ A is an ideal, let

n ΓI = {a ∈ A; I a = 0 for some n>0}

Let A be a commutative ring and let I be an ideal of A. From algebraic geometry we recall the deﬁnition of the variety corresponding to I, V (I):={p ∈ Spec A|I ⊂ p}.Let

n a ∈ ΓI .Ifp 6∈ V (I), then there is an x ∈ I \ p such that x a = 0 for some positive integer n. However xp is a unit in Ap, and therefore it follows that ap =0inAp.Conversely assume that a ∈ A is such that ap =0inAp for all p 6∈ V (I). Choose an x ∈ I. Then for all p ∈ Spec A \ V (Ax), we have by assumption that ap = 0, and hence it follows that n a(x) =0inA(x). This implies that x a = 0 for some positive integer n. This shows that if A is noetherian, then ΓI is the ideal of regular functions on Spec A that are supported on V (I).

For a prime ideal p ∈ Spec A, it follows right from the deﬁnitions that Γp =0ifand6 only if p ∈ Ass(A). Using lemma 3.1.1 we can characterize Γp when p is an associated prime ideal of a commutative noetherian reduced ring:

Lemma 3.1.2 Let A be a commutative, noetherian and reduced ring, and let p ∈ Ass(A). Then T (i) Γp = q∈Ass(A) q, q=6 p

(ii) p =ΓΓp

n Proof: Let x ∈ Γp.Thenp x = 0 for some integer n. Since all the associated primes are minimal, for all q ∈ Ass(A), with p =6 q, we can ﬁnd an y ∈ p \ q.Thenynx =0∈ q, so x ∈ q.Converselyifx is contained in all associated primes of A except p,thenpx is contained in all the associated primes of A.SinceA is reduced, it follows that px =0.

Hence x ∈ Γp.Thisprovespart(i).

71 Since the intersection of all the associated primes of a reduced ring is zero, part (i) ⊂ ∈ of the lemma implies immediately that p ΓΓp . Conversely assume that x ΓΓp .Then

for each q ∈ Ass(A), q =6 p choose an yq ∈ q \ p (possible since all the associated primes are minimal), and let y be the multiple of these elements. Then by lemma 3.1.2 we have

n that y ∈ Γp, and hence it follows that y x =0∈ p for some integer n ≥ 0. Since y 6∈ p and p is a prime ideal, this implies that x ∈ p. 2

The sum of all the Γp for p ∈ Ass(A), will play an important role in the sequel. Thus we give it a name:

Deﬁnition Let A be a commutative, noetherian and reduced ring. We let

X ΓA := Γp p∈Ass(A)

By using the basic properties of varieties and some elementary set theory, it follows easily that

\ [ V (ΓA)= V (q) p∈Ass(A) q∈Ass(A) 6 [ q=p = V (p) ∩ V (q) p,q∈Ass(A) q=6 p

That is, the variety corresponding to ΓA is the closed subset of Spec A consisting of all points that lie in at least two diﬀerent irreducible components of Spec A. We will now

give some descriptions of ΓA. First we have:

Lemma 3.1.3 Let A be a commutative, noetherian and reduced ring. Then ΓA = L p∈Ass(A) Γp and it contains a regular element of A.

72 P Proof: Assume that xp ∈ Γp are elements such that xp =0.Chooseap ∈ Ass(A). P − Then xp = q=6 p xq, and by lemma 3.1.2 we have that the right hand side of this equation lies in p.Thereforewehavethatxp ∈ p ∩ Γp = 0. This shows that ΓA is equal to the direct sum of all the Γp for p ∈ Ass(A).

It is clear from the deﬁnition that Γp =6 0 for all p ∈ Ass(A). Thus we may choose P 0 =6 xp ∈ Γp for all p ∈ Ass(A). Let x = xp. From lemma 3.1.2 it then follows that x is not contained in any associated prime ideal of A, and then lemma 3.1.1 implies that x is regular. 2

Let A be a commutative, noetherian and reduced ring and write Ass(A)={p1,...,pn} (we recall that this is also the set of minimal prime ideals of A). For i =1,...,n we Q → ¯ n → ¯ let Ai := A/pi, πi : A Ai the natural projections, A := i=1 Ai and λi : Ai A the natural injections. Finally we let ι : A → A¯ be the map a 7→ (πi(a)), and for all i we let ei := λi(1). It is clear that each ei is idempotent, eiej = 0 whenever i =6 j and that P n i=1 ei =1. T n Since A isreduced,wehavethat i=1 pi = 0 and hence it follows that ι is injective. WemaythusviewA as a subring of A¯ via this map. Then K, the total ring of fractions Q ¯ n of A, is also the total ring of fractions of A and it is equal to the direct product i=1 Ki where Ki is the field of fractions of Ai, i =1,...,n. Nowfixaninteger1≤ i ≤ n. From the definitions and identifications above, and lemma 3.1.2 it then follows that Γpi =(0,...,0,πi(Γpi ), 0,...,0), which is clearly an ideal of A¯. Hence under the identification A ⊂ A¯ we may identify ΓA with the ideal

(π1(Γp1 ),...,πn(Γpn )). We have more, namely:

Proposition 3.1.4 Let A be a commutative, noetherian and reduced ring, and let A¯ be as above. Then ΓA is equal to the conductor of A¯ into A.

73 Proof: Since ΓA is an ideal of A¯ and contained in A,itfollowsimmediatelythatitis contained in the conductor. Conversely let a ∈ A be such that aA¯ ⊂ A. Fix an integer

1 ≤ i ≤ n. We have by assumption that aei =(0,...,0,πi(a), 0,...,0) ∈ A,andit 6 ∈ follows that πj(aei) = 0 for all j = i. This, and lemma 3.1.2 then implies that aei Γpi .

This holds for all i and then we have by deﬁnition that a = ae1 + ...+ aen ∈ ΓA. 2

We end this section by giving a certain combinatorial description of ΓA,whichwe need when we discuss ´etale homomorphisms.

Definition Let U be a finite set. A partition of U is a finite family S = {Si} of subsets S ∩ ∅ 6 of U such that i Si = U and Si Sj = whenever i = j.

We will study partitions of the set of associated primes in a reduced noetherian ring. Hencewemake:

Definition Let A be a commutative, noetherian and reduced ring. A partition S = {Si} of the finite set Ass(A)issaidtobefine if p + p0 = A whenever p and p0 lie in the same 0 Si and p =6 p .

Remark First of all note that we allow some of members of a partition to be the empty set. Also we observe that by this deﬁnition, any partition of Ass(A) into singletons and empty sets is automatically ﬁne.

Lemma 3.1.5 Let {I1,...,Ir}, r ≥ 2 be a set of pairwise comaximal ideals of a commutative ring A. Then Xr \r A = ( Iν) ν0=1 ν=1 ν=6 ν0

Proof: We have that

P T \r [r r r V ( ν0=1 ( ν=1 Iν)) = V (Iν) 0 ν=6 ν ν0=1 ν=1 ν=6 ν0

74 [ = (V (Iν) ∩ V (Iν0 )) 1≤ν<ν0≤r = ∅

0 since by our assumption, V (Iν) ∩ V (Iν0 )=∅ whenever ν =6 ν . This proves the lemma. 2

Proposition 3.1.6 Let A be a commutative, noetherian and reduced ring, and let S = { }n Si i=1 be a ﬁne partition of Ass(A). Then

Xn \ ΓA = IS := ( p) i=1 p∈Ass(A)\Si

{ }m Proof: Let T = Tj j=1 be a partition of Ass(A) into singletons. By deﬁnition of ΓA

and lemma 3.1.2 we have that ΓA = IT .Moreover,sinceS is a partition of Ass(A)it

follows that for every j there is an i such that Tj ⊂ Si. From this we get that IT ⊂ IS. We prove the other inclusion. By symmetry it clearly suﬃces to show that

\ p ⊂ IT p∈Ass(A)\S1

If S1 = ∅, the statement holds trivially since the intersection of all the associated primes of a reduced ring is equal to zero. Let us then assume that S1 contains only one element,

say S1 = {q}.SinceT is a partition of Ass(A), there is an j such that q ∈ Tj. It clearly follows that \ \ p ⊂ p ⊂ IT . p∈Ass(A)\{q} p∈Ass(A)\Tj

75 If S1 contains more than one element, we use the assumption that S is a ﬁne partition of Ass(A) and lemma 3.1.5 to write

X \ A = ( q) 0∈ q S1 q∈S1 q=6 q0

andthenwecanwrite

\ X \ \ p = ( q)( p) 0∈ p∈Ass(A)\S1 q S1 q∈S1 p∈Ass(A)\S1 q=6 q0 X \ ⊂ ( p) 0 0 q ∈S1 p∈Ass(A)\{q }

FinallywecanusethecasewhenS1 contains only one element to show that each term

of the last sum is contained in IT . 2

3.1.2 Geometrical unibranches

Recall our deﬁnition of a partition from section 3.1.1. We start this section by proving a lemma on the behavior of minimal prime ideals under ´etale homomorphisms:

Lemma 3.1.7 Let A be a commutative noetherian ring and let A → B be an ´etale homomorphism. For all p ∈ Min(A) let Sp := {P ∈ Spec B|P ∩ A = p}. Then

(i) the family {Sp}p∈Min(A) is a partition of Min(B); (ii) for all p ∈ Min(A) we have that

\ pB = P P ∈Sp

(where the right hand side is deﬁned to be equal to B if Sp = ∅).

76 Proof: (i). Flat local homomorphisms are faithfully ﬂat. Hence if P ∈ Spec B and

p = P ∩ A, then the morphism Spec BP → Spec Ap is surjective. If P ∈ Min(B), this

implies that Ap has only one prime ideal, and thus p must be a minimal prime of A.

Conversely let p be a minimal prime of A.IfSp = ∅ there is nothing to prove (recall that we allow members of a partition to be the empty set). Hence we may assume that p = P ∩ A for some P ∈ Spec B.ThenP contains a minimal prime P 0 of B,and by the minimality of p it follows that p = P 0 ∩ A also. By [9, Ex. II.3.10] we have

that the ﬁber over p is homeomorphic to Spec B ⊗A κ(p), and since A → B is ´etale it

follows from lemma 2.1.3 that B ⊗A κ(p) is a ﬁnite direct product of ﬁelds, and hence of krull dimension 0. This implies that P 0 = P and thus P ∈ Min(B). This shows that

Sp ⊂ Min(B). (ii). By base extension, the ring B/pB is an ´etale A/p-algebra. Hence it is reduced by lemma 2.1.6, so the intersection of all the minimal (= associated) primes of B/pB is equal to zero. This means precisely that the intersection of all the minimal elements of the set S := {P ∈ Spec B|p ⊂ P } is equal to pB. Thus it only remains to show that the minimal elements of S are actually minimal primes of B. Assume that p ⊂ P 0 for some P 0 ∈ Spec B.Thenp ⊂ p0 = P 0 ∩ A and since the going-down theorem holds for ﬂat homomorphisms, we have that there is a P ∈ Spec B such that P ⊂ P 0 and p = P ∩ A. Thus P ∈ S and part (i) of the lemma implies that P ∈ Min(B). 2

Let A be a commutative noetherian local ring and let ι : κ(A) → Ω be a separable closure of κ(A). As in [7, (IV,18.8)], all diagrams of the type

B - Ω 6 6

can. A - κ(A)

77 where A → B is a local étale homomorphism and B → Ω is a local homomorphism, form a direct system (observe that our definition of “local étale homomorphism” coincides with the definition of “essentiellement étale” in [7, (IV, 18.6.1)]).

Deﬁnition Let A be a local noetherian ring. The direct limit of the direct system deﬁned above is called the strict henselization of A, and it is denoted by Ash.

If A is a local noetherian ring, then Ash is also a local noetherian ring and A → Ash is a local homomorphism (cf. [7, (IV, 18.8.8)]). This morphism is faithfully flat since directs limit of flat modules are flat and flat local homomorphisms are faithfully flat. This implies that A → Ash is injective and Spec Ash → Spec A is surjective. Finally [7, (IV, 18.8.13)] implies that Ash is reduced if and only A is reduced.

Remark We should note that the deﬁnition of the strict henselization of the local ring A depends on the choice of the separable closure ι : κ(A) → Ω. Hence if A0 is the strict henselization of A with respect to some other separable closure ι0 : κ(A) → Ω0 of κ(A), then Ash and A0 are not canonically isomorphic. Nevertheless as A-algebras, these strict henselizations are isomorphic (cf. [7, (IV, 18.8.8)]). For our purposes we may ﬁx a separable closure once and for all.

Lemma 3.1.8 Let A be a noetherian and reduced local ring. Every local ´etale A-algebra is isomorphic to an A-subalgebra of Ash, and every element of Ash is contained in a local ´etale A-subalgebra of Ash.

Proof: Fix a separable closure κ(A) → Ω. Let A → B be a local étale homomorphism. Then by lemma 2.1.3, κ(B) is a finite separable field extension of κ(A), and thus there is a κ(A)-linear morphism α : κ(B) → Ω. By [7, (IV,18.8.4)], this morphism corresponds

78 to a unique local homomorphism B → Ω making the diagram

B - Ω 6

A commutative. From the deﬁnition of Ash we thus have a local A-algebra homomorphism B → Ash,andAsh is also the strict henselization of B. Injectivity follows directly. The second claim follows directly from the deﬁnitions of Ash and basic properties of direct limits. 2

Lemma 3.1.9 Let A be a local noetherian ring. (i) A prime ideal P ∈ Spec Ash is in Min(Ash) if and only if P ∩ A is in Min(A). (ii) Let p ∈ Min(A). Then \ pAsh = P P ∈Min(Ash) P ∩A=p

sh Proof: (i). By [7, (IV, 18.8.12)] we have for all p ∈ Spec A that A ⊗Aκ(p) is a finite direct product of fields. Since A → Ash is flat, we can thus proceed precisely as in the proof of part (i) of lemma 3.1.7. (ii). We already pointed out that Ash is reduced if and only A is. Then by using part (i), we can proceed as in the proof of part (ii) of lemma 3.1.7. 2

Remark By [7, IV, corollaire (18.8.14)] we also have that a prime ideal P ∈ Spec Ash is in Ass(Ash) if and only if P ∩ A is in Ass(A). If A is reduced, this fact implies the lemma above.

79 The lemma, combined with the earlier fact that Spec Ash → Spec A is surjective, implies that we have a surjective morphism Min(Ash) → Min(A); P 7→ P ∩ A.Weuse this in the next deﬁnition:

Deﬁnition We say that a noetherian ring A is a componentwise geometrical unibranch ∈ sh → at p Spec A if the the map Min(Ap ) Min(Ap) is injective (and hence bijective). We say that A is a componentwise geometrical unibranch if A is a componentwise geometrical unibranch at all p ∈ Spec A.

Let A be a noetherian local ring and assume that A is irreducible (i.e., that it has only one minimal prime ideal). Then the definition above clearly implies that A is a componentwise geometrical unibranch at its maximal ideal if and only if Ash has only one minimal prime ideal. This shows that for such a ring, the definition of a geometrical unibranch in [7, 21] is similar to our definition of a componentwise geometrical unibranch at the maximal ideal. However:

Important Warning We must note that a local noetherian ring may be a componentwise geometrical unibranch at its maximal ideal without being a componentwise geometrical unibranch (at all points).

We should also point out that the definition above is not quite in line with the usual definition of geometrical unibranches (cf. e.g. [21, définition IX.2] or [7, (0IV , 23.2.1) and (IV, 18.8.15)]), since usually they are irreducible by definition. The next lemma shows though, that being a componentwise geometrical unibranch is in fact a condition on the irreducible components:

Lemma 3.1.10 Let A be a noetherian ring. It is a componentwise geometrical unibranch if and only if all its irreducible components are geometrical unibranches.

Proof: Assume that A is a componentwise geometrical unibranch. Let p be a minimal prime ideal of A and let q ∈ Spec A be such that p ⊂ q. By assumption there is a unique

80 sh ∩ minimal prime ideal of Aq ,sayP , such that p = P Aq. Then lemma 3.1.9 implies that sh sh ∼ sh sh sh P = pAq , and thus we have by [7, (IV, 18.8.10)] that (A/p)q = Aq /pAq = Aq /P . sh This shows that (A/p)q is a domain and hence A/p is a geometrical unibranch at q. This holds for all minimal prime ideals p of A and all prime ideals q containing p,and therefore by deﬁnition all the irreducible components are geometrical unibranches. 0 ∈ sh For the converse let q be a prime ideal of A and assume that P, P Min(Aq )are 0 such that p := P ∩ Aq = P ∩ Aq. By lemma 3.1.9, p is necessarily a minimal prime ideal sh ⊂ ∩ 0 of Aq (which we may take to be a minimal prime ideal of A). Then pAq P P ,and 0 sh sh so P and P are both minimal prime ideals of Aq /pAq which in turn is isomorphic to sh (A/p)q , again by [7, (IV, 18.8.10)]. Since we are assuming that A/p is a geometrical sh sh 0 2 unibranch, this implies that Aq /pAq is a domain, and hence P = P .

Next lemma (cf. [21, corollaire IX.1, définition IX.2]) gives a different description of geometrical unibranches:

Lemma 3.1.11 Let A be a noetherian local integral domain. The following two conditions on A are equivalent: (i) Ash, the strict henselization of A, has a unique minimal prime. (ii) Ae, the normalization of A, is a local ring and the corresponding extension of residue ﬁelds has separable degree 1.

For the next result we recall the deﬁnition of ﬁne partitions from section 3.1.1:

Proposition 3.1.12 Let A be a componentwise geometrical unibranch, let A → B be an

´etale homomorphism and for p ∈ Min(A) let Sp := {P ∈ Min(B)|P ∩ A = p}. Then the

family {Sp}p∈Min(A) is a ﬁne partition of Min(B).

Proof: Let M be a maximal ideal of B and let m := M ∩ A. Then the local morphism

Am → BM is also ´etale by the results in section 2.1.2, so by lemma 3.1.8 we have that BM

81 sh is isomorphic to a subalgebra of Am.SinceA is a componentwise geometrical unibranch,

this implies that the morphism Min(BM ) → Min(Am) is injective (in fact bijective). Now let P, P 0 ∈ Min(B) such that P ∩A = p = P 0 ∩A where p ∈ Min(A)andP =6 P 0. If P + P 0 =6 B, then there is a maximal ideal M ∈ Spec B containing both the ideals, 0 and hence P and P are distinct minimal prime ideals of BM lying over the same minimal

prime ideal of Am,wherem = M ∩ A. This is a contradiction to the ﬁrst paragraph of the proof. 2

For a reduced noetherian ring A we recall the deﬁnition of ΓA from section 3.1.1. Having the result above we can now prove:

Theorem 3.1.13 Let A be a reduced componentwise geometrical unibranch and let A → ∼ B be an ´etale homomorphism. Then there is a canonical isomorphism B ⊗A ΓA = ΓB.

Proof: Since A is reduced we have that Ass(A)=Min(A). The same is true for B

by lemma 2.1.6. For p ∈ Min(A)weletSp := {P ∈ Min(B)|P ∩ A = p} and we let S

denote the family {Sp}p∈Min(A). This family is a ﬁne partition of Min(B)=Ass(B)by proposition 3.1.12. For p ∈ Ass(A)wehavebylemma3.1.2that

\ Γp = q q∈Ass(A) q=6 p

Thus, since B is a ﬂat A-module, it follows from [14, Theorem 7.4] that

\ B ⊗A Γp = B ⊗A q q∈Ass(A) q=6 p

82 By lemma 3.1.7 we have for all q ∈ Ass(A) a canonical isomorphism

∼ \ B ⊗A q = Q Q∈Sq

and thus it follows that

X B ⊗A ΓA = B ⊗A Γp p∈Ass(A) X \ = B ⊗A q p∈Ass(A) q∈Ass(A) q=6 p ∼ X \ \ = Q p∈Ass(A) q∈Ass(A) Q∈Sq q=6 p

Here we pause for a moment to observe the elementary set theoretic fact that since

S = {Sq}q∈Ass(A) is a partition of Ass(B) , we have

\ \ \ Q = Q. ∈ q∈Ass(A) Q Sq Q∈Ass(B)\Sp q=6 p

This, along with the previous formula, implies that

∼ X \ B ⊗A ΓA = Q = IS p∈Ass(A) Q∈Ass(B)\Sp

where IS is deﬁned as in proposition 3.1.6. Since S is a ﬁne partition of Ass(B),the result now follows directly from proposition 3.1.6. 2

83 3.2 Minimal twosided ideals and minimal left submodules

3.2.1 Diﬀerential operators and supports

We will now relate supports of regular functions on spectra of commutative rings to their rings of diﬀerential operators. For that we will need the following easy lemma from the folklore. We include a proof here for the convenience of the reader.

Lemma 3.2.1 Let A be a commutative k-algebra and let I be an ideal of A.Theideal

ΓI is a left D(A)-submodule of A. That is, D(A) ∗ ΓI ⊂ ΓI.

Proof: Let Dd denote the set of diﬀerential operators on A of order ≤ d. We claim that

In+dDd ⊂DdIn for all integers d ≥ 0andn ≥ 0.

The lemma follows immediately: Let a ∈ ΓI. Then there is an integer n ≥ 0 such that In a = 0. Hence if D ∈D(A), say of order ≤ d, the claim implies

In+dD ∗ a ⊂DdIn ∗ a = Dd ∗ (In a)=0

so by deﬁnition D ∗ a ∈ ΓI. We now prove the claim. First we note that for n =0ord = 0 the statement is trivial. We then proceed by a double induction. We prove the claim for a pair (n, d) where both n and d are greater than zero. By induction we may assume that the statement is true for all pairs (n0,d0) where either n0

84 Let x ∈ Id+n−1, y ∈ I and D ∈Dd. Our task is to show that xyD ∈DdIn. However, by our induction hypothesis we have that xD ∈DdIn−1 and x[D, y] ∈Dd−1In (since [D, y] ∈Dd−1), and then

xyD = xDy − x[D, y] ∈DdIn−1I + Dd−1In = DdIn

We already pointed out that ΓI is the ideal of regular functions on Spec A whose support is contained in the closed subvariety V (I). Hence in terms of geometry, this lemma says that diﬀerential operators preserve supports of regular functions.

Now let A be a commutative, noetherian and reduced k-algebra. By the deﬁnition

of ΓA in section 3.1.1, the lemma above implies immediately that it is always a left D(A)-submodule of A. Using this we can sometimes construct less obvious left D(A)- submodules of A. But to simplify our notation we ﬁrst make a deﬁnition:

Definition If A → B is a faithfully flat étale homomorphism, we say that B is an étale covering of A.

This definition is justified by the fact that if A → B is a faithfully flat homomorphism of commutative rings, then the corresponding map Spec B → Spec A is surjective (cf. [14, theorem 7.3]). We have:

Theorem 3.2.2 Let A be a commutative, noetherian and reduced k-algebra, and assume that A has an ´etale covering B which is a componentwise geometrical unibranch. Then ∼ (i) there is a unique ideal χA of A such that B ⊗A χA = ΓB via the multiplication map;

(ii) this ideal is an invariant of A, i.e., χA is independent of the choice of a componentwise geometrical unibranch ´etale covering;

85 (iii) if A is a k-algebra, then χA is a left D(A)-submodule of A.

Proof: Fix an ´etale covering f : A → B where B is a componentwise geometrical

unibranch. Let fi : B → B ⊗A B, i =1, 2 denote the base extensions of f, b 7→ 1 ⊗ b and b 7→ b ⊗ 1 respectively. These morphisms are ´etale and hence, by our assumption on B, theorem 3.1.13 applies. We recall the notation of section 2.2.4, and we have by theorem 3.1.13 canonical isomorphisms

⊗ ∼ ∼ ⊗ B A ΓB = f1∗ΓB = ΓB⊗AB = f2∗ΓB =ΓB A B

which correspond to the restriction of the identity on B ⊗A B. The existence and unique-

ness of the ideal χA follows now directly from the theorem on faithfully flat descent (cf. [12]) Now assume that A → B0 is another étale covering of A which is a componentwise 0 geometrical unibranch. Let χA be the ideal defined by that covering. By base extension, 0 we get an étale covering B → B ⊗A B , and then by theorem 3.1.13 there is a canonical isomorphism 0 ∼ 0 ∼ ⊗ ⊗ ⊗ 0 ΓB A B = (B A B ) B ΓB = ΓB⊗AB .

The same argument, with the roles of B and B0 reversed, shows that

∼ ⊗ 0 0 B A ΓB = ΓB⊗AB

0 Hence by the deﬁnitions of χA and χA we have that

0 ∼ 0 ∼ ∼ 0 0 (B ⊗A B ) ⊗A χA = ΓB ⊗A B = B ⊗A ΓB0 = (B ⊗A B ) ⊗A χA

0 0 and since A → B ⊗A B is faithfully flat, this implies that χA = χA . Finally assume that A is a commutative algebra over some ring k, and we have an étale covering f : A → B as above. We define a k-algebra structure on B via f.Bythe

86 theorem on faithfully ﬂat descent we have that

f(χA)={b ∈ ΓB|f1(b)=f2(b)}.

By a remark after corollary 2.2.6 we have an exact sequence

f −→f1 0 −→ D (A) −→ D(B) −→ D(B ⊗A B). f2

Let a ∈ χA and D ∈D(A). Then f(a) ∈ ΓB, and since ΓB is a left D(B)-module, this implies that f(D ∗ a)=f(D) ∗ f(a) ∈ ΓB. Furthermore

f1(f(D ∗ a)) = f1(f(D) ∗ f(a)) = (f1 ◦ f)(D) ∗ (f1 ◦ f)(a)

=(f2 ◦ f)(D) ∗ (f2 ◦ f)(a)

= f2(f(D ∗ a))

and since f : A → B is injective, this and the description of f(χA) above, implies that

D ∗ a ∈ χA.ThereforewehavethatχA is a left D(A)-submodule of A. 2

n Remarks 1. We observe that if A is a k-algebra such that the modules PA are ﬁnitely presented for all n ≥ 0, and the conditions in the theorem hold, then the existence and uniqueness of the left D(A)-module χA also follows from proposition 2.2.15. 2. If A, as in the theorem, is already a componentwise geometrical unibranch, then obviously χA =ΓA. This is deﬁnitely no longer true if A is not a componentwise geometrical unibranch.

87 3.2.2 Essential submodules

Deﬁnition Let R be a ring and let M be an R-module. An R-submodule N ⊂ M is said to be essential if for any R-submodule L ⊂ M, N ∩ L = 0 implies that L =0.We let Ess(M) denote the set of all essential R-submodules of M.

It is clear that Ess(M)isnotempty(M ∈ Ess(M)). It also follows directly from the deﬁnition that if N,N0 ∈ Ess(M), then N ∩ N 0 ∈ Ess(M). Thus if we order Ess(M)by inclusion, this implies that Ess(M) has at most one minimal element. For a general ring R and an R-module M, the family Ess(M) need not have a minimal element. However if Ess(M) has a minimal element N, we will say that N is the minimal essential submodule of M. Having this, we can identify essential twosided ideals of some rings of diﬀerential operators:

Lemma 3.2.3 Let A be a reduced noetherian k-algebra and let I be a twosided ideal of D(A). Then I is and essential ideal of D(A) if and only if I contains a regular element of A.

Proof: Assume that I is an essential ideal of D(A). For p ∈ Ass(A), we recall the deﬁnition of Γp from section 3.1.1. We then have that D(A, Γp) is a nonzero twosided ideal of D(A), and hence is I ∩D(A, Γp) =6 0. Lemma 1.2.3 now implies that I ∩D(A, Γp) contains a nonzero element of A, and subsequently any such element must be in Γp.This holds for all p ∈ Ass(A) and therefore, as in the proof of lemma 3.1.3, we have that I contains a regular element of A. Conversely assume that I contains a regular element a ∈ A,andlet0=6 J be a twosided ideal of D(A). By lemma 1.2.3 it contains a nonzero element b ∈ A, and hence 0 =6 ab ∈ I ∩ J. 2

88 Let A be a commutative k-algebra. Besides essential twosided ideals of D(A), we will also study essential left D(A)-submodules of A. They are also ideals of A,anda priori it is not clear whether essential left D(A)-submodules of A are essential ideals of A. Sometimes this is indeed the case as th following lemma shows:

Lemma 3.2.4 Let A be a reduced noetherian k-algebra and let I ⊂ A be an ideal of A. The following two statements are equivalent: (i) I is an essential ideal of A; (ii) I contains a regular element of A. If I is also a left D(A)-submodule of A, then the statements above are equivalent to the following two statements: (iii) I is an essential left D(A)-submodule of A; (iv) D(A, I) is an essential twosided ideal of D(A).

Proof: It is clear that (ii) implies (i). Conversely, a proof similar to that of lemma 3.2.3 shows that (i) implies (ii). Since the ideals Γp, p ∈ Ass(A)areleftD(A)-submodules of A, the same proof also shows that (iii) implies (ii). Conversely, since all left D(A)- submodules of A are ideals, it is clear that (i) implies (iii). Finally, since I = D0(A, I) ⊂ D(A, I), lemma 3.2.3 and (ii) imply that (iii) is equivalent to (iv). 2

Remark From this and lemma 3.1.3 it follows that if A is a commutative, noetherian

and reduced k-algebra, then ΓA is an essential D(A)-submodule of A,andD(A, ΓA)isan essential twosided ideal of D(A). The lemma shows also that we can talk about essential D(A)-submodules of A without a danger of confusion.

Next we will show how essential ideals behave under faithfully étale homomorphisms. Before stating the result, we recall some properties of étale homomorphisms and differential operators. Let A be a commutative, noetherian and reduced k-algebra such that

89 n ≥ → the modules PA are finitely presented for all n 0. Let f : A B be an étale homomorphism of k-algebras. Then we showed in section 2.2.3 that D(B)=B ⊗A D(A) as B-D(A)-bimodules, and hence it followed that B ⊗A M is a left D(B)-module in a canonical way whenever M is a left D(A)-module. Furthermore, under the same assumptions we showed in section 2.3.3 that B ⊗A I is a twosided ideal of D(B)ifI is a twosided ideal of D(A). We use these facts in the statement and proof of the next proposition. Butfirstweneedalemma:

Lemma 3.2.5 Let f : A → B be an ´etale homomorphism of commutative, noetherian and reduced rings. If a ∈ A is a regular element of A, then f(a) is a regular element of B.

Proof: We recall that for a reduced noetherian ring A we have Min(A)=Ass(A). By lemma 3.1.1 we have that the set of zero-divisors in a noetherian ring is precisely the union of all the associated primes. Thus if f(a) is a zero-divisor of B ,thenf(a) ∈ P for some P ∈ Ass(B). Hence a ∈ f −1(P ). Finally by lemma 3.1.7 we have f −1(P ) ∈ Min(A)=Ass(A) and thus a is a zero-divisor of A. 2

Proposition 3.2.6 Let A be a commutative noetherian reduced ring, and let f : A → B be a faithfully ´etale homomorphism. Let I be an ideal of A. Then

(i) I is an essential ideal of A if and only if B ⊗A I is an essential ideal of B. n Assume furthermore that f is a morphism of k-algebras, the modules PA are ﬁnitely presented for all n ≥ 0, and I is a left D(A)-submodule of A. We have:

(ii) If B ⊗A I is the minimal essential D(B)-submodule of B, then I is the minimal essential D(A)-submodule of A. We have similar statements for ideals of D(A). Namely, if I is a twosided ideal of D(A) and all the conditions above hold, then we have:

90 (iii) I is an essential twosided ideal of D(A) if and only if B ⊗A I is an essential twosided ideal of D(B);

(iv) if B ⊗A I is the minimal essential twosided ideal of D(B), then I is the minimal essential twosided ideal of D(A).

Proof: If I is an essential ideal of A, it contains a regular element a ∈ A, by lemma 3.2.4.

Then f(a) is regular by lemma 3.2.5, and it is contained in B ⊗A I. Hence B ⊗A I is essential, again by lemma 3.2.4. For the converse let 0 =6 J be an ideal of A.SinceB is faithfully ﬂat over A,wehave

that B ⊗A J =0and6 B ⊗A (J ∩ I)=(B ⊗A J) ∩ (B ⊗A I) by [14, Theorems 7.2, 7.4].

If B ⊗A I is an essential ideal of B,wethushavethat(B ⊗A J) ∩ (B ⊗A I) =0which6 in turn implies that J ∩ I =6 0. This shows that I is an essential ideal of A and we have proved (i). For (ii), let J be an essential D(A)-submodule of A such that J ⊂ I. Then by the

ﬁrst part of the proposition and lemma 3.2.4, B ⊗A J is an essential D(B)-submodule of

B, contained in B ⊗A I. Hence if B ⊗A I is minimal, we have that B ⊗A J = B ⊗A I, and since B is faithfully ﬂat over A, this implies that J = I. The proofs of (iii) and (iv) are completely analogous, once we recall that under our

assumptionswehavethatD(B)=B ⊗A D(A), and that B ⊗A I is a twosided ideal of D(B)ifI is a twosided ideal of D(A). 2

3.2.3 Minimal nonzero ideals in rings of diﬀerential operators

Let A be a commutative, noetherian and reduced k-algebra. Let p be an element of Ass(A). We recall that by lemmas 3.1.2 and 3.2.1 we have indeed that p is a D(A)-submodule of A. This implies that D(A, p) (which we may identify with the set {D ∈D(A)|D ∗ A ⊂ p}) is an ideal of D(A). We have:

91 Lemma 3.2.7 Let A be a commutative, noetherian and reduced k-algebra, and let p be an associated prime ideal of A.Letπ : A → A/p be the natural projection. The image π(Γp)

is an ideal of A/p and π restricted to Γp is an isomorphism of A-modules Γp → π(Γp).

Proof: Since π is surjective it is clear that the image of Γp under π is an ideal. The

second statement follows directly from the fact that p ∩ Γp =0. 2

It was pointed out in section 1.2.1 that π : A → A/p extends to a homomorphism π : D(A) →D(A/p) of ﬁltered rings, and it is characterized by the property that π(D) ∗ π(a)=π(D ∗ a) for all D ∈D(A), a ∈ A. We use these observations in the statement of next result:

Proposition 3.2.8 Let A be a commutative, noetherian and reduced k-algebra, and let p be an associated prime ideal of A.Letπ : A → A/p be the natural projection, and denote

the corresponding map D(A) →D(A/p) also by π. The space π(D(A, Γp)) ⊂D(A/p) is

equal to the right ideal D(A/p, π(Γp)). Moreover π restricted to D(A, Γp) is injective.

Proof: If D ∈D(A, Γp)andπ(D) = 0 then for all a ∈ A

0=π(D) ∗ π(a)=π(D ∗ a).

Therefore D ∗ A ⊂ p ∩ Γp = 0. This proves the second statement.

For the ﬁrst statement let D ∈D(A, Γp). Then for all a ∈ A

π(D) ∗ π(a)=π(D ∗ a) ∈ π(Γp)

and thus π(D) ∗ (A/p) ⊂ π(Γp). That is, π(D) ∈D(A/p, π(Γp)).

92 Conversely let D ∈D(A/p, π(Γp)). We recall that by lemmas 1.2.1 and 1.2.6 we have the following natural isomorphism

∼ D(A/p, π(Γp)) = DA/p(A/p, π(Γp)) = DA(A/p, π(Γp)) and so D is a diﬀerential operator between A-modules. By abuse of language we let

−1 π : π(Γp) → Γp denote the inverse to the natural A-linear isomorphism in lemma 3.2.7. −1 Then the map π Dπ : A → Γp is the composite of diﬀerential operators (two of which −1 ∼ are A-linear) between A-modules. Therefore π Dπ ∈DA(A, Γp) = D(A, Γp), again by lemma 1.2.1. Finally we have for all a ∈ A that

π(π−1Dπ) ∗ π(a)=π(π−1(D ∗ π(a))) = D ∗ π(a)

−1 since D ∗ π(a) ∈ π(Γp). This shows that D = π(π Dπ) ∈ π(D(A, Γp)). 2

We can now prove:

Theorem 3.2.9 Let A be a commutative noetherian and reduced k-algebra. If p ∈

Ass(A) is such that D(A/p) is a simple ring, then D(A, Γp) is a minimal nonzero twosided ideal of D(A)(i.e. it does not strictly contain any nontrivial twosided ideal of D(A)). Moreover, if D(A/p) is a simple ring for all p ∈ Ass(A), then

(i) the set {D(A, Γp)|p ∈ Ass(A)} is precisely the set of all minimal nonzero twosided ideals of D(A), (ii) we have an equality M D(A, ΓA)= D(A, Γp), p∈Ass(A) and

(iii) the ideal D(A, ΓA) is the minimal essential twosided ideal of D(A).

93 Proof: Let I ⊂D(A, Γp) be a nonzero twosided ideal of D(A). Then clearly I ⊃

D(A, Γp)ID(A, Γp) and hence

π(I) ⊃ π(D(A, Γp))π(I)π(D(A, Γp)).

Since I is a nonzero twosided ideal, by lemma 1.2.3 it must contain a nonzero element of A,

and since I ⊂D(A, Γp), such an element must lie in Γp. It follows that π(I)π(D(A, Γp)) =6

0. Last proposition implies that π(D(A, Γp)) is a right ideal of D(A/p), and since we assume that D(A/p) is a simple ring we then have that

π(I) ⊃ π(D(A, Γp))π(I)π(D(A, Γp))

=(π(D(A, Γp))D(A/p))π(I)(π(D(A, Γp))D(A/p))

= π(D(A, Γp)).

Since π is injective on D(A, Γp) it follows that D(A, Γp) ⊂ I. This proves the ﬁrst part of the theorem.

For the remainder write Ass(A)={p1,...,pn} and for the moment let J be the ideal P n D D ∈ i=1 (A, Γpi ). We assume that (A/p) is simple for all p Ass(A). For i =1,...,n P P ∈D n n ∈ let Di (A, Γpi ) and assume that i=1 Di =0.ThenD1 = i=2 Di, so for every a A ∗ ∈ ∩ D we have that D1 a p Γp = 0. This shows that the sum of the (A, Γpi )s is in fact direct. Let D be any nonzero diﬀerential operator of D(A)and let a ∈ A be such that D ∗a =6 6 ∈ 0. Then by taking x = x1 + ...+ xn with 0 = xi Γpi ,wehavethatx is a regular

element of A, and therefore xD ∗ a =0.Thisimpliesthat6 xD = x1D + ...+ xnD =0,6 ∈D 6 and since xiD (A, Γpi ) for all i,wehavethatxiD = 0 for some i. Thisimpliesthat D D ∩ 6 if I is a nonzero twosided ideal of (A), then there is an i such that (A, Γpi ) I = D ⊂ 0. The ﬁrst part of the theorem then implies that (A, Γpi ) I. This proves that

{D(A, Γp)|p ∈ Ass(A)} is the set of all nonzero minimal twosided ideals of D(A). This also shows that J is the minimal essential twosided ideal of D(A).

94 By deﬁnition it is clear that J,deﬁnedasabove,iscontainedinD(A, ΓA). For the → other inclusion, let πi :ΓA Γpi denote the natural projection. It is A-linear, and D →D 7→ ◦ thus it induces a natural morphism (A, ΓA) (A, Γpi ), D πi D. It is clear that

π1 +...+πn is the identity on ΓA, and therefore it follows immediately that D(A, ΓA) ⊂ J. 2

From this theorem, a similar statement for left D(A)-submodules of A follows easily:

Corollary 3.2.10 Let A be a commutative, noetherian and reduced k-algebra. If p ∈

Ass(A) is such that D(A/p) is simple, then Γp is a minimal nonzero left D(A)-submodule of A. Moreover, if the rings D(A/p) are simple for all p ∈ Ass(A), then

(i) the set {Γp|p ∈ Ass(A)} is precisely the set of all minimal nonzero left D(A)- submodules of A, and

(ii) the ideal ΓA is the minimal essential left D(A)-submodule of A.

Proof: If 0 =6 I is a left D(A)-submodule of A,thenD(A, I) is a nonzero twosided ideal of D(A). From this observation, the corollary follows. 2

We observe that the theorem and its corollary say nothing about integral domains since the assumptions become the implications. In order to deal with domains, we hence must prove a more general version of these results. For that we will use étale coverings (definedinsection3.2.1).LetA be a commutative, noetherian and reduced k-algebra. Assume that there is a faithfully étale homomorphism f : A → B of k-algebras (i.e., an étale covering) where B is a componentwise geometrical unibranch. We recall that B is reduced, by lemma 2.1.6. Then we can define the left D(B)-submodule ΓB of B,and

theorem 3.2.2 implies the existence and uniqueness of χA,aleftD(A)-submodule of A

such that ΓB = B ⊗A χA. Finally we have of course that D(A, χA) is a twosided ideal n ≥ and if PA is ﬁnitely presented for all n 0, then by the remark following theorem 2.2.10

it follows that D(B,ΓB)=B ⊗A D(A, χA).

95 We can now state and prove the generalization we want of theorem 3.2.9 and its corollary:

Theorem 3.2.11 Let A be a commutative, noetherian and reduced k-algebra such that n ≥ → PA is finitely presented for all n 0. Assume that A has an étale covering f : A B where B is a componentwise geometrical unibranch and the irreducible components of B have rings of differential operators that are simple. ∼ (i) Then A has a minimal essential D(A)-submodule χA, and B ⊗A χA = ΓB via the natural multiplication map.

(ii) Moreover, the ideal D(A, χA) is the minimal essential twosided ideal of D(A),

and B ⊗A D(A, χA)=D(B,ΓB).

Proof: We have already constructed χA in theorem 3.2.2. This ideal and D(A, χA)are essential by proposition 3.2.6. Thus it remains only to show that these objects are in fact minimal. For that let I be an essential twosided ideal of D(A). Then by corollary 2.3.13

and a remark following proposition 3.2.6, we have that B ⊗A I is an essential twosided

ideal of D(B). Hence it follows from theorem 3.2.9 that B ⊗A D(A, χA)=D(B,ΓB) ⊂

B ⊗A I, and then faithful ﬂatness implies that D(A, χA) ⊂ I. We get the minimality of

χA in a similar way from corollary 3.2.10. 2

3.2.4 Applications of theorem 3.2.11

Theorems 3.2.2 and 3.2.11 raise two questions: 1. Do all commutative noetherian reduced rings have an ´etale covering which is a componentwise geometrical unibranch? 2. When is it true for a commutative, noetherian algebra A that D(A)isasimple ring? In this section we will reﬂect on these two questions.

96 The answer to the first question is “no”, even with the additional assumption that the ring is local and essentially of finite type over an algebraically closed field of characteristic zero. The problem lies in the fact that even if a local ring is a componentwise geometrical unibranch at its maximal ideal, it may not be so at other prime ideals. However if we dispose of that problem, we have the following:

Proposition 3.2.12 Let A be a commutative noetherian local ring such that A is a componentwise geometrical unibranch at all prime ideals except possibly the maximal ideal. Then A has a local ´etale covering B which is a componentwise geometrical unibranch (at all prime ideals).

Proof: In section 3.1.2 we pointed out that Ash, the strict henselization of A,isnoethe- rian. Therefore it has finitely many minimal prime ideals by lemma 3.1.1. Assume that P, P 0 ∈ Min(Ash) such that P =6 P 0. Then there is an x ∈ P \P 0, so by lemma 3.1.8 there is a local étale A-subalgebra A0 of Ash such that x ∈ A0. Then obviously P ∩A0 =6 P 0 ∩A0. We can repeat this procedure for all pairs of minimal prime ideals of Ash, and thus there exists a local étale A-subalgebra B of Ash such that P ∩ B =6 P 0 ∩ B whenever P and P 0 are two distinct minimal prime ideals of Ash. This implies that B is a componentwise geometrical unibranch at its maximal ideal. Finally if q is a prime ideal of B distinct from the maximal ideal, then p := q ∩ A is distinct from the maximal ideal of A (fibers of étale homomorphisms have krull dimension

0) and Ap → Bq is a local ´etale homomorphism. Then, by assumption we have that the sh → map Min(Ap ) Min(Ap) is injective, and since Bq is a local ´etale extension of Ap, the same is obviously true for that ring. This shows that B is also a componentwise geometrical unibranch at all prime ideals distinct from the maximal ideal. 2

All ﬁelds are automatically geometrical unibranches, and hence by using lemma 3.1.10 we can apply the proposition to rings of krull dimension 1:

97 Corollary 3.2.13 Let A be a commutative noetherian local ring of krull dimension 1. Then A has an ´etale covering B which is a componentwise geometrical unibranch.

Proof: If p ∈ Spec A is not the maximal ideal, then p is a minimal prime ideal and

Ap/p is a ﬁeld. 2

Proposition 3.2.12 implies also the following:

Proposition 3.2.14 Let A be a commutative noetherian reduced ring, and a k-algebra. If D(A) is a simple ring, then A is a domain and a geometrical unibranch.

Proof: If A is not a domain and p is one of the minimal prime ideals of A, it follows

directly from the deﬁnition that Γp is neither A nor 0. Hence D(A, Γp) is a nontrivial twosided ideal of D(A). We may now assume that A is a domain. Assume that A is not a geometrical unibranch, and let p be a minimal element of the set

{q ∈ Spec A|A is not a geometrical unibranch at q}.

Then Ap is a geometrical unibranch at all prime ideals not equal to p, and proposi-

tion 3.2.12 implies that Ap has a local ´etale covering B which is a componentwise geo-

metrical unibranch. Since Ap is not a geometrical unibranch at p it follows that B has

more than one minimal prime ideal, and therefore is ΓB a nontrivial left D(B)-submodule

of B. By theorem 3.2.2 it then follows that Ap has a nontrivial left D(Ap)-submodule, and by intersecting this module with A, we get a nontrivial left D(A)-submodule of A. This shows that the ring D(A)isnotsimple. 2

98 Remark This proposition shows that the condition on geometrical unibranchedness in theorem 3.2.11 is in fact redundant. We should also note that for reduced algebras, essentially of ﬁnite type over a ﬁeld of characteristic zero, proposition 3.2.14 is implied by the results of [4].

We now turn to the second question stated at the beginning of this section. Propo- sition 3.2.14 gives a necessary condition for D(A)tobesimpleifA is a commutative noetherian and reduced k-algebra. That is, A must be a domain and a geometrical unibranch. Furthermore it is not hard to see that A must also be equal to the intersection

of all its height one localizations (i.e., A must satisfy the condition S2. See chapter 4 for the definition). This follows also from [4]. On the other hand, these conditions are not sufficient. The well known example of [1] shows that D(A)neednotbesimpleevenif A is a normal domain, finitely generated over the complex numbers. However, if A is a regular domain, essentially of finite type over a field of characteristic zero, it was pointed out in the beginning that D(A)=∆(A), and in that case it follows, for example from [15, Chapter 15], that D(A) is simple. Furthermore, if krull dimension of A is equal to 1, we have:

Theorem 3.2.15 Let A be a local domain of krull dimension 1, a localization of a ﬁnitely generated k-algebra, where k is a ﬁeld of characteristic zero. If A is a geometrical unibranch, then A has no proper nonzero D(A)-submodules, and D(A) is a simple ring.

Proof: This is [4, corollary 3.6]. 2

From this and theorem 3.2.11 we get an immediate corollary:

Corollary 3.2.16 Let k be a field of characteristic zero and let A be a local, reduced k-algebra of krull dimension 1 and essentially of finite type over k. Then χA, as defined

99 in theorem 3.2.2, is the minimal essential left D(A)-submodule of A, and D(A, χA) is the minimal essential twosided ideal of D(A).

Remark We must note that the existence of such minimal modules was already known, at least if A is the localization of an aﬃne reduced curve over a ﬁeld of characteristic zero (cf. [17, 25]).

There are other examples of algebras A for which D(A) is simple (cf. [13]), but in general the problem is quite open. In next chapter, however, we will show that if we restrict the class of twosided ideals, then it is possible to prove a minimality condition which holds quite generally.

100 Chapter 4

Reﬂexive modules and rings of diﬀerential operators

In [4] it was proved that if A is a domain, essentially of finite type over a field of characteristic zero, then A satisfies the condition S2 and is a geometrical unibranch if and only if D(A) is a krull domain with no proper reflexive ideals. The goal of this chapter is to prove a related result. Namely, if A is a domain, essentially of finite type over a field of characteristic zero, and satisfies the condition S2, then the set

{J|J is a reflexive nonzero ideal of A and a D(A)-submodule of A }, ordered by inclusion, has a unique minimal element. In fact if we let X denote the set of all height one prime ideals of A, it follows from section 3.2.4 that the ideals χAp can be T ∈X constructed for all p . We show that I := p∈X χAp is the unique minimal element we are seeking. Moreover, in that case the ideal D(A, I)ofD(A) is reflexive on the right and it contains all nonzero twosided ideals of D(A) which are reflexive on either side.

101 4.1 Some noncommutative ring theory

4.1.1 Rings of quotients

We refer to [15, Section 2.1] for the general deﬁnition of rings of fractions of noncommutative rings (also called “rings of quotients”). We will only need the following special case:

Deﬁnition Let R be a ring. The total right (resp. left) ring of fractions of R is a ring Q containing R such that the following conditions hold: (i) Every regular element of R has an inverse in Q; (ii) Every element of Q may be written as rs−1 (resp. s−1r)wherer, s ∈ R and s is a regular element.

If R is commutative, this is the classical definition, and we know that the ring Q exists. However if R is noncommutative, the existence of its total (left or right) ring of fractions is by no means guaranteed. However if it exists, it is unique up to an isomorphism by [15, Corollary 2.1.4], and that result also shows that if R has both a left and a right total rings of fractions, they must be isomorphic. Therefore if R has both left and right ring of fractions, we will call it the ring of fractions of R (without the qualifiers “left” and “right”). In [15, Section 3.1] we have also the following definitions, related to that above:

Deﬁnition AringQ is called a quotient ring if every regular element of Q is invertible. Given a quotient ring Q, a subring R is called a right (left) order in Q if every q ∈ Q has the form rs−1 (s−1r) for some r, s ∈ R. A left and a right order is called an order.

Note that if R has a total ring of fractions Q,thenQ is a quotient ring and R is an order in Q.

102 In this section we will show that if k is a ﬁeld of characteristic zero, and A is a reduced k-algebra, essentially of ﬁnite type over k,thenD(A) has a total ring of fractions. We carry out some of the proofs for the sake of completeness, but essentially the same result and proof may be found in [17].

Theorem 4.1.1 Let k be a field of characteristic zero, and let k → K be a field extension of finite transcendence degree. Then D(K) is a simple noetherian integral domain.

Proof: This is [15, Proposition 15.3.2(ii), Theorem 15.5.5(ii)] 2

The following is essentially proved in [17]:

Theorem 4.1.2 Let k be a ﬁeld of characteristic zero, and let A be a commutative k- algebra, reduced and essentially of ﬁnite type over k.LetK be the total ring of fractions of A. Then D(K) has a total ring of fractions QD(K), and we have a chain of inclusions

D(A) ⊂D(K) ⊂ QD(K)

Moreover, QD(K) is also the total ring of fractions of D(A).

∼ Proof: Since A is reduced and essentially of ﬁnite type over k, it follows that K =

K1 × ...Kn where each Ki is a ﬁeld of ﬁnite transcendence degree over k. This implies especially that K is reduced, so D(K) is semiprime by lemma 1.2.4. Lemma 1.2.5 implies

that D(K)=D(Ki) × ...D(Kn), and therefore we have by theorem 4.1.1 that D(K)is noetherian (a direct product of ﬁnitely many noetherian rings is noetherian). Noetherian rings are Goldie (cf. [15, Section 2.3.1]), and hence [15, Theorem 2.3.6] implies that D(K) has a total ring of fractions.

103 By a remark following theorem 2.2.5 we have that D(A) ⊂D(K). Moreover, since A is essentially of ﬁnite type over k, propositions 2.2.11 and 2.2.12 imply that theorem 2.2.10 applies for the extension A ⊂ K. Therefore every element of D(K) may be written as s−1D where s is a regular element of A and D ∈D(A). The same is true for the other side: We may write s−1D = Ds−1 − [D, s−1], and by induction on order of diﬀerential operators, we may assume that [D, s−1]=D0t−1 for some D0 ∈D(A)andt ∈ A,aregular element. Therefore we have that s−1D = Ds−1 − D0t−1 =(Dt − D0s)(st)−1 (this proof comes from [17, Proposition 1.8]). From this last result it now follows that regular elements of D(A) are also regular

elements of D(K) and hence invertible in QD(K). To show that QD(K) is the total ring of fractions of D(A), we copy the proof of [17, Lemma 3.2(c)]. Let DD0−1 be an element of

0 0 00 −1 0 000 −1 QD(K) with D, D ∈D(K)andD regular. We may write D = D s and D s = D t with D00,D000 ∈D(A)ands, t regular elements in A.Then

DD0−1 = D00(D0s)−1 = D00(D000t−1)−1 =(D00t)D000−1.

− It follows in a similar way that DD0−1 = D000 1D00 for some D00, D000 ∈D(A)whereD000 is regular. 2

Corollary 4.1.3 Let k and A be as in the theorem. Then D(A) is a semiprime (right and left) Goldie ring.

Proof: This follows directly from the theorem and [15, Theorem 2.3.6]. 2

104 4.1.2 Reﬂexive modules

In this section we will introduce reflexive modules, and show how they apply to rings of differential operators. We refer to [15] for most of the general theory, but some of the commutative theory may also be found in [2, chapter VII.4]. The applications to rings of differential operators can be found in [4].

Let R be a ring and let M be a right (left) R-module. The dual of M is deﬁned as ∗ the left (right) R-module M := HomR(M,R). The doubledual is again a right (left) ∗∗ R-module, and there is an obvious R-module homomorphism M → M , m 7→ φm,where ∗ ∗∗ φm : M → R is the map α 7→ α(m). The module M is said to be torsionless if M → M is injective and in that case we identify M with its image in M ∗∗.Tosaythatamodule M is torsionless is equivalent to saying that M embeds in some direct product of copies of R (cf. [15, sections 5.1.7, 3.4.2]).

Deﬁnition Let R be a ring and let M be a torsionless (left or right) R-module. We say that M is reﬂexive if M = M ∗∗.

We quickly observe that for any R-module M, its dual M ∗ is torsionless and in fact reflexive [15, sections 3.4.2, 5.1.7]. The definition of reflexive modules we have just stated is rather hard to work with, and we will restrict our attention to a certain class of modules where duals can be described in somewhat more concrete terms. From section 4.1.1 we recall the definition of orders. The following definition appears in [15, section 3.1.11].

Deﬁnition Let R be an order in a quotient ring Q.LetI be a right (left) R-submodule of Q.WesaythatI is a fractional right (left) R-ideal if there are units a and b of Q such that aI ⊂ R (Ia ⊂ R)andbR ⊂ I (Rb ⊂ I).

We immediately get another description of duals:

105 Proposition 4.1.4 Let R be an order in a quotient ring Q, and let I be a fractional ∗ ∼ right (left) R-ideal. Then we have an isomorphism of left (right) R-modules I = ∗ ∼ {q ∈ Q|qI ⊂ R} (I = {q ∈ Q|Iq ⊂ R}).

Proof: This is [15, Proposition 3.1.15(iv)] 2

Having this proposition, we will from now on identify the objects therein. We use it in the next lemma

Lemma 4.1.5 Let R be an order in a quotient ring Q.IfI is a fractional right (left) R-ideal, then I∗ is a fractional left (right) R-ideal.

Proof: We carry the proof out for fractional right R-ideals. Then by deﬁnition there are units a, b of Q such that aI ⊂ R and bR ⊂ I. This implies that b ∈ I, and thus I∗b ⊂ R. Since I∗ = {q ∈ Q|qI ⊂ R} it follows especially that a ∈ I∗. We recall (or observe) that I∗ is a left R-module and hence Ra ⊂ I∗. 2

This lemma, along with the previous proposition applied again, shows that doubleduals of fractional (left or right) R-ideals may also be identiﬁed with fractional R-ideals, and from now on we will do so without further comment. Moreover the results show directly that I ⊂ I∗∗ whenever I is a fractional R-ideal. We should note that the deﬁnition of fractional ideals implies immediately that they are torsionless R-modules.

At this point we issue a warning. On several occasions in the sequel we will be considering fractional twosided R-ideals, i.e., twosided R-submodules of Q that are fractional ideals on both sides. When taking duals of such fractional ideals, we must be careful to specify on which side we are doing it, since we have no guarantee that the end result is the

106 same. (However compare [15, Proposition 5.1.8] for a case where there is no ambiguity.) If there is a danger of confusion, we will thus follow the example of [4, 15] and write

∗ RI = {q ∈ Q|qI ⊂ R} and ∗ { ∈ | ⊂ } IR = q Q Iq R .

At last we warn that a fractional twosided R-ideal may well be reﬂexive on one side without being so on the other side.

We now apply the general results above to rings of differential operators. Let k be a field of characteristic zero, and A be a commutative k-algebra, reduced and essentially of finite type over k.LetK be the total ring of fractions of A. From theorem 4.1.2 we then have a chain of inclusions D(A) ⊂D(K) ⊂ QD(K),whereQD(K) is the total ring of fractions of D(K). Moreover we have that D(A)isanorderinQD(K). We now recall the notion of essential submodules from section 3.2.2. The next lemma shows that many ideals are in fact fractional:

Lemma 4.1.6 Essential twosided ideals of D(A) are fractional twosided D(A)-ideals.

Proof: Let I be an essential twosided ideal of D(A). By lemma 3.2.3 it contains a regular element of A, which in turn is a regular element of D(A). Let a ∈ A be such an element. Then aD(A), D(A)a ⊂ I and hence we have shown that I is a fractional twosided D(A)-ideal. 2

ThereforewehavethatallessentialidealsofI are fractional, and we may assume that their duals and doubleduals have the same form as in proposition 4.1.4:

107 Proposition 4.1.7 Let I be an essential twosided ideal of D(A). Then I∗ is a twosided ∗ ∗∗ D(A)-submodule of QD(K) such that D(A) ⊂ I ⊂D(K). Moreover I is a twosided ideal of D(A).

∗ Proof: We do the proof for the right side. Then I = {q ∈ QD(K)|qI ⊂D(A)}.SinceI is a twosided ideal of D(A), it follows that D(A) ⊂ I∗ and that I∗ is a twosided submodule of QD(K). By lemma 3.2.3 we have that I contains a regular element a ∈ A.Thusif q ∈ I∗,wehavethatqa ∈D(A), so q ∈D(A)a−1 ⊂D(K). Finally, since 1 ∈D(A) ⊂ I∗, ∗∗ ∗ ∗∗ and I = {q ∈ QD(K)|I q ⊂D(A)}, it follows immediately that I ⊂D(A). 2

This proposition is what we have been aiming for. It tells us that if I is an essential twosided ideal of D(A), then the right and left duals and doubleduals of I are contained in D(K), and hence we have the formulas

I∗ = {D ∈D(K)|DI ⊂D(A)} and I∗∗ = {D ∈D(A)|I∗D ⊂D(A)}

(and similarly for the left side).

We note that the general theory at the beginning of this section applies especially to commutative rings. We will establish some connections between reflexive ideals of commutative algebras and reflexive twosided ideals of their rings of differential operators. To do so we need to show some additional properties of reflexive modules over commutative rings. The definitions of duals and reflexive modules are as before. First we need a result from homological algebra:

108 Lemma 4.1.8 Duals commute with flat extensions on finitely presented modules. That is, if A is a commutative ring, f : A → B is a flat homomorphism and M is a finitely presented A-module, then there is a natural isomorphism

∗ τM ∗ B ⊗A M −→ (B ⊗A M)

0 0 given by the formula τM (b⊗φ)(b ⊗m)=bb f(φ(m))

Proof: In order to simplify the notation, we let F (−)andG(−) denote the two functors

B ⊗A HomA(−,A)andHomB(B ⊗A −,B)fromA − Mod into B − Mod.SinceB is ﬂat over A,wehavethatF and G are both left exact contravariant. Moreover if we deﬁne

τM as in the statement of the lemma it is trivial to verify that τ : F → G is a natural transformation of functors. If M is ﬁnitely generated free A-module, say M = Ar for some integer r,thenbothF (M)andG(M) are isomorphic to Br in an obvious way, and

τM becomes the identity under these identiﬁcations. Finally if M is ﬁnitely presented, we have an exact sequence 0 ←− M ←− Ar ←− As

for some integers r and s, and then the result follows from the diagram

0 - F (M) - F (Ar) - F (As)

∼ ∼ τM = = ? ? ? 0 - G(M) - G(Ar) - G(As)

For a commutative, noetherian and reduced ring A, the the ideal ΓA was deﬁned in section 3.1.1. If furthermore A has an ´etale covering B which is a componentwise

geometrical unibranch, then by theorem 3.2.11 we have that A contains an ideal χA such

109 that B ⊗A χA =ΓB.Wehave:

Lemma 4.1.9 Let A be a commutative, noetherian and reduced ring. Then ΓA is a reﬂexive ideal of A.

Proof: Write A ⊂ A¯ ⊂ K, as at the end of section 3.1.1. Since ΓA contains a regular

element of A, it is an essential ideal of A, and thus we have that ΓA is a fractional ideal of A. Hence by proposition 4.1.4 and lemma 4.1.5 it follows immediately that ∗ ∗∗ ∗ ΓA = {x ∈ K|xΓA ⊂ A}⊃A and ΓA = {x ∈ K|xΓA ⊂ A}⊂A. By lemma 3.1.4 we ∗ have that ΓA is the conductor of A¯ into A, and hence it follows that A¯ ⊂ ΓA .Thusif ∗∗ ¯ ∗ a ∈ ΓA ⊂ A,thenaA ⊂ aΓA ⊂ A,soa ∈ ΓA, again by lemma 3.1.4. This shows that ∗∗ ΓA ⊂ ΓA . The other inclusion is always true. 2

Proposition 4.1.10 Let A be a commutative, noetherian and reduced ring, and assume that A has an ´etale covering B which is a componentwise geometrical unibranch. Then the ideal χA is reﬂexive.

Proof: We have by deﬁnition that ΓB = B ⊗A χA, and by lemma 3.2.5 it follows that

χA contains a regular element of A. Therefore is χA a fractional ideal of A.SinceB is flat over A, lemma 4.1.8 implies that duals on finitely presented modules commute with the extension A → B, and since ideals of noetherian rings are finitely presented, ∗ ∗ this implies that ΓB = B ⊗A χA . Fractional ideals of commutative noetherian rings are obviously isomorphic (as modules) to ideals, and are hence finitely presented. Therefore ∗ ∗∗ ∗∗ we can use lemma 4.1.8 on χA to get that ΓB = B ⊗A χA . The ring B is noetherian

and reduced, so the ideal ΓB is reﬂexive by lemma 4.1.9. Hence we get that B ⊗A χA = ∗∗ ∗∗ ΓB = B ⊗A χA .SinceA → B is faithfully ﬂat, we thus have that χA = χA. 2

110 4.2 Rings satisfying the condition S2

4.2.1 Minimal reﬂexive modules

We now introduce the condition S2 for commutative noetherian domains. If A is a commutative noetherian domain, it is clear that we may identify A with a subring of K,its ﬁeld of fractions. Moreover, if S is any multiplicatively closed subset of A (not containing zero), then we may identify S−1A with a subring of K such that the localization map A → S−1A is simply the inclusion. We use this convention in what follows:

Deﬁnition Let A be a commutative noetherian domain and let X denote the set of all

height 1 prime ideals of A.WesaythatA satisﬁes the condition S2(or the condition 2 T of Serre) if A = p∈X Ap.

An immediate task is to show that if A is a k-algebra which satisﬁes the condition

S2,thenD(A) possesses a similar property (cf. [4]):

Lemma 4.2.1 Let A be a commutative noetherian k-algebra. Assume that A is a domain which satisﬁes the condition S2, and let K denote the ﬁeld of fractions of A. Then as subrings of D(K), \ D(A)= D(Ap) p∈X

Proof: From a remark following theorem 2.2.5 it follows that

D(A)={D ∈D(K)|D ∗ A ⊂ A}.

Let p ∈ Spec A.SinceK is also the ﬁeld of fractions of Ap, a similar formula holds for

D(Ap). Furthermore, since A → Ap is injective and formally ´etale, we have by the same

111 remark that D(A) ⊂D(Ap). This is true for all p ∈X, and hence the inclusion “⊂” follows. T ∈ D ∈ ∗ ∈ Conversely assume that D p∈X (Ap). Then if a A,wehavethatD a Ap for all p ∈X, and since A satisﬁes the condition S2, this implies that D ∗ a ∈ A. Hence is D ∗ A ⊂ A andwehavethatD ∈D(A). 2

Next we show a connection between reflexive ideals of certain commutative algebras, and reflexive ideals in their rings of differential operators:

Lemma 4.2.2 Let k be a field of characteristic zero and let A be a commutative, reduced k-algebra, and essentially of finite type over k.IfI ⊂ A is an essential D(A)-submodule of A, and a reflexive ideal of A, then D(A, I) is a reflexive right ideal of D(A).

Proof: Lemma 3.2.4 implies that D(A, I) is an essential twosided ideal of D(A). Hence by the remarks following proposition 4.1.7 we have that

∗ D(A, I) = {D ∈D(K)|DD(A, I) ⊂D(A)}, and ∗∗ ∗ D(A, I) = {D ∈D(K)|D(A, I) D ⊂D(A)}.

By proposition 4.1.4 we have that I∗ = {x ∈ K|xI ⊂ A}. We also recall that in this case we can identify D(A) with the ring {D ∈D(K)|D ∗ A ⊂ A}. From all this it follows ∗ ∗∗ immediately that I∗ ⊂D(A, I) .ThusifD ∈D(A, I) ,thenwehaveespeciallythat I∗D ⊂D(A) and hence I∗(D ∗ A) ⊂ A.SinceI is reﬂexive, this implies that D ∗ A ⊂ I, that is, D ∈D(A, I). 2

112 Lemma 4.2.3 Let k be a field of characteristic zero, let A be a commutative algebra, essentially of finite type over k, and assume that A is a domain which satisfies the condition

S2.IfJ is a twosided nonzero ideal of D(A), then

\ \ ∗ ∗ ∗ J = (J )p = (Jp) p∈X p∈X where −∗ means the dual either as a right or as a left D(A)-module.

Proof: Let K denote the ﬁeld of fractions of A. We may assume that all localizations of A are subrings of K, and their respective rings of diﬀerential operators are contained in D(K). This implies especially that if I is any left D(A)-submodule of D(K), and p ∈ Spec A, then we may identify the left D(Ap)-module D(Ap)⊗D(A)I with the set

Ip = ApI⊂D(K). Since J is a twosided essential ideal of D(A), we may use the description of J ∗ following proposition 4.1.7. We will prove the theorem by showing a chain of inclusions: \ \ ∗ ∗ ∗ ∗ J ⊂ (J )p ⊂ (Jp) ⊂J p∈X p∈X

∗ ∗ From the comment above it is clear that J ⊂ (J )p for all prime ideals p of A,so T ∈ J ∗ J⊂ the ﬁrst inclusion certainly holds. Likewise assume that D p∈X ( p) .ThenD T J ⊂D ∈X J⊂ D D D p (Ap) for all p , and hence D p∈X (Ap)= (A) by lemma 4.2.1. The other side is similar. Therefore the last inclusion also holds. ∗ ∗ Thus it only remains to show that (J )p ⊂ (Jp) for all p ∈X.Fixap ∈ Spec A. ∗ By proposition 2.3.10, Jp is a twosided ideal of D(Ap), so (Jp) is a twosided D(Ap)- ∗ ∗ ∗ ∗ submodule of D(K). Since (J )p = ApJ it therefore suﬃces to show that J ⊂ (Jp) .

I.e., we must show that if D ∈D(K) is such that DJ⊂D(A), then DJp ⊂D(Ap), and the same for the other side. Since Jp is a twosided ideal of D(Ap), and Jp =

ApJ , it follows immediately that J D ⊂D(A) implies that JpD ⊂D(Ap). Similarly if −1 0 −1 0 DJ⊂D(A)ands D is an element of Jp (with s ∈ Ap and D ∈J), then we use proposition 2.3.4 to move D to the right of s−1 and the result follows. 2

113 Remark We have not been able to prove that duals commute with localizations in ∗ ∗ D(A). We only have that (J )p ⊂ (Jp) . However if A is also a geometrical unibranch, equality follows from [4].

Theorem 4.2.4 Let A be a local domain of krull dimension 1, a localization of a ﬁnitely generated k-algebra, where k is a ﬁeld of characteristic zero. Then

(i) A has a unique minimal nonzero left D(A)-submodule χA,

(ii) The ideal D(A, χA) is the unique minimal nonzero twosided ideal of D(A),

(iii) χA is reﬂexive as an A-module,

(iv) D(A, χA) is a reﬂexive right ideal of D(A).

Proof: Parts (i)–(ii) are just restatements of corollary 3.2.16, part (iii) follows directly from proposition 4.1.10 combined with corollary 3.2.13, and part (iv) follows from part (iii) and lemma 4.2.2. 2

Remark The existence of a submodule and ideal as in parts (i) and (ii) in the theorem was previously known, at least in the special case when A is the localization of an aﬃne curve (cf. [17, 25]). However, our description of these objects and the fact that they are reﬂexive is new.

We use this to get a result which holds for domains that satisfy the condition S2:

Theorem 4.2.5 Let k be a field of characteristic zero, and let A be a commutative k- algebra, essentially of finite type over k.LetX denote the set of all height 1 prime ideals of A.IfA is a domain which satisfies the condition S2, then T

(i) the ideal I = p∈X χAp , where the ideals χAp are as in theorem 4.2.4, is the unique minimal element of the set

{J|J is a reﬂexive nonzero ideal of A and a D(A)-submodule of A };

114 (ii) The twosided ideal D(A, I) is reﬂexive as a right D(A)-module and it is contained in all twosided nonzero ideals of D(A) which are reﬂexive on either side.

Proof: (i) For every p ∈X, Ap satisfies the conditions of theorem 4.2.4 and hence it has D aleft (Ap)-submodule χAp as prescribed there. Since χAp = Ap for every p at which A is T 6 ∈X regular, χAp = Ap for at most finitely many p . ThereforewehavethatI := p∈X χAp is a nonzero ideal of A. Moreover, the description of D(A) in lemma 4.2.1 implies that I D ∈X ⊂ is a left (A)-submodule of A. For every p we then have that Ip χAp is a nonzero D left (Ap)-submodule of Ap, and hence Ip = χAp by minimality. Every χAp is reflexive, so by [2, theorem VII.4.1] it follows that

\ \ \ ∗∗ ∗∗ ∗∗ I = Ip = χAp = χAp = I p∈X p∈X p∈X

I.e., I is a nonzero reﬂexive ideal of A.

Now let 0 =6 J ⊂ A be a D(A)-submodule of A. Then for every p ∈X, Jp is a D ⊂ ∗ ⊂ ∗ nonzero (Ap)-submodule of Ap, so by minimality, Ip = χAp Jp. Hence is Jp Ip

for all p ∈X.SinceA satisfies the condition S2, we have, again by [2, theorem VII.4.1], T ∗ ∗ ∗ ⊂ ∗ that J = p∈X Jp , and therefore J I . By taking duals again we thus have that I = I∗∗ ⊂ J ∗∗. ThisimpliesthatifJ is a nonzero left D(A)-submodule of A,anda reflexive ideal, then I ⊂ J. (ii). Lemma 4.2.2 implies that D(A, I)isareflexiveright ideal of D(A). Let J be a

nonzero twosided ideal of D(A). Then for all p ∈X, Jp is a nonzero twosided ideal of

D(Ap) and hence by corollary 3.2.16 and the remark after theorem 2.2.10 we have that D D D ⊂J (A, I)p = (Ap,Ip)= (Ap,χAp ) p. Taking duals and using lemma 4.2.3 we thus have that \ \ J ∗ J ∗ ⊂ D ∗ D ∗ = ( p) ( (A, I)p) = (A, I) p∈X p∈X

115 By taking duals again we then have in any case that

∗∗ D(A, I) ⊂D(A, I) ⊂J∗∗.

∗∗ Hence if J is reﬂexive on either side, it contains D(A, I) , which in turn is equal to D(A, I) if it is the doubledual of D(A, I) as a right ideal. 2

116 Chapter 5

Algebras related to rings of diﬀerential operators

Let A be a commutative, noetherian and reduced k-algebra and assume that A has an ´etale covering which is a componentwise geometrical unibranch. By theorem 3.2.2 it is

then possible to deﬁne the essential D(A)-submodule χA of A, and the essential twosided

ideal D(A, χA)ofD(A). In particular, χA is an ideal of A, and it is not contained in any of the minimal prime ideals of A. Hence if A is of ﬁnite krull dimension it follows that

A/χA is of krull dimension strictly less than A. This implies especially that if k is a field, and A is a finitely generated k-algebra of krull dimension 1 which satisfies the conditions above, then A/χA is a finite dimensional vector space over k. ThesameistrueifwereplaceA with its completion at a maximal ideal. From this, the definition of rings of differential operators, and lemma 1.2.2 it

follows that the ring D(A)/D(A, χA) is isomorphic to a subalgebra of Mn(k), the algebra

of all n × n-matrices over k,wheren =dimk(A/χA). In the case that k is an algebraically closed ﬁeld of characteristic zero and A is

reduced and a ﬁnitely generated k-algebra of krull dimension 1, the algebras A/χA and

D(A)/D(A, χA) have been studied in [3, 17, 23, 24, 25] and they have been given the

117 names

C(A):=A/χA and

H(A):=D(A)/D(A, χA)

We use these names whenever χA can be defined, i.e., when A is a commutative, noetherian and reduced k-algebra which has an étale componentwise geometrical unibranch covering. In this chapter we prove two results. We give a formula for the dimension of C(A) if A is the completion at a maximal ideal of an affine plane curve over an algebraically closed field of characteristic zero. This is a partial answer to a question asked in [24].

We also identify the canonical image of the algebra H(A) in the ring D(A/χA)ifA is a reduced algebra, essentially of ﬁnite type over a ﬁeld of characteristic zero, and with irreducible components that are regular. This is an analog of [3, theorems 4.5, 4.6], and the proof is similar.

5.1 The algebra C(A)

5.1.1 A dimension formula related to aﬃne plane curves

Let A be a commutative, noetherian and reduced k-algebra. For such an algebra we defined the ideal ΓA in section 3.1.1. It is clear from the definition that ΓA is not contained in any of the minimal prime ideals of A.Thusifk is a field and A is a finitely generated and reduced k-algebra of krull dimension 1, or a completion thereof, it follows that A/ΓA is a finite dimensional vector space over k. In this section we will give a formula for the dimension of this vector space (in terms of intersection multiplicities) in some special cases.

118 For power series f1, ..., fn in k[[X, Y ]] and i =1,..., n we let

b fi := f1 ···fi−1fi+1 ···fn, and for 0 ≤ i

b fij := f1 ···fi−1fi+1 ···fj−1fj+1 ···fn

(where the product of zero number of factors is deﬁned as 1).

Lemma 5.1.1 Let k be a ﬁeld and let f1,..., fn be in k[[X, Y ]]. Assume that for all i, b the power series fi and fi have no common components. Then there is an exact sequence

b b ψ b b 0 −→ k[[X, Y ]]/(f1,n,...,fn−1,n) −→ k[[X, Y ]]/(f1,...,fn) π b −→ k[[X, Y ]]/(fn, fn) −→ 0

where the map ψ is deﬁned by ψ(z):=fnz, and π is the natural projection (we note that b b b (f1,...,fn) ⊂ (fn, fn)).

Proof: First of all we observe that the map ψ is indeed well deﬁned. The map π is surjective, and exactness in the middle follows immediately from the deﬁnitions of the maps. Hence we only need to show that ψ is injective. For that we imitate the proof b b of [5, p. 76]. Namely, let z ∈ k[[X, Y ]] be such that ψ(z) = 0, i.e., fnz = u1f1 + ...+ unfn for some u1,...,un ∈ k[[X, Y ]]. Then

b b b unfn = fnz − u1f1 − ...− un−1fn−1 b b = fn(z − u1f1,n − ...− un−1fn−1,n)

119 b and since by assumption fn and fn have no common factors, and k[[X, Y ]] is a unique b b factorization domain, this implies that fn divides the power series z − u1f1,n − ...− b un−1fn−1,n,say b b b ufn = z − u1f1,n − ...− un−1fn−1,n where u ∈ k[[X, Y ]]. Hence

b b b z = ufn + u1f1,n + ...+ un−1fn−1,n b b ∈ (f1,n,...,fn−1,n)

For an algebraically closed ﬁeld k and power series f and g in k[[X, Y ]], we let i(f ∩ g) denote the intersection multiplicity of f and g at the maximal ideal, deﬁned as in [5, 6]. If gcd(f,g)=1wehavethat

i(f ∩ g)=dimk(k[[X, Y ]]/(f,g))

Lemma 5.1.2 Let k be an algebraically closed ﬁeld and let f1, ..., fn ∈ k[[X, Y ]] be such that gcd(fi,fj)=1whenever i =6 j. Then

X b b dimk(k[[X, Y ]]/(f1,...,fn)) = i(fi ∩ fj) 1≤i

Proof: First we observe that the assumptions in this lemma imply the assumptions of lemma 5.1.1. We prove the lemma by induction on n, and note that for n = 2 this is just the formula b given above. From the deﬁnition of fn and the properties of intersection multiplicities

120 (cf. [6, examples 1.1.1, 1.2.5d]) we have that

nX−1 b dimk(k[[X, Y ]]/(fn, fn)) = i(fi ∩ fn). i=1

b b By the induction hypothesis we may assume that k[[X, Y ]]/(f1,n,...,fn−1,n) is a ﬁnite dimensional vector space over k and that

X b b dimk(k[[X, Y ]]/(f1,n,...,fn−1,n)) = i(fi ∩ fj). 1≤i

Finally by using the exact sequence of k-vectorspaces from lemma 5.1.1 we deduce that

b b dimk(k[[X, Y ]]/(f1,...,fn)) b b b =dimk(k[[X, Y ]]/(f1,n,...,fn−1,n)) + dimk(k[[X, Y ]]/(fn, fn)) X nX−1 = i(fi ∩ fj)+ i(fi ∩ fn) ≤ ≤ − 1 Xi

With this we can now prove:

Theorem 5.1.3 Let A be the k-algebra

A := k[[X, Y ]]/(f1f2 ···fn)

where k is an algebraically closed ﬁeld, and f1, ..., fn are irreducible power series such that gcd(fi,fj)=1whenever i =6 j. Then

X dimk(A/ΓA)= i(fi ∩ fj) 1≤i

121 Proof: Let π denote the natural projection k[[X, Y ]] → A. The minimal prime ideals of A are precisely the ideals (π(fi)), i =1,..., n, and then from lemma 3.1.2 we have T b that Γ(π(fi)) = j=6 i(π(fj)) = (π(fi)) (where the last equality follows from the fact that k[[X, Y ]] is a unique factorization domain). This, and the deﬁnition of ΓA implies that b b ΓA =(π(f1),...,π(fn)),andthenwehavethat

∼ b b A/ΓA = k[[X, Y ]]/(f1,...,fn)

Thus the result follows directly from lemma 5.1.2. 2

Let A be a complete local ring with maximal ideal m and residue field k. Assume that A is noetherian, reduced, of krull dimension 1 and of embedding dimension 2 (i.e., m is generated by two elements). Furthermore assume that k is algebraically closed and ∼ of characteristic 0. Then A = k[[X, Y ]]/I for some ideal I of k[[X, Y ]] (cf. [14, §29]). T n Since we assume that A reduced and of krull dimension 1, it follows that I = i=1 pi where p1, ..., pn are (distinct) height 1 prime ideals of k[[X, Y ]]. By [14, theorem 20.1] we have that all the ideals pi are principal, say pi =(fi) for some irreducible power series T ∈ n ··· fi k[[X, Y ]]. This shows that I = i=1(fi)=(f1f2 fn), and therefore theorem 5.1.3 applies to the ring A. We observe especially the following. Let R bethecoordinateringofareducedaffine plane curve over an algebraically closed field k of characteristic 0, i.e., R = k[X, Y ]/I where I is a radical ideal. If A = Rˆ, the completion of R at some maximal ideal, then A is as above (A is reduced since R is excellent, cf. [7]) and hence theorem 5.1.3 applies.

122 5.2 The algebra H(A)

5.2.1 A description when the irreducible components of A are regular

Let A be a commutative noetherian and reduced k-algebra. By the remark following lemma 3.2.4 we have that ΓA is an essential D(A)-submodule of A,andsoD(A, ΓA)is an essential ideal of D(A). By lemma 1.2.2 we then have an injective morphism

φ : D(A)/D(A, ΓA) −→ D (A/ΓA).

In general we do not expect this morphism to be surjective. For example we observe that by lemmas 3.1.2 and 3.2.1 every p ∈ Ass(A)isaD(A)-submodule of A, and thus it

follows that φ(D) ∗ (p +ΓA)/ΓA ⊂ (p +ΓA)/ΓA. I.e., we have that

Im φ ⊂{D ∈D(A/ΓA)|D ∗ ((p +ΓA)/ΓA) ⊂ (p +ΓA)/ΓA for all p ∈ Ass(A)}

We will show that sometimes this inclusion is in fact an equality (cf. [3, 25]).

Let I be an ideal of a commutative k-algebra A.IfI is not a left D(A)-submodule of A,thenD(A, I) is only a right ideal of D(A) but not a left ideal. Hence we introduce the following notation. Let ID(A)(D(A, I)) denote the idealizer of D(A, I)inD(A). That

is, ID(A)(D(A, I)) is the largest subring of D(A)inwhichD(A, I) is a twosided ideal. It follows immediately that

ID(A)(D(A, I)) = {D ∈D(A)|D ∗ I ⊂ I}

123 and it is clear that order of differential operators defines a filtration on ID(A)(D(A, I)). If I is a left D(A)-submodule, by lemma 1.2.2 the natural projection A → A/I induces an homomorphism π : D(A) →D(A/I) of filtered rings, and its kernel is equal to D(A, I). If I is not a D(A)-submodule of A, this is of course no longer true. However, by using the idealizer, we can state a similar result which holds for all ideals of A.

Lemma 5.2.1 Let A be a commutative k-algebra, and let I be an ideal of A.Letπ :

A → A/I be the natural projection. Then there is a homomorphism π : ID(A)(D(A, I)) →

D(A/I) of ﬁltered rings, deﬁned for D ∈ ID(A)(D(A, I)) by the formula π(D) ∗ π(a):= π(D ∗ a) for all a ∈ A. The kernel of this map is precisely D(A, I), so we have an exact sequence π 0 −→ D (A, I) −→ ID(A)(D(A, I)) −→ D(A/I)

Proof: Same as the proof of lemma 1.2.2, after recalling that ID(A)(D(A, I)) is equal to {D ∈D(A)|D ∗ I ⊂ I}. 2

Let A be a commutative, noetherian and reduced k-algebra, and as in section 3.1.1, Q let A := p∈Ass(A) A/p. For simplicity of notation we will identify A with its image in A, and by lemma 3.2.7 we may also identify all the ideals Γp with their canonical images Q ⊂ in the rings A/p. With these identiﬁcations we have that ΓA = p∈Ass(A) Γp A.This Q shows that A/ΓA = p∈Ass(A)(A/p)/Γp, and thus

Y D(A)= D(A/p) p∈Ass(A) Y D(A, ΓA)= D(A/p, Γp) p∈Ass(A) Y D(A/ΓA)= D((A/p)/Γp). p∈Ass(A)

With this description it is now clear that

Y I D I D D(A)( (A, ΓA)) = D(A/p)( (A/p, Γp)), p∈Ass(A)

124 I D →D and that the canonical morphism D(A)( (A, ΓA)) (A/ΓA) of lemma 5.2.1 is equal to the direct product of the canonical morphisms

ID(A/p)(D(A/p, Γp)) −→ D ((A/p)/Γp).

Using this we can now prove:

Lemma 5.2.2 Let k be a ﬁeld of characteristic zero, and let A be a commutative k- algebra, essentially of ﬁnite type and reduced. Assume that A/p is regular for all p ∈ Ass(A). Then the natural morphism

I D −→ D D(A)( (A, ΓA)) (A/ΓA) is surjective.

Proof: By using the descriptions above, it is suﬃcient to prove that the morphisms

ID(A/p)(D(A/p, Γp)) −→ D ((A/p)/Γp) are surjective for all p ∈ Ass(A). However, since we assume that the rings A/p are regular for all p ∈ Ass(A), this follows directly from [25, proposition 1.6]. 2

We also recall that the inclusion A ⊂ A induces an inclusion D(A) ⊂D(A)insucha way that D(A)={D ∈D(A)|D ∗ A ⊂ A}.Wehave:

Lemma 5.2.3 Let A be a commutative, noetherian and reduced k-algebra. Then

D(A, A)=D(A, ΓA)=D(A, ΓA)

125 Proof: We identify A and D(A) with their images in A and D(A). As before proposition 3.1.4, we let ei be the indecomposable idempotents of A corresponding to the minimal prime ideals of A.SinceA is a direct product of rings, D(A) is also a direct product of rings by lemma 1.2.5, and then it is clear that all the eis commute with all P D elements of (A). We recall also that A = i eiA and that ΓA is in fact the conductor of A into A. Let D be in D(A). Having the above it follows that

D ∈D(A, A) ⇔ D ∗ eiA ∈ A for all i

⇔ eiD ∗ A ∈ A for all i

⇔ D ∗ A ⊂ ΓA

⇔ D ∈D(A, ΓA).

This shows that D(A, A)=D(A, ΓA). For the third equality we ﬁrst observe that

D(A, ΓA) is obviously contained in D(A, ΓA). Conversely let D ∈D(A, ΓA). Since

ΓA is an ideal of A,wehavethatD ∗ eiA = eiD ∗ A ⊂ eiΓA ⊂ ΓA for all ei and hence it follows that D ∈D(A, ΓA). 2

Having established these lemmas, we can now prove the main result of this section:

Theorem 5.2.4 Let k be a ﬁeld of characteristic zero, and let A be a commutative noetherian and reduced k-algebra, essentially of ﬁnite type. Assume that A/p is regular for all p ∈ Ass(A). Then there are canonical isomorphisms

H(A):=D(A)/D(A, ΓA) ∼ = {D ∈D(A/ΓA)|D ∗ (A/ΓA) ⊂ A/ΓA} ∼ = {D ∈D(A/ΓA)|D ∗ ((p +ΓA)/ΓA) ⊂ (p +ΓA)/ΓA for all p ∈ Ass(A)}

126 Proof: Let I D T := D(A)( (A, ΓA)).

It is clear that D(A)={D ∈ T |D ∗ A ⊂ A},andD(A, ΓA) is an ideal of T by deﬁnition.

We let π : A → A/ΓA denote the natural projection, and for D ∈D(A), we let D denote its image in H(A). Having set this notation it now follows from our assumptions on A andallthelemmasinthissectionthat

∼ D(A)/D(A, ΓA) ⊂ T/D(A, ΓA)=T/D(A, ΓA) = D(A/ΓA)

Furthermore it is clear that the (injective) morphism D(A)/D(A, ΓA) →D(A/ΓA)is given by D 7→ π(D). From this it follows that if D ∈D(A), then π(D) ∗ (A/ΓA)=

π(D) ∗ π(A)=π(D ∗ A) ⊂ π(A)=A/ΓA. This shows that we have a well deﬁned and injective canonical homomorphism

H(A) −→ { D ∈D(A/ΓA)|D ∗ (A/ΓA) ⊂ A/ΓA}.

For surjectivity let D ∈D(A/ΓA) be such that D ∗ (A/ΓA) ⊂ A/ΓA. By lemma 5.2.2 there is a D0 ∈ T with π(D0)=D, and then we have for all a ∈ A that

0 0 π(D ∗ a)=π(D ) ∗ π(a)=D ∗ π(a) ∈ A/ΓA.

This shows that D0 ∗ A ⊂ A,soD0 ∈D(A), and hence D is the image of D0 ∈

D(A)/D(A, ΓA). This proves the ﬁrst isomorphism To prove the second isomorphism, let

S := {D ∈D(A/ΓA)|D ∗ ((p +ΓA)/ΓA) ⊂ (p +ΓA)/ΓA for all p ∈ Ass(A)} and 0 S := {D ∈D(A/ΓA)|D ∗ A ⊂ A}

127 As mentioned earlier, we have the canonical injective homomorphism

φ : D(A)/D(A, ΓA) → S.

We have already established the isomorphism H(A) → S0, so in order to prove that φ is an isomorphism, it suﬃces to provide an injective morphism S → S0 making the diagram

∼ = - 0 H(A) S @ 6 @ φ @ @@R S commutative.

It is clear that A/ΓA ⊂ A/ΓA, so our first problem is to show that differential operators in S extend to differential operators on A/ΓA. ∈ 0 → Fix a p Ass(A)andletπp : A/ΓA A/(p +ΓA) denote the canonical projection. By definition of S it is contained in the ring

I D D(A/ΓA)( (A/ΓA, (p +ΓA)/ΓA)), so it follows immediately from lemma 5.2.1 that we have a canonical ring homomorphism →D ∗ 0 0 ∗ ∈ ψp : S (A/(p +ΓA)) such that ψp(D) πp(a)=πp(D a) for all D S and all a ∈ A/ΓA.

We now observe that (p +ΓA)/p =Γp, so it follows immediately that A/(p +ΓA)is 0 → canonically isomorphic to (A/p)/Γp. We therefore get a ring homomorphism ψp : S D 00 → → ((A/p)/Γp), and if we let πp denote the composite A/ΓA A/(p +ΓA) (A/p)/Γp, 0 ∗ 00 00 ∗ then it is trivial to check that ψp(D) πp (a)=πp (D a). Q ∈ The above is true for all p Ass(A), and since ΓA = p∈Ass(A) Γp, we get a morphism Q 0 → D ∼ D ψ : S p∈Ass(A) ((A/p)/Γp) = (A/ΓA). Furthermore we observe that under the

128 identiﬁcation Y ∼ (A/p)/Γp = A/ΓA, p∈Ass(A) 00 → the map (πp )p∈Ass(A) : A/ΓA A/ΓA is simply the natural inclusion. Therefore it follows 0 0 that ψ (D) ∗ a = D ∗ a for all D ∈ S and all a ∈ A/ΓA.Thuswehavethatψ is injective, 0 and Im ψ0 ⊂ S0. Finally it is easy to verify that the composite H(A) → S →ψ S0 is equal ∼ to the isomorphism H(A) →= S0 constructed in the ﬁrst half of the proof. 2

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