Descent for differential modules and skew fields

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Journal of Algebra 296 (2006) 18–55

www.elsevier.com/locate/jalgebr

Descent for differential modules and skew field

Mark van Hoeija,∗,1, Marius van der Putb

a Department of Mathematics, Florida State University, Tallahassee, FL 32306, USAb Department of Mathematics, University of Groningen, PO Box 800, 9700 AV Groningen, The Netherl

Received 13 March 2004

Communicated by Eva Bayer-Fluckiger

Abstract

In this paper we study rationality questions for differential modules and differential operIf a differential operatorL is equivalent to its conjugates overk, is it then equivalent to an operatdefined overk? We will show how counterexamples to this question correspond to skew fields, awill make this correspondence explicit in both directions. Similar questions are studied for proequivalence of differential operators. The main tool is the study of differential modules overfields. 2005 Elsevier Inc. All rights reserved.

1. Introduction

1.1. Examples of descent phenomena

Let K/k be a Galois extension of differential fields of characteristic 0. The skew rindifferential operators overK is denoted byD := K[∂]. An elementσ in the Galois groupGal(K/k) acts onD by σ(

∑ai∂

i) = ∑σ(ai)∂

i . Theconjugatesof an elementL ∈D arethe operatorsσ(L). Theorder of L is the degree in∂ . If L ∈ k[∂] thenL is calledrational.

* Corresponding author.E-mail addresses:[email protected] (M. van Hoeij), [email protected] (M. van der Put).

1 Supported by NSF grant 0098034.

0021-8693/$ – see front matter 2005 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2005.09.041

M. van Hoeij, M. van der Put / Journal of Algebra 296 (2006) 18–55 19

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Two operatorsL1, L2 are calledequivalent(or of thesame type) if the two differentialmodulesD/DLi , i = 1,2 are isomorphic. The operators are calledprotectively equivalenif there exists aD-moduleE of dimension 1 overK such thatD/DL1 ∼= E ⊗D/DL2.

Abbreviations:

R Rational, i.e. element ofk[∂].ER Equivalent to an element ofk[∂].EC Equivalent to all its conjugates overk.

EpR Projectively equivalent to an element ofk[∂].

Ep

C Projectively equivalent all its conjugates overk.

The following implications are obvious:

R ⇒ ER ⇒ (EC and Ep

R

),

(EC or Ep

R

) ⇒ Ep

C.

These implications leave seven possible combinations for the truth-values of R, ER, EC,Ep

R, Ep

C. In casek = Q(x) andK = Q(x), all seven cases occur. The examples, giveTable 1, are irreducible inD.

The cases of interest are 3, 4, and 6. An example of a second order operator thaCbut not ER was already give in [H, pp. 101–102] using computer computations. Tacan also be verified by computer computations, however, this would not explain thelying mathematics nor where these examples come from. That is the main themepaper.

DenoteCK respectivelyCk as the field constants ofK respectivelyk. Suppose thatK =CK(x) andk = Ck(x) with differentiation d

dx. Then the descent phenomenon “EC without

ER” (case 3 or 4) is related to skew fieldsF 0 of finite dimension over their centerCk .The main result in Section 2 is that any such skew field yields examples for “EC withoutER.” Case 6 (a descent phenomenon up to projective equivalence) also corresponskew fieldF , this time of finite dimension over its centerk. For second-order equationthis skew field is a quaternion field and corresponds to a conic overk. The main resulin Section 3 is that all skew fields of this type produce examples for case 6. For mour constructions it will be more convenient to use modules instead of operators. Woperators in Table 1 for compactness of notation.

Table 1

Case R ER EC EpR Ep

C Example

1 Y Y Y Y Y ∂

2 N Y Y Y Y ∂ − 1/(x − i)

3 N N Y Y Y (x2 + 1)∂2 + (2x + i)∂ + 14 N N Y N Y ∂2 + x2 + 1+ i

5 N N N Y Y ∂ + i

6 N N N N Y 4x∂2 − 2∂ + x2 + x + i

7 N N N N N ∂2 + i

20 M. van Hoeij, M. van der Put / Journal of Algebra 296 (2006) 18–55

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1.2. Descent problems

In Section 2, one considers a differential moduleM overK which is isomorphic to alits conjugatesσ M under the Galois group ofK/k. The question is whetherM descendsto k, i.e.,M ∼= K ⊗k N for some differential moduleN overk. The obstruction to desceis a 2-cocycle which corresponds to a skew field, see Theorem 2.10 and Propositiofor a general differential fieldk, with additional results fork = Ck((x)) in Theorem 2.4 andfor k = Ck(x) in Theorem 2.8. In Section 2.4 we show how all possible examplesk = Ck(x) can be constructed. For completeness, Amitsur’s completely different contion (which makes a very special case of the descent problem explicit, namely partProposition 2.11) is presented in Section 2.5.

In Section 3 we study the problem whether a 3-dimensional differential moduleM isthe second symmetric power of a 2-dimensional differential moduleN . At the heart of therationality issues in this problem is a conic, and we show that every conic overk = Ck(x)

occurs. A surprising result is Theorem 4.7 in Section 4, which states that the isomorclass of a solutionN of this problem is unique up to tensoring with 1-dimensional moduif the field of constants is algebraically closed. This implies that the rationality issuSection 3 can also be viewed as a “projective descent problem,” which is the subTheorem 4.4. In this type of descent problem we consider modules that are not neceisomorphic to their conjugates, but only isomorphic up to tensoring with 1-dimensmodules.

1.3. Motivation and overview of the main results

In Section 2.3 we follow standard Galois cohomology techniques to describe dephenomena for differential modules. This way one can explain the descent phenin terms of 2-cocycles or skew fields, and classify descent phenomena in one dir“a differential module that is isomorphic to its conjugates⇒ a skew field” but not in theopposite direction: Which skew fields occur this way? Our goal is a thorough classificand hence we study this question in detail for several common differential fields. Inrem 2.4 in Section 2.1 we show that over the field of formal Laurent series only thecase occurs (there is no obstruction to descent) whereas in Section 2.4 we show ththe rational functions every skew field occurs. Each of these results require compdifferent techniques. The key ingredient in our construction “skew field⇒ differentialmodule” is the introduction of differential modules over skew fields, which are then vieas differential modules over a commutative subfield. To illustrate this idea we give exexamples. We then give an construction that we prove to be complete (it provides eexamples for every skew field, and every example can be constructed this way). Oustruction also implies an interesting complexity result, namely that factoring fourth-operators inQ(x)[∂] is at least as hard as finding rational points on a conic overQ. Thelatter problem involves factoring integers, which is generally assumed to be very ha

One of the motivations to study the descent problem is the question which exteof the constants need to be considered in algorithms for solving differential equationthe results in Section 2 are of interest for factoring differential operators.


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Section 3 is motivated by the problem of trying to reduce a third-order linear diffetial equation to a second-order equation by determining if the corresponding 3-dimendifferential module is the symmetric square of a 2-dimensional module. This problequires finding a point on a conic that is defined over the differential fieldk. Such a coniccorresponds to a skew field of dimension 4 over its centerk. A natural question is: Caevery conic occur? For the formal Laurent series again only the trivial case occurs,the rational functionsk = Ck(x) we show explicitly in Section 3.2 that every conic ovekoccurs. This result implies that if an algorithm attempts to reduce a third-order differequation by trying to write the corresponding module as a symmetric square, thendegree extension of the constants could be necessary, see Section 3.2.2. Our conin Section 3 differs from the one in Section 2, it uses skew fields overk instead of overCk

(the field of constants ofk).In Theorem 4.7 in Section 4 we show that if a 3-dimensional differential module i

symmetric square of a 2-dimensional module, then this 2-dimensional module is uup to projective equivalence, provided that the field of constants is algebraically cCorollary 4.2 shows that this is then also true overCk(x) whenCk is not algebraicallyclosed except in the imprimitive case, and to complete the result we give counterexafor the imprimitive case in Examples 4.3. The conic occurs in Sections 3 and 4dimensional submodule of the symmetric square of a 3-dimensional module. How tothe skew field that corresponds to this conic in terms of Galois cohomology is shoTheorem 4.4 combined with Theorem 4.7.

2. The descent problem

2.1. Twists and descent data

Let K ⊃ k denote two fields and letM be some object overK . Thedescent problemasks for an objectN overk such thatK ⊗k N is isomorphic toM . We are interested in thcase whereM andN are differential modules. We start with definitions and notation.

Definition 2.1. The twistσ V of a vector spaceV .

Let K be a field andσ an automorphism ofK . For any vector spaceV over K

one associates a vector spaceσ V which is equal toV as additive group and has a nescalar multiplication defined byλ ∗ v := σ−1(λ)v for all λ ∈ K and v ∈ V . One hasσ1(σ2V ) = σ1σ2V . For anyK-linear mapf :V → W betweenK-vector spaces one dnotes byσ f : σ V → σ W the same mapf , which is K-linear for the new structures. Ithis way, one has defined a functor from the category of theK-vector spaces to itself. Thfunctor commutes with tensor products, exterior powers, symmetric powers, etc. Ifb is thematrix of f with respect to a basis ofV and a basis ofW , thenσ(b) is the matrix ofσ f

with respect to the same bases. Hereσ acts onb by acting on the entries.Let V,W beK-vector spaces and letσ be an automorphism ofK . The mapf :V → W

is calledσ -linear if f is additive andf (λv) = σ(λ)f (v) for all λ ∈ K , v ∈ V . To a linear


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mapf : σ V → W we associate aσ -linear mapF :V → W by composing the set-theoretidentityV → σ V with f .

Observation. Let K ⊃ k be a finite Galois extension with Galois groupG. Let V be avector space overK . Then there exists a natural linear map fromK ⊗k V to σ V , and anatural isomorphism fromK ⊗k V to

⊕σ∈G

σ V .

The maps are given bya ⊗ v �→ a ∗ v = σ−1(a)v. The map to⊕

σ∈Gσ V is one-to-one

(hence onto by comparing dimensions) because of the linear independence of auphisms.

Definition 2.2. Descent data and the descent problem.

(1) k is a differential field of characteristic zero. Its algebraic closure will be denotedk.Let K ⊂ k be a Galois extension (finite or infinite) ofk. ThenK is also a differentiafield. The action of the Galois groupG of K/k commutes with differentiation onK .Let D := K[∂] denote the skew ring of the differential operators over the fieldK .A differential moduleoverK is a leftD-module, of finite dimension as vector spaover K . Let M be a differential module overK . For σ ∈ G one defines the twisσ M of M as follows: TheD-moduleσ M is M as additive group, has a new scamultiplication as defined in Definition 2.1 and has the same operator∂ .The action ofG on K is extended to an action onD by imposingσ(∂) = ∂ for allσ ∈ G. As usual, one associates to a monic differential operatorL ∈ D the differentialmoduleM = D/DL. Forσ ∈ G one has thatσ M is the differential module associateto σ(L).

(2) Descent datafor M are given by:(i) for eachσ ∈ G an isomorphismφ(σ) : σ M → M of differential modules,

(ii) satisfyingφ(σ) σ φ(τ) = φ(στ) for all σ, τ ∈ G.Let Φ(σ) :M → M be theσ -linear map associated toφ(σ). Then the two conditioncan be formulated as follows:(i) Φ(σ) :M → M is aσ -linear bijection, commuting with∂ ,

(ii) satisfyingΦ(σ)Φ(τ) = Φ(στ) for all σ, τ ∈ G.The above definition coincides with the one used in algebraic geometry. IndeeK/k be a finite Galois extension. The algebraK ⊗k K has a left and a rightK-algebrastructure. LetNl := (K ⊗k K)⊗K M andNr := (K ⊗k K)⊗K M denote the two tensoproducts w.r.t. the left and the right structure. The first part (I) of the descent daK ⊗k K-linear isomorphismNl → Nr which commutes with∂ . The second part (II) othe descent data is a relation between the various structures on(K ⊗k K ⊗k K)⊗K M .Now Nl is isomorphic with the direct sum of[K : k] copies ofM andNr

∼= K ⊗k M

is isomorphic to⊕

σ∈Gσ M . Therefore (I) is equivalent to (i). Furthermore, one c

show that (II) is equivalent with (ii).(3) One says thatM descends tok if there exists a differential moduleN overk such that

M is isomorphic toK ⊗k N . If M descends tok, thenM has obviously descent data


s

cents

ws:

nt

fields

-

-

On the other hand, let descent data forM be given. Choose a basism1, . . . ,mn of M

over K . This also provides a basis forσ M using the set-theoretic identity. Letb(σ )

denote the matrix ofφ(σ) w.r.t. this basis. Then condition (ii) translates into:(ii) b(στ) = b(σ )σ (b(τ )),in other wordsσ �→ b(σ ) is a 1-cocycle forG and GLn(K). Hereσ ∈ G acts on amatrix by acting on its entries. It is well known thatH 1(G,GLn(K)) is trivial (see [S])and that this implies that there exists aK-basism1, . . . ,mn of M such that the matriceb(σ ) of φ(σ) w.r.t. this new basis are 1. NowN := km1 ⊕ · · · ⊕ kmn = {m ∈ M | ∀σ

Φ(σ)(m) = m} is a differential module overk because theΦ(σ) commute with∂ . Thenatural mapK ⊗k N → M is an isomorphism of differential modules, soM descendsto k.

(4) In the sequel we will study the situation where only the first part of the desdata is given, i.e., for a differential moduleM over K a collection of isomorphism{φ(σ) : σM → M} is given. Thedescent problemis to decide whetherM descendsto k or, more generally, to find the differential fields� (say with K ⊃ � ⊃ k) suchthatM descends to�. This problem can be stated for differential operators as folloAssume thatL is EC (see abbreviations in Section 1), isL also ER?

(5) For any algebraic extension of differential fieldsK ⊃ k one can also define descedata and a weak form of descent data, as in (4), for a differential module overK . LetKc

denote the normal closure ofK . Then these data translate into data forN := Kc ⊗K M

w.r.t. the Galois extensionKc/k and with the additional information thatN descendsto K . Therefore we will restrict ourselves to Galois extensions.

We start with a useful lemma. Consider an algebraic extension of differentialk ⊂ K , obtained byextension of constants,and a differential moduleM over K whichdescends tok. Then, by part (a) of the lemma, the moduleN overk with K ⊗k N ∼= M isunique up to isomorphism.

Lemma 2.3. Consider an algebraic extension of differential fieldsk ⊂ K , with fields ofconstantsCk , CK . Suppose thatK = CK · k.

(a) LetM1, M2 be differential modules overK , which descend to modulesN1, N2 overk.Then

HomK[∂](M1,M2) = CK ⊗CkHomk[∂](N1,N2).

Furthermore, ifM1 ∼= M2, thenN1 ∼= N2.(b) LetN be a differential module overk and putM = K ⊗k N . If the groupAutk[∂](N) of

differential automorphisms ofN isC∗k , thenEndk[∂](N) = Ck , andEndK[∂](M) = CK .

Moreover,AutK[∂](M) = C∗K .

Proof. (a) LetW := Homk[∂](N1,N2) denote theCk-vector space of the differential homomorphisms between the two differential modules overk. The naturalCK -linear mapCK ⊗ W → HomK[∂](M1,M2) is a bijection. Indeed, let1 denote the trivial differential module of dimension one overk. Then Homk[∂](N1,N2) ∼= Homk[∂](1,N∗

1 ⊗ N2)

and the latter is the space{a ∈ N∗ ⊗ N2 | ∂a = 0} of the “rational solutions” of the
1


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differential moduleN∗1 ⊗ N2. Similarly, HomK[∂](M1,M2) is the space of the ratio

nal solutions ofM∗1 ⊗ M2 = K ⊗k (N∗

1 ⊗ N2). The bijectivity follows now from [P-SChapter 4, Proposition 4.3]. Consider a basisw1, . . . ,wd of W . There is a polynomiah ∈ K[X1, . . . ,Xd ] (the determinant of

∑Xiwi w.r.t. any bases forN1, N2) such that the

set{(λ1, . . . , λd) ∈ CdK |∑i λiwi is an isomorphism} is given byh(λ1, . . . , λd) �= 0. This

implies that somew ∈ W induces an isomorphism betweenM1 andM2. Thenw is anisomorphism betweenN1 andN2.

(b) PutE := Endk[∂](N) = Homk[∂](N,N). This is a finite-dimensional algebra ovCk with basise1, . . . , ed . For anye = ∑

λiei one writes det(e) for the determinant of thematrix of left multiplication bye on N . This is a polynomial inλ1, . . . , λd . The automor-phisms ofN are the elementse ∈ E with det(e) �= 0. The assumption that this group isC∗

k

implies thatE = Ck . Then, as in (a), EndK[∂](M) = CK and AutK[∂](M) = C∗K . �

If K �= CK · k then lemma does not hold in general. See also [Ka, Lemma 2.7.1similar results.

Theorem 2.4. Let CK/Ck be a finite Galois extension with groupG. Consider the dif-ferential fieldsk = Ck((x)) and K = CK((x)) with differentiationx d

dx. Suppose that th

differential moduleM overK has the property: σ M is isomorphic toM for all σ ∈ G.ThenM descends tok.

Proof. First we treat the case thatM is irreducible. The proof is based on the classificatof irreducible differential modules overK as given in the thesis [So] of R. Sommeling. Trelevant information is the following:

(1) One associates toQ ∈ ⋃m�i CK [x−1/m] the one-dimensional differential modu

K(Q)e over the finite field extensionK(Q) of K , given by∂e = Qe (recall that∂ nowrefers tox d

dxinstead of d

dx). ThenK(Q)e, viewed as a differential module overK , will be

denoted asE(Q). It is irreducible and has dimension[K(Q) : K].(2) Every irreducible differential module overK is obtained in this way.(3) E(Q1) ∼= E(Q2) if and only if there exists aτ in the Galois group ofK/K such

thatτ(Q1)−Q2 ∈ 1rZ, wherer is the ramification index ofK(Q1)/K (or equivalently the

ramification index ofK(Q2)/K).One associates toQ (as above) its monic minimal polynomialfQ ∈ K[T ] overK . Then

(3) translates into [So, Proposition 3.3.6]:(3′) E(Q1) ∼= E(Q2) if and only if there is aλ ∈ 1

rZ, wherer is the ramification index

of K(Q1)/K , such thatfQ1(T ) = fQ2(T + λ).Now suppose thatσ M is isomorphic toM for all σ . Choose aQ ∈ CK [x−1/r ] with

r � 1 minimal, such thatM ∼= K(Q)e. Let fM = fQ ∈ K[T ] be the minimal monicpolynomial ofQ over K . Takeσ ∈ Gal(CK/Ck) = Gal(K/k) and extendσ to an auto-morphismσ of K/k. Thenσ M is seen to be associated toσ (Q). Thereforefσ M = σ(fM).Thusσ M ∼= M translates intoσ(fM)(T ) = fM(T +λ) for someλ ∈ 1

rZ. Sinceσ has finite

order, one hasλ = 0. HencefM = fQ ∈ k[T ]. Now Q defines also an irreducible differe


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tial moduleN := k(Q)e with ∂e = Qe overk of dimension[k(Q) : k]. ClearlyK ⊗k N isisomorphic toM . This completes the proof in caseM is irreducible.

We now sketch the proof that the assumption thatM is irreducible can be omittedAny differential moduleM can be written as a finite direct sum

⊕i E(Qi) ⊗ Ri , where

E(Qi) �∼= E(Qj) for i �= j and where eachRi has the property that the matrix of∂ w.r.t. abasis is nilpotent. This decomposition is unique (see [L]) and theQi are unique up to thequivalence stated in the proof of Theorem 2.4. One observes that eachRi descends tok.This information suffices to prove the theorem without assuming thatM is irreducible. �Remark 2.5.

(1) The following example illustrates that Theorem 2.4 does not hold for positiveacteristic. LetF2 respectivelyF4 = F2(α) denote the fields with 2 respectivelyelements. Letσ be the nontrivial automorphismσ(α) = α2 = α + 1. Let k = F2((x))

andK = F4((x)) with differentiationx′ = 1. LetM = Ke with ∂e = αxe. Thene �→ xe

defines an isomorphismσ M ∼= M , but M does not descend tok. In the rest of thispaper we will only consider characteristic 0.

(2) LetK be any finite (Galois) extension ofCk((x)). SuchK has the formCK((t)) wheret has the propertytm = cx with c ∈ C∗

K . The above theorem and its proof remain vafor this K and any differential moduleM overK . However, in casem > 1, the differ-ential moduleN overk with K ⊗k N ∼= M is no longer unique up to isomorphism.

(3) Theorem 2.4 and (1), (2) above, answer the descent problem for formal Laureries fields. For convergent Laurent series field, the situation is quite different, sand [P2]. In the sequel of this paper we will study the descent problem for the gcase, i.e., for differential fields which are function fields in one variable over theof constants. The situationk = Ck(x) ⊂ K = CK(x), whereCk ⊂ CK is an algebraicextension, is easier to deal with (see Theorem 2.8) than the general case. The foexample illustrates this.

(4) Consider the differential fieldsk = Q(x) andK = Q(t) with x = t − 1t

and with differ-entiation′ given byx′ = 1. Letσ denote the nontrivial automorphism ofK/k. We notethatσ t = −1

t. Define the 1-dimensional differential moduleM = Ke with ∂e = t ′

2te.

Thenσ M ∼= Ke with ∂e = − t ′2t

e and thusM is isomorphic toσ M . We will prove that

there does not existh ∈ K∗ such thata := t ′2t

+ h′h

∈ k. This implies thatM does not

descend tok. Suppose thath exists. Thenσa = a and consequentlyt′t

= σh′σh

− h′h

.Thent = α σh

hfor someα ∈ Q. This yields the contradiction−1= σ(t)t = α2.

2.2. Semi-simple modules and semi-simple algebras

In this subsection some known facts are collected that are useful for the descenlem. It seems that Proposition 2.7 is not available is the literature. For the first staresult we omit the proof.

Lemma 2.6. LetM be a differential module overK . The following are equivalent.


s ofetric

es

l

(1) M is a sum of irreducible submodules.(2) M is a direct sum of irreducible submodules.(3) Every submoduleN ⊂ M is a direct summand, i.e., there exists a submoduleN ′ with

M = N ⊕ N ′.

A differential moduleM , having the equivalent properties of Lemma 2.6, is calledsemi-simpleor completely reducible.

Proposition 2.7. LetK ⊃ k be an algebraic extension of differential fields.

(1) LetN be a differential module overk. ThenN is semi-simple⇔ the differential moduleK ⊗k N overK is semi-simple.

(2) Suppose that[K : k] < ∞ and thatM is a semi-simple differential module overK .ThenM considered as a differential module overk is semi-simple.

(3) LetA,B be semi-simple differential modules overk. ThenA ⊗k B is semi-simple.(4) The collection of semi-simple differential modules is stable under all “operation

linear algebra,” i.e., submodules, quotients, direct sums, tensor products, symmpowers,. . . .

Proof. (1′). We suppose first thatK ⊃ k is a finite Galois extension with Galois groupG.(1′) (⇒) We may suppose thatN is irreducible. Take an irreducibleK-submoduleD

of K ⊗k N . Everyσ ∈ G acts onK ⊗k N in the obvious way and this action commutwith ∂ . Then eachσ(D) is also an irreducibleK-submodule ofK ⊗k N . ThereforeE =∑

σ∈G σ(D) is a semi-simpleK-submodule. Moreover,σ(E) = E for all σ ∈ G. HenceE = K ⊗k F for some submodule ofN . SinceN is irreducible, one has thatF = N andE = K ⊗k N .

(1′) (⇐) Let F ⊂ N be a submodule. ThenK ⊗k F is aK-submodule ofK ⊗k N . Thereexists aK-linearP :K ⊗k N → K ⊗k N with the properties imP = K ⊗k F , P 2 = P andP∂ = ∂P . DefineQ = 1

|G|∑

σ∈G σPσ−1. ThenQ∂ = ∂Q, Qm = m for m ∈ K ⊗k F ,

imQ = K ⊗k F andτQ = Qτ for all τ ∈ G. ThenQ2 = Q. Let us identify anyn ∈ N

with 1⊗ n ∈ K ⊗k N . For anyn ∈ N one has thatQ(n) is invariant underG and thereforebelongs toN . The restrictionQ of Q to N has the properties imQ = F , Q∂ = ∂Q andQ2 = Q. ThenN = F ⊕ kerQ. ThusN is semi-simple.

(1′′). Consider any finite extension of differential fieldsk ⊂ K . LetKc denote its normaclosure. By (1′), N is semi-simple overk if and only if Kc ⊗k N is semi-simple overKc.But, again by (1′), Kc ⊗k N = Kc ⊗K K(K ⊗k N) is semi-simple overKc if and only ifK ⊗k N is semi-simple overK .

(1) Let K ⊃ k be an arbitrary algebraic extension. Suppose thatN is semi-simple. AnyK-submoduleD of K ⊗k N comes (by tensoring) from aK-submoduleD of K ⊗k N

for some fieldK ⊂ K which is finite overk. SinceD is a direct summand,D is a directsummand, too.

Suppose thatK ⊗k N is semi-simple. LetF ⊂ N be a submodule. ThenK ⊗k F is adirect summand ofK ⊗k N . For a suitable finite extensionK1 of k, contained inK , alsoK1⊗k F is a direct summand ofK1⊗k N . One replacesK1 by its normal closureK2 ⊃ K1.


l,

urther

nd ther

e

als

ThenK2 ⊗k F is a direct summand ofK2 ⊗k N . As in the proof of (1′) (⇐), it follows thatF is a direct summand.

(2) LetKc be the normal closure ofK . ThenKc ⊗K M is semi-simple overKc. FurtherM is ak-submodule ofKc ⊗K M . Therefore it suffices to consider the case whereK is aGalois extension ofk with Galois groupG. As in Definition 2.1 one hasK ⊗k M ∼= σM .EachσM is semi-simple and thusK ⊗k M is a semi-simple differential module overK .By (1), M is semi-simple as ak-differential module.

(3) We use the notation:Ck is the field of constants ofk, Ck is an algebraic closure ofCk

andk = Ck · k. By (1), A := k ⊗k A andB := k ⊗k B are semi-simple. Using differentiaGalois theory, see [P-S, Exercise 2.38(4)], one has thatA ⊗k B is semi-simple. By (1)A ⊗k B is semi-simple.

(4) follows from (3). �Let M be a semi-simple differential module overk. Theisotypical decompositionM =

M1 ⊕ · · · ⊕ Ms is defined by: eachMi is a direct sum of, say,ni copies of an irreducibledifferential moduleDi . Moreover theD �≡Dj for i �= j . PutFi = Endk[∂](Di). Then onehas the following obvious results:

(a) The isotypical decomposition ofM is unique.(b) Fi is a (skew) field of finite dimension over the field of constantsCk of k.(c) Endk[∂](M) is isomorphic to the product of the algebras Matr(ni,Fi).

Here Matr(n,F ) denotes the ring ofn by n matrices with entries inF . Let C be anyfield. The algebrasA that we consider here are supposed to have a neutral element, fC lies in the center ofA, andA as vector space overC has finite dimension.A is calledsemi-simpleif for every two-sided idealI , there is a two-sided idealJ with A = I ⊕ J .The semi-simple algebras are the algebras of the form

∏i Matr(ni,Fi) where theFi are

(skew) fields of finite dimension overC, with C in the center ofFi . The algebraA is calledsimpleif A has no two-sided ideals other than 0 andA. Equivalently,A ∼= Matr(n,F ) forsomen and some (skew) fieldF of finite dimension overC, with C in the center ofF .

In the remainder of this section we recall some standard facts on skew fields aBrauer group. For more information we refer to [Bl,Bo,Rei,S]. LetC be any field. Considea skew fieldF of finite dimension over its centerC. Then the dimension ofF overC isa square, sayn2. A field extensionC′ ⊃ C is called asplitting field forF if C′ ⊗ F isisomorphic to the matrix algebra Matr(n,C′). Any maximal commutative subfieldC′ ⊃ C

of F satisfies[C′ : C] = n and is moreover a splitting field forF of minimal degree overC.This is illustrated by the example of Hamilton’s quaternion field overQ, namelyH =Q+Qi+Qj +Qk. The minimal splitting fields areQ(

√−m) for every squarefree positivintegerm that can be written as the sum of three squares inQ.

A central simplealgebraA over the fieldC is an algebra whose only two-sided ideare A and {0} and which has finite dimension over its centerC. Every central simplealgebra overC has the form Matr(d,F ), whereF is a (skew) field with centerC. TheBrauer group Br(C) of a field C consists of the equivalence classes[A] of the centralsimple algebras overC. Two such algebrasAi = Matr(di,Fi) are called equivalent,[A1] =


g

t

a

-

-

[A2], if the (skew) fieldsF1, F2 are isomorphic. The group structure on Br(C) is inducedby the tensor product.

If CK/Ck is a finite Galois extension with groupG then one defines the followinsubgroup of Br(Ck)

Br(CK/Ck) = {[A] ∈ Br(Ck) | A is split byCK

}.

Let c be a 2-cocycle with values inC∗K , i.e.c(σ, τ ) ∈ C∗

K for all σ, τ ∈ G andc satisfiesthe 2-cocycle relation. Nowc is called normalized ifc(1, σ ) = c(σ,1) = 1 for all σ . Theimage ofc in H 2(G,C∗

K) is trivial, i.e. c = 1, if and only if there exists a mapf :G → C∗K

such thatc(σ, τ )f (στ) = f (σ )σ (f (τ)) for all σ , τ . Now take theCK -vector spaceA withbasis{bσ | σ ∈ G} whereb1 = 1. One turnsA into an algebra (a so-calledcrossed-producalgebra) with multiplication rules:bσ · bτ = c(σ, τ )bστ , andbσ · λ = σ(λ)bσ for λ ∈ CK .The 2-cocycle relation guarantees thatA is associative, in fact,A is a central simple algebroverCk . Now c → [A] defines an isomorphism

H 2(G,C∗K

) ∼= Br(CK/Ck).

Taking the limit over all finite Galois extensions one finds

H 2(Gal(Ck/Ck), C∗k

) ∼= Br(Ck).

The inflation homomorphismH 2(G,C∗K) → H 2(Gal(Ck/Ck), C

∗k ) is injective, and corre

sponds to the embedding of Br(CK/Ck) in Br(Ck).

2.3. The associated two-cocycle and skew fields

Theorem 2.8. The associated2-cocycle for a special case.Consider the differential fieldsk = Ck(x) and K = CK(x) whereCK/Ck is a finite

Galois extension with groupG. Assume that the differential moduleM of dimensionnoverK has the properties:

(i) The group of the differential automorphism ofM is C∗K .

(ii) For everyσ ∈ G one hasσ M ∼= M .

These assumptions define a2-cocycle classc ∈ H 2(G,C∗K) which has the following prop

erties:

(a) c = 1⇔ M descends tok.(b) The orderd of c dividesn and[K : k].(c) c ∈ H 2(G,C∗

K) ∼= Br(CK/Ck) ⊂ Br(Ck) determines a(skew) fieldF 0 with centerCk .(d) E := Endk[∂](M) is a simple algebra, of dimension[CK : Ck]2 over its centerCk ,

having imagec ∈ Br(Ck). In particular, E is isomorphic toMatr(m,F 0) for somem � 1. Moreover,CK is a maximal commutative subfield ofE.


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(e) A finite field extension� ⊃ Ck is a splitting field forF 0 if and only ifM descends to�(x).

Proof. Choose an isomorphismφ(σ) : σ M → M for eachσ ∈ G. By assumption (i),φ(σ)

is unique up to an element inC∗K . Let Φ(σ) :M → M denote the associatedσ -linear map.

For σ, τ ∈ G one definesc(σ, τ ) := Φ(σ)Φ(τ)Φ(στ)−1. This is aK-linear bijection onM and commutes with∂ . By assumption (i),c(σ, τ ) ∈ C∗

K . Furtherc(σ, τ ) satisfies theusual 2-cocycle relation. The imagec ∈ H 2(G,C∗

K) is independent of the choices for t{φ(σ)}. Part (a) follows at once from Definitions 2.2 part (3).

(b) Consider the case whereM has dimension 1. WriteM = Ke and write∂e = ae.For anyσ one normalizesΦ(σ) :M → M by Φ(σ)e = b(σ )e such thatb(σ ) ∈ CK(x)

has the formb1(σ )/b2(σ ) with bi(σ ) monic elements ofCK [x]. In other words, the firscoefficient in the series expansion ofb(σ ) atx = ∞ is 1. Thenc(σ, τ ) ∈ C∗

K has this formtoo, which impliesc(σ, τ ) = 1.

Let M of dimensionn induce the 2-cocycle classc. Then the 2-cocycle class of th1-dimensional differential modulesΛnM is easily seen to becn. Since the latter is trivialthe orderd of c is a divisor ofn. Moreover, every element inH 2(G,C∗

K) is annihilated bythe order ofG, which is[K : k]. This proves (b). The statement in (c) is a standard facBrauer groups, see the previous section.

(d) According to Lemma 2.3(a), EndK[∂](K ⊗k M) ∼= CK ⊗CkEndk[∂](M). By the ob-

servation in Definition 2.1,K ⊗k M ∼= ⊕σ∈G

σM , which by assumption (ii) is isomorphto the direct sum of[CK : Ck] copiesM . Now EndK[∂](M) = CK , so EndK[∂](K ⊗k M) isisomorphic to the matrix algebra Matr([CK : Ck],CK). It follows that the dimension ofEoverCk is [CK : Ck]2.

Consider the algebraA ⊂ E consisting of theL ∈ E of the formL = ∑σ∈G cσ Φ(σ)

with cσ ∈ CK . It is easily verified thatL = 0 if and only allcσ are 0.A is a crossed-producalgebra with multiplication rules (a factor set) given by thec(σ, τ ). SoA is a simple algebrathat represents the image ofc in Br(Ck). By comparing dimensions one findsA = E. Oneconcludes thatE ∼= Matr(m,F 0) for somem whereF 0 is the (skew) field associated toc.FurtherCK = EndK[∂](M) consists of all elements ofE which commute withCK . HenceCK is a maximal commutative subfield ofE.

If Ck ⊂ � ⊂ CK then part (e) follows from the statement that� ⊃ Ck is a splitting fieldfor F 0 if and only if the image ofc in H 2(Gal(CK/�),C∗

K), under the restriction map, isIf � is not a subfield ofCK , then replaceCK by the normal closure of� ·CK . Then the sameproof applies; assumption (i) still holds by Lemma 2.3(b), assumption (ii) is clear.�Corollary 2.9. We keep the notation and assumptions of Theorem2.8.

(1) Suppose thatM is irreducible as aK-differential module. ThenM viewed ask-differential module is the direct sum ofm copies of an irreducible differential moduN overk with Endk[∂](N) = F 0.

(2) Let be given:∗ a skew fieldF 0 of finite dimension over its centerCk ,∗ m � 1 and a maximal commutative subfieldCK of Matr(m,F 0), which is Galois

overCk ,


e

0

n

-

[Bl,

ma 2.3g thepart

∗ an irreduciblek-differential moduleN with Endk[∂](N) = F 0.ThenM , the direct sum ofm copies ofN , has a natural structure of an irreduciblK = CK(x)-differential module that satisfies the assumptions of Theorem2.8.

Proof. (1) Let Pi ∈ Matr(m,F 0) = E = Endk[∂](M) denote the matrix with all entrieswith the exception of an entry 1 at the position(i, i). Put Ni = PiM . ThenNi is a k-submodule ofM andM = ⊕

i Ni . An elementL ∈ Endk[∂](Ni) can be extended to aelementL ∈ E by prescribingL = 0 on eachNj with j �= i. The structure ofE impliesthatL is a diagonal matrix with zeros on the diagonal, except for the position(i, i). HenceL ∈ F 0. Since the matricesPi are conjugated in Matr(m,Ck), thek-differential modulesNi are all isomorphic. Moreover, sinceNi is semi-simple and Endk[∂](Ni) = F 0 one hasthatNi is irreducible.

(2) M = N ⊕ · · · ⊕ N has Endk[∂](M) = Matr(m,F 0). The embedding ofCK inMatr(m,F 0) makesM into a K-differential module. EndK[∂](M) consists of the elements in Matr(m,F 0) commuting withCK . Thus EndK[∂](M) = CK . FurtherM is asemi-simpleK-differential module since it is a semi-simplek-differential module. ThusM is an irreducibleK-differential module. Finally, by the Skolem–Noether theorem (Théorème III-4], or [Rei, (7.21)]) for everyσ in the Galois group ofCK/Ck , there is aninvertible elementΦ(σ) ∈ Matr(m,F 0) with σ(λ) = Φ(σ)λΦ(σ)−1 for all λ ∈ CK . ThenΦ(σ) :M → M is aσ -linear map which commutes with∂ . �

For a general finite Galois extension of differential fieldsk ⊂ K with groupG, the situa-tion is more complicated; e.g., one-dimensional modules need not descend, and Lemno longer applies. Parts (a) and (b) of the following proposition can be proved alonlines of the proof of Theorem 2.8. Part (c) follows from Proposition 2.11(a) below, and(d) follows from Proposition 2.11(c).

Theorem 2.10. The associated2-cocycle in the general case.LetK/k be a Galois extension of differential fields with Galois group G. LetCK denote

field of constants ofK . Let a differential moduleM overK of dimension n satisfy:

(i) the group of the differential automorphisms ofM is C∗K , and

(ii) for everyσ ∈ G one hasσ M ∼= M .

(a) These data determine a2-cocycle classc ∈ H 2(G,C∗K) of an orderd , dividing[K : k]

if K is finite overk. MoreoverM descends tok if and onlyc = 1.(b) The tensor productM ⊗ · · · ⊗ M of d copies ofM has associated2-cocyclecd and

therefore this tensor product descends tok. The same holds forsymd M , the symmetrictensor product ofd copies ofM .

(c) The image ofc under the mapH 2(G,C∗K) → H 2(G,K∗) defines a(skew) fieldF with

centerk. Let k ⊂ � ⊂ K . Then� is a splitting field forF if and only if there exists amoduleL of dimension1 overK such thatσ L ∼= L for all σ ∈ Gal(K/�) andL ⊗ M

descends to�.(d) If K = k, andF , � as in part(c), then� is a splitting field if and only ifM descends

to �.


ules

-

of

ialte

The conditionσ L ∼= L in (c) is irrelevant whenM is not cyclic-imprimitive (for a defi-nition see Section 4, see also Theorem 4.4).

Consider differential modulesM of dimension 1 overK such thatσ M ∼= M for allσ ∈ G. The isomorphism classes of these modules form an abelian groupI with the tensorproduct as multiplication. LetI0 denote the subgroup of the classes of differential modwhich descend tok.

Proposition 2.11. LetCK denote the constant field ofK . Then

(a) I/I0 is naturally isomorphic to the kernel ofH 2(G,C∗K) → H 2(G,K∗).

(b) Let M = Ke represent an element ofI . ThenE := Endk[∂](M) is a semi-simple algebra overCk .

If K = CK · k, thenE is a central simple algebra overCk . Moreover,M descends to� withk ⊂ � ⊂ K if and only if the field of constantsC� is a splitting field forE.

If K �= CK · k, then, in general, the2-cocycle inH 2(G,C∗K) attached toM , does not

describe a central simple algebra overCk or k. Moreover,E is (in general) not a centralsimple algebra overCk .

(c) If K = k thenI = I0.

Proof. (a) For any differential fieldL one writesQ(L) for the isomorphism classesthe 1-dimensional differential modules overL. The tensor product makesQ(L) into acommutative group and there is an exact sequence

0→ L∗/C∗L → L → Q(L) → 0,

whereCL denotes the field of constants ofL; the first nontrivial arrow is defined byf �→ f ′f

and the second nontrivial arrow mapsf ∈ L to the isomorphism class of the differentmodule(Le, ∂) with ∂e = f e. We consider this exact sequence withL = K and the exacsequence 0→ C∗

K → K∗ → K∗/C∗K → 0. These sequences ofG-modules induce th

usual long exact sequences. Using thatH 1(G,K) = 0 andH 1(G,K∗) = 0 one obtainsexact sequences

0→ H 1(G,C∗K

) →Q(k) → Q(K)G → H 1(G,K∗/C∗K

) → 0,

0→ H 1(G,K∗/C∗K

) → H 2(G,C∗K

) → H 2(G,K∗).ClearlyI = Q(K)G andI0 is the image ofQ(k) →Q(K)G.

(c) If K = k, then the groupK∗/C∗K is a vector space overQ and henceH 1(G,K∗/

C∗K) = 0.(b) The algebra{∑σ∈G mσ Φ(σ) | mσ ∈ K} is equal to the algebra of allk-linear endo-

morphisms ofM . This follows from theK-linear independence of the maps{σ :K → K |σ ∈ G}. One derives from this thatE = {∑ cσ Φ(σ) | cσ ∈ CK}.
σ∈G


isal

ne

trivial

int

ion.

ore

felds.

from

r

r

d ex-

If K = CK · k, thenG is the Galois group ofCK/Ck . Again, by [Bl, Proposition IV-1],one has thatE is a central simple algebra overCk of dimension[CK : Ck]2. Consider anintermediate field�. ThenK = CK ·�. SoK/� andCK/C� have the same Galois groupH .Now M descends to� if and only if the restriction of the 2-cocyclec to H is trivial, if andonly if C� is a splitting field forE.

Suppose thatK �= CK · k. ThenCK/Ck is still a Galois extension but its Galois groupdifferent fromG. The 2-cocycle inH 2(G,C∗

K), attached toM , need not describe a centrsimple algebra. Consider the example (4) of Remark 2.5. The algebraE for this exampleis isomorphic toQ(i). It is also interesting to make part (a) explicit for this example. Ohas thatI/I0 is isomorphic to the kernel ofH 2({1, σ },Q∗) → H 2({1, σ },Q(t)∗). One canverify that this kernel is the group of two elements. The example describes the nonelement in this kernel. �Remark. AssumeK = CK ·k. Then Theorem 2.8 is valid for a differential fieldk preciselywhenI = I0. The proof of Theorem 2.8 part (b) showsI = I0 for k = Ck(x). A similarargument shows that ifk is the function field of a nonsingular algebraic curve with a podefined overCk thenI = I0.

In Section 2.5 we will show that Proposition 2.11 is related to Amitsur’s construct

2.4. Differential modules over a skew differential field

In the sequel of this section we will produce explicit examples for Corollary 2.9. Mprecisely, for a given finite Galois extensionCk ⊂ CK and a skew fieldF 0 of finite dimen-sion over its centerCk such thatCK is a maximal commutative subfield of Matr(m,F 0),we will produce an irreducible differential moduleM overK = CK(x) which satisfies (i)and (ii) of Theorem 2.8 and such that Endk[∂](M) = Matr(m,F 0). The basic feature othe construction is the introduction of differential modules over skew differential fiBy Corollary 2.9,M corresponds to ak[∂]-moduleN with Endk[∂](N) = F 0. SoN is ak[∂]-module as well as anF 0-module. ThenN is ank[∂]-module whereF is defined inProposition 2.13 below. Hence, every example for Theorem 2.8/Corollary 2.9 comesa differential module over a skew field.

Definitions 2.12. Let k be a differential field andF a skew field of finite dimension oveits centerk. A differentiationf �→ f ′ on F is an additive map fromF to itself such that(fg)′ = f ′g + fg′ for all f,g ∈ F and such that the restriction off �→ f ′ to k ⊂ F isthe differentiation ofk. A differential moduleM overF is a finite-dimensional left vectospace overF , equipped with an additive map∂ :M → M satisfying∂(f m) = f ′m + f ∂m

for all f ∈ F andm ∈ M .

Let e1, . . . , en be a basis ofM overF . Then∂ is determined by the elements∂e1, . . . ,

∂en. Moreover, these elements inM can be chosen arbitrarily.The next proposition together with part (2) of Corollary 2.9 provides the require

amples.


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Proposition 2.13. Let F 0 be a skew field of finite dimension over its centerCk . On theskew fieldF = F 0(x) := F 0 ⊗Ck

Ck(x) with centerk = Ck(x), a differentiation is givenby (f ⊗ a)′ = f ⊗ a′ for all f ∈ F 0 anda ∈ k. Let N be a finite-dimensional left vectospace overF . ThenN can be given the structure of differential module overF such thatN is irreducible as ak-differential module andEndk[∂](N) = F 0.

Proof. Choose a maximal commutative subfieldC of F 0 containingCk . ThenN is alsoa vector space overC(x). Write C = Ck(α) and letP ∈ Ck[x] denote the monic minimapolynomial ofα. The completion of the local ringCk[x](P ) is denoted byOL. The residuefield of OL is C. There is a unique subfield ofOL containingCk , which maps bijectively tothe residue fieldC. We will identify C with this subfield ofOL. After this identification, theelementt = x −α ∈ OL is a generator of the maximal ideal. Thus we can identifyOL withC[[t]]. The embeddingCk[x] → OL has dense image. LetL denote the field of fractionof OL. ThenL is the completion of the fieldk = Ck(x) w.r.t. the valuation associatedthe irreducible polynomialP .

Let V = EndF (N) and W = EndC(x)(N). The naturalk-algebra homomorphismφ :C(x) ⊗k V → W is bijective. Indeed, the first algebra is simple and the two algehave the same dimension overk. Consider a property(∗) of elementsB ∈ W which is pre-served for allB such thatB is close toB w.r.t. a metric onW induced by the embeddinW ⊂ EndL(L ⊗C(x) N). Then, sincek is dense inC(x), there exists an element inV withproperty(∗).

We identify N with Fa and define the standard differentiation onN by n =(n1, . . . , na) �→ n′ = (n′

1, . . . , n′a). A structure∂ of N as a differential module overF

has the form∂n = n′ + A(n) with A ∈ V . A structure∂ onN as a differential module oveC(x) = C(t) has the form∂n = n′ + B(n) with B ∈ W . Property(∗) is defined by: theNewton polygon of a cyclic element of the moduleC((t)) ⊗C(t) N has slope1

r, wherer is

the dimension ofN asC(x)-vector space. This property implies thatN := C((t)) ⊗C(t) N

is irreducible and EndC((t))[∂](N) = C. ThusN is an irreducible differential module oveC(x) and EndC(x)(N) = C. Clearly (∗) is preserved under a small perturbation w.r.metric onW induced by the embeddingW ⊂ EndL(L ⊗C(x) N). We conclude that:

N can be given the structure of a differential module overF such thatN andC(x) ⊗C(x) N are irreducible differential modules overC(x) and C(x). MoreoverEndC(x)[∂](C(x) ⊗C(x) N) = C.

Let C ⊃ C be the normal closure ofC ⊃ Ck . PutN = C(x)⊗C(x) N . As ak-differentialmodule,N is isomorphic to a direct sum of[C : C] copies ofN . Therefore, Endk[∂](N)

is equal to Matr([C : C],Endk[∂](N)) and contains the simple algebraA := Matr([C : C],F 0) with centerCk . FurtherC is a maximal commutative subfield ofA (since it has thecorrect dimension). By the Skolem–Noether theorem, everyσ ∈ Gal(C/Ck) extends toan inner automorphism ofA. Thus theC(x)-moduleN satisfies conditions (ii) of Theorem 2.8. FurtherN is irreducible and End

C(x)[∂](N) = C. Condition (i) of Theorem 2.8

holds and we can apply the first part of Corollary 2.9. One has Endk[∂](N) = Matr(b,F 0)

for someb � 1. Now C is also a maximal commutative subfield of Endk[∂](N). Therefore


y (atl

r

o-ds tour

ra

thise.

the latter algebra coincides withA and Endk[∂](N) = F 0. Finally, sinceN is semi-simpleas a differential module overk, it follows thatN is irreducible. �

We now give asecond construction of examples for Corollary2.9 with m = 1, whichwill produce nicer examples because there will be only one irregular singularitx = ∞). Given is a skew fieldF 0 of finite dimension over its centerCk and a maximacommutative subfieldCK of F 0 which is a Galois extension ofCk . As beforeF = F 0(x) =F 0 ⊗Ck

Ck(x) is made into a skew differential field by the formula(f ⊗ a)′ = f ⊗ a′ forall f ∈ F 0 anda ∈ Ck(x). ThenK = CK(x) is a maximal commutative subfield ofF andthe restriction of the differentiation ofF to K is the obvious differentiation ofK .

Proposition 2.14. We use the above notation. LetM be a finite-dimensional left vectospace overF . ThenM can be given a structure of differential module overF such thatMis an irreducibleK-differential module with the properties:

(a) EndK[∂](M) = CK .(b) σ M ∼= M for all σ ∈ Gal(K/k).(c) Endk[∂](M) = F 0 andM is irreducible ask-differential module.

(d) M := CK(x) ⊗K M is an irreducible differential module overCK(x).(e) EndCK (x)[∂](M ) = CK .

Proof. As in the proof of Proposition 2.13, we identifyM with Fa for somea � 1 anddefine the operation′ on Fa by (f1, . . . , fa)

′ = (f ′1, . . . , f

′a). A structure∂ of differential

module overF onM is given by∂m = m′ + mD, with D ∈ Matr(a,F ).ForD we make the choiceD = ∑N

i=0 Dixi , with:

(1) DN is a diagonal matrix with distinct coefficients inCk on the diagonal,(2) {D0, . . . ,DN−1} are generators of Matr(a,F 0) overCk ,(3) M is irreducible as a differential module overF .

For a = 1, property (3) is obvious. Moreover in that case we may replace(1) + (2) by:D0, . . . ,DN generateF 0 over Ck . If a > 1, then one could apply the method of Propsition 2.13 in order to obtain (3) and (a)–(e), but we will avoid this because it leaexamples more complicated than

∑Ni=0 Dix

i , and because (3) is easy to verify for oexplicit examples.

(i) Endk[∂](M) = F 0.

For any algebraB we writeBopp for the opposite algebra. For any algebraB we writeMatr(a,B) for the algebra of the(a × a)-matrices with coordinates inB. Now F andMatr(a,F )opp are central simple algebras overk. By [Ren, Corollaire 4, p. 107], the algebF ⊗k Matr(a,F )opp is again simple. This algebra is mapped to Endk(F

a), the algebra of thek-linear endomorphisms ofFa , by the following formula(f ⊗ B)(v) = f vB for f ∈ F ,v ∈ Fa , B ∈ Matr(a,F )opp. Since the first algebra has only trivial two-sided ideals,map is injective. By counting dimensions overk, one concludes that the map is bijectiv


n

left

tsthat

ow

t

Let b1, . . . , bd2 denote a basis ofF 0 overCk . From the above it follows that everyk-linear

mapL :M → M can uniquely be written asL(v) = ∑d2

i=1 bivAi , with all Ai ∈ Matr(a,F ).

For L as above one calculates that(∂L − L∂)v = ∑d2

i=1 bivBi , whereBi equals thematrix A′

i + AiD − DAi . HenceL commutes with∂ if and only if A′i + AiD − DAi = 0

for everyi.Suppose thatA ∈ Matr(a,F ) is a nonzero solution ofA′ = [D,A]. Let q ∈ Ck[x]

denote the monic polynomial of minimal degree such thatqA ∈ Matr(a,F 0[x]). WriteA = q−1B, then −q ′q−2B + q−1B ′ + q−1BD − q−1DB = 0. After multiplying thisidentity with q one finds thatq ′A ∈ Matr(a,F 0[x]). One concludes thatq = 1. HenceA ∈ Matr(a,F 0[x]).

The setV of all solutionsA ∈ Matr(a,F 0[x]) of A′ = [D,A] is an algebra overCk .Moreover,V is a finite-dimensional vector space overCk sinceA′ = [D,A] can be in-terpreted as a linear differential equation over the differential fieldk. Any A ∈ V , A �= 0can be written in the formA = ∑e

i=0 Aixi with all Ai ∈ Matr(a,F 0) andAe �= 0. One

finds that[DN,Ae] = 0 and thusAe is a diagonal matrix (with coefficients inF 0). Theelements1,A,A2,A3, . . . all belong toV . In particular, there is a nontrivial relatioλ01 + λ1A + · · · + λsA

s = 0 with all λi ∈ Ck andλs �= 0. It e > 0, then, sinceAe is anonzero diagonal matrix, the termxes is present inAs and not in theAi with i < s. Thiscontradiction implies thate = 0. FurtherA = A0 lies in the center of Matr(a,F 0), sinceA0 commutes with allDi . HenceA = λ1 for someλ ∈ Ck . Hence Endk[∂](M) = F 0.

(ii) EndK[∂](M) = CK andM is an irreducible differential module overK .

The first statement holds becauseCK is a maximal commutative subfield ofF 0. Let0 �= N ⊂ M be an irreducibleK-differential module. For anyf ∈ F 0, f �= 0, alsof N

is an irreducibleK-differential module. The sum∑

f N , wheref runs in a basis ofF 0

overCK , is a semi-simpleK-differential module. Since this object is invariant undermultiplication byF , andM is an irreducibleF -differential module, one has

∑f N = M .

SoM is semi-simple overK and since EndK[∂](M) contains only the trivial idempotenit follows thatM is irreducible overK . The same argument (or Proposition 2.7) showsM is semi-simple and irreducible overk.

(iii) σ M ∼= M as differential modules overK .

The Skolem–Noether theorem asserts that forσ ∈ G there exists nonzerofσ ∈ F 0 suchthatσ(λ) = fσ λf −1

σ for everyλ ∈ CK . One definesΦ(σ)(m) = fσ m. ClearlyΦ(σ) :M →M is aσ -linear bijection, commuting with∂ . This proves (b). Statements (d) and (e) follfrom Proposition 2.7 and Lemma 2.3.�Example 2.15. Skew fields overQ.

(1) LetH = Q+Qi+Qj +Qk denoteHamilton’s quaternion field overQ. We considera maximal commutative subfieldCK = Q(i) and the fieldsK := CK(x), k = Q(x). Oneprovides the 1-dimensional left vector spaceM = H(x)e over H(x) with ∂ defined by∂e = de for somed ∈ H(x). According to Proposition 2.14, the choiced = i + jx makesM into an example for Theorem 2.8 and Corollary 2.9. Letσ be the nontrivial elemenin Gal(K/k). ThenΦ(σ), defined byΦ(σ)he = jhe for all h ∈ H(x), is a good choice


a

s

for the σ -linear bijection commuting with∂ . We note that the 2-cocyclec has the formc(1,1) = c(1, σ ) = c(σ,1) = 1 andc(σ,σ ) = −1.

(i) Explicit formulas.

e ∈ M is cyclic for M as a differential module overk = Q(x). The minimal monicoperatorL4 ∈ k[∂] with L4e = 0 has the form

L4 = ∂4 + (2+ 2x2)∂2 + 4x∂ + (

4+ 2x2 + x4).By Proposition 2.14(c),L4 is irreducible as an element ofQ(x)[∂]. Moreover,e ∈ M isalso cyclic forM as Q(i)(x)[∂]-module. The minimal monic operatorL2 ∈ Q(i)(x)[∂]has the form

L2 = ∂2 − x−1∂ + (ix−1 + 1+ x2).

L2 must be a right-hand factor ofL4 in Q(i)(x)[∂]. Indeed,

L4 = (∂2 + x−1∂ + (−x−2 − ix−1 + 1+ x2)) · L2.

By Proposition 2.14(d), the operatorL2 is irreducible as an element ofQ(x)[∂].As we knowL2 is equivalent to its conjugate∂2 − x−1∂ + (−ix−1 + 1 + x2) andL2

is not equivalent to any second-order differential operator inQ(x)[∂]. Now Q(√−m) is a

splitting field ofH whenm > 0 is the sum of three squares inQ. Hence for every suchm,L2 must be equivalent to an operator inQ(

√−m)(x)[∂], andL4 can be factored asproduct of two irreducible operators inQ(

√−m)(x)[∂].(ii) The associated third-order operatorL3 ∈ Q(x)[∂].According to Theorem 2.10(b), thesymmetric squareN = sym2 M of M asQ(i)(x)[∂]-

module descends toQ(x). We want to make this explicit for our example. Theσ -linear mapΨ (σ) : sym2 M → sym2 M is defined by

Ψ (σ)(m1 ⊗ m2) = Φ(σ)m1 ⊗ Φ(σ)m2 = jm1 ⊗ jm2.

In particular,Ψ (σ)2 is the identity. LetN0 denote the set of the elements ofN invariantunderΨ (σ). GiveM the basise, je. ThenN hasQ(i)(x)-basise⊗e, je⊗e, je⊗je. Onefinds thatN0 hasQ(x)-basise. ⊗ e + je ⊗ je, ie ⊗ e − ije ⊗ je, ije ⊗ e. FurtherN0 is adifferential module overQ(x), sinceΨ (σ) commutes with∂ . ThusN = Q(i)(x)⊗Q(x) N

0.We takeije⊗e as cyclic element ofN0 and letL3 be its minimal operator:L3(ije⊗e) = 0.A calculation shows that

L3 = ∂3 − 2x−1∂2 + (2x−2 + 4+ 4x2)∂ + 4x.

L3 must be equivalent to the symmetric square of the operatorL2 above. The latter doenot have coefficients inQ(x) because it corresponds to the cyclic vectore ⊗ e, which isnot invariant underΨ (σ).


ak-s

o

fmod-

. Nowheonen,s onfactor-nly if

ing

SinceN is a symmetric square, we see thatN0 becomes a symmetric square after ming a suitable algebraic extensionC of Q. The fieldsC are precisely the splitting fieldfor H. We will return to this in the next section.

(2) Example for a general quaternion field overQ.

If we repeat the computation withi replaced byb1, j replaced byb2, with the followingrules of multiplication:b2

1 = A1, b22 = A2, b2b1 = −b1b2, replacingH by F 0 = Q+Qb1+

Qb2 + Qb1b2, andd by b1 + b2x then we find the followingL2

L2 = ∂2 − x−1∂ + √A1x

−1 − A1 − A2x2.

Now the splitting fields ofF 0 are precisely the fieldsC for which the conic

A1X2 + A2Y

2 − 1= 0

has a point(X,Y ) ∈ C2. In particular,F 0 is a skew field if and only if this conic has nQ-rational point. For example, ifA1 = 2 andA2 = 3, thenF 0 is a skew field with centerQ.We note that the following operator

L′2 = ∂2 − A2x

2 − A1 + √A2

is equivalent toL2. Hence it defines the same descent problem, andL′2 descends toC(x)[∂]

if and only if C is a splitting field ofF 0. The symmetric square ofL2 is equivalent to

L3 = ∂3 − 2x−1∂2 + (2x−2 − 4A1 − 4A2x

2)∂ − 4A2x

which is equivalent to

∂3 − 4(A1 + A2x

2)∂ − 12A2x.

SupposeA1, A2 are integers, and that the conic has a rational point(X,Y ) ∈ Q2. ThenL2 descends toQ(x) and the cocycle classc in Theorem 2.8 is 1. If both parts (i) and (ii) othe descent data are explicitly known, then one can explicitly calculate descent: Theule N in Definition 2.2 part (3) can be found as{∑σ∈G Φ(σ)(m) | m ∈ M}. Conversely,if one knows an explicit descent, then descent data can also be explicitly calculatedsuppose thatA1, A2 are given, butX, Y are not. It is easy to calculate part (i) of tdescent data for the exampleL2. However, to calculate part (ii) of the descent data,must multiplyφ(σ) by a suitable element ofCK . This means solving a norm equatiowhich is equivalent to finding a rational point on the conic. Finding rational pointa conic requires computing square roots modulo integers, which in turn requiresing those integers. So finding a rational point can be computationally hard (but oA1,A2,−A1A2 are not squares, and at least oneAi is hard to factor inZ, for an exam-ple see http://www.math.fsu.edu/~hoeij/files/conic). For suchA1, A2, finding part (ii) ofthe descent data forL2, or equivalently, finding descent, is computationally hard. Find


a

er

r

an irreducible submodule ofM , viewed as a differential module overQ(x), is then alsocomputationally hard, so factoring

L4 = ∂4 − 2(A2x

2 + A1)∂2 − 4A2x∂ + A2

1 − 3A2 + A22x

4 + 2A1A2x2

in Q(x)[∂] is hard. Indeed, one can parametrize all monic second-order factors

∂2 − s/(t + sx)∂ − A2x2 − A1 − u/(t + sx)

of L4 in terms of points(s : t : u) on the conicA1s2 + A2t

2 − u2 = 0, and hence findingsecond-order factor is equally hard as finding a point on the conic.

(3) Example withM of dimension2 overF .

Consider the quaternion fieldF 0 = Q + Qb1 + Qb2 + Qb1b2, with b21 = 2, b2

2 = 3,b2b1 = −b1b2. Let F = F 0(x) andM = F 2 be the 2-dimensional differential module ovF given by the following action of∂ (recall that we are using row notationv = (v1, v2))

∂v = v′ + v

(0 1

b1 + b2x 0

).

EndQ(x)[∂](M) = −F 0, see Proposition 2.14.M is irreducible asQ(√

2)(x)-differentialmodule. The cyclic vector(1,0) gives the following operator:

∂4 − 2x−1∂3 + 2x−2∂2 + 2√

2x−1∂ − 2√

2x−2 − 2− 3x2.

This operator is irreducible (even overQ) and descends toC(x)[∂] if and only if C is asplitting field forF 0.

(4) A skew fieldF 0 of dimension9 overQ.

Let α be a solution ofα3 − 3α − 1= 0. ThenQ(α) is Galois overQ with Galois groupgenerated byσ , whereσ mapsα to 2− α2. Now takeb such thatbα = σ(α)b andb3 = 2.Let F 0 be the skew field generated byα andb, taked = b + αx, let M = F 0(x), and∂v = v′ + vd . Then we find the following operator

∂3 + (α2 − 2α − 2− 3x2)∂ − x3 − (

1+ α + α2)x − 2∈ Q(α)[∂].

It is irreducible, even as an element ofQ(x)[∂], and descends toC(x)[∂] if and only if C

is a splitting field forF 0.

2.5. Amitsur’s construction

Let k be a differential field of characteristic 0, havingCk as field of constants. Amitsuconsiders an irreducible differential moduleM of dimensionn over k with the propertythat M∗ ⊗ M is a trivial differential module. In other words, Hom(M,M) is a triv-ial differential module and the ring of endomorphismsE := Endk[∂](M), which equals


ule

m

itself

em,its

or

ntial

l

ker(∂,Hom(M,M)), has dimensionn2 over Ck . Now k ⊗CkE is thek-algebra of allk-

linear mapsM → M . It follows thatE is a skew field with centerCk and thatk is a splittingfield for E.

If CK ⊃ Ck is a finite extension and a splitting field forE, andK = CK · k, then the dif-ferential moduleK ⊗k M is a direct sum of copies of a 1-dimensional differential modoverK . Indeed, EndK[∂](K ⊗k M) is isomorphic to the matrix algebra Matr(n,CK).

One of the main results, Theorem 16 of [A] is:

• Any skew fieldE of dimensionn2 over its centerCk that hask as splitting field isobtained in this way.

We sketch the proof of this result. PutM = kn. By definition there is a homomorphisa ∈ E → Pa ∈ Endk(M). One defines∂0 :M → M by ∂0(f1, . . . , fn) = (f ′

1, . . . , f′n). For

any k-linear mapL :M → M one defines thek-linear mapL′ by L′ = ∂0L − L∂0. The

two representationsE → Endk(M ⊕ M), given bya �→ ( Pa 00 Pa

)anda �→ ( Pa P ′

A

0 Pa

), are iso-

morphic. Indeed, every finitely generated left module over a semi-simple algebra issemi-simple. Using this one obtains aQ ∈ Endk(M) with the propertyP ′

a = QPa − PaQ

for all a ∈ E. Define now∂ :M → M by ∂ = ∂0−Q. This makesM = (M,∂) into a differ-ential module overk and there is a homomorphismE → Endk[∂](M). Hence Hom(M,M)

is a trivial differential module and Endk[∂](M) = E. This is Amitsur’s construction.The differential fieldk = Ck(x) does not produce an example for Amitsur’s theor

since it is not a splitting field for any nontrivial skew field of finite dimension overcenterCk . Consider the differential fieldk = Q(s, t) with s2 + t2 = −1 ands′ = 1, t ′ =−st−1. Thenk is a splitting field for Hamilton’s quaternionsH = Q + Qi + Qj + Qk

over Ck = Q. One definesa ∈ H �→ Pa ∈ Endk(M) with M = k2 by Pi = ( 0 −11 0

)and

Pj = ( s tt −s

). One calculates thatQ = ( 0 1/2t

−1/2t 0

)and that the minimal monic operat

that annihilates the cyclic vector(10

) ∈ M is L = (s2 + 1)∂2 + s∂ − 14.

Let m be a positive squarefree integer andK = k(√−m). From the above it follows

thatL is irreducible as an element ofk[∂], and thatL factors inK[∂] if and only if m isthe sum of� 3 squares inQ.

Note that Amitsur’s theorem follows from Proposition 2.11(a). Consider a differefield k with field of constantsCk and a skew fieldE of dimensionn2 over its centerCk ,such thatk is a splitting field forE.

Let CK be a splitting field forE, and a finite Galois extension ofCk . Let G be theGalois group ofCK/Ck . Then[E] ∈ Br(CK/Ck) ∼= H 2(G,C∗

K). Now G is also the Galoisgroup ofK/k, whereK = CK · k, soH 2(G,K∗) ∼= Br(K/k). Sincek is a splitting field,[E ⊗ k] is trivial, hence[E] is in the kernel ofH 2(G,C∗

K) → H 2(G,K∗). According toProposition 2.11(a),[E] is the image of someN ∈ I = Q(K)G. ThisN is a 1-dimensionadifferential module overK such thatσ N ∼= N for all σ ∈ G. Let M be N consideredas a differential module overk. Explicit construction of the mapQ(K)G → H 2(G,C∗

K)

shows that[Endk[∂](M)] = [E]. ThenM has a submoduleM ′ of dimensionn over k (ifCK is a maximal commutative subfield ofE thenM = M ′), and Endk[∂](M ′) = E. SinceN is unique up toI0, see Proposition 2.11(a), the isomorphism class ofM ′ is unique up


’s

-r ofositive.l fields

ren--

atcond

sand weof the

rm

l

f differ-ds

mal

to tensoring with 1-dimensional modules overk. Thus it must correspond to Amitsurconstruction up to this equivalence.

3. Differential modules of dimension 3

In this section we consider a differential operatorL3 ∈ K[∂] of order 3 over a differential fieldK . The first question is whetherL3 is equivalent to the second symmetric powea differential operator of order 2. Singer (see [Si]) showed that the question has a panswer if and only if one can produce a certain conic and aK-rational point on this conicThis raises a second question: Which conies can occur? We will use skew differentiato answer this.

The first question translates in terms of differential modules as follows. A diffetial module B of dimension 3 overK is given. IsB isomorphic to the second symmetric power sym2K A of a differential moduleA of dimension 2 overK? We recallthat sym2

K A is defined as theK-vector spaceAs⊗K A with ∂ given by the formula

∂(a1 ⊗ a2) = (∂a1) ⊗ a2 + a1 ⊗ (∂a2). Our interest in this question lies in the fact tha differential moduleB may not be a second symmetric power, but could become a sesymmetric power after enlarging the field of constants ofK . Example 2.15(1)(ii) has thifeature. There is again a 2-cocycle responsible for this phenomenon (see Section 4)will construct examples using quaternion fields. First we investigate some propertiessecond symmetric power.

3.1. Properties of the second symmetric power

Proposition 3.1. LetB be a differential module of dimension3 over a differential fieldK .The following are equivalent:

(1) B ∼= sym2K A for some differential moduleA of dimension2 overK .

(2) sym2K B has a1-dimensional submoduleL with the following property. Letb1, b2, b3

be any basis ofB over K . Then L is generated by an element of the fo∑1�i�j�3 ci,j bi ⊗ bj with ci,j ∈ K and the quadric

∑1�i�j�3 ci,jXiXj = 0 in

P2K is nondegenerate and has aK-rational point.

Proof. (1) ⇒ (2) Let a1, a2 be a basis ofA overK . Putb1 = a ⊗ a1, b2 = a2 ⊗ a2, b3 =a1 ⊗ a2. Then the 1-dimensional subspaceL of sym2

K B with generatorb1 ⊗ b2 − b3 ⊗ b3

is easily seen to be a differential submodule. MoreoverX1X2 − X23 = 0 has a nontrivia

solution inK3.The reasoning above is based on the observation that the canonical morphism o

ential modulesφ : sym2K B → sym4

K A is surjective. Comparing the dimensions, one finthat the kernel ofφ is a 1-dimensional submodule of sym2

K B.(2) ⇒ (1) The assumptions on the quadric

∑1�i�j�3 ci,jX1Xj = 0 imply that there

exists a linear change of the variablesX1,X2,X3 which transforms the quadratic forinto a multiple ofX1X2 − X2. ThusB has a basisb1, b2, b3 such that the 1-dimension
3


.s

flinear

--

e, then

le

t

al

e

f

K-vector space generated byb1 ⊗ b2 − b3 ⊗ b3 of sym2K B is a differential submodule

One considers a 2-dimensional vector spaceA over K with basisa1, a2 and one definetheK-linear bijectionφ : sym2

K A → B by φ(a1 ⊗ a1) = b1, φ(a2 ⊗ a2) = b2 andφ(a1 ⊗a2) = b3. Let the∂ on B be given by the formula∂bi = ∑

j ej,ibj for i = 1,2,3. TheK-vector spaceA is made into a differential module by putting∂ai = ∑

j dj,iaj withd1,1 = e1,1/2, d2,1 = e3,1/2, d1,2 = e3,2/2, d2,2 = e2,2/2. Using that∂(b1 ⊗ b2 − b3 ⊗ b3)

is equal tof (b1 ⊗ b2 − b3 ⊗ b3) for somef ∈ K , one can verify thatφ is an isomorphismof differential modules. �

Some observations. Let C be an algebraically closed field of characteristic 0. LetK =C(x) be the differential field with differentiationf �→ df

dx. Since this fieldK is aC1-field,

one can omit in part (2) of Proposition 3.1 the assumption that the quadric has aK-rationalpoint (see also [F]). The Tannakian equivalence between differential modules overK andfinite-dimensionalC-linear representations of the universal differential Galois group oK

leads to the following translation of Proposition 3.1 in terms of representations ofalgebraic groups overC:

Let G be a linear algebraic subgroup ofGL(W), whereW is a vector space of dimension3 overC. Suppose thatsym2 W contains aG-invariant line that defines a nondegenerate quadric. Then there exists a linear algebraic groupH ⊂ GL(V ), whereV hasdimension2 overC, such thatH/(H ∩ {±1}) ∼= G and the twoG-modulessym2 V andW are isomorphic.

A similar result holds for Galois representations.

Lemma 3.2. Let B be a differential module of dimension3 over K . If sym2K B has a

1-dimensional submodule such that the corresponding quadratic form is degeneratB is reducible.

Proof. Suppose that the quadratic form associated to the 1-dimensional submoduL isdegenerate. This degenerate quadratic form has rank 1 or 2. There existsc1, c2 ∈ CK anda basisb1, b2, b3 of B such thatL is generated byb1 ⊗ b1 (for rank 1) or byc1b1 ⊗ b1 +c2b2 ⊗ b2 (for rank 2). ThenKb1 or Kb1 + Kb2 is a differential submodule ofB. �Lemma 3.3. LetA be an irreducible differential module of dimension2. PutB = sym2

K A.ThenB is reducible if and only if there is a field extensionK ⊃ K of degree two such thaK ⊗K A is reducible.

Proof. Because of Proposition 2.7(4), ifB is reducible thenB has a one-dimensionsubmoduleL. One can choose a basisa1, a2 of A such that a generator ofL has one of thefollowing forms:a1 ⊗ a1, a1 ⊗ a2 or a1 ⊗ a1 −f a2 ⊗ a2 wheref ∈ K is not a square. Thfirst two cases are excluded sinceA is irreducible. In the last case one putsK = K(

√f )

andK ⊗K A = K(a1 + a2√

f ) ⊕ K(a1 − a2√

f ) and thusK ⊗ A is reducible.Conversely, letK = K(t) with t2 = f ∈ K and writeσ for the nontrivial element o

Gal(K/K). Let Ke be a submodule ofK ⊗ A. If Kσe = Ke, thenσe = ge for some


s.tants

e

a

le,

ed

wer

m

s.

g ∈ K∗ andgσ(g) = 1. Theng = hσh

for someh ∈ K∗ and soσ(he) = he. It follows thathe ∈ A andKhe is a submodule ofA. This is a contradiction sinceA is irreducible. ThusK ⊗A is the direct sum of the submodulesKe andKσe. Thena1 = e +σe, a2 = te − tσ e

is a basis ofA. Finally a1 ⊗ a1 − f −1a2 ⊗ a2 generates a submodule ofB. �Remark. An irreducible differential moduleA of dimension 2 will be calledimprimitiveifthere exists a quadratic extensionK of K such thatK ⊗K A is reducible. OtherwiseA willbe calledprimitive. A differential moduleA can be imprimitive for two different reasonIt is possible thatA becomes reducible after a quadratic extension of the field of consof K . In the second case,A remains irreducible after replacingK by CKK . The differentialGalois groupG is defined for the differential moduleCKK ⊗A. Imprimitive is now equiva-lent to: the action ofG on the two-dimensional solution spaceV is irreducible and there arlinesL1,L2 ⊂ V such thatV = L1 ⊕L2 and anyg ∈ G permutes the linesL1,L2. In otherwords,G is contained in the infinite dihedral group{g ∈ GL2 | {gL1, gL2} = {L1,L2}}.Lemma 3.4. Suppose that the two-dimensional differential moduleA is irreducible andprimitive. PutB = sym2

K A. There are two possibilities:

(1) sym2K B has three distinct submodules of dimension1. Each one of them determines

nondegenerate quadric having aK-rational point.(2) sym2

K B has a unique submodule of dimension1.

Proof. After replacingA by L ⊗ A for a suitable 1-dimensional differential moduwe may suppose that the differential Galois groupG of A is an irreducible, primitivesubgroup of SL2(CK). We note that the differential Galois group is only well definover the differential fieldK := CKK . Let A andB denoteK ⊗ A andK ⊗ B. If G isnot the groupASL2

4 (see Remark 4.6(3) for a list of groups), then the symmetric posym4

K A has no 1-dimensional submodule and so the same holds for sym4K A. It follows

that the only 1-dimensional submodule of sym4K B is the kernel of the canonical morphis

sym4K B → sym4

K A.

Suppose thatG is ASL24 . Then by differential Galois theory:

(a) The differential Galois group ofB is A4.(b) sym4

K B ∼= M0 ⊕ M1 ⊕ M2 ⊕ B, whereM0,M1,M2 are 1-dimensional submoduleThe differential Galois group ofM0 is trivial andM0 = L0, whereL0 is the kernel ofsym2

K B → sym4K A.

(c) Mi ⊗ B ∼= B for all i.(d) The differential Galois group ofMi , i = 1,2 is the quotientC3 of A4.

Thus sym3K Mi is a trivial module. Moreover sym2K M1 ∼= M2.

We assume that sym2K B has a 1-dimensional submoduleL1 �= L0. ThenL1 is eitherM1

or M2. We may suppose thatL1 = M1. From the above one concludes that sym2K(L1 ⊗A)

is isomorphic toB and that the kernel of sym2K B → sym2K(L1 ⊗ A) is sym4

K L1 = M1.From Lemma 2.3(a) one concludes that all isomorphisms are defined over the fieldK . As a


1

n

sen

er

al

consequenceM2 = K ⊗L2, whereL2 := sym2K L1. In the same way,L2 is the kernel of the

morphism sym2K B ∼= sym2

K(L2 ⊗B) → sym4(L2 ⊗A). An application of Proposition 3.ends the proof. �3.2. Examples obtained from quaternion fields

Notation and assumptions.F is a quaternion algebra overk = Ck(x) with basisb0, . . . , b3. The multiplication is given byb0 = 1, b2

i = Ai ∈ k = Ck(x) for i = 1,2,3,b1b2 = −b2b1 = b3, A3 = −A1A2. ThenF is a skew field if and only if the equatioA1X

2 + A2Y2 − Z2 = 0 has only the trivial solution(0,0,0) in the field k (see [Bl,

Théorème I-5]). We will assume that this is the case. PutFi = k(bi) for i = 1,2,3. Theseare maximal commutative subfields ofF . As before, a differentiation′ onF will be a mapf �→ f ′ such that(fg)′ = f ′g + fg′ for all f,g ∈ F andf ′ = df

dxfor anyf ∈ Ck(x) ⊂ F .

We note that in general forf ∈ F , the elementsf andf ′ need not commute, in which cak(f ) is not a differential subfield ofF . Differentiations onF are not unique, but we cachoose one as follows.

Lemma 3.5. There is a unique differentiation′ on F such that the fieldsF1,F2 are stableunder differentiation. The fieldF3 is also stable under this differentiation.

Proof. The condition prescribesb′0 = 0 andb′

i = A′i

2Aibi for i = 1,2. Fromb3 = b1b2 one

deduces thatb′3 = A′

32A3

b3. Thus(∑3

i=0 aibi)′ = ∑3

i=0(a′i + ai

A′i

2Ai)bi , whereA0 := 1. On

the other hand, one can easily verify that this formula defines a differentiation onF . Thelast statement holds becauseb′

3 ∈ F3. �A 1-dimensional moduleM = Fe overF is described by∂e = de, whered ∈ F is an

arbitrarily chosen element. One can seeM as a two-dimensional differential module ovthe differential fieldF1. Let σ denote the nontrivial element of the Galois group ofF1/k.One considers theσ -linear bijectionA(σ) :M → M , defined byA(σ)m = b2m. One has:

∂A(σ) = A(σ)(∂ + A′2

2A2), or in a shorter notation∂b2 = b2(d + A′

22A2

).

Intermezzo. One hasσ M ∼= L ⊗F1 M for the one-dimensional differential moduleL =F1b overF1 with ∂b = A′

22A2

b. FurtherL ⊗ L is isomorphic to the trivial one-dimension

module. As a consequence the 3-dimensional differential moduleN := sym2F1

M descendsto k. We will make this explicit by a calculation.

On the differential moduleN =: sym2F1

M of dimension 3 overF1 we define an additive

operatorB(σ) :N → N by the formulaB(σ)(m1 ⊗m2) = 1A2

b2m1 ⊗b2m2. The following

properties are easily verified:B(σ) is σ -linear, commutes with∂ , andB(σ)2 is the identity.Let N0 denote the subset ofN consisting of the elements invariant underB(σ). An

explicit calculation shows thatN0 is a vector space of dimension 3 overk with basisn0

1 = e ⊗ e + 1A2

b2e ⊗ b2e, n02 = b1e ⊗ e − b1

A2b2e ⊗ b2e, n0

3 = b2e ⊗ e. SinceB(σ) and∂

commute,N0 is a differential module overk and moreoverN = F1 ⊗k N0. In other wordsN descends tok.


it-

edm

me

over

e are

lution

M , as a differential module overF1, has basise, b2e. The differential moduleN hasbasisn1 = e ⊗ e, n2 = b2e ⊗ b2e, n3 = e ⊗ b2e. By Proposition 3.1, sym2F1

N has a1-dimensional submodule, generated byn1 ⊗ n2 − n3 ⊗ n3. This expression can be rewrten in the basisn0

1, n02, n0

3 and reads

A2

4n0

1 ⊗ n01 − A2

4A1n0

2 ⊗ n02 − n0

3 ⊗ n03.

Thus sym2k N0 contains a one-dimensional submoduleL and the quadratic form associat

to L has the formA24 X2

1 − A24A1

X22 − X2

3. This form is equivalent to the quadratic for

A1X2 + A2Y

2 − Z2 associated to the quaternion fieldF .

Theorem 3.6. LetF , F1, k, M , N0 be as above and letK be an algebraic extension ofk.

(1) If K is a splitting field forF thenK ⊗k N0 is the second symmetric power of sodifferential module overK .

(2) Assume thatK ⊗k N0 is the second symmetric power of some differential moduleK and thatKF1 ⊗k N0 is irreducible. ThenK is a splitting field forF .

Proof. (1) Apply Proposition 3.1 to the 1-dimensional submoduleK ⊗k L of sym2(K ⊗k

N0). By assumption the quadratic form associated toL has a nonzeroK-rational point.(2) By Proposition 3.1, sym2(K ⊗k N0) contains a 1-dimensional submoduleZ such

that the associated quadratic form is nondegenerate and has a nonzeroK-rational point. Ingeneral, this does not imply thatK is a splitting field forF sinceZ need not beK ⊗k L.

The assumption thatKF1 ⊗k N0 is irreducible is equivalent, by Lemma 3.3, toM :=KF1 ⊗F1 M is irreducible and primitive. One applies Lemma 3.4 toN := sym2 M . Thussym2 N satisfies (1) or (2) of Lemma 3.4. Then the same holds for sym2(K ⊗k N0). Inparticular the quadratic form associated toL has a nonzeroK-rational point and thusK isa splitting field forF . �3.2.1. An example withA1 = x andA2 = s0 − s1x

We keep the notation of Section 3.2. Assume thats0, s1 ∈ Q and s0 �= 0, s1 �= 0,1.The central simple algebraF over Q(x) need not be a skew field. In fact,K ⊃ Q(x) isa splitting field forF if and only if the quadratic equationA1X

2 + A2Y2 − Z2 = 0 has

a solution(X,Y,Z) �= (0,0,0) in K3. For K = CK(x) this condition is equivalent tos0ands1 being squares inCK . Indeed, ifs0, s1 are squares, then one easily sees that thera, b ∈ CK such thatxa2 + (s0 − s1x)b2 −1= 0. On the other hand, let(X,Y,Z) ∈ CK(x)3

be a nontrivial solution of the quadratic equation. Then one can normalize this sosuch thatX,Y,Z ∈ CK [x] and the g.c.d. ofX,Y,Z is 1. Substitution ofx = 0 andx = s0

s1

in the equationxX2 + (s0 − s1x)Y 2 − Z2 = 0 yields thats0 ands1 are squares inCK .

Now we suppose that not boths0 and s1 are squares inQ. Then F is a quater-nion field. Examples of splitting fields forF are: Q(

√s0,

√s1 )(x) or F1 = Q(

√x ) or

Q(√

s0 − s1x + m2x ) for any m ∈ Q. The F -vector spaceM = Fe is made into a dif-


6

-merc

isthedter

re-

,

ator

ferential module overF by ∂e = de andd = b1 + xb2. We want to apply Theorem 3.with K = CK(x) where CK is any algebraic extension ofQ. Thus we have to showthat N := Q(

√x ) ⊗Q(x) N0 is irreducible, since the composite of the fieldsQ(x) and

F1 is Q(√

x ). Suppose thatM contains a one-dimensionalF1-vector spaceV , invariantunder∂ . Thenb2V is also a 1-dimensionalF1-vector space, invariant under∂ (indeed,

b1b2 = −b2b1 andb′2 = A′

22A2

b2) andM = V ⊕b2V . ThusM is always semi-simple overF1.

Then alsoN is semi-simple. Suppose thatN is reducible. ThenN has a direct sum decomposition. The Galois group of the extensionQ(

√x ) ⊃ Q(x) acts on these direct su

decompositions. From this one concludes thatQ(x) ⊗ N0 is also reducible and moreovthat Q(x) ⊗ N0 contains a submodule of dimension 2. LetL3 denote the minimal monioperatorL3 for the cyclic vectorb2e ⊗ e of N0. We will prove thatN is irreducible byshowing thatL3 has no right-hand factor of order 1 inQ(x)[∂]. One calculates thatL3 isequal to

x(s0 − s1x)∂3 + 3(s0/2− s1x)∂2 + (4(s1 − 1)x − 3s1/4− 4s0

)∂ + 6(s1 − 1).

Suppose that∂–u with u ∈ Q(x) is a right-hand factor ofL3. We make now a local analysat the singular pointsx = 0, s0/s1, ∞ of L3. The first two points are regular singular wilocal exponents 0,1,1/2. The point∞ is irregular singular and has only one “generalizlocal exponent” which is unramified, i.e., does not involve a root of the local parame1

x

at∞. This exponent is 3/2 and gives a local right-hand factor of the form∂ +3/2x−1+· · · .Thenu has the formu = l0

x+ l1

x−s0/s1+ f ′

fwith f ∈ Q[x], wherel0, l1 ∈ {0,1,1/2}. This

cannot produce the prescribed local right-hand factor at∞.Let K = CK(x). From Theorem 3.6 one concludes that there existsL2 ∈ K[∂] whose

symmetric square is equivalent toL3, if and only if the equationA1X2 + A2Y

2 − Z2 = 0has a nontrivial solution inK3. The latter is equivalent to

√s0,

√s1 ∈ CK .

Observation. One can compute a monic operatorL2 ∈ Q(√

s0,√

s1 )(x)[∂] with Sym2 L2equivalent toL3 above. Suppose thatQ(

√s0,

√s1 ) has degree 4 overQ. Let σ be a non-

trivial automorphism ofQ(√

s0,√

s1 ). Thenσ(L2) is not equivalent toL2. But accordingto Theorem 4.7 it must be protectively equivalent (see Section 4) toL2. We verified bycomputer computation the following:

Let σi interchange√

si and −√si , and leave

√sj invariant wherej �= i. Let t0 =√

A1A2 and let t1 = √A2. Thenσ0(L2) is equivalent toL2 ⊗ (∂ − t ′0

t0) and σ1(L2) is

equivalent toL2 ⊗ (∂ − t ′1t1

). If one of these operators is equivalent toL2 then (the modulecorresponding to)L2 is imprimitive by Lemma 4.1. The latter is excluded by the irducibility of L3.

3.2.2. Quaternion fields with generalA1 andA2Consider elementsA1,A2 ∈ Q(x) with A′

1 �= 0 �= A2. Even if F is not a skew fieldone can define∂ on M = Fe by ∂e = de whered = b1 + xb2. Again M is a differentialmodule overF1 and sym2

F1M = F1 ⊗Q(x) N0. Moreoverb2e ⊗ e is a cyclic vector forN0.

Let L3 denote the monic operator of order 3 withL3(b2e ⊗ e) = 0. In the following wewill make the equivalence betweenL3 and the second symmetric power of some oper


t

one

elle:

e

p of

ed

t

L2 over some fieldK ⊃ Q(x) explicit. For this purpose we introduce an operatorR oforder < 3, with coefficients inK ⊃ Q(x). One describesR by: the least common lefmultiple LCLM(R,L3) of R andL3 has the form Sym2 L · R = B · L3 for someL2 ∈K[∂] of order 2. In other words,R provides the isomorphismK[∂]/K[∂]Sym2 L2 →K[∂]/K[∂]L3 of differential modules:

L3 = 4A2∂3 − 2(A′

1A2/A1 + 2A′′1A2/A

′1 − 2A′

2)∂2

− (16A1A

′21 A2 + A′2

2 /A2 + A′1A

′2/A1 + 2A′′

1A′2/A

′1 − 2A′′

2 + 16)∂

+ 8(A′2/A2 + A′

1/A1 + 2A′′1/A

′1),

R = uA2∂2 + (2A′

1A2w + A′2u/2)∂ + 4A′

1A2v − 4u,

where(u, v,w) is a nonzero point on the conic

A1u2A2v

2 − w2 = 0. (1)

Assuming thatK ⊗Q(x) sym2 N0 has only one submodule of dimension 1, namely thewith generator

A2

4n0

1 ⊗ n01 − A2

4A1n0

2 ⊗ n02 − n0

3 ⊗ n03,

we have a one-to-one correspondence between all nonzero points(u, v,w) ∈ K3 on theconic, and all operatorsR ∈ K[∂] of order< 3 with the required property LCLM(R,L3) =Sym2 L2 · R for someL2 of order 2. We note thatQ(x) is a C1-field, and hence theris always a fieldK of the formCK(x) with [CK : Q] < ∞ such that (1) has a nontriviasolution. The degree[CK : Q] can be arbitrarily high as is shown in the following examp

A1 = x, A2 = (x − 2)(x − 3)(x − 5) · · · (x − pn),

wherepn is thenth prime number. The smallest field extensionCK of Q for which (1) hasa nontrivial solution inK = CK(x) is Q(

√2, . . . ,

√pn,

√(−1)n ).

Remark. For any specific choice, say ofA1,A2 ∈ Q[x] with A′1 �= 0 �= A2, we need to

verify thatL3 has the desired properties (i.e.,K ⊗Q(x) sym2 N0 has only one submodulof dimension 1). Suppose that the differential Galois groupG of M := QF1⊗F1 M satisfiesG ⊃ SL2. Then the differential Galois group ofF1 ⊗F1 M is G0 and contains SL2. Thenfor any algebraic extensionK of Q(x), the second symmetric power ofK ⊗Q(x) N0 has aunique submodule of dimension 1.

If at a pointp formal solutions involve logarithms, then the differential Galois grouM contains SL2. Indeed, the differential Galois group is a reductive group (recall thatM issemi-simple) and contains the additive groupGa . Examples with this feature are obtain

by a different choice for∂e, namely∂e = (A′

1 b0 + A′1 b1 + A′

1 b2)e. Suppose tha
A1−1 A1(A1−1) A2


at

thaterh

rsalons

:

-

the pointp satisfiesA1(p) = 1 andA2(p) �= 0, then one can verify that local solutionsp contain logarithms (it is sufficient to check this for the caseA1 = x, A2 ∈ Q(x) becauseone can then generalize the result by applying a pullbackx �→ A1). Thus logarithms willappear in local solutions ifA2 is not a multiple ofA1 − 1.

The operator obtained this way is:

L3 = 4∂3 + (4B2 − 12A′′

1/A′1 + (6A1 − 2)A′

1/B1)∂2

+ (2A′′

2/A2 + (3A1 − 1)A′1B2/B1 − 6A′′

1B2/A′1 − 4A′′′

1 /A′1

− B22 + 12(A′′

1/A′1)

2 − 2(3A1 − 1)A′′1/B1 − 16A′2

1 /(A1(A1 − 1)2)

− 16A′21 /A2

)∂ + 8

(A′2

1 B2 − (3A1 − 1)A′31 /A2

)/A2,

whereB1 = A1(A1 − 1) andB2 = A′2/A2.

This operator has the same conic (1). After, if necessary, replacingA1 with c2A1 forsome nonzeroc ∈ Q, we obtain thatA2 is not a multiple ofA1 − 1 (if A1 ∈ Q, then takec ∈ Q(x) instead of inQ). The conic (1) changes into an equivalent one. We concludefor any algebraic extensionK of Q(x), the operatorL3 is equivalent to a symmetric powof someL2 ∈ K[∂] if and only if (1) has a nonzero solution inK3. Furthermore, sucexamples exist for every nondegenerate conic overQ(x).

4. Projective equivalence

4.1. Some notation and definitions

Let k be a differential field with field of constantsCk . Put k = Ckk, whereCk is thealgebraic closure ofCk . The trivial 1-dimensional differential module is denoted by1. Thedeterminant det(M) of a differential moduleM is the 1-dimensional moduleΛnM , wheren is the dimension ofM . For a 1-dimensional differential moduleL, one writesL⊗n forthe tensor productL ⊗ · · · ⊗ L of n copies ofL.

Two differential modulesM1,M2 will be calledprotectively equivalentif there existsa differential moduleL of dimension 1 such thatM2 is isomorphic toL ⊗ M1. SupposethatCk is algebraically closed, thenM1,M2 correspond to representations of the univedifferential Galois groupU on finite-dimensionalCk-vector spaces. These representatiare projectively equivalent if and onlyM1 andM2 are projectively equivalent.

The translation in terms of monic differential operatorsL1,L2 ∈ k[∂] reads as followsL1 and L2 are projectively equivalent if there existsf ∈ k such that thek-algebra au-tomorphismφf : k[∂] → k[∂], given byφf (∂) = ∂ + f , has the property thatφf (L1) isequivalent toL2.

The problem. Let M be an irreducible differential module overK of dimensionn, whereK is a Galois extension ofk. Suppose thatσ M is projectively equivalent toM for everyσ ∈ Gal(K/k). The problem is to find the fieldsk ⊂ � ⊂ K for which there exists a differential moduleN , projectively equivalent toM , such thatN descends to�. The main caseof interest isK = k and� = C�k.


-

an

)

roup-

Lemma 4.1. Suppose thatM is irreducible, of dimensionn > 1 overk and thatk containsa primitiventh-root of unity. The following are equivalent:

(1) There exists a1-dimensionalL �∼= 1 with L ⊗ M ∼= M .(2) There exists a cyclic field extensionk ⊂ k(t) with equationtd = f ∈ k of degreed �= 1

dividing n, such that[k(t) : k ] = d and k(t) ⊗ M is a direct sum ofd irreducibledifferential submodules overk(t).

Proof. (1) ⇒ (2) det(M) ∼= det(L ⊗ M) ∼= L⊗n ⊗ det(M) and soL⊗n = 1. Let d be theminimal divisor ofn such thatL⊗d = 1. Put L = ke with ∂e = be. There exists an elementf ∈ k∗ with f ′

f= db. The isomorphismM → M ⊗ ke can be written asm �→

φ(m)⊗e, whereφ :M → M is a bijectivek-linear map. One finds thatφ∂φ−1 = ∂ +b andφd∂φ−d = ∂ +db = f −1∂f . Thenf φd commutes with∂ and thusf φd ∈ C∗

k . After chang-ing f we may suppose thatf φd = 1. One considers the field extensionk(t) ⊃ k, given bytd = f . It is easily seen that[k(t) : k ] = d . Defineψ := tφ : k(t) ⊗ M → k(t) ⊗ M . Thenψ∂ = ∂ψ andψd = 1. Let N0, . . . ,Nd−1 ⊂ N =: k(t) ⊗ M denote the eigenspaces ofψ

corresponding to the eigenvaluesζ id , i = 0, . . . , d − 1, whereζd denotes a primitived th-

root of unity. ThenN = ⊕i Ni and theNi are submodules ofN . We will use a cyclic

notation for theNi , e.g.,Nd = N0.Let σ ∈ Gal(k(t)/k) be the generator of this Galois group, given byσ(t) = ζ−1

d t .Thenσ acts also onN and one hasσ(Ni) = Ni+1 for all i. Indeed, forni ∈ Ni one hasψ(σni) = tφσni = ζdσ tφni = ζdσζ i

dni = ζ i+1d σni . Now suppose that (say)N0 has a non-

trivial submoduleN ′0. ThenN ′ := N ′

0 ⊕ σN ′0 ⊕ · · · ⊕ σd−1N ′

0 is a nontrivial submoduleof N , stable under the action ofσ . Then the set of theσ -invariant elements ofN ′ forms anontrivial submodule ofM . This contradicts the assumptions.

(2)⇒ (1) Write againN = k(t)⊗M and letσ have the same meaning as before. Takeirreducible submoduleN0 of N . PutNi := σ iN0 for i = 0, . . . , d −1. ThenN ′ := ∑d−1

i=0 Ni

is a differential submodule ofN , invariant under the action ofσ . The set of theσ -invariantelements ofN ′ forms a nonzero submodule ofM . HenceN ′ = N . The assumption in (2implies thatN = ⊕d−1

i=0 Ni . Define thek(t)-linear mapψ :N → N , by ψni = ζ idni for

any i = 0, . . . , d − 1 and anyni ∈ Ni . Thenψ∂ = ∂ψ andψσ = ζdσψ . Define thek(t)-linearφ :N → N by φ = t−1ψ . Thenφ∂φ−1 = ∂ + f ′

dfandσφ = φσ . The last equality

implies thatφ(M) = M and that the restrictionφ′ of φ to M is a k-linear bijection. Therelationφ′∂ = (∂ + f

df)φ′ implies thatL⊗M is isomorphic toM , whereL = ke is given by

∂e = f ′df

e. The assumption[k(t) : k ] = d implies thatd is minimal such thatL⊗d ∼= 1. �Remark. A differential moduleM of dimensionn > 1 will be calledcyclic-imprimitiveif M satisfies property (1) of Lemma 4.1. We recall that an irreducible action of a gG on a finite-dimensional vector spaceV is calledimprimitive if V has a direct sum decompositionV = V1 ⊕ · · · ⊕ Vd (with d > 1) such thatG permutes theVi . This induces ahomomorphismG → Sd which has as image a transitive subgroup ofSd since the actionof G is irreducible. The action ofG is calledcyclic-imprimitiveif the image ofG → Sd

is a cyclic group of orderd . If Ck is algebraically closed then the solution spaceV of theirreducible differential moduleM and the action of the differential Galois groupG on V


isrt

s

se

s

e

ts can

s

thm 4.7.

t

l

is well defined. In this situation, it is easily seen thatM is cyclic-imprimitive if and only ifthe action ofG onV is cyclic-imprimitive. An example of a cyclic-imprimitive operatorgiven in case 3 in Table 1 in Section 1 (ifk = Ck(x), K = CK(x) then case 3 implies pa(1) of Lemma 4.1).

Corollary 4.2. Let k = Ck(x) and k = Ck(x) and let M1, M2 be differential moduleover k, which descend to differential modulesN1,N2 over k. Suppose thatM1 and M2are protectively equivalent and thatM1 is not cyclic-imprimitive. ThenN1 and N2 areprotectively equivalent.

Proof. There is a one-dimensional moduleL over k, unique up to isomorphism becauM1 does not satisfy Lemma 4.1(1), such thatL⊗M1 ∼= M2. For anyσ in the Galois groupof k/k one hasσ L ⊗ σ M1 ∼= σ M2. Sinceσ Mi

∼= Mi for i = 1,2, one hasσ L ∼= L. ByTheorem 2.8, there exists a one-dimensionalL0 overk with L ∼= k ⊗L0. The two modulesL0 ⊗ N1 andN2 are isomorphic after tensoring withk overk. By Lemma 2.3,L0 ⊗ N1 isisomorphic toN2. �Example 4.3. Corollary 4.2 is no longer valid ifM1 is cyclic-imprimitive. We give two ex-amples. LetMi = Q(x)⊗Q(x)Ni for i = 1,2, whereN1,N2 denote the differential moduleoverQ(x) defined by the differential operatorsL1,L2. In the first example one chooses

L1 = ∂2 + 3(5x4 − 2)

2x2(x4 − 2)2and L2 = ∂2 − 3x2(x4 − 10)

4(x4 − 2)2.

One can verify the following. The differential Galois group ofM1 is DSL22 . Moreover

sym2 N1 ∼= sym2 N2 and the modulesN1,N2 are not projectively equivalent. They becomprojectively equivalent after extending the constants with a solution of the equationt4 +2 = 0 (not t4 − 2 as the denominators would suggest). The proof of these statemenbe deduced from part (2d) of the proof of Theorem 4.7.

In the second example one fixesc ∈ Q, not a square, an integern > 2, and considermonic operatorsL1,L2 ∈ Q(x)[∂] of degree 2, defined by the data:

– three regular singularities√

c,−√c,∞,

– exponent-difference 1/2 at±√c for L1,L2,

– L1 respectivelyL2 have exponent-difference1n

respectively 1+ 1n

atx = ∞.

The differential Galois group ofM1 is DSL2n , n > 2. The modulesN1,N2 are not projec-

tively equivalent, but become projectively equivalent after extending the constants wi√

c.The proof of these statements can be deduced from part (2b) of the proof of Theore

Theorem 4.4. Let K be Galois overk andG = Gal(K/k). Let M be an irreducible, nocyclic-imprimitive, differential module of dimensionn > 1 over K such thatdet(M) de-scends tok and EndK[∂](M) = CK . Suppose that for anyσ ∈ G the twisted differentiamoduleσ M is projectively equivalent toM . Then:


s

at

ng

pro-

me

(1) M induces a2-cocycle classc ∈ H 2(G,K∗), having an orderd with d | n. Let F

denote the(skew) field with centerk associated toc.(2) A fieldk ⊂ � ⊂ K is a splitting field forF if and only ifM is projectively equivalent to

a moduleN that descends to�.

Proof. (1) For σ ∈ G there is an isomorphismφ(σ) : σ M → M ⊗ L(σ) of differentialmodules. Then one has isomorphisms:

στMσ φ(τ)−→ σ

(M ⊗ L(τ)

) = σ M ⊗ σ L(τ)φ(σ )⊗1−→ M ⊗ L(σ) ⊗ σ L(τ)

andφ(στ) : στM → M ⊗ L(ατ). The assumption onM implies thatL(στ) ∼= L(σ) ⊗σ L(τ). From det(M) = det(σ M) = det(M) ⊗ L(σ)⊗n one concludes thatL(σ)⊗n = 1.Fix a basise(σ ) for eachL(σ). Let A(σ) be theσ -linear mapM → M for which m �→A(σ)(m) ⊗ e(σ ) is theσ -linear mapM → M ⊗ L(σ) associated toφ(σ). ThenA(σ)∂ =(∂ + a(σ ))A(σ) with a(σ ) ∈ K given by∂e(σ ) = a(σ )e(σ ).

One hasc(σ1, σ2)A(σ1σ2) = A(σ1)A(σ2) with c(σ1, σ2) ∈ K∗. Moreovera(σ1σ2) =a(σ1) + σ1(a(σ2)) + c(σ1,σ2)

′c(σ1,σ2)

. The collection{c(σ1, σ2)} defines a 2-cocycle with clas

c ∈ H 2(G,K∗). The 2-cocycle classcn is associated to the differential module det(M). Byassumption, det(M) descends tok and thuscn = 1.

(2) It suffices to consider the case� = k. Suppose thatc = 1, then one may suppose thc(σ1, σ2) = 1 for all σ1, σ2. Now {a(σ )} is a 1-cocycle with values inK . Such a 1-cocycleis trivial and thus has the forma(σ ) = σ(b)−b for a certainb ∈ K . The differential moduleN = M ⊗ Ke, with ∂e = be, descends tok. Indeed, forN one has that the correspondimapsA(σ) commute with∂ andA(σ1)A(σ2) = A(σ1σ2) holds for allσ1, σ2.

On the other hand, ifN with property (2) is given, then clearlyc = 1. �The condition that det(M) descends tok is not a serious restriction because everyM is

projectively equivalent to a module with this property. Note that ifK = k thenL(σ)⊗n = 1impliesL(σ) = 1, so the theorem follows from Theorem 2.10(d) in this case.

Example 4.5. A skew fieldF of dimension 9 overQ(x).Let α be a root of the polynomialz3 − 3z − 1. ThenQ(α) is Galois overQ. Let σ be

the automorphism that sendsα to 2− α2. Let F be the skew field with centerQ(x), withbasis 1, b, b2 asQ(α, x)-vector space, and multiplication rulesb3 = x andbf = σ(f )b.

We turn the vector spaceFe into aQ(α, x)[∂]-module by definingα′ = 0, b′ = b3x

and∂e = α+b

xe. Taking a cyclic vector then leads to the following operator:

L = L0L1L2 − x whereLi = x∂ − σ i(α) − i/3.

Now L, σ(L), σ 2(L) are not equivalent but are projectively equivalent. They are notjectively equivalent to an element ofQ(x)[∂].

If we increase our differential field toQ(α, x1/3) then these three modules becoisomorphic and descend to a module overQ(x1/3) given by the following operator


at

edle

s ofll

sym-

on-seuishesith theics,d dis-

n is as

9x3(x + 1)∂3 + 3x2(9− x1/3 + x2/3 + 8x)∂2 − 3x

(6+ 2x1/3 − x2/3 + 7x + x4/3)∂

− 9+ 5x1/3 − 11x2/3 − 10x − 3x4/3 − 3x5/3 − 9x2.

This operator was obtained through a cyclic vector computation ofFe, but this time viewedasQ(b)[∂]-module.

Remark 4.6. (1) Let the differential moduleM over K of dimensionn > 1 satisfy theassumptions of Theorem 4.4. Then symn

K M is projectively equivalent to a module thdescends tok. Indeed, the 2-cocycle associated to this module iscn = 1.

(2) If K = k andM is of dimensionn = 2 overK then a converse for (1) can be obtainfrom Theorem 4.7 below, and one finds: If sym2 M is projectively equivalent to a moduthat descends tok thenM is projectively equivalent to its conjugates overk.

(3) We recall the classification of irreducible algebraic subgroups of SL2(C), whereC

is an algebraically closed field of characteristic 0

DSL2∞ ,DSL2n ,A

SL24 , S

SL24 ,A

SL25 ,SL2(C).

The first five groups are the pre-images in SL2(C) of the subgroupsDpr∞, Dn, A4, S4 and

A5 of PSL2(C). FurtherDpr∞ is the projective dihedral group consisting of the element

PSL2(C) which stabilize the subset{0,∞} of P1(C). For more details see [Ko]. We wiuse this classification to prove Theorem 4.7.

(4) Theorem 4.7 below does not hold for higher symmetric powers nor for secondmetric powers of modules of dimension> 2. We give three examples whereM1,M2are not projectively equivalent, and symi M1 ∼= symi M2 with i = 3 for (a), (b) andi = 2 for (c). Let E(Q) be the one-dimensional module given by∂e = Q/xe. ThenE(Q1) ⊗ E(Q2) ∼= E(Q1 + Q2) andE(1) ∼= E(0).

(a) M1 = E(0) ⊕ E(15) andM2 = E( 1

15) ⊗ (E(0) ⊕ E(25)).

(b) M1 = E(0) ⊕ E( 110) ⊕ E( 3

10) andM2 = E(15) ⊗ (E(0) ⊕ E(− 1

10) ⊕ E(− 310)).

(c) M1 = E(0) ⊕ E(17) ⊕ E(3

7) andM2 = E( 114) ⊗ (E(0) ⊕ E(1

7) ⊕ E(57)).

Theorem 4.7. LetM1,M2 denote differential modules overk of dimension2. If sym2k M1 ∼=

sym2k M2, then there exists a one-dimensional differential moduleL overk such thatM2 ∼=

L ⊗ M1.

Remark. Corollary 4.2 implies that over rational functions this result is also valid for nalgebraically closed field of constantsexcept(see Example 4.3) in the imprimitive ca(cases (2b) and (2d) in the proof below). The proof below is long because it distingmany cases. Bas Edixhoven informed us that a shorter proof can be obtained wfollowing ideas: replace the groupsGi by the largest group that stabilizes both quadrconsider the intersectionR of these quadrics as a closed subscheme of length 4, antinguish cases based on the structure ofR.

Proof. The theorem is in fact a statement concerning representations. The translatiofollows. LetU denote the universal differential Galois group of the fieldk. The category


s

ted

how

nce

rential

f

sub-

e that

f

hen

-i-

of the differential modules overk is equivalent to the category of the finite-dimensionalC-linear representations of the affine group schemeU . HereC = Ck is the field of constantof k. One associates to a differential moduleM overk its solution spaceV equipped withthe action ofU . Let ρi :U → GL(Vi) for i = 1,2, denote the representations associato Mi . PutWi = sym2 Vi equipped with the induced representation sym2 ρi . There is givenan isomorphismB :W1 → W2 between the two representations. Now it suffices to sthat there exists aC-linear bijectionA :V1 → V2 such that sym2 A = B. Indeed, for anyg ∈ U one has thatAρ1(g)A−1 andρ2(g) have the same second symmetric power. Hethere exists aχ(g) ∈ {±1} such thatAρ1(g)A−1 = χ(g)ρ2(g). The mapg ∈ U �→ χ(g) ∈{±1} is a one-dimensional representation and corresponds to a one-dimensional diffemoduleL overk. It follows thatM1 ∼= L ⊗ M2.

Since we are dealing with only two differential modulesM1 andM2, we may replace inthe sequel the (somewhat fancy) affine group schemeU by the differential Galois group oM1 ⊕ M2, which is an ordinary linear algebraic group overC.

As before, one considers the canonical surjective map sym2C Wi → sym4 Vi and

its one-dimensional kernelKi for i = 1,2. One observes that aC-linear bijectionB :W1 → W2 has the form sym2 A for someC-linear bijectionA :V1 → V2 if and onlyif sym2 B : sym2

C W1 → sym2C W2 mapsK1 to K2.

Let Gi denote the image ofρi . The isomorphism betweenW1 andW2 implies that theinduced morphismsU → Gi/{±1} ∩ Gi , i = 1,2 coincide.

(1a) Suppose thatV1 is reducible and that it has precisely one proper invariantspaceL1. It easily follows that the only nontrivial invariant subspaces ofW1 areL1 ⊗ L1

and L1 ⊗ V1. Also V2 has a unique proper invariant subspaceL2, sinceG1/{±1} ∼=G2/{±1}. ThusL1 ⊗ V1 is isomorphic toL2 ⊗ V2. HenceV2 ∼= L1 ⊗ L−1

2 ⊗ V1.(1b) Suppose thatV1 has precisely two proper invariant subspaces. We may assum

the representationV1 is the sum of the trivial character (denoted by1) and a nontrivialcharacterχ1. BecauseG1/{±1}∩G1 ∼= G2/{±1}∩G2 one has thatV2 is the direct sum otwo distinct charactersχ2, χ3. The two sequences of characters 1, χ1, χ

21 andχ2

2 , χ23 , χ2χ3

are equal up to their order. Suppose thatx22 �= 1 �= x2

3. Then we may suppose thatχ2χ3 = 1,χ2

2 = χ1, χ23 = χ2

1 . Thenχ31 = 1, χ2 = χ2

1 , χ3 + χ1 andV2 = χ1 ⊗ V1.If say χ2

3 = 1, then after changingV2 into χ3 ⊗ V2. Thus we may suppose thatV2 =1 ⊕ χ2. If χ1 = χ2, thenV1 ∼= V2. If χ1 �= χ2, thenχ1 = χ2

2 , χ21 = χ2 andχ3

1 = 1. ThusV1 = 1 ⊕ χ1 = χ1 ⊗ (1 ⊕ χ2) = χ1 ⊗ V2.

(1c) Suppose thatV1 has more than two invariant subspaces of dimension 1. TG1/{±1} = {1}. Also G2/{±1} = {1}. HenceV1 andV2 differ by a character.

(2) Now we consider the case whereG1 is irreducible. After multiplyingρ1 by a charac-ter, one may assume thatG1 ⊂ SL2. The isomorphism betweenW1 andW2 implies that theimageG2 of the second representation lies in{Z ∈ GL2 | det(Z)3 = 1}. Using the abovenotation, we will show that there is a choice of the isomorphismB betweenW1 andW2

such that sym2(B) mapsK1 to K2.(2a) If G1 ∈ {SSL2

4 ,ASL25 ,SL2(C)}, then sym4 Vi , i = 1,2 has no invariant one

dimensional subspace. Then sym2 Wi , i = 1,2 has onlyKi as invariant subspace of dmension 1. HenceB(K1) = K2 and we are done.


per-

onsns

.-ble

get

s

nn

alter

ker-nel

nta-

(2b) If G1 = DSL2∞ , then we have to make a more detailed calculation. Let{v1, v2} be

a basis ofV1 such thatG1 consists of the transformations with determinant 1 whichmute the two linesCv1,Cv2 in V1. Thenb1 = v1 ⊗ v1, b2 = v2 ⊗ v2, b3 = v1 ⊗ v2 is abasis ofW1 = sym2

C V1. The spaceW1 is the direct sum of the irreducible representatiCb1 ⊕ Cb2 andCb3. The space sym2C W1 is the sum of the irreducible representatioCb1 ⊗ b1 + Cb2 ⊗ b2, Cb1 ⊗ b2, Cb1 ⊗ b3 + Cb2 ⊗ b3 andCb3 ⊗ b3. The family ofall invariant lines in sym2C W1 is C(λb1 ⊗ b2 + µb3 ⊗ b3) with λ,µ ∈ C, not both zeroEach member of the family defines a quadric inW1. The family has two special members namelyCb1 ⊗ b2 andCb3 ⊗ b3, where the quadric consists of two lines or a douline. The situation is similar forW2 and sym2 W2 (with adapted notationv′

1, v′2, b

′1, b

′2, b

′3).

Thus the isomorphismB :W1 → W2 sends the family of quadrics ofW1 to the one ofW2.The special quadrics are mapped to the special quadrics and one concludes thatBb3 is amultiple of b′

3 and{Bb1,Bb2} are multiples of{b′1, b

′2}. One has the freedom to chan

the isomorphismB by prescribingBb3 = λb′3 for anyλ ∈ C∗. In this way one obtains tha

sym2 B maps the lineK1 with generatorb1 ⊗ b2 − b3 ⊗ b3 to the lineK2 with generatorb′

1 ⊗ b′2 − b′

3 ⊗ b′3. Thus the changedB is a sym2 A. ForG1 = D

SL2n with n > 2 the same

proof works.Note that sym2 B depends quadratically onλ, which explains why

√c is needed in

Example 4.3.(2c) Assume now thatG1 = A

SL24 . Then W1 is the irreducible representationD of

G1/{±1} = A4 of dimension 3. Further sym2 W1 is the direct sum of three invariant lineL1(0),L1(1),L1(2) and D. The groupG1/{±1} = A4 acts trivially onL1(0) which isthe kernelK1 of sym2 W1 → sym4 V1. The action ofG1 on the other two lines is giveby the two nontrivial characters ofA4. The space sym2 W2 has a similar decompositioL2(0) ⊕ L2(1) ⊕ L2(2) ⊕ D. The notation is chosen such thatL2(0) is the kernelK2 ofsym2 W2 → sym4 V2. Again, L2(0),L2(1),L2(2) correspond to the three 1-dimensioncharacters ofG2/{±1} = A4. However,L2(0) need not correspond to the trivial characof A4 andB need not satisfy sym2(B)K1 = K2.

Now we replaceV2 by V ′2 := L2(0)⊗2 ⊗ V2. This new representation has the same

nel asρ2. Put W ′2 = sym2 V ′

2 = L2(0)⊗4 ⊗ W2. This representation has the same keras sym2 ρ2. The image of sym2 ρ2 is identified withA4. As representations ofA4 the twoobjectsL2(0)⊗4 ⊗D andD are isomorphic, since there is only one irreducible represetion of A4 with dimension 3. Thus sym2 V2 and sym2 V ′

2 are isomorphic. Then sym2 V1 andsym2 V ′

2 are isomorphic. LetB ′ denote the isomorphism. The decomposition of sym2 W ′2 is

L2(0)⊗9 ⊕ (L2(1) ⊗ L2(0)⊗8) ⊕ (

L2(2) ⊗ L2(0)⊗8) ⊕ D,

whereL2(0)⊗9 is the kernelK ′2 of sym2 W ′

2 → sym4 V ′2. Now, as required, sym2(B ′)K1 =

K ′2 and we conclude thatV1 andV ′

2 differ by a character. HenceV1 andV2 differ by acharacter.

(2d) Assume thatG1 = DSL22 . Consider a two-dimensional vector spaceV and a rep-

resentationρ :U → SL(V ) with imageDSL22 . For a suitable basisv1, v2 of V the space

W = sym2 V is a direct sum of the three invariant linesL1,L2,L3 with generatorse1 := v1 ⊗ v1 + v2 ⊗ v2, e2 := v1 ⊗ v1 − v2 ⊗ v2 and e3 := v1 ⊗ v2. They correspond


har-

t

er-

dy

he

iffer-

to the three characters ofD2 having order two. Then sym2 W is a direct sum of the3-dimensional spaceT spanned byei ⊗ ei , i = 1,2,3, having trivialD2-action, and thethree linesL1 ⊗ L2, L1 ⊗ L3, L2 ⊗ L3. These lines correspond again to the three cacters ofD2 of order two. The lineK , kernel of sym2 W → sym4 V , lies in T and isgenerated by an element

∑3i=1 ciei ⊗ ei with nonzeroc1, c2, c3 ∈ C. For any other elemen∑3

i=1 diei ⊗ ei in T with all di �= 0, there is an automorphismD of the representationWsuch that sym2 D maps the given element

∑3i=1 ciei ⊗ ei to

∑3i=1 diei ⊗ ei . Indeed, take

D of the formDei = λiei for i = 1,2,3 whereλ2i = di/ci .

Consider, as before, two representations(Vi, ρi) of U having dimension 2, such that thimageG1 of ρ1 is D

SL22 . Let an isomorphismB :W1 → W2 be given. Using the automo

phismD above, one changesB into B ′ such that sym2 B ′ mapsK1 to K2.In the first example of Example 4.3, the three linesL1,L2,L3 are defined over the fiel

of constantsQ(√

2). The above proof then shows thatN1,N2 must become projectivelequivalent if we extend the constants toC = Q(

√2, λ1

λ3, λ2

λ3). In the example,C turns out

to be the splitting field ofx4 + 2, which has degree 8 overQ. In general, ifN1,N2 aredifferential modules overQ(x) that become projectively equivalent overQ(x), and havedifferential Galois groupDSL2

n , then they become projectively equivalent overC′(x) forsome field extensionC′ of Q of degree� 4 whenn = 2, and degree� 2 whenn > 2. Inthe example one can show that the two subfields ofC that contain a solution ofx4 + 2= 0are the smallest fields of constants over whichN1,N2 become projectively equivalent.�

The following explains the constructions with quaternions in Section 3.

Corollary 4.8. Suppose thatN is an irreducible differential module overk of dimension3,which descends tok. Assume thatsym2 N has a1-dimensional submodule, such that tcorresponding quadratic form has ak-rational point. Then:

(1) There exists a differential moduleM over k with sym2 M ∼= N . For every σ ∈Gal( k/k) the modulesσ M andM are projectively equivalent.

(2) Let c ∈ H 2(Gal( k/k), k∗) denote the2-cocycle of order1 or 2, associated toM . LetF be the quaternion algebra overk associated toc. An algebraic extensionC

k⊃ Ck

yields a splitting fieldk = Ckk for F if and only if there exists a differential moduleM

over k such thatk ⊗k

sym2kM is isomorphic toN .

Proof. Apply Lemma 3.2, Proposition 3.1, Theorems 4.7 and 4.4 (observe that the dential Galois group ofM is irreducible and primitive sinceN is irreducible). �

Acknowledgments

We thank J.-A. Weil and E. Compoint for sharing their ideas in [C-W].


, 1972.

lge-

en Fun-

6.Univ.

mbolic

975)

plexe,96 (2004)

(2004)

Wiss.,

rential

, 1993.

References

[A] A.S. Amitsur, Differential polynomials and division algebras, Ann. of Math. (2) 59 (1954) 245–278.[Bl] A. Blanchard, Les corps non commutatifs, Collection SUP, vol. 9, Presses universitaires de France[Bo] N. Bourbaki, Élements de mathématiques, in: Algebres, Hermann, Paris, 1958, Chapitre 8.[C-W] E. Compoint, J.-A. Weil, Absolute reducibility of differential operators and Galois groups, J. A

bra 275 (1) (2004) 77–105.[F] G. Fano, Über lineare homogene Differentialgleichungen mit algebraischen Relationen zwischen d

damentallösungen, Math. Ann. 53 (4) (1900) 493–590.[H] M. van Hoeij, Factorization of linear differential operators, PhD thesis, University of Nijmegen, 199[Ka] N. Katz, Exponential Sums and Differential Equations, Ann. of Math. Stud., vol. 124, Princeton

Press, Princeton, NJ, 1990.[Ko] J. Kovacic, An algorithm for solving second order linear homogeneous differential equations, J. Sy

Comput. 2 (1986) 3–43.[L] A.H.M. Levelt, Jordan decomposition for a class of singular differential operators, Ark. Mat. 13 (1

1–27.[P] M. van der Put, Skew differential fields, differential and difference equations, in: Analyse com

systèmes dunamiques, sommabilité des séries divergentes et théories galoisiennes. I, Astérique 2191–205.

[P2] M. van der Put, Galois properties of linear differential equations, Ann. Fac. Sci. Toulouse 13 (3)459–475.

[P-S] M. van der Put, M.F. Singer, Galois Theory of Linear Differential Equations, Grundlehren Math.vol. 328, Springer-Verlag, Berlin, ISBN 3-540-44228-6, 2003.

[Ren] G. Renault, Algèbre non commutative, Gauthier–Villars, Paris, 1975.[Rei] I. Reiner, Maximal Orders, Academic Press, London, 1975.[S] J.-P. Serre, Corps locaux, Hermann, Paris, 1968.[Si] M.F. Singer, Solving homogeneous linear differential equations in terms of second order linear diffe

equations, Amer. J. Math. 107 (1985) 663–696.[So] R. Sommeling, Characteristic classes for irregular singularities, PhD thesis, University of Nijmegen

Descent for differential modules and skew fields

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