Non-linear Di erential Galois Theory · 1 Introduction Given a eld K, algebraic Galois theory allows one to associate a ( nite) group to any polynomial f2K[x]. From here, we can begin

Non-linear Differential Galois Theory

Alex Wood

Supervised by Dr. Robert A. Van Gorder

September 27, 2017

Contents

1 Introduction 2

2 Linear Differential Galois Theory 3

2.1 Differential Rings and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2 Differential Equations and Picard-Vessiot Extensions . . . . . . . . . . . . . . . . . . . . . . . 4

2.2.1 Continuing the Scalar Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.2 Aside: Introducing the Matrix Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3 The Differential Galois Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3.1 Setting the Scene . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3.2 Some Concrete Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3.3 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4 Liouvillian Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4.1 A Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4.2 Solvability by Quadratures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4.3 Relating Solvability by Quadratures to the Differential Galois Group . . . . . . . . . . 13

2.5 Comparing the Galois Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Introducing a Galois Theory of Non-linear Differential Equations 16

3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.1.1 The Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.1.2 Clarifying the Goal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.1.3 Some Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2 The Differential Galois Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2.1 First Integrals and the General Group of Analytic Series . . . . . . . . . . . . . . . . 18

3.2.2 Admissible Differential Isomorphisms and the Differential Galois Group . . . . . . . . 19

3.3 The Structure of the Differential Galois Group . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.3.1 Generalised Differential Polynomials and the Regular Prime Ideals . . . . . . . . . . . 20

3.3.2 The Fundamental Ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.4 Solvability Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4.1 A Clutch of Elementary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4.2 The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.4.3 Some More Concrete Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 Conclusion 31

4.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1

1 Introduction

Given a field K, algebraic Galois theory allows one to associate a (finite) group to any polynomial f ∈ K[x].From here, we can begin to resolve prima facie difficult questions regarding the properties of this polynomialin terms of concrete group-theoretic facts. In particular, the solvability of a polynomial by radicals may becharacterised in terms of the solvability of the associated Galois group.

One might wonder whether a similar procedure could be carried out with a differential equation; thisis the domain of differential Galois theory. For ordinary homogeneous linear differential equations, there isa well-understood existing theory, introduced by Emile Picard and Ernest Vessiot at the end of the 19thcentury, but formalised and polished by Ellis Kolchin around 50 years later. In order to give the reader aflavour of the rudiments of differential Galois theory, I will be reviewing the main results in this area at thebeginning of this paper. In my exposition of this material, I am indebted to both [2] and [9]. Any imprecisionhere can undoubtedly be clarified by their papers.

In contrast, little has been done vis-a-vis the Galois theory of ordinary non-linear differential equations.In this paper, I study the non-linear differential Galois theory of [5], focusing on the general polynomialfirst order non-linear differential equation. Jinzhi Lei has developed the theory to such an extent that wemay recover a result, analogous to the aforementioned highlight of algebraic Galois theory, regarding thesolvability by quadratures of this narrow class of equations. Furthermore, he provides criteria to demonstrateboth the solvability and insolvability of such equations. I intend to reconstruct Lei’s work in a more reader-friendly manner, providing examples to elucidate proceedings where necessary.

To conclude, I will give a brief discussion of the material covered here, with a view to outlining somepossible avenues of research where Lei’s theory might be generalised and extended.

∗ ∗ ∗

I’d like to thank my supervisor Robert for his guidance and probing questions every week. I’d also like tothank my friends and family for having the patience to let me jabber on about maths every day. Finally, I’dlike to thank Chris Rokos, for without his generous donation (in the form of the Rokos Awards) this paperand my accompanying experience of academic research would not have been possible.

2

2 Linear Differential Galois Theory

2.1 Differential Rings and Fields

We begin by giving the fundamental definitions setting apart the differential Galois theory from its algebraiccounterpart.

Definition 1. Let R be a ring. A derivation on R is a map δ : R 7→ R satisfying

δ(r + s) = δ(r) + δ(s), δ(rs) = δ(r)s+ rδ(s) ∀r, s ∈ R

Definition 2. A differential ring is a commutative ring R (containing a multiplicative identity) togetherwith a derivation δ on R. A differential ring is a differential field if R is a field.

Of course, the conditions on δ in Definition 1 are designed to algebraically mimic our usual notion ofthe derivative from differential calculus. As such, we encounter a first, almost canonical, example of adifferential ring: the ring of all infinitely differentiable functions on the real line with the aforementioned‘typical’ derivative.

In fact, any commutative ring R can be made into a differential ring, albeit a rather boring one, bytaking the trivial derivation δ : R 7→ R given by δ(r) = 0 ∀r ∈ R. Actually, for Z this is the only possiblederivation. Other, more substantial, examples include the ring of analytic functions in the complex planewith the ‘typical’ derivative and any abelian Lie algebra g with the derivation ad(a) for any a ∈ g.

Example 1. As one might expect, there are various ways one can build new differential rings from existingones. I emphasise a handful here:

1. The following proposition, which is easily proven, is particularly useful for construction purposes:

Proposition 1. Let R be a differential ring. If R is a domain, then the derivation δ extends toFrac(R), the fraction field of R, uniquely. Moreover, we have that

δ(r

s) =

δ(r)s− rδ(s)s2

∀r ∈ R ∀s ∈ R \ 0

Consequently, given a differential ring R without zero divisors, we can consider Frac(R) as a differentialfield.

2. In a similar fashion, if Y is a multiplicatively closed subset of a differential domain R such that 1 ∈ Ybut 0 /∈ Y , then δ extends uniquely to

Y −1R := ry−1 | r ∈ R, y ∈ Y ⊆ Frac(R)

Consequently, given a differential domain R, we can consider RP , the localisation of R at a prime idealP / R as a differential ring.

3. If R is a differential ring, then δ can be extended to a derivation on R[x] by assigning δ(x) an arbitraryvalue in R[x]. Consequently, given a differential ring R, we can consider the polynomial ring in onevariable (or more, by iteration) over R as a differential ring.

4. The final tool for forming new differential rings from old ones (that I will mention, at least), firstrequires an adaptation of a familiar definition:

Definition 3. Let R be a differential ring and I E R an ideal. I is a differential ideal of R ifδ(a) ∈ I ∀a ∈ I.

From here it is straightforward to check that Definition 3 allows us to define a derivation δ : R/I 7→ R/Iby setting δ(r+ I) = δ(r) + I. Consequently, given a differential ring R and a differential ideal I of R,we can consider the quotient ring R/I as a differential ring.

3

Definition 3 might make one wonder if any other basic notions in ring theory carry over to a differentialtheory. Indeed, let us first make the idea of a structure preserving map precise in this context:

Definition 4. Let R and S be differential rings, with derivations δR and δS and φ : R 7→ S a homomorphismof rings. Then φ is a differential homomorphism if φ(δR(r)) = δS(φ(r)) ∀r ∈ R.

Note: Clearly there is the potential for the subscript notation used above to become messy, and the use ofthe symbol δ only serves to complicate such expressions. Henceforth I primarily write r′ for δ(r), and theambient ring will usually be clear from the context. The condition in Definition 4 can thus be rewritten asφ(r′) = (φ(r))′.

Definition 4 enables us to recover differential versions of the classic cohort of isomorphism theoremsfamiliar from ring theory, whose statements and proofs proceed in the usual way.

Finally recall that a key concept in algebraic Galois theory is that of a field extension L/K. An analogousnotion plays an equally crucial role in the differential theory:

Definition 5. An injection K −→ L of fields is a differential field extension if the derivation of Lrestricts to that of K.

One type of differential field extension, known as a Picard-Vessiot extension, is of particular impor-tance; its definition and role in differential Galois theory will be made explicit with the introduction of thehomogeneous linear differential equations in the next section.

2.2 Differential Equations and Picard-Vessiot Extensions

Henceforth let K be a differential field of characteristic 0. Consider the following homogeneous lineardifferential equation over K:

L(y) = y(n) + a(n−1)y(n−1) + · · ·+ a2y

′′ + a1y′ + a0y = 0 ai ∈ K (1)

Note that, of course, the monicness of 1 can always be arranged since K is a field. In order to properlyunderstand the solutions to this differential equation, we first make the following important definition:

Definition 6. Let R be a differential ring. Then r ∈ R is a constant if r′ = 0. The set of constants in Ris denoted CR.

Easy checks show that CR is a subring of R, and in particular, that CK is a subfield of K. The notionof constants allows us to give some structure to the solutions of (1):

Proposition 2. Let V = f ∈ K | L(f) = 0. V is a vector space over CK . Also we have dimCK (V ) ≤ n.

If dimCK (V ) = n and we can exhibit concretely a basis for V then there isn’t much more left to askabout (1); we’d have a thorough understanding of its solutions in a similar way to having written down all ofthe roots of an algebraic polynomial. However, if the solution space is too small, i.e dimCK (V ) < n, there isstill work to do. We must construct a differential field extension L/K in which the solution space of (1) overCK has dimension n without containing any extraneous elements. This extension is the natural analogue ofthe splitting field of a polynomial from algebraic Galois theory and is known as a Picard-Vessiot extension.

Before making this notion precise, note that there are two ways of expressing homogeneous linear dif-ferential equations in the literature, and consequently there is a duality of perspective when it comes toPicard-Vessiot extensions. I presented (1) in what is known as scalar form, in the style of [2], and I willcontinue to favour this notation. There is, however, a matrix form of (1) described in [9]. Each method offersa distinct definition of a Picard-Vessiot extension; I will describe both and show how they marry together.

Note: In any case, the construction of a Picard-Vessiot extension for (1) requires that CK is algebraicallyclosed, which we will assume from now on. A good example to keep in mind then, is the differential fieldC(x) = Frac(C[x]) with the usual derivation. Here CC(x) = C, which is certainly algebraically closed.

4

2.2.1 Continuing the Scalar Approach

We first give a name to the situation in which we have found a sufficiently large differential extension L/Kfor (1):

Definition 7. Let L be a differential extension of K, and f1, . . . , fn ∈ L be solutions of (1) which arelinearly independent over CK . Then f1, . . . , fn is a fundamental set of solutions of (1) in L.

We are now able to give the (first) definition of a Picard-Vessiot extension:

Definition 8. (P-V 1): L is a Picard-Vessiot extension of (1) over K if:

1. L admits a fundamental set of solutions f1, . . . , fn of (1).

2. L is differentially generated over K by f1, . . . , fn. This coincides with our usual understanding of‘generation by a subset’ from ring theory, only that we also allow iterated applications of the derivation.We hereby denote this L = K〈f1, . . . , fn〉.

3. CK = CL

Roughly then, the first condition, as we have already noted, ensures that the differential extensionis sufficiently large to contain all of the solutions of (1), while the latter two conditions ensure that theextension is as small as possible. In this subsection, all references to Picard-Vessiot extensions will mean anextension satisfying (P-V 1).

The logic of the second criterion is clear, but the third might seem a bit obscure. The following example,courtesy of [2] ought to demystify it:

Example 2. Let K be a differential field and consider the homogeneous linear differential equation L∗(y) =y′−y = 0. Let L = K〈f〉, where f ′ = f , so that f is a solution of L∗(y) = 0. Bearing in mind that a Picard-Vessiot extension is analogous to a splitting field, we require that the Picard-Vessiot extension of L∗(y) = 0over L be trivial, since L is already large enough to provide us with all the solutions to L∗(y) = 0. Takingg to be a differential indeterminate, as discussed in Example 1, with g′ = g, then the extension M = L〈g〉clearly satisfies the first two conditions in Definition 8 but is a non-trivial extension of L. However, notethat g

f ∈M \ L is a constant, since, thanks to Proposition 1:

(g

f)′ =

g′f − gf ′

f2=gf − fgf2

= 0

Consequently CL 6= CM , and so the third condition prevents this situation, of repeatedly being able toconstruct bigger Picard-Vessiot extensions from old ones, from arising. In short, it guarentees the minimalityof the Picard Vessiot extension of a differential equation.

Before we begin our existence proof of the Picard-Vessiot extension of (1) over K, there is a final device thatwe must introduce:

Definition 9. Let R = K[g01, g02, . . . , g0n, g11, . . . , g1n, . . . . . . . . . , g(n−1)n] be a polynomial ring in n2 inde-terminates. We can extend the derivation of K to R by setting, with ai as in (1):

g′ij =

g(i+1)j if 0 ≤ i ≤ n− 2

−a(n−1)g(n−1)j − · · · − a0g0j if i = n− 1(2)

Now let Y be the multiplicatively closed subset (det(g(i−1)j))k | k ∈ N ⊆ R. Then the derivation of R (and

hence of K) extends to A := Y −1R by Example 1. A is called the full universal solution algebra of (1).

Observant readers will recognise that det(g(i−1)j) is in fact the wronskian of g01, g02, . . . , g0n.Now we are ready to construct the Picard-Vessiot extension L of (1) over K.

5

Note: One might wonder why it doesn’t suffice to take L = Frac(A), since g01, g02, . . . , g0n would be afundamental set of solutions for (1) and clearly we would have L = K〈g01, g02, . . . , g0n〉. Unfortunately wecannot guarentee that CK = CL, so we have to be more delicate than this.

Theorem 1. There exists a Picard-Vessiot extension L of (1) over K.

Proof. As indicated by the above note, the full universal solution algebra will at least play some role infinding such an extension. First recall two facts, both proved in [2]

Fact 1. Let R/K be a differential extension of rings, and I be a maximal element in the set of properdifferential ideals of R, which exists by Zorn’s Lemma. Then I is a prime ideal of R. (This is non-trivial:maximal differential ideals are not necessarily maximal ideals).

Fact 2. Let R/K be a differential extension of rings, where R is a domain, finitely-generated as a K-algebra,with no proper differential ideals. Then CK = CFrac(R).

Consider the full universal solution algebra A of (1). A is certainly a differential extension of K byconstruction, so letting J denote a maximal element in the set of differential ideals of A, we have that J is aprime ideal of A by Fact 1. Consequently, A/J is a domain, which is finitely generated as a K-algebra sinceA is finitely-generated as a K-algebra by g01, g02, . . . , g0n, g11, . . . , g1n, . . . . . . . . . , g(n−1)n, (det(g(i−1)j))

−1.If A/J had a proper differential ideal then this would contradict the maximality of J by the CorrespondenceTheorem on ideals, so we may apply Fact 2 to deduce that CK = CFrac(A/J).

Now take L = Frac(A/J). By the above discussion, CK = CL, so the final condition in (P-V 1) issatisfied. Clearly L = K〈g01 + J, g02 + J, . . . , g0n + J〉, and so it remains to show that this constitutes afundamental set of solutions of (1) to satisfy the first two criteria of (P-V 1). Indeed, det(g(i−1)j) ∈ A∗,the set of units of A, so det(g(i−1)j) /∈ J , because J was assumed to be a proper ideal of A. Thus

det(g(i−1)j + J) = det(g(i−1)j) + J 6= 0 + J

As we noted earlier, this states that the wronskian of g01 + J, g02 + J, . . . , g0n + J is non-zero, so g01 +J, g02 + J, . . . , g0n + J are linearly independent, as required.

In fact, we can do better:

Theorem 2. Suppose that M is also a Picard-Vessiot extension of (1). Then L and M are differentiallyK-isomorphic.

Proof. Again, we begin with a fact proved in [2]:

Fact 3. Let L1, L2 be Picard-Vessiot extensions of K. Let N/K be a differential field extension such thatCK = CN and let φi : Li 7→ N be differential K-homomorphisms, for i = 1, 2. Then φ1(L1) = φ2(L2)

Consider the ring S := (A/J) ⊗K M This is clearly finitely generated as an M -algebra by (g01 +J) ⊗K 1, (g02 + J) ⊗K 1, . . . , (g0n + J) ⊗K 1. Define a derivation δ on S by setting δ((a + J) ⊗K m) =δ(A/J)(a+J)⊗K m+ (a+J)⊗K δM (m). Denote by P a maximal proper differential ideal of S, which againexists by Zorn’s Lemma.

Consider the natural differential K-homomorphism ψ1 : A/J 7→ S. ψ−11 (P ) is an ideal of A/J , whichhas no proper differential ideals, and ψ−11 (P ) 6= A/J since this would force P = S when P was assumed

to be a proper ideal of S. Consequently, ψ−11 (P ) = 0, so A/JQψ1−−−→ S/P , where Q : S 7→ S/P is the

natural quotient map. Clearly, this means that L −→ Frac(S/P ). Symmetrically, by considering the natural

differential K-homomorphism ψ2 : M 7→ S, observe that MQψ2−−−→ S/P , and so M −→ Frac(S/P ).

By Fact 1, P is a prime ideal of S, so S/P is a domain and it is a finitely generated M -algebra since Sis a finitely generated M -algebra. Applying Fact 2, CM = CFrac(S/P ). But since M/K is a Picard-Vessiotextension, we must have CK = CM , and so CK = CFrac(S/P ).

6

Write φ1 and φ2 for the differential K-homomorphisms L −→ Frac(S/P ) and M −→ Frac(S/P ) re-spectively. Now we can easily obtain the required differential K-isomorphism between L and M , sinceφ1(L) = φ2(M) by Fact 3.

Thus we have shown that a Picard-Vessiot extension for (1) exists, and that it is unique, up to differentialK-isomorphism.

2.2.2 Aside: Introducing the Matrix Approach

Something about the introduction of the full universal solution algebra for (1) might have felt rather un-satisfactory; why take a polynomial ring in n2 variables when we could streamline the whole process byusing matrices? I now briefly outline this alternative formulation of the previous section, paying particularattention to the ‘new’ definition of a Picard-Vessiot extension.

Recall (1):L(y) = y(n) + a(n−1)y

(n−1) + · · ·+ a2y′′ + a1y

′ + a0y = 0 ai ∈ K

We can rewrite this as follows, to produce a matrix differential equation, where the derivation actscomponentwise:

yy′

...y(n−1)

′

=

0 1 0 0 · · · 00 0 1 0 · · · 00 0 0 1 · · · 0...

......

.... . .

...0 0 0 0 · · · 1−a0 −a1 −a2 −a3 · · · −a(n−1)

yy′

...y(n−1)

(3)

To save space, I will often write y′ = Ty for (3). Following [9], we can rephrase Definition 7 in terms ofmatrices.:

Definition 10. Let R/K be a differential extension of rings with CK = CR. Then B ∈ GLn(R) is afundamental matrix for (3) if B′ = TB. (Note the stipulation on the constants at this stage.)

Now we can introduce our second definition of a Picard-Vessiot extension:

Definition 11. (P-V 2) A Picard-Vessiot ring for (3) is a differential ring extension R of K such that:

1. R is a simple differential ring.

2. There is a fundamental matrix B ∈ GLn(R) for (3).

3. R is generated as a K-algebra by the coefficients of B and (det(B))−1.

L is a Picard-Vessiot extension of (3) over K if L = Frac(R).

As one would hope, there is an equivalence of definitions (stated below), so that the use of the term‘Picard-Vessiot extension’ is completely unambiguous. Given that we have seen the construction of a Picard-Vessiot extension in Theorem 1, one can imagine how this proof would go - see [9] for the details.

Proposition 3. Let L/K be a differential field extension. Then L satisfies (P-V 1) if and only if L satisfies(P-V 2).

7

2.3 The Differential Galois Group

2.3.1 Setting the Scene

The first two definitions will come as no surprise:

Definition 12. Let N/K be a differential field extension. The differential Galois group of N/K, writtenGal(N/K) is the set of all differential automorphisms of N fixing K pointwise, with composition of maps asthe binary operation.

Definition 13. The differential Galois group of (1) is Gal(L/K), where L is the Picard-Vessiot exten-sion of (1) over K.

Of major importance in the algebraic theory was the notion of a Galois extension: recall that a field ex-tension M/K is Galois if MGal∗(M/K) := m ∈M | σ(m) = m ∀σ ∈ Gal∗(M/K) = K, where Gal∗(M/K)refers to the algebraic Galois group. One reason this stipulation is so crucial is that it allows us to harnessthe full power of the fundamental Theorem. Since we will be proving a differential analogue of this soon,it makes sense to check that the same holds for Picard-Vessiot extensions vis-a-vis their differential Galoisgroup.

Theorem 3. Let L be the Picard-Vessiot extension of (1) over K. Then LGal(L/K) = K

Proof. It is immediate from Definition 12 that K ⊆ LGal(L/K).

The reverse inclusion follows at once from the fact proved in [2]:

Fact 4. If x ∈ L \K, then there exists a differential K-automorphism σ of L such that σ(x) 6= x.

What does the differential Galois group of (1) actually look like? Stop for a moment and recall the algebraicGalois group of a polynomial with degree n. The action of the group on the splitting field of the polynomialis determined by the way it permutes its roots (which is a finite set of n elements), since the roots generatethe splitting field. As such, the algebraic Galois group can naturally be identified with a subgroup of Sn.

Likewise, the action of the differential Galois group of (1) on the Picard-Vessiot extension is determinedby the way it moves the fundamental solutions (which now form a vector space of dimension n over CK),since the fundamental solutions generate the Picard-Vessiot extension. In parallel to the algebraic case, then,the differential Galois group can naturally be identified with a subgroup of GLn(CK).

Explicitly, let f1, . . . , fn be a fundamental set of solutions of (1) in L and σ ∈ Gal(L/K). Since σ is adifferential K-automorphism of L, we see that σ(fj) is a solution of (1) for each 1 ≤ j ≤ n. But f1, . . . , fnis a CK-basis of the solution space of (1) in L, so we must have that

σ(fj) =

n∑i=1

cijfi (1 ≤ j ≤ n)

where cij ∈ CK . Now the identification is completed by simply considering the map φ : Gal(L/K) 7→GLn(CK) given by φ(σ) = (cij).

Before we consider some concrete examples of differential Galois groups, there is a final question one mightask before vacating the abstract setting: how can we attach a notion of size to such a group? In the algebraictheory, we had the useful fact that, for a field extension M/K, |Gal∗(M/K)| = |M : K|. But since CK isalgebraically closed (and thus infinite), a non-trivial differential Galois group will often itself be infinite - wethus adopt a different perspective.

We start by introducing some algebraic geometry. Recall that in the affine space Am over CK , an affinevariety is a set of points in Am on which a set of polynomials in the ring CK [x1, . . . , xm] simultaneously

8

vanish. A typical exercise is to show that Am admits a topology (the Zariski topology) consisting of thealgebraic varieties as closed sets. We aim to view the differential Galois group as an algebraic variety; wewill then have the option of utilising the tools of Krull dimension.

Naıvely, one might identify Mn(CK), the set of all n×n matrices, with An2

, and use the non-vanishing ofthe determinant (a polynomial in the entries of a matrix) to define the invertible matrices. This would,however, result in GLn(CK) becoming an open set in the Zariski topology, when we require it to beclosed. To avoid this, introduce another variable z and identify GLn(CK) with the variety defined by

det(x11, . . . , xnn)z − 1. We can then move back into An2

by ‘forgetting’ the last variable. As a result,GLn(CK) satisfies the criterion for an algebraic group over CK , defined here:

Definition 14. An algebraic group over an algebraically closed field F of characteristic zero is an algebraicvariety G over F such that the multiplication and inversion operations are morphisms of varieties.

Note that matrix multiplication and inversion are clearly morphisms of varieties because of the polynomialexpressions describing them, as you will have seen in any introductory linear algebra course.

Now we turn to the differential Galois group itself. How do we know that Gal(L/K) is also an algebraicvariety? Thankfully we have the following statement from [2]:

Fact 5. There exists a set S of polynomials in n2 variables such that the elements (cij) ∈ GLn(CK) onwhich the polynomials in S vanish simultaneously are precisely those identified with Gal(L/K).

This shows that Gal(L/K) is a Zariski closed subgroup of GLn(CK). Moreover we can offer, with somework (see [2] again), a result regarding the Krull dimension of Gal(L/K):

Corollary 1. dim(Gal(L/K)) = tr.degKL

2.3.2 Some Concrete Examples

In this section, all differential equations will be defined over the differential field K = C(x) with the usualderivation, and σ denotes an element of Gal(Li/K).

Example 3. Consider the homogeneous linear differential equation L1(y) = y′ − y = 0. Clearly ex is afundamental set of solutions of L1 and adding it to K does not create any new constants, so L1 = C(x, ex)is a Picard-Vessiot extension for L1 over K.

By earlier discussion, we must have σ(ex) = cex for some c ∈ C∗K = C∗. Note that ex is transcendentalover K, so there are no other algebraic restrictions on our choice of c ∈ C∗. Since this choice whollydetermines the action of σ, we have Gal(L1/K) ∼= C∗

Example 4. Similarly, let L2(y) = y′ − 12xy = 0. In this case, L2 = C(x,

√x) is a Picard-Vessiot extension

for L2 over K. Again, we must have σ(√x) = c

√x for some c ∈ C∗K = C∗. However, since

√x is algebraic

over K, we have the additional restriction that c2 = 1, because

c2x = (c√x)2 = (σ(

√x))2 = σ(x) = x

Therefore, Gal(L2/K) ∼= ±1 ∼= C2. Evidently, differential Galois groups need not be infinite!

Example 5. Another instructive example can be found by taking L3(y) = y′′ + 1xy′ = 0. Here 1, ln(x)

gives a fundamental set of solutions of L3, and L3 = C(x, ln(x)) is a Picard-Vessiot extension for L3 overK. This time σ(1) = 1 since 1 ∈ CK , while σ(ln(x)) = c1 + c2 ln(x) for some c1, c2 ∈ CK . A differentialrestriction is obtained by noticing that

c2x

= (σ(ln(x))′ = σ(ln(x)′) = σ(1

x) =

1

x

Therefore c2 = 1. Since ln(x) is transcendental over K there are no further algebraic restrictions on ourchoice of c1, so σ(ln(x)) = c+ ln(x) for c ∈ CK = C. This choice wholly determines the action of σ; we haveGal(L3/K) ∼= (C,+), the additive group C.

9

Example 6. Finally, consider the equation L4(y) = y′′ + y = 0. By checking the wronskian, it isclear that sin(x), cos(x) is a fundamental set of solutions of this equation, adding no constants, soL4 = C(x, sin(x), cos(x)) is a Picard-Vessiot extension for L4 over K.

Suppose σ(sin(x)) = c1 sin(x)+c2 cos(x) and σ(cos(x)) = c3 sin(x)+c4 cos(x) for some c1, c2, c3, c4 ∈ CK .Then there is a differential restriction from:

c3 sin(x) + c4 cos(x) = σ(cos(x)) = σ(sin(x)′) = (σ(sin(x))′ = (c1 sin(x) + c2 cos(x))′ = c1 cos(x)− c2 sin(x)

As a result, we must have that c1 = c4 and c3 = −c2; we obtain the same relations by consideringσ(cos(x)′). There is also an algebraic restriction originating from the equation sin2(x) + cos2(x) = 1, sincewe have:

c21 + c22 = (c21 + c22)(sin2(x) + cos2(x))

= c21 sin2(x) + c22 cos2(x) + c21 cos2(x)) + c22 sin2(x)

= (c1 sin(x) + c2 cos(x))2 + (c1 cos(x)− c2 sin(x))2

= (σ(sin(x))2 + (σ(cos(x))2

= σ(sin2(x) + cos2(x))

= σ(1)

= 1

Therefore we see that Gal(L4/K) ∼=(

a −bb a

)∈ GL2(C) | a2 + b2 = det

(a −bb a

)= 1

.

2.3.3 The Fundamental Theorem

We move on to the jewel in the crown of linear differential Galois theory. As in the algebraic theory, thereis a correspondence between the differential subfields of L containing K and the Zariski closed subgroups ofGal(L/K). Thus, in analogue with algebraic Galois theory, we can begin to understand fairly intractableproblems regarding intermediate differential subfields in terms of more approachable questions in the theoryof groups. We state the result here for completeness, but the proof is fairly arduous - interested readers cancheck [2], [9] or any text dedicated to the linear differential Galois theory for that matter:

Theorem 4. Let Gal(L/K) be the differential Galois group of (1).

1. There exists an inclusion reversing correspondence between the intermediate differential field extensionsL/M/K and Zariski closed subgroups H ≤ Gal(L/K), given by the mutually inverse bijections:

M 7→ Gal(L/M) H 7→ LH

2. The intermediate field M is a Picard-Vessiot extension of K if and only if Gal(L/M) E Gal(L/K).In this case,

Gal(M/K) ∼= Gal(L/K)/Gal(L/M)

As an illustration of the power of this theorem, I present two examples, the first of which I owe to [7]:

Example 7. Recall Example 3. The equation L1(y) = y′−y = 0, defined over K = C(x), has Picard-Vessiotextension L1 = C(x, ex), and so the differential Galois group of L1 is Gal(L1/K) ∼= C∗.

By considering elementary properties of the modulus and argument of the complex numbers, the Zariskiclosed subgroups of C∗ are precisely the groups of units µn := e 2kπ

n i ∈ C | k = 1, . . . , n for each n ∈ N -note that µn is, of course, the set of simultaneous zeroes of xn − 1.

10

Thus, by the fundamental theorem, the intermediate differential fields L1/Mn/K are given by Mn =Lµn1 = K(enx). To see this, identify σ with an element of µn, then:

σ(enx) = (σ(ex))n = (cex)n = cnenx = enx

since c ∈ µn. In the special case that m | n, then we can denote the correspondence of the resulting inclusionreversed towers of subfields and subgroups in the familiar way:

L e

K(emx) µm

K(enx) µn

K C∗

Finally, note that C∗ is abelian, so µn E C∗ for each n ∈ N. Therefore, by the fundamental theorem,K(enx)/K is a Picard-Vessiot extension (for the equation y′ − ny = 0) for each n ∈ N, with differentialGalois group Gal(K(enx)/K) ∼= C∗/µn.

Example 8. Consider L5(y) = y′ − 16xy = 0 defined over K = C(x). In this case, L5 = C(x, x

16 ) is a

Picard-Vessiot extension for L5 over K. We must have σ(x16 ) = cx

16 for some c ∈ C∗K = C∗. However, since

x16 is algebraic over K, we have the additional restriction that c6 = 1, since

c6x = (cx16 )6 = (σ(x

16 ))6 = σ(x) = x

Therefore, Gal(L5/K) ∼= µ6∼= C6.

Since we have obtained a finite group we are back in familiar territory from correspondences seen inalgebraic Galois theory. We present another diagram with the details:

L e

K(x13 ) K(x

12 ) C2 C3

K C6

Of course, C2, C3 E C6. Therefore, K(x13 )/K is a Picard-Vessiot extension (for the equation y′− 1

3xy = 0),

with differential Galois group Gal(K(x13 )/K) ∼= C6/C2

∼= C3. Likewise K(x12 )/K is a Picard-Vessiot

extension (for the equation y′ − 12xy = 0), with differential Galois group Gal(K(x

12 )/K) ∼= C6/C3

∼= C2; wesaw this in Example 4.

2.4 Liouvillian Extensions

2.4.1 A Recap

Let’s remind ourselves of some of the results proved in algebraic Galois theory. Consider f ∈ Q[x]. Informally,we say that f is solvable by radicals if every root of the polynomial can be generated from Q using the

11

operations +,−,×,÷, and the taking of nth roots for any n ∈ N. As a, rather convoluted, example, thepolynomial g(x) = x3 + 2x2 + 3x+ 4 is solvable by radicals; its roots are:

x1 =1

3

(3

√5(3√

6− 7)− ( 3√

5)2

3√

3√

6− 7

)x2, x3 = −2

3+

( 3√

5)2(1± i√

3)

63√

3√

6− 7− 1

6(1∓ i

√3)

3

√5(3√

6− 7)

We can translate this rather intuitive notion of solvability by radicals into the language of fields. Moreformally, then, we say that a field extension F/Q is radical if there is a sequence of elements α1, . . . , αn ∈ Fand m1, . . . ,mn ∈ Z+ such that F = Q(α1, . . . , αn) and αmii ∈ Q(α1, . . . , αi−1) for each 1 ≤ i ≤ n. Then fis solvable by radicals precisely if f splits in a radical extension F of Q (in other words, the splitting field off embeds into a radical extension F of Q).

In practice, this is a very awkward characterisation to deal with, particularly when it comes to workingwith a polynomial that isn’t solvable by radicals - how would one go about showing there doesn’t exist aradical extension of Q in which h(x) = x5 − 6x + 3 splits? This is where algebraic Galois theory comes in,with the following theorem, proved by Galois himself:

f is solvable by radicals if and only if the Galois group of f is solvable.

Now the task of demonstrating that h(x) is not solvable by radicals becomes far easier to perform.Explicitly, h(x) is irreducible by Eisenstein’s criterion, and using differential calculus, we can show that ithas precisely two real roots. From here, we know that the Galois group of f contains a 2-cycle and a 5-cycle,which by elementary group theory suffices to show that it is isomorphic to S5. Of course, S5 is not solvable,since it contains the non-solvable A5 as a subgroup, which completes, by Galois’ theorem, our proof thath(x) is not solvable by radicals.

It makes sense for the differential Galois theorist to ask a similar question. That is, can we tell whetherour homogeneous linear differential equation (1) admits a ‘nice’ fundamental set of solutions, just by lookingat some property of its differential Galois group?

2.4.2 Solvability by Quadratures

Firstly, we must clarify what we mean by ‘nice’ solutions in a differential context - what will play thedifferential analogue of radical solutions to a polynomial equation? Let us make a couple of definitions:

Definition 15. An elementary function is a function of one variable which is the (finite) composition ofthe operations +,−,×,÷, exponentials, logarithms, constants and solutions of algebraic equations.

Remark 1. This class of functions is wider than it might prima facie appear. For instance, it includes thetrigonometric functions since sin(x) = 1

2i (eix − e−ix) and similarly arccos(x) = −i ln(x+ i

√1− x2).

Definition 16. The homogeneous linear differential equation (1) is solvable by quadratures if everyfundamental solution of (1) is formed from elementary functions, allowing the operations +,−,×,÷, com-position, integration and taking solutions of algebraic equations.

Warning: In the literature, the inclusion of ‘solutions of algebraic equations’ in the above definitions some-times results in the use of the term ‘generalised elementary function’ and ‘solvable by generalised quadratures’respectively. We do not make this distinction here.

We can see then that L4(y) = y′′+y = 0 from Example 6 is solvable by quadratures, for example. In fact,all second-order homogeneous linear differential equations with constant coefficients are solvable by radicals;an easy observation demonstrated in any elementary differential calculus course. Again, we can translatethis intuitive notion of solvability by quadratures into the language of differential fields, in parallel with theearlier algebraic discussion.

Definition 17. Let L/K be a differential field extension and α ∈ L. α is primitive over K if α′ ∈ K, and

α is exponential over K if α′

α ∈ K.

12

Remark 2. The relationship between the above definition and quadratures ought to be clear, but let usmake it explicit. Concretely, take K = C(x). Of the list of legitimate methods to form our quadratures,the arithmetic operations and constants are taken care of by the fact we are working in the differential fieldK. Solutions of algebraic equations will be dealt with separately in the next definition. This leaves us withintegration, logarithms and exponentiation:

• Primitive elements clearly provide us with a notion of integration in K. Take L = K〈arctan(x)〉, forexample; note that arctan(x) is primitive over K since arctan(x)′ = 1

1+x2 . Consequently we can write

arctan(x) =∫ x 1

1+t2 dt. Handily, the fact that we have introduced integration also gives us the notion

of logarithm, since ln(x) =∫ x 1

t dt.

• Exponential elements (unsurprisingly) provide us with a notion of exponentiation in K, now that we

have integration and logarithms. Let g(x) ∈ L/K be such that g(x)′

g(x) = f ∈ K. Then ln(g(x)) =∫ xf(t)dt, which implies that g(x) = e

∫ x f(t)dt.Definition 18. A differential field extension L/K is Liouvillian if there is a sequence of elementsα1, . . . , αm ∈ L such that L = K〈α1, . . . , αm〉, with CK = CL, and for each 1 ≤ i ≤ m, either:

1. αi is primitive over K〈α1, . . . , αi−1〉.

2. αi is exponential over K〈α1, . . . , αi−1〉.

3. αi is algebraic over K〈α1, . . . , αi−1〉.

Therefore, we see that (1) is solvable by quadratures precisely if the Picard-Vessiot extension of (1)embeds into a Liouvillian extension of K. Once more, though, this is a very awkward characterisation todeal with, particularly when it comes to working with a homogeneous linear differential equation that isn’tsolvable by quadratures. What we need is a result in analogue with Galois’ theorem above, relating thenotion of solvability by quadratures to a property of the differential Galois group.

2.4.3 Relating Solvability by Quadratures to the Differential Galois Group

Recall that the differential Galois group Gal(L/K) of (1) is an algebraic group, so in particular, it is analgebraic variety. Therefore, we can write Gal(L/K) as a finite union of irreducible algebraic varieties. Thefollowing is a straightforward observation:

Proposition 4. The identity element e ∈ Gal(L/K) belongs to a unique irreducible component of Gal(L/K).

Proof. Let V1, . . . , Vm be the distinct irreducible components of Gal(L/K) containing e. Then W := V1 ×. . . × Vm is an irreducible variety, and so V1 . . . Vm (the image of W under the product morphism) is anirreducible subset of Gal(L/K) containing e. Consequently, V1 . . . Vm is contained in some Vi. Therefore wehave that

Vj ⊆ V1 . . . Vm ⊆ Vi (1 ≤ j ≤ m)

and since V1, . . . , Vm are irreducible, it follows that V1 = . . . = Vm. The only way to avoid a contradictionis to conclude m = 1.

This ensures that the following notion is well-defined.

Definition 19. The identity component of Gal(L/K) is the unique irreducible component of Gal(L/K)containing the identity element e. It is denoted Gal(L/K)0.

The aim of this section, then, is to prove the following result:

Equation (1) is solvable by quadratures if and only if the identity component Gal(L/K)0 of the differentialGalois group of (1) is solvable.

13

We shall proceed by proving each direction separately, making liberal use of results from the theory ofalgebraic groups - all of the necessary arguments are demonstrated fully in the appendix of [2].

Theorem 5. Suppose that (1) is solvable by quadratures. Then the identity component Gal(L/K)0 of thedifferential Galois group of (1) is solvable.

Proof. By definition, the Picard-Vessiot extension L of (1) over K embeds into a Liouvillian extension Mof K. Therefore, write M = K〈α1, . . . , αm〉, with the sequence α1, . . . , αm ∈ M as in Definition 18. Weperform induction on m, so we may assume Gal(L〈α1〉/K〈α1〉)0 is solvable. Now we rely on a result provedin [2]:

Fact 6. Let N be a Picard-Vessiot extension of K, and N〈z〉 be an extension of N without new constants.Then N〈z〉/K〈z〉 is Picard-Vessiot and Gal(N〈z〉/K〈z〉) ∼= Gal(N/(N ∩K〈z〉)).

Thus we have Gal(L〈α1〉/K〈α1〉) ∼= Gal(L/(L∩K〈α1〉)). Since M/K is Liouvillian, we can make a casedistinction:

• Assume α1 is primitive or exponential over K. By generalising Examples 3, 4 and 5, we see thatK〈α1〉 is a Picard-Vessiot extension of K with differential Galois group isomorphic to one of C∗K , afinite cyclic group, or (CK ,+). Since Gal(K〈α1〉/K) is, in any case, commutative, any subgroup ofit must be normal, so by the fundamental theorem, all differential fields between K〈α1〉 and K arePicard-Vessiot extensions of K. In particular, L ∩ K〈α1〉 is a Picard-Vessiot extension of K withGal((L ∩K〈α1〉)/K) ∼= Gal(K〈α1〉/K)/Gal(K〈α1〉/(L ∩K〈α1〉)) a commutative group.

This means that Gal(L/(L ∩K〈α1〉)) is normal in Gal(L/K) and Gal(L/K)/Gal(L/(L ∩K〈α1〉)) iscommutative, which, by general theory of algebraic groups, suffices to show that Gal(L/K)0 is solvable.

• Assume α1 is algebraic over K, so that Gal(L/(L ∩ K〈α1〉)) has finite index in Gal(L/K). By gen-eral theory of algebraic groups, this suffices to show that Gal(L/K)0 = Gal(L/(L ∩ K〈α1〉))0 ∼=Gal(L〈α1〉/K〈α1〉)0, which we have already seen to be solvable above.

Therefore, whichever case we find ourselves in, we must have that Gal(L/K)0 is solvable.

Theorem 6. Suppose that the identity component Gal(L/K)0 of the differential Galois group of (1) issolvable. Then (1) is solvable by quadratures.

Proof. Let F = LGal(L/K)0 . By general theory of algebraic groups, Gal(L/K)0 is normal in Gal(L/K), withfinite index.

Firstly, by the fundamental theorem, this means that F/K is a finite Picard-Vessiot extension, and so bythe primitive element theorem (since charK = 0), F = K(α) for some α ∈ F algebraic over K. A fortiori,F = K〈α〉. Therefore, F/K is Liouvillian.

Secondly, we know that Gal(L/F ) ∼= Gal(L/K)0; note that L/F is clearly a Picard-Vessiot extension.Now we require a result attributed to Lie and Kolchin:

Fact 7. Let G be a connected solvable subgroup of GLn(CK). Then G is triangularisable.

In particular, this gives us that Gal(L/F ) is triangularisable. That is, there exists a fundamental setv1, . . . , vk ∈ L such that for each σ ∈ Gal(L/F ):

σ(vj) = a1jv1 + . . .+ ajjv1 (1 ≤ j ≤ k) (4)

where aij ∈ CF depend on σ. We claim, by induction on k, that L = F 〈v1, . . . , vk〉 is a Liouvillian extensionof F .

Taking j = 1 in (4) gives σ(v1) = a11v1, from which we obtain σ(v′1) = (σ(v1))′ = a11v′1. Clearly,

v′1v1

isinvariant under each σ ∈ Gal(L/F ). Now L/F is a Picard-Vessiot extension, so recall that by Theorem 3,

we have LGal(L/F ) = F . This means thatv′1v1∈ F , and so v1 is exponential over F .

14

For the remaining k− 1 equations in (4), we divide each by the equation σ(v1) = a11v1 and differentiate,so we arrive at:

σ(vjv1

)′ =a2ja11

(v2v1

)′ + . . .+ajja11

(vjv1

)′ (2 ≤ j ≤ k) (5)

By induction, F 〈( v2v1 )′, . . . , ( vkv1 )′〉 is a Liouvillian extension of F . But clearly, this means that F 〈 v2v1 , . . . ,vkv1〉 is

a Liouvillian extension of F since we are merely adding elements primitive over F 〈( v2v1 )′, . . . , ( vkv1 )′〉. Finally,since v1 was shown to be exponential over F , this demonstrates that F 〈v1, v2v1 , . . . ,

vkv1〉 = F 〈v1, . . . , vk〉 = L

is a Liouvillian extension of F .

Therefore we have shown that both F/K and L/F are Liouvillian, which implies that L/K is Liouvillian.Trivially then, the Picard-Vessiot extension of (1) over K embeds into a Liouvillian extension of K. As such,(1) is solvable by quadratures.

2.5 Comparing the Galois Theories

As one will have surely noticed by now, there are considerable analogies between algebraic Galois theory andlinear differential Galois theory. To summarise this exposition of linear differential Galois theory, we presenta table to document the similarities between the features of the two theories.

Algebraic Galois Theory Linear Differential Galois Theory

Field Differential fieldHomomorphism Differential homomorphism

Polynomial Homogeneous linear differential equationRoots Fundamental set of solutions

Splitting field Picard-Vessiot extensionGalois group Differential Galois group

Solvable by radicals Solvable by quadraturesRadical extension Liouvillian extension

15

3 Introducing a Galois Theory of Non-linear Differential Equa-tions

We have seen that there is a fairly complete Galois theory for linear differential equations which, as we notedat the conclusion of the previous section, mimics many of the constructions encountered in algebraic Galoistheory. The situation is different for non-linear differential equations. Not only has there been a relativescarcity of literature in this area, but publications have tended to focus on specific classes of non-lineardifferential equations, rather than treating them in generality as we saw in the previous section. Of course,this is no surprise, given the heterogeneity of the class of non-linear differential equations in comparison totheir linear counterparts.

In this section, we outline the non-linear differential Galois theory introduced in [5]. Any proofs, eitherpresented or omitted here, can, unless explicitly noted, be found in Lei’s paper.

3.1 Preliminaries

3.1.1 The Equation

We focus now on the following first-order non-linear differential equation:

dy

dx=p(x, y)

q(x, y)(6)

where p and q are polynomials. Corresponding to this this equation, we have the differential form F , givenby:

F(ω) = (q(x, y)δx + p(x, y)δy)(ω) = 0 (7)

where here and henceforth we use δx and δy to denote ∂∂x and ∂

∂y , respectively. Now, (7) may look even more

unwieldy than (6), but to motivate its use, suppose we have a non-constant solution ω(x, y) of (7), analyticat a point (x0, y0). Then on a solution to (6), we have:

q(x, y)dω

dx= q(x, y)(ωx +

dy

dxωy)

= q(x, y)(ωx +p(x, y)

q(x, y)ωy)

= q(x, y)ωx + p(x, y)ωy

= (q(x, y)δx + p(x, y)δy)(ω)

= F(ω)

= 0

Since we are assuming q(x, y) does not vanish on the domain, we must have dωdx = 0, which means that we

have arrived at the implicit solution ω(x, y) = C of (6), where C is some constant. Therefore, non-trivialsolutions to (7) are invaluable, because they allow us to extract information about our chosen non-lineardifferential equation. Thankfully, the above situation is not a rare one, with the following theorem due to[3]:

Theorem 7. Let (x0, y0) ∈ C2 be a non-critical point. Then there exists a non-constant solution ω(x, y) of(7) that is analytic at (x0, y0).

The uniformity of this situation allows us to make a definition:

Definition 20. Let ω(x, y) be the non-constant solution of (7) which is analytic at the non-critical point(x0, y0). We call ω(x, y) the first integral of (6) at (x0, y0).

16

Henceforth, for ease of notation, we will assume that the origin, (0, 0), is a regular point, and thatq(0, 0) 6= 0. Any mention of a ‘first integral’ will implicitly refer to a first integral at the origin. All of theresults hold equally well for some other regular point (x0, y0) provided that q(x0, y0) 6= 0. Meanwhile, K willalways mean the differential field of rational functions in x and y, with derivations δx and δy, containing Cas the field of constants.

The first integrals of (6) form a crucial part of this non-linear theory. As we will see, where the algebraicand linear differential Galois groups act on roots and fundamental sets of solutions respectively, the non-lineardifferential Galois group defined later in this section will act on the first integrals.

3.1.2 Clarifying the Goal

In the linear theory, we sought to characterise the solvability of a linear differential equation by quadraturesin terms of a property of the differential Galois group. Here we have an analogous goal, but some minoramendments need to be made to take into account the fact that we now have two derivations, and that weare working with first integrals rather than fundamental sets of solutions.

Definition 21. A differential field extension L/K is Liouvillian if there is a sequence of elementsα1, . . . , αm ∈ L such that L = K〈α1, . . . , αm〉, with CK = CL, and for each 1 ≤ i ≤ m, either:

1. δj(αi) ∈ K〈α1, . . . , αi−1〉 for j = x, y. That is, αi is primitive over K〈α1, . . . , αi−1〉 under both δx andδy.

2.δjαiαi∈ K〈α1, . . . , αi−1〉 for j = x, y. That is, αi is exponential over K〈α1, . . . , αi−1〉 under both δx and

δy.

3. αi is algebraic over K〈α1, . . . , αi−1〉.

Definition 22. Equation (6) is solvable by quadratures at (0, 0) if there exists a first integral ω of (6)such that K〈ω〉 embeds into a Liouvillian extension of K.

Remark 3. Note that if (6) is solvable by quadratures at (0, 0), then the first integral described in Definition22 is expressible by quadratures in the way we described in the previous section. From here, one can showthat this first integral is in fact analytic on a dense open set in C2 - see [8] - and so (6) is solvable byquadratures on a dense open set in C2. We can thus say that (6) is solvable by quadratures.

As hinted at above, our ultimate aim is to be able to relate to solvability of (6) by quadratures to someproperty of the differential Galois group of (6), which we are yet to define. For completeness, we state themain result of this section here:

Equation (6) is solvable by quadratures if and only if the differential Galois group of (6) over K is solvable.

3.1.3 Some Notation

Throughout this paper, the notion of analyticity arises so regularly that it is worth bringing in some notation.In particular, we write:

• A0 for the set of all functions f(z) of one variable which are analytic at z = 0,

• A00 = f(z) ∈ A0 | f(0) = 0,

• A10 = f(z) ∈ A0

0 | f ′(0) 6= 0.

In this way, we also write Ω(0,0)(F) for the set of all first integrals of (6) and also

Ω0(0,0)(F) = ω ∈ Ω(0,0)(F) | ω(0, 0) = 0, Ω1

(0,0)(F) = ω ∈ Ω0(0,0)(F) | δyω(0, 0) 6= 0.

17

3.2 The Differential Galois Group

3.2.1 First Integrals and the General Group of Analytic Series

Let’s begin by elaborating on the set Ω(0,0)(F) of first integrals of (6). One might wonder, having seenTheorem 7, whether the notion of a first integral is quite a rigid one. Due to a result in [1], we see that thisis far from the case:

Proposition 5. Let f(y) ∈ A0 be a function. Then there exists a unique ω(x, y) ∈ Ω(0,0)(F) such thatω(0, y) = f(y) for all y in a neighbourhood of 0.

This shows us that the set of all first integrals of (6) is at least as big as the set of functions in onevariable which are analytic at the origin. Equally, this prima facie raises the worry that Ω(0,0)(F) is so bigthat it starts to become intractable, resisting any meaningful sort of analysis. Thankfully, given that we areonly really interested in the solvability of (6) by quadratures, we can restrict our attention to the narrowerclass Ω1

(0,0)(F) of first integrals, by the following proposition:

Proposition 6. Equation (6) is solvable by quadratures if and only if there exists a first integral ω ∈ Ω1(0,0)(F)

such that K〈ω〉 embeds into a Liouvillian extension of K.

Proof. Suppose that (6) is solvable by quadratures, so that there is some u ∈ Ω(0,0)(F) such that K〈u〉embeds into a Liouvillian extension M of K. Without loss of generality, we may assume that u ∈ Ω0

(0,0)(F),

since we can always subtract the constant u(0, 0) from u without vacating the class of first integrals. Ifδyu(0, 0) 6= 0 then we are done. Otherwise, write u(0, y) =

∑∞i=k aiy

i for some k ≥ 2 such that ak 6= 0.

Now we can take f(y) = k√u(0, y) = y(

∑∞i=k aiy

i−k)1k , so that clearly f(y) ∈ A1

0. Taking advantage ofproposition 5, there is a unique ω ∈ Ω(0,0)(F) such that ω(0, y) = f(y) ∈ A1

0. By construction, we have

ωk = u, so M〈ω〉 is a Liouvillian extension of K, and trivially K〈ω〉 embeds into M〈ω〉.The reverse direction is simply a special case of the definition of (6) being solvable by quadratures.

Now that we need only concentrate on the first integrals in Ω1(0,0)(F), we can see some structure on the

first integrals beginning to form. In particular, by proposition 5, there is a 1 − 1 correspondence betweenΩ1

(0,0)(F) and A10. Since A1

0 is clearly a group under composition, we can let it act on Ω1(0,0)(F) in the

canonical way to obtain a rather useful characterisation of the first integrals in this set:

Proposition 7. Let ω ∈ Ω1(0,0)(F). Then Ω1

(0,0)(F) = f(ω) | f ∈ A10. Denote this set by A1

0(ω).

Proof. Let f ∈ A10. It is clear that f(ω) ∈ Ω0

(0,0)(F), and δyf(ω)(0, 0) = f ′(0)δyω(0, 0) 6= 0, which completesthe reverse inclusion of the statement.

Let u ∈ Ω1(0,0)(F), and write g(y) = ω(0, y) and h(y) = u(0, y), so that g, h ∈ A1

0. Then, since A10 is a

group, we can take f = h g−1 ∈ A10, and deduce that u = f(ω) by proposition 5. This proves the forward

inclusion.

We shall not stop, however, at letting A10 act on Ω1

(0,0)(F). In fact, we can construct an even larger

group, subsuming A10, which allows us to include (roughly) the action of the broader class of functions A0 on

the first integrals in Ω1(0,0)(F). To accomplish this, we first adjoin an infinitesimal variable ε to the constant

field C, and make the following definition.

Definition 23. Let f(z, ε) =∑∞i=0,j=0 fijz

iεj ∈ C[[z, ε]]. The series f(z, ε) is analytic if it converges when(z, ε) takes values in a neighbourhood of (0, 0). Denote the set of all analytic series as A0[[ε]].

Remark 4. Another way of expressing an analytic series is by writing f(z, ε) =∑∞i=0 fi(z)ε

i where fi(z) ∈A0. Not only does it look cleaner, but it elucidates our choice of notation for the set of all analytic series.

18

Definition 24. Let G[[ε]] = f(z, ε) =∑∞i=0 fi(z)ε

i ∈ A0[[ε]] | f0(z) ∈ A10. We can define a multiplication

on G[[ε]] by setting f(z, ε) ∗ g(z, ε) = f(g(z, ε), ε). The group (G[[ε]], ∗) is called the general group ofanalytic series.

The details of showing that (G[[ε]], ∗) is a bona fide group are straightforward, and can be found in [5],but note that the condition on f0(z) is crucial, because the non-vanishing of the first derivative ensures thatthe criterion regarding inverses in G[[ε]] is satisfied.

The introduction of the general group of analytic series marks the beginning of our construction of adifferential Galois group for (6). Given some f(z, ε) ∈ G[[ε]], we can define an action on ω ∈ Ω1

(0,0)(F) by

setting f(z, ε)(ω) = f(ω, ε). As one might hope, f(ω, ε) is also a first integral of (6), which we can verify bychecking:

F(f(ω, ε)) = F(

∞∑i=0

fi(ω)εi)

=

∞∑i=0

F(fi(ω)εi)

= F(f0(ω)) +

∞∑i=1

F(fi(ω)εi)

= F(f0(ω)) +

∞∑i=1

F(fi(ω))εi since δxε = 0 = δyε

=

∞∑i=0

F(fi(ω))εi

=

∞∑i=0

(q(x, y)f ′i(ω)ωx + p(x, y)f ′i(ω)ωy)εi

= F(ω)

∞∑i=0

f ′i(ω)εi

= 0

Before we move on, note that the multiplication ∗ extends partially to A0[[ε]]. That is, if h(z, ε) ∈ A0[[ε]]and f(z, ε) ∈ G[[ε]], then h(z, ε) ∗ f(z, ε) = h(f(z, ε)ε) ∈ A0[[ε]] is well defined. This will be useful in someproofs of later results, once we introduce the notion of an admissible differentiable isomorphism.

3.2.2 Admissible Differential Isomorphisms and the Differential Galois Group

So far, our approach to the non-linear differential equation (6) has been largely analytic. To recover apowerful Galois theory, we need to reintroduce the algebraic notion of a differential isomorphism. This willallow us to pick out a narrower subset of G[[ε]] with a clearer structure, making the properties of the (6)more transparent.

The following definition, which may seem obscure at first, will turn out to be critical in our developmentof the theory:

Definition 25. Let ω ∈ Ω1(0,0)(F) and write M = K〈ω〉. An admissible differential isomorphism of

M/K with respect to F at (0, 0) is a differential K-isomorphism φ : K〈ω〉 7→ K〈f(ω, ε)〉, where f(z, ε) ∈ G[[ε]]and φ(ω) = f(ω, ε), satisfying the following condition:

• Let h1(z, ε), . . . , hm(z, ε) ⊆ A0[[ε]], where m ∈ Z+. Then φ extends to a differential K-isomorphismψ : K〈ω, h1(ω, ε), . . . , hm(ω, ε)〉 7→ K〈f(ω, ε), h1(f(ω, ε), ε), . . . , hm(f(ω, ε), ε)〉, where ψ(hi(ω, ε)) =hi(f(ω, ε), ε) for each 1 ≤ i ≤ m.

19

Remark 5. Since an admissible differential isomorphism is determined by the element f(z, ε) ∈ G[[ε]] whichacts on the first integral ω, we can naturally identify the set of all admissible differential isomorphisms ofM/Kwith respect to F at (0, 0), denoted henceforth by Ad(M/K,F)(0,0), with a subset of G[[ε]]. Furthermore,we can also endow Ad(M/K,F)(0,0) with the multiplication ∗ defined earlier on G[[ε]]. Simple checks, whichare again demonstrated in [5], show that (Ad(M/K,F)(0,0), ∗) is a group, and hence a subgroup of (G[[ε]], ∗).

By considering Ad(M/K,F)(0,0) as a subgroup of the general group of analytic series in this way, wearrive at some rather nice properties. Firstly, we will show in the upcoming proposition that the group ofadmissible differential isomorphisms is independent, up to isomorphism, of our choice of first integral for (6):

Proposition 8. Let u ∈ Ω1(0,0)(F) and write N = K〈u〉. Then Ad(M/K,F)(0,0) ∼= Ad(N/K,F)(0,0)

Proof. From proposition 7, there is some f ∈ A10 such that u = f(ω). Write σ = f(z, 0) ∈ G[[ε]], so that

ω = σ−1u.

Now let φ ∈ Ad(M/K,F)(0,0). We claim that σ ∗ φ ∗ σ−1 ∈ Ad(N/K,F)(0,0); it is clearly a differentialK-isomorphism as required, since φ ∈ Ad(M/K,F)(0,0), so it suffices to prove the additional condition indefinition 25. To this end, let h ∈ A0[[ε]]. Since φ ∈ Ad(M/K,F)(0,0) and σ, h ∗ σ ∈ A0[[ε]], we can extendφ to K〈ω, σ(ω), h ∗ σ(ω)〉, so that it sends σ(ω) = u and h ∗ σ(ω) = h(u) to σ ∗ ψ(ω) = σ ∗ φ ∗ σ−1(u) andh ∗ σ ∗ φ(ω) = h ∗ σ ∗ φ ∗ σ−1(u) respectively. Clearly, ψ |K〈u,h(u)〉 is the required extension of σ ∗ φ ∗ σ−1,proving our claim.

Now consider the map π : Ad(M/K,F)(0,0) 7→ Ad(N/K,F)(0,0), given by π(φ) = σ ∗ φ ∗ σ−1, whichwe now know to be well-defined. π is clearly an injective group homomorphism. Note, by adapting theprevious argument, that if ψ ∈ Ad(N/K,F)(0,0), then σ−1 ∗ ψ ∗ σ ∈ Ad(M/K,F)(0,0), so we have thatπ(σ−1 ∗ ψ ∗ σ) = ψ. This proves that π is also surjective, and consequently an isomorphism.

We will also see, later, that the group Ad(M/K,F)(0,0) is independent of our choice of (0, 0) as theregular point; in short, Ad(M/K,F)(0,0) is uniquely determined by our differential equation (6). As such,we make the following definition:

Definition 26. The differential Galois group of (6) over K at (0, 0) is defined to be Ad(M/K,F)(0,0),and is denoted Gal(F)(0,0).

3.3 The Structure of the Differential Galois Group

At this point, as in the linear exposition, we ask ourselves what such a Galois group might look like. Insome sense, because we are only interested in the solvability of (6) by quadratures, we are not particularlyworried about being able to write the group down explicitly, but rather we are more concerned with certainproperties of the group, namely (as we will see) its solvability. As such, we are motivated to take quite aspecific approach in our investigation of the group.

3.3.1 Generalised Differential Polynomials and the Regular Prime Ideals

Of pivotal structural importance to the differential Galois group is the fundamental ideal. This ideal and itsorder, both to be defined, suffice to characterise completely the solvability of (6) by quadratures. In order toproperly introduce it, we need the concept of a differential polynomial, together with some related notions.Most, if not all, of these ideas were developed by one of the founding fathers of differential algebra, JosephRitt. Consequently, our review of this material here was inspired, and indeed can be supplemented, by [6].

We begin by taking t to be an indeterminate over K, and writing A0(t) = f(t) | f ∈ A0. Then elementsof the polynomial ring K[A0(t)] can be written in the form

∑ni=0 aifi(t) where ai ∈ K and fi ∈ A0. We can

consider K[A0(t)] as a differential ring, with derivations δx and δy, by setting:

δif(t) = f ′(t)δit (i = x, y), δj(δkxδly)t =

δk+1x δlyt if j = x

δkxδl+1y t if j = y

20

where f ∈ A0. Denote this differential ring by KA0(t); clearly, by the above definitions of the actions ofδx and δy the elements are polynomials in the derivatives δkxδ

lyt with coefficients in K[A0(t)] (by noting that

f ′ ∈ A0 is implied by f ∈ A0).

Remark 6. Readers with a rudimentary knowledge of differential algebra will recognise that the elementsof KA0(t) in which the aforementioned ‘coefficients’ are all polynomials in t are called differential poly-nomials. However, since Lei’s theory uses analytic functions in the Galois group to act on the first integrals,we need a broader notion of differential polynomials, which we will encounter in the next few definitions.

Definition 27. Let α ∈ KA0(t), such that α is polynomial in the derivatives δkxδlyt whose coefficients

consist of a combination of polynomials in t and f(t) with f ∈ A0). Then α is a quasi-differentialpolynomial (henceforth written QDP). If, in addition, α involves at least one proper derivative of t, thenwe say α is a proper quasi-differential polynomial (henceforth written PQDP).

It is possible, and important for future results, to give the set of QDP an ordering. To do this, we start byintroducing a lexicographic ordering on the derivatives of t. Any mention of ‘higher’ vis-a-vis the derivativesof t will henceforth refer to this lexicographic order. If α is a QDP, then we will call the highest derivativeof t in α the ‘leader’.

This allows us to create a weak total order on the set of QDP:

Definition 28. Let α1 and α2 be QDP. Then say that α2 has higher rank than α1 if either:

1. α2 has a higher leader than α1.

2. α2 has the same leader as α1, but the degree of the leader in α2 exceeds the degree of the leader in α1.

We are almost ready to connect our equation (6) to the QDP outlined above. We need a final definition:

Definition 29. A regular prime ideal of KA0(t) is a prime ideal P of KA0(t) containing exclusivelyPQDP.

Now let P / KA0(t) be a regular prime ideal containing F(t). Write F(t) for the differential ringgenerated by F(t) and consider α(t) ∈ P , an irreducible element with lowest rank. By the primality of P ,there are two cases: Either F(t) 6⊆ P or F(t) = P . For now, suppose the former. α(t) does not involveδxt or any of its derivatives, since this would imply it had higher rank than F(t). Therefore we can writeδµy for the leader of α(t). The value of µ is of critically important to the theory. To this end, it deserves itsown definition.

Definition 30. Let P / KA0(t) be a regular prime ideal containing F(t) and let µ be as constructed inthe above discussion. Then µ is called the order of P , denoted ord(P ) = µ. In the case that F(t) = P ,we say that ord(P ) =∞.

We now have all of the tools to introduce the most important object in Lei’s differential Galois theory:the fundamental ideal. The exposition of this material will be dealt with in the next section.

3.3.2 The Fundamental Ideal

We begin by proving a useful result.

Lemma 1. Suppose there exists α(t) ∈ K[A0(t)], which is not identically zero, and ω ∈ Ω1(0,0)(F) such that

α(ω(x, y)) = 0 for all (x, y) in a neighbourhood of (0, 0). Then K contains a first integral of (6).

Proof. Henceforth, for ease of notation, write α(ω) = 0 for the hypotheses in the statement.

Let Σ = α(t) ∈ K[A0(t)] | α(ω) = 0 and α(t) 6≡ 0. Then by assumption, Σ 6= ∅. Recall from ourearlier discussion that elements in Σ ⊆ K[A0(t)] can be written in the form

∑ni=0 aifi(t) where ai ∈ K and

21

fi ∈ A0. Pick such an element of Σ where n is as small as possible. Write α0(t) for this element, and n0 forits additive length.

Suppose n0 = 1. Then α0(t) = a1f1(t) with a1 6= 0. We are assuming a1f1(ω) = 0, which impliesf1(ω) = 0. This is only possible if ω is a constant, but first integrals are non-constant by definition.Therefore n > 1.

Now write α0(t) =∑n0

i=1 aifi(t), so that α0(ω) =∑n0

i=1 aifi(ω) = 0. From this, with some familiarcalculations, we deduce F(α0(ω)) =

∑n0

i=1 F(ai)fi(ω) = 0. Taking an appropriate linear combination of theprevious two equations, we see :

a1F(α0(ω))−F(a1)α0(ω) =

n0∑i=2

(a1F(ai)−F(a1)ai)fi(ω) = 0

Assume that a1F(ai) − F(a1)ai 6= 0 for some 2 ≤ i ≤ n. Then β(t) =∑n0

i=2(a1F(ai) − F(a1)ai)fi(t) isan element of Σ with smaller additive length than α0, which is a contradiction.

In particular, a1F(a2)−F(a1)a2 = 0, which suffices to show that F(a2a1 ) = 0. Clearly a2a1

is non-constant,since otherwise we would again have contradicted the minimality constraint on n0. Therefore a2

a1∈ K is a

first integral of (6).

What is the relevance of this lemma to our discussion? Well, if K does not contain a first integral of (6),and α(t) ∈ KA0(t) such that α(ω) = 0 for some ω ∈ Ω1

(0,0)(F), then by contraposition, we must conclude

that α(t) 6∈ K[A0(t)]. This means α(t) must involve some proper derivatives of t, which is the same as sayingthat α(t) is a PQDP.

Most importantly, however, it allows us to give a proof of the existence of what will come to be knownas the fundamental ideal. Before we state this theorem, we first need another definition.

Definition 31. Let P / KA0(t) be a regular prime ideal, ω ∈ Ω1(0,0)(F) and f(z, ε) ∈ G[[ε]]. Then

f(ω(x, y), ε) satisfies P if for any α(t) ∈ P , α(f(ω(x, y), ε)) = 0 for all x, y, ε small enough.

The following is proved fully in [5]:

Theorem 8. Let K be a differential field containing no first integral of (6) and ω ∈ Ω1(0,0)(F). Then there

exists a unique regular prime ideal ξ / KA0(t), such that:

1. For every φ ∈ Gal(F)(0,0), let φ(ω) = fφ(ω, ε) where fφ(z, ε) ∈ G[[ε]]. Then fφ(ω(x, y), ε) satisfies ξ.

2. Let f(z, ε) ∈ G[[ε]] be such that f(ω(x, y), ε) satisfies ξ. Then there exists φ ∈ Gal(F)(0,0) such thatφ(ω) = f(ω, ε).

In other words, the elements of G[[ε]] satisfying ξ are precisely those which are associated to elements inthe differential Galois groups by the natural identification mentioned in Remark 5.

This result ensures that the key object in Lei’s theory is well-defined:

Definition 32. Let ξ be the ideal with the properties as stipulated in Theorem 8. Then ξ is called thefundamental ideal of (6).

Remark 7. Recall the notion of the order of a regular prime ideal introduced in Definition 30. Of course,in the case where K does contain a first integral of (6), it currently doesn’t make sense to consider ord(ξ),since we cannot appeal to Theorem 8 to ensure that the fundamental ideal exists. To maintain a uniformityof language, if K contains a first integral of (6), we henceforth write ord(ξ) = 0.

We will now see that the fundamental ideal, in particular its order, provides us with all of the informationwe need about (6). The next theorem is the zenith of Lei’s theory, and all subsequent results are effectivelycorollaries from this point onwards. The proof is lengthy, with many laborious calculations, so we leave theinterested reader to review [5].

22

Theorem 9. Let K be a differential field and ξ as in Theorem 8 (noting the previous remark). Then0 ≤ ord(ξ) ≤ 3 or ord(ξ) =∞. Furthermore, we have:

ord(ξ) Status of First Integral Gal(F)(0,0)

0 there exists ω ∈ Ω1(0,0)(F) such

that ω ∈ Ke

1 there exists ω ∈ Ω1(0,0)(F) such

that (δyω)n ∈ K for some n ∈ Nf(z, ε) ∈ G[[ε]] | f(z, ε) = λz + c(ε), c(0) = 0, λn = 1

2 there exists ω ∈ Ω1(0,0)(F) such

thatδ2yω

δyω∈ K

f(z, ε) ∈ G[[ε]] | f(z, ε) = a(ε)z + c(ε), c(0) = 0

3 there exists ω ∈ Ω1(0,0)(F) such

that2δyωδ

3yω−3(δ

2yω)

2

(δyω)2∈ K

f(z, ε) ∈ G[[ε]] | f(z, ε) = a(ε)z1+b(ε)z + c(ε), c(0) = 0

∞ there exists ω ∈ Ω1(0,0)(F) G[[ε]]

In particular, if Gal(F)(0,0) is solvable, then ord(ξ) ≤ 2.

Evidently, the power of this theorem cannot be overstated. It will, of course, be invaluable when it comesto our characterisation of the solvability of (6) by quadratures in terms of the solvability of the differentialGalois group, which we will outline in the next section.

However, the result can do more for us than this. Pedantic readers may have been troubled by oursteadfast loyalty to the regular point (0,0) with respect to the differential Galois group throughout thisexposition of Lei’s work. What can we say about the equation (6), one might wonder, if picking a differentregular point led to a different differential Galois group? Thankfully, Theorem 9 is the key to showing thatthis won’t happen.

We start by stating, what was at the time, a rather technical proposition from earlier in Lei’s paper. Bypostponing it until now, one can more easily appreciate the significance of the result, since it is at its mostpotent when working in tandem with Theorem 9.

Proposition 9. Assume ω ∈ Ω1(0,0)(F). Then we have the following:

1. If (δyω)n ∈ K for some n ∈ N, then φ(ω) = µnω + c(ε), ∀φ ∈ Gal(F)(0,0), where µn is an nth root ofunity.

2. Ifδ2yω

δyω∈ K, then φ(ω) = a(ε)ω + c(ε), ∀φ ∈ Gal(F)(0,0).

3. If2δyωδ

3yω−3(δ

2yω)

2

(δyω)2∈ K, then φ(ω) = a(ε)ω

1+b(ε)ω + c(ε), ∀φ ∈ Gal(F)(0,0).

Here a(ε), b(ε), c(ε) ∈ A0[[ε]] ∩ C[[ε]] and c(0) = 0. Furthermore, let (x0, y0) be such that q(x0, y0) 6= 0, ω isanalytic at (x0, y0) and δy(x0, y0) 6= 0. Note that if we set u = ω−ω(x0, y0), we clearly have u ∈ Gal(F)(x0,y0).Then the above results 1− 3 are valid, ∀φ ∈ Gal(F)(x0,y0).

This proposition may seem quite obscure in isolation, but now that we have Theorem 9, it ought to beclear what can be said. In the cases where Gal(F)(0,0) is non-trivial, by reading down the middle columnof the above, we will always find ourselves in one of the three cases stipulated by Proposition 9. Now theaddendum to this proposition is music to our ears. In these cases, the structure of the differential Galoisgroup (determined, as we noted earlier in Remark 5, by the action on ω), is independent of our choice ofregular point. That is, we have demonstrated the following:

Corollary 2. Let (x0, y0) be a regular point. Then Gal(F)(0,0) ∼= Gal(F)(x0,y0)

To summarise, we have, in separate parts of this exposition, shown that the differential Galois group of(6) is independent of both our choice of regular point and of first integral for (6). As such, we can now simplywrite Gal(F) in place of Ad(K〈ω〉/K,F)(0,0), where ω ∈ Ω1

(0,0)(F). We have shown this to be well-defined.

23

3.4 Solvability Revisited

Now that the concept of the fundamental ideal and its significance have been established, we are well-placedto return to the matter at hand. Recall that our original goal was to characterise the solvability of (6) interms of the solvability of its differential Galois group. To do this, we first require a clutch of elementaryresults to ease the process.

3.4.1 A Clutch of Elementary Results

We first reintroduce the analogue of a Galois extension from algebraic Galois theory:

Definition 33. Let M/K be a differential field extension and G be a group of differential K-isomorphismsof M . Then say M/K is differentially Galois with respect to G if MG = K.

Now would be a disturbing time to discover that the proper differential extensions of K involved whenGal(F) is non-trivial did not actually exhibit this property, given our coining of the phrase ‘differentialGalois group’ long ago. Thankfully this is not the case; the following (surprisingly technical) proposition isproved fully in [5]:

Proposition 10. Suppose K does not contain a first integral of (6). Then for any ω ∈ Ω1(0,0)(F), K〈ω〉/K

is differentially Galois over Gal(F).

This is, naturally, quite reassuring, because it creates parallels with both the algebraic Galois theory andthe linear differential Galois theory we have seen thus far; the elements of Gal(F) fix the base field andnothing more, allowing us to obtain maximal information about the solution to (6) itself.

The following result is, once again, quite straightforward but please check Lei’s paper for the details. Tostate it properly, we need to fix some notation.

Definition 34. Let M/K be a differential field extension and G be a group of differential K-isomorphismsof M . Then for any intermediate differential subfield M/L/K, we write L′ = φ ∈ G | φ(a) = a ∀a ∈ L.

Proposition 11. Let ω ∈ Ω1(0,0)(F) and write M = K〈ω〉. Suppose we have M/N/L/K with N = L〈u〉 and

that K,L,N do not contains first integrals of (6). Then:

• If u is algebraic over L, then |L′/N ′| ≤ [N : L].

• If δiu ∈ L or δiuu ∈ L (i = x, y), then N ′ E L′ and L′/N ′ is abelian.

There is a final, simple lemma that will clearly be handy when it comes to proving the prima facie moredifficult direction of the upcoming theorem.

Lemma 2. Let ω ∈ Ω1(0,0)(F) and write M = K〈ω〉. Suppose that M/K is differentially Galois with respect

to Gal(F).

1. If for every φ ∈ Gal(F), there exists c(ε) ∈ C[[ε]] such that φ(ω) = ω + c(ε), then M/K is Liouvillian.

2. If for every φ ∈ Gal(F), there exist c(ε), a(ε) ∈ C[[ε]] such that φ(ω) = a(ε)ω + c(ε), then M/K isLiouvillian.

Proof. 1. We can easily see that for any φ ∈ Gal(F), we have φ(δiω) = δiω for i = x, y. In other words,δiω ∈ MGal(F). But since we are assuming that M/K is differentially Galois with respect to Gal(F),this means δiω ∈ K for i = x, y, which suffices to show that M/K is Liouvillian.

2. By elementary calculations, we note that for any φ ∈ Gal(F), we have φ(δ2iωδiω

) =δ2iωδiω

for i = x, y. In

other words,δ2iωδiω∈MGal(F). But since we are assuming that M/K is differentially Galois with respect

to Gal(F), this meansδ2iωδiω

= ai ∈ K for i = x, y.

24

Now note that, since q(x, y)δxω + p(x, y)δyω = 0, there exists γ ∈ M such that δxω = γp(x, y) andδyω = −γq(x, y). We thus have a tower M/K〈γ〉/K, and since δiω ∈ K〈γ〉 for i = x, y, M/K〈γ〉 isobviously Liouvillian.

To show that M/K is Liouvillian, then, it suffices to demonstrate that K〈γ〉/K is Liouvillian. Butthis is almost immediate, because we can calculate:

δxγ

γ=δx( δxω

p(x,y) )

δxωp(x,y)

=δ2xωp(x, y)− δxωδxp(x, y)

δxωp(x,y)

=δ2xω

δxω− δxp(x, y)

p(x, y)

= ax −δxp(x, y)

p(x, y)

∈ K

Similarly, we can see thatδyγγ = ay − δyq(x,y)

q(x,y) ∈ K and so K〈γ〉/K is certainly Liouvillian.

3.4.2 The Main Result

Without further ado, let’s state, again, the headline result of Lei’s paper.

Equation (6) is solvable by quadratures if and only if the differential Galois group of (6) over K is solvable.

Thanks to our work in the previous section, we essentially have all of the requisite tools at our disposalto give a proof of this result. We proceed, as in the linear case, by proving each direction separately. Theforward direction is fairly simple at this stage.

Theorem 10. Suppose that (6) is solvable by quadratures. Then Gal(F) is solvable.

Proof. If K contains a first integral of (6), then, trivially, we must have Gal(F) ∼= e, which is certainlysolvable.

Assume, then, that K does not not contain a first integral of (6), and that (6) is solvable by quadratures.Then there exists ω ∈ Ω1

(0,0)(F) such that M = K〈ω〉 embeds into a Liouvillian extension of K. Note that,

without loss of generality, we may actually assume that M/K is Liouvillian, since M/K is differentiallyGalois by Proposition 10 and quotients of solvable groups are solvable. This means that there is a sequenceof elementsα1, . . . , αn ∈ M such that M = K〈α1, . . . , αn〉, and for each 1 ≤ i ≤ n, either δj(αi) ∈ K〈α1, . . . , αi−1〉,δjαiαi∈ K〈α1, . . . , αi−1〉 or αi is algebraic over K〈α1, . . . , αi−1〉 for j = x, y. Now, for 0 ≤ i ≤ n, set

Gi = K〈α1, . . . , αi〉′, so that we have a sequence:

e = Gn ⊆ Gn−1 ⊆ · · · ⊆ G1 ⊆ G0 = Gal(F)

Now we can appeal to Proposition 11. If αi is algebraic over K〈α1, . . . , αi−1〉, then |Gi−1/Gi| ≤[K〈α1, . . . , αi〉 : K〈α1, . . . , αi−1〉]. But [K〈α1, . . . , αi〉 : K〈α1, . . . , αi−1〉] is finite, because αi is alge-braic over K〈α1, . . . , αi−1〉, and so |Gi−1/Gi| is also finite. Alternatively, if δj(αi) ∈ K〈α1, . . . , αi−1〉 orδjαiαi∈ K〈α1, . . . , αi−1〉 for j = x, y, then Proposition 11 tell us that Gi E Gi−1 and Gi−1/Gi is abelian.

Together, the above observations say precisely that Gal(F) is solvable.

25

As one would expect, the backwards direction is more difficult, but, as we will see, the fact that we haveTheorem 9 in our arsenal cuts down the workload significantly.

Theorem 11. Suppose Gal(F) is solvable. Then (6) is solvable by quadratures.

Proof. By Theorem 9, we must have ord(ξ) ≤ 2. We treat each case individually.

If ord(ξ) = 0, then K contains a first integral of (6). Trivially, this implies that (6) is solvable byquadratures, since K is vacuously a Liouvillian extension of itself.

Suppose ord(ξ) = 1; there exists ω ∈ Ω1(0,0)(F) and write M = K〈ω〉. By Theorem 9, Gal(F) = f(z, ε) ∈

G[[ε]] | f(z, ε) = λz + c(ε), c(0) = 0, λn = 1. Let E = φ ∈ Gal(F) | φ(ω) = ω + c(ε), c(0) = 0. ThenE is a subgroup of Gal(F), and clearly |Gal(F)/E| = n. Now we need a result from [4]:

Fact 8. Let J ⊂ H be subgroups of Gal(F), with J of index r in H. Then [MJ : MH ] ≤ r.

This immediately gives us that [ME : K] = [ME : MGal(F)] ≤ n, since Proposition 10 showed that M/K isdifferentially Galois with respect to Gal(F), which implies that ME/K is an algebraic (and hence Liouvillian)extension. Also, it is clear from the definition that M is differentially Galois over ME with respect to E,and so Lemma 2 yields that M/ME is Liouvillian. Given that we have a Liouvillian tower M/ME/K, wededuce that M/K is Liouvillian. M embeds trivially into itself, so (6) is solvable by quadratures.

Suppose ord(ξ) = 2; there exists ω ∈ Ω1(0,0)(F) and write M = K〈ω〉. By Theorem 9, Gal(F) = f(z, ε) ∈

G[[ε]] | f(z, ε) = a(ε)z + c(ε), c(0) = 0. As before, M/K is differentially Galois with respect to Gal(F)by Proposition 10. Now Lemma 2 yields that M/K is Liouvillian. M embeds trivially into itself, so (6) issolvable by quadratures.

Therefore, whichever case we find ourselves in, we must have that (6) is solvable by quadratures.

3.4.3 Some More Concrete Examples

To demystify Lei’s main result, I will demonstrate how it can be used in practice. Encoded in Lei’s proof ofTheorem 9 is a way to determine the solvability by quadrature of many differential equations in the form of(6). As it turns out, we will not need to bother with the stress of writing the differential Galois group downexplicitly; we only need to carry out some quite simple tests.

Before we get into any examples, we establish a couple of lemmas from [5]. First we need to set out somenotation:

Definition 35. The ith differential intermediary of (6), denoted Bi, is defined by

Bi = −q(x, y)δi+1y (p(x,y)q(x,y) ), for 0 ≤ i ≤ 2.

Lemma 3. Suppose that some non-zero u satisfies F(u) = B0u. Then there exists a first integral ω of (6)such that δyω = u.

Proof. Suppose that u 6= 0 satisfies F(u) = B0u. Using our definitions of the differential form F and the 0th

differential intermediary B0, this means that qδxu+ pδyu = −qδy(pq )u, which implies that:

δxu = −δy(p

q)u− p

qδyu = δy(−p

qu)

Now set v = −pqu, so that δxu = δyv, and define ω(x, y) = 12

∫ (x,y)

(0,0)vdx+ udy. We check that ω is indeed a

first integral of (6):

26

F(ω) = (qδx + pδy)(ω)

= qδxω + pδyω

=q

2(2v) +

p

2(2u)

= −pu+ pu

= 0

as required. We can also see clearly from the above working that δyω = u.

Lemma 4. Suppose that some u satisfies F(u) = B0u+B1. Then there exists a first integral ω of (6) such

thatδ2yω

δyω= u.

Proof. Suppose that u satisfies F(u) = B0u + B1. Using our definitions of the differential form F and thedifferential intermediaries B0 and B1, this means that qδxu + pδyu = −qδy(pq )u − qδ2y(pq ) = −qδy(pq )u +

qδy(B0

q ), which implies that:

δxu = −δy(p

q)u− p

qδyu+ δy(

B0

q) = δy(−p

qu+

B0

q)

Now set v = −pqu+ B0

q , so that δxu = δyv, and define β(x, y) = e12

∫ (x,y)

(0,0)vdx+udy

, which is clearly non-trivial.Working in a similar fashion to before, note that:

F(β) = (qδx + pδy)(β)

= qδxβ + pδyβ

= (q

2(2v) +

p

2(2u))β

= ((−pu+B0) + pu)β

= B0β

This means we can apply the previous result to obtain that there exists a first integral ω of (6) such that

δyω = β. Now we can deduce thatδ2yω

δyω=

δyββ = uβ

β = u, by the above working.

We’ll start with some easy examples of equations which are solvable by quadratures. Indeed, as a check,we will give these solutions by working in the usual ways seen in any undergraduate differential calculuscourse.

Example 9. Consider the equation dydx = y2 + 1, which is clearly non-linear. By easy calculations, we have

that B0 = −2y. We want to apply Lemma 3, which means that we need to find some non-zero u satisfying(y2 + 1)δyu+ δxu = −2yu. Evidently, this becomes significantly more straightforward if we allow ourselvesto assume that u is a function of y alone; one can then easily verify that the resulting expression does indeedsatisfy the initial partial differential equation. This reduces our problem to solving the linear differentialequation (y2 + 1)dudy = −2yu. We can proceed by separating the variables as usual, and integrating with

respect to y produces log |u| = − log |y2 + 1|, from which we deduce u = 1y2+1 . Now, as we’d hoped, Lemma

3 applies, so there is some first integral ω such that δyω = u. In particular, note that δyω = u ∈ K, whichmeans that we can appeal to Proposition 9, in order to say that φ(ω) = ω + c(ε), ∀φ ∈ Gal(F), wherec(ε) ∈ A0[[ε]] ∩ C[[ε]] and c(0) = 0. We saw in Lemma 2 that this is enough to ensure the solvability ofdydx = y2

1+xby quadratures.

27

Remark 8. Solvability by quadratures means that there is a first integral of dydx = y2 + 1 expressible by

quadratures, which implies, by our discussion in the preliminary part of this chapter, that this equation hasa quadrature solution. Of course, we can see this last observation directly, because dy

dx = y2 + 1 is itselfseparable.

Note first that∫

1y2+1dy = arctan(y) +C, by making the substitution y = tan(θ). Now, returning to the

equation at hand, we separate the variables, integrate with respect to y, and having lumped our constantswe ought to obtain arctan(y) = x+D. Therefore y = tan(x+D) is our quadrature solution whose existencewe secured as a consequence of the previous example.

Example 10. Consider the equation dydx = y2

1+x , which is clearly non-linear. By easy calculations, we havethat B0 = −2y. Again, we want to apply Lemma 3, which means that we need to find some non-zerou satisfying y2δyu + (1 + x)δxu = −2yu. Evidently, this becomes significantly more straightforward ifwe allow ourselves, as we did previously, to assume that u is a function of y alone; one can then easilyverify that the resulting expression does indeed satisfy the initial partial differential equation. This reducesour problem to solving the linear differential equation y2 dudy = −2yu. We can proceed by separating the

variables as usual, and integrating with respect to y produces log |u| = −2 log |y|, from which we deduceu = 1

y2 . Now, as we’d hoped, Lemma 3 applies, so there is some first integral ω such that δyω = u. Inparticular, note that δyω = u ∈ K, which means that we can appeal to Proposition 9, in order to say thatφ(ω) = ω + c(ε), ∀φ ∈ Gal(F), where c(ε) ∈ A0[[ε]] ∩ C[[ε]] and c(0) = 0. We saw in Lemma 2 that this is

enough to ensure the solvability of dydx = y2

1+x by quadratures.

Remark 9. Solvability by quadratures means that there is a first integral of dydx = y2

1+x expressible byquadratures, which implies, by our discussion in the preliminary part of this chapter, that this equation

has a quadrature solution. Once more, we can see this last observation directly, because dydx = y2

1+x is itselfseparable.

We separate the variables, integrate with respect to y, and having lumped our constants we ought toobtain −1y = log |1 + x| + C. Therefore y = 1

D−log |1+x| , where D = −C, is our quadrature solution whose

existence we secured as a consequence of the previous example.

So far, both of our examples have been separable differential equations. Just to make sure I’m keepingnothing up my sleeves, here is a fairly basic case of a non-separable differential equation unto which we mayapply Lei’s theory.

Example 11. Consider the equation dydx = x + y. This equation is linear, but still falls within the scope

of our main result and so will prove to be an instructive example. By easy calculations, we have thatB0 = −1. Again, we want to apply Lemma 3, which means that we need to find some non-zero u satisfying(x + y)δyu + δxu = −u. This time, we cannot simply assume the u is a function of y alone, due to to theoccurence of x in the coefficient of δyω. Equally, with hindsight, we cannot assume that u is a function ofx alone, since this produces a function which lies outside K, which would force us to abandon Proposition9. Therefore it remains to solve this PDE. We might use the method of characteristics, although this seemsagainst the spirit of this discussion, since it would involve solving the equation dy

dx = x + y. It is, in fact,fairly straightforward to see by inspection that u = 1

x+y+1 is a solution of this PDE. Now, as we’d hoped,Lemma 3 applies, so there is some first integral ω such that δyω = u. In particular, note that δyω = u ∈ K,which means that we can appeal to Proposition 9, in order to say that φ(ω) = ω+ c(ε), ∀φ ∈ Gal(F), wherec(ε) ∈ A0[[ε]] ∩ C[[ε]] and c(0) = 0. We saw in Lemma 2 that this is enough to ensure the solvability ofdydx = x+ y by quadratures.

Remark 10. Solvability by quadratures means that there is a first integral of dydx = x + y expressible by

quadratures, which implies, by our discussion in the preliminary part of this chapter, that this equation hasa quadrature solution. Once more, we can see this last observation directly, because dy

dx = x + y is a lineardifferential equation, admitting an integrating factor.

Multiplying through by e−x and subtracting y from both sides, we can see (by the product rule) thatd(ye−x)dx = xe−x. Now integrating with respect to x, the right-hand side by parts, we obtain ye−x =

28

C − (x + 1)e−x. Therefore y = Cex − (x + 1) is our quadrature solution whose existence we secured as aconsequence of the previous example.

Finally, we outline an example using an equation which is neither separable, nor linear. Here, we canbegin to appreciate the potential importance of Lei’s theory, as the equations become harder to solve by eye.

Example 12. Consider the equation dydx = y2− y

x = xy2−yx , which is clearly non-linear. By easy calculations,

we have that B0 = 1 − 2xy. Again, we want to apply Lemma 3, which means that we need to find somenon-zero u satisfying (xy2− y)δyu+xδxu = (1− 2xy)u. Again, we cannot simply assume the u is a functionof y alone, due to to the occurence of x in the coefficient of δyω. Likewise, we cannot assume that u is afunction of x alone, due to the occurence of y on the right-hand side. Therefore it remains to solve thisPDE; one can check that u = 1

xy2 is a solution of this PDE. Now, as we’d hoped, Lemma 3 applies, so thereis some first integral ω such that δyω = u. In particular, note that δyω = u ∈ K, which means that we canappeal to Proposition 9, in order to say that φ(ω) = ω+ c(ε), ∀φ ∈ Gal(F), where c(ε) ∈ A0[[ε]]∩C[[ε]] andc(0) = 0. We saw in Lemma 2 that this is enough to ensure the solvability of dy

dx = y2 − yx by quadratures.

Remark 11. Solvability by quadratures means that there is a first integral of dydx = y2 − y

x expressible byquadratures, which implies, by our discussion in the preliminary part of this chapter, that this equationhas a quadrature solution. Once more, we can see this last observation directly, because dy

dx = y2 − yx is a

Bernoulli differential equation.

Begin by making the substitution v = y−1 and multiplying through by v2, we obtain that dvdx −

vx = −1.

We now have a linear differential equation which admits the integrating factor 1x , and so we deduce that

d( vx )

dx = −1x . Now integrating with respect to x,we arrive at v

x = C − log |x|. Therefore y = 1x(C−log |x|) is our

quadrature solution whose existence we secured as a consequence of the previous example.

One ought to have seen a general method to demonstrate the solvability of an equation of form (6)emerging. Let’s formalise this in the next proposition.

Proposition 12. Suppose that some non-trivial u ∈ K satisfies either F(u) = B0u or F(u) = B0u + B1.Then (6) is solvable by quadratures.

Proof. Suppose we can find some non-trivial u ∈ K satisfying F(u) = B0u, then we can apply Lemma 3 tosay that there exists a first integral ω of (6) such that δyω = u ∈ K. By Proposition 9, φ(ω) = ω+c(ε), ∀φ ∈Gal(F), where c(ε) ∈ A0[[ε]] ∩ C[[ε]] and c(0) = 0. But, by Lemma 2, this suffices for us to conclude thatK〈ω〉/K is Liouvillian. That is, (6) is solvable by quadratures.

Similarly, suppose we can find some non-trivial u ∈ K satisfying F(u) = B0u + B1, then we can apply

Lemma 4 to say that there exists a first integral ω of (6) such thatδ2yω

δyω= u ∈ K. By Proposition 9,

φ(ω) = a(ε)ω + c(ε), ∀φ ∈ Gal(F), where a(ε), c(ε) ∈ A0[[ε]] ∩ C[[ε]] and c(0) = 0. But, by Lemma 2, thissuffices for us to conclude that K〈ω〉/K is Liouvillian. That is, (6) is solvable by quadratures.

Of course, not all equations of form (6) are solvable by quadratures. One might be wondering how wetest for such an eventuality, given the concrete method we have just described for proving that an equationis solvable by quadratures. We will illuminate the situation by giving an example, courtesy of [5]. First weneed another fact from Lei’s work, which plays a role in the proof of Theorem 9.

Lemma 5. Consider equation (6). Then:

1. If ord(ξ) = 0, then K contains a first integral of (6).

2. If ord(ξ) = 1, then there exists u ∈ K \ 0, and n ∈ N such that F(u) = nB0u.

3. If ord(ξ) = 2, then there exists u ∈ K such that F(u) = B0u+B1.

4. If ord(ξ) = 3, then there exists u ∈ K such that F(u) = 2B0u+B2.

29

Example 13. Consider the van der Pol equation given by dydx = −x

y−µ( 13x

3−x) for some µ ∈ C\0. Certainly K

does not contain a first integral of this equation, and one can check that F(u) = nB0u and F(u) = B0u+B1

have no solutions in K. Consequently, ord(ξ) > 2 by Lemma 5. However, by Theorem 9, the differentialGalois group of the van der Pol equation must be non-solvable. This implies, with our main result, that thevan der Pol equation is not solvable by quadratures.

In fact, we can do better than this. Lei shows that F(u) = 2B0u+ B2 also has no solutions in K. Thismeans, due to Lemma 5, that ord(ξ) =∞, and so the differential Galois group of the van der Pol equationis actually G[[ε]], by Theorem 9.

30

4 Conclusion

4.1 Summary

We began by introducing the Galois theory of ordinary homogeneous linear differential equations. Havingprovided the elementary definition of a differential field, we were able to recover many of the key devicesfrom standard ring theory in a differential context, such as ideals and homomorphisms. We saw that linearODEs are particularly convenient to work with because, of course, their solution space can be viewed as avector space over the field of constants of our ambient differential field. The crucial step at this point wasto secure the existence and uniqueness of the Picard-Vessiot extension for such an equation, which plays ananalogous role to the splitting field of a polynomial in algebraic Galois theory. We noted that there was aduality of perspective surrounding this notion, arising from different presentations in the literature - mostnotably the scalar approach given in [2] and the matrix approach outlined in [9]. Despite this, we emphasisedthat there is no ambiguity here: a differential field extension is Picard-Vessiot in the former sense if andonly if it is Picard-Vessiot in the latter sense also. This meant that we could properly define the differentialGalois group of an ordinary homogeneous linear differential equation.

Unlike the Galois group of a polynomial, which, since it permutes the roots of the polynomial, can beidentified with a subgroup of some symmetric group, the differential Galois group acts on the vector spaceof solutions to the linear ODE, and consequently can be identified with a subgroup of some general lineargroup. In order to visualise clearly what such a Galois group might look like, however, we had to delveinto some algebraic geometry. It turns out that the differential Galois group is in fact an algebraic group,which meant we could use tools from commutative algebra (such as Noether’s normalisation lemma) to givea concrete notion of size, in the form of Krull dimension, to these groups. We then offered some explicitexamples of differential Galois groups for specific equations, before recovering the fundamental theorem ofGalois theory in a differential context. We concluded our discussion of the linear theory by recapping theconcept of solvability by radicals for a polynomial, before detailing the corresponding idea of solvability byquadratures for a differential equation, noting that both notions could be defined in a formal manner using(differential) field theory. Of course, Galois proved that a polynomial is solvable by radicals precisely whenthe Galois group of the polynomial is solvable; we proved the analogous result for ordinary homogeneouslinear differential equations regarding their solvability by quadratures. Ultimately, the Galois theory for theseequations is, to all intents and purposes, complete. Having followed a broadly similar path to an expositionof algebraic Galois theory, we exhibited a table to document the parallels between the two theories.

We then turned to non-linear ordinary differential equations. In particular, we followed the work of Lei,who focussed on the general polynomial first order non-linear differential equation. Having introduced adifferential form, we saw that the main objects of the discussion were in fact first integrals: solutions tothis form which are analytic at some non-critical point. Such first integrals are abundant, but thankfullywe could restrict ourselves to considering those which exhibited special properties - those which vanished atthe origin, but have non-vanishing derivative there. We then defined the general group of analytic series,demonstrating that it was indeed a bona fide group, and noted that we could let this group act on ourchosen class of first integrals. For our purposes, however, the general group of analytic series is too broad analgebraic structure. We picked out a subgroup, the admissible differential isomorphisms, and, having shownthat it was independent of our choice of first integral, this became our differential Galois group of the generalpolynomial first order non-linear differential equation.

To analyse the structure of this Galois group, we had to develop several new concepts. Primary amongthem was the notion of a (proper) quasi-differential polynomial, which allowed us to translate our working,which until that point had been relatively analytic, into the more algebraic language which is usually asso-ciated with Galois theory. A key development was the definition of regular prime ideals: ideals in the ringof generalised differential polynomials consisting purely of PQDP. One regular prime ideal is of particularimportance. I coined it the ‘fundamental ideal’; the elements of the general group of analytic series satisfyingthis ideal are precisely those in the differential Galois group. Ultimately, this was the reason for the primafacie convoluted definition of an admissible differential isomorphism. The fundamental ideal, or more specif-ically its order, facilitates a complete classification of the differential Galois groups, which is a key result of

31

Lei’s paper. As a corollary, we deduced that the differential Galois group is also independent of our choice ofregular point, so our earlier choice of the origin was ultimately irrelevant in our development of the theory.Once again, we concluded our discussion by revisiting the notion of solvability and most notably identifiedthe solvability of the general polynomial first order non-linear differential equation with the solvability of itsassociated Galois group, in analogue with the corresponding results in the algebraic and linear differentialtheories. Having described a method for demonstrating the solvability of such an equation, we offered someconcrete examples.

4.2 Concluding Remarks

Where can we go from here? One might wonder whether we could utilise the final proposition to providesufficient conditions on p(x, y) and q(x, y) for (6) to be solvable by quadratures. Unfortunately, this hopeis a naıve one. As we saw in the final set of examples, unless we can solve either of the equations in thehypotheses of Proposition 12 by inspection, we must resort to using the method of characteristics; it isimpossible to avoid this technique in the general case. But to ask for the solution to these general PDEsusing characteristics is to beg the question: it relies on us being able to solve (6), which would immediatelysettle the issue of solvability by quadratures. We cannot, it seems, use Lei’s results to immediately judgewhether any given general polynomial first order non-linear differential equation is solvable by quadratures- more work must be done to provide a non-circular criterion of solvability.

Two other avenues of research, however, might be of interest. Firstly, one could consider broadening thescope of Lei’s more theoretical results on differential Galois groups by considering equations of the sameform of (6), but that also allow other functions, such as sines, exponentials and logarithms. Doing so wouldseemingly require the extension of the concept of quasi-differential polynomials to quasi-differential formalseries, which would bring sweeping changes to the foundations of Lei’s theory. In particular, the notion oforder for a regular prime ideal, which plays such an important role in Theorem 9, would have to be revised.Another possibility, which doesn’t drift as far from Lei’s original work, is to consider general polynomialnon-linear differential equations of higher orders. This is due to the fact that we may write such an equationas a system of first order polynomial non-linear differential equations, unto which we can apply Lei’s originalwork. Of course, even if we could extend Lei’s theoretical result to this wider class of equations, we wouldstill struggle to provide sufficient conditions for the solvability by quadratures of a given equation, due tothe circularity worries in the first order case mentioned above. As such, it is not immediately clear whetherdifferential Galois theory, at least in the style of Lei, can offer any further insight into non-linear differentialequations in the near future.

32

References

[1] AD Bruno. Local method in nonlinear differential equations. part i the local method of nonlinear analysesof differential equations, part ii the sets of analyticity of a normalizing transformation, 1988.

[2] Teresa Crespo, Zbigniew Hajto, and Juan Jose Morales Ruiz. Introduction to differential Galois theory.Wydawnictwo PK, 2007.

[3] W Kaplan. Ordinary differential equations (reading, 1958.

[4] Irving Kaplansky. An introduction to differential algebra, hermann, paris, 1957. MR, 57:297, 2014.

[5] Jinzhi Lei. Nonlinear differential galois theory. arXiv preprint math/0608492, 2006.

[6] Joseph Fels Ritt. Differential algebra, volume 33. American Mathematical Soc., 1950.

[7] Bruce Simon. An introduction to differential galois theory, 2015.

[8] Michael F Singer. Liouvillian first integrals of differential equations. Transactions of the AmericanMathematical Society, 333(2):673–688, 1992.

[9] Marius van der Put. Galois theory of differential equations, algebraic groups and lie algebras. Journalof symbolic computation, 28(4-5):441–473, 1999.

33