Introduction to Differential Galois Theory

Teresa Crespo and Zbigniew Hajto

IMUJ PREPRINT 2007/11

Contents

1 Introduction 5

2 Differential rings 7 2.1 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Differential rings . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.3 Differential extensions . . . . . . . . . . . . . . . . . . . . . . 10 2.4 The ring of differential operators . . . . . . . . . . . . . . . . 11

3 Picard-Vessiot extensions 13 3.1 Homogeneous linear differential equations . . . . . . . . . . . . 13 3.2 Existence and uniqueness of the Picard-Vessiot extension … 14

4 Differential Galois group 21 4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2 The differential Galois group as a linear algebraic group …. 23

5 Fundamental theorem 31

6 Liouville extensions 41 6.1 Liouville extensions . . . . . . . . . . . . . . . . . . . . . . . . 41 6.2 Generalized Liouville extensions . . . . . . . . . . . . . . . . . 42

7 Appendix on algebraic varieties 45 7.1 Affine varieties . . . . . . . . . . . . . . . . . . . . . . . . . . 45 7.2 Abstract affine varieties . . . . . . . . . . . . . . . . . . . . . 50 7.3 Auxiliary results . . . . . . . . . . . . . . . . . . . . . . . . . 52

8 Appendix on algebraic groups 55 8.1 The notion of algebraic group . . . . . . . . . . . . . . . . . . 55 8.2 Connected algebraic groups . . . . . . . . . . . . . . . . . . . 56 8.3 Subgroups and morphisms . . . . . . . . . . . . . . . . . . . . 58 8.4 Linearization of affine algebraic groups . . . . . . . . . . . . . 60 8.5 Homogeneous spaces . . . . . . . . . . . . . . . . . . . . . . . 62 8.6 Decomposition of algebraic groups . . . . . . . . . . . . . . . . 63 8.7 Solvable algebraic groups . . . . . . . . . . . . . . . . . . . . . 66 8.8 Characters and semi-invariants . . . . . . . . . . . . . . . . . . 70 8.9 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9 Suggestions for further reading 75 10 Bibliography 77

1 Introduction

Some classical methods used to solve certain differential equations can be unified by associating to the equation a group of transformations leaving it invariant. This idea, due to Sophus Lie, is at the origin of differential Galois theory. The group associated to the differential equation gives then information on the properties of the solutions. However, most differential equations do not admit a nontrivial group of transformations. In the case of ordinary homogeneous linear differential equations, there exists a satisfactory Galois theory introduced by Émile Picard and Ernest Vessiot. The group associated to the differential equation is in this case a linear algebraic group and a characterization of equations solvable by quadratures is given in terms of the Galois group. In the middle of the 20th century, Picard-Vessiot theory was clarified by Ellis Kolchin, who also built the foundations of the theory of linear algebraic groups. Kolchin used the differential algebra developed by Joseph F. Ritt and established the Fundamental Theorem of Picard-Vessiot theory, which is the counterpart of its homonymous theorem in polynomial Galois theory.

Our lecture notes develop Picard-Vessiot theory from an elementary point of view, based on the modern theory of algebraic groups. They are mainly aimed at graduate students with a basic knowledge of abstract algebra and differential equations. The necessary topics of algebraic geometry and linear algebraic groups are included in the appendices.

In chapter 2, we introduce differential rings and differential extensions and consider differential equations defined over an arbitrary differential field. In chapter 3, we prove that we can associate to an ordinary linear differential equation defined over a differential field K, of characteristic 0 with algebraically closed field of constants, a uniquely determined minimal extension L of K containing the solutions of the equation, the Picard-Vessiot extension. In chapter 4, we introduce the differential Galois group of an ordinary linear differential equation defined over the field K as the group of differential K-automorphisms of its Picard-Vessiot extension L and prove that it is a linear algebraic group. In chapter 5, we prove the fundamental theorem of Picard-Vessiot theory, which gives a bijective correspondence between intermediate fields of a Picard-Vessiot extension and Zariski closed subgroups of its Galois group. In chapter 6, we give the characterization of homogeneous linear differential equations solvable by quadratures in terms of their differential Galois group.

These lecture notes are based on the courses on Differential Galois Theory given by the authors at the University of Barcelona and the Cracow University of Technology. Some parts of them were presented at the Differential Galois Theory Seminar at the Mathematical Institute of the Cracow University of Technology during the academic year 2006-2007. The authors would like to thank the members of the DGT Seminar, especially Dr. Marcin Skrzyński and Dr. Artur Piȩkosz for his useful remarks on the previous version of these notes.

During the work on this monograph both authors were supported by the Polish Grant N20103831/3261 and the Spanish Grant MTM2006-04895. During her stay at the Cracow University of Technology, Teresa Crespo was supported by the Spanish fellowship PR2006-0528.

2 Differential rings

2.1 Derivations

Definition 2.1 A derivation of a ring A is a map d : A → A such that

d(a + b) = da + db , d(ab) = d(a) b + a d(b).

We write as usual a′ = d(a) and a′′, a′′′, …, a^{(n)} for successive derivations. By induction, one can prove Leibniz’s rule

(ab)^{(n)} = a^{(n)} b + ⋯ + \binom{n}{i}a^{(n−i)}b^{(i)} + ⋯ + a b^{(n)}.

If a′ commutes with a, we have (a^{n})′ = na^{n−1}a′. If A has an identity element 1, then necessarily d(1) = 0, since d(1) = d(1.1) = d(1).1 + 1.d(1) ⇒ d(1) = 0. If a ∈ A is invertible with inverse a^{−1}, we have a.a^{−1} = 1 ⇒ a′a^{−1} + a(a^{−1})′ = 0 ⇒ (a^{−1})′ = −a^{−1}a′a^{−1}. Hence, if a′ commutes with a, we get (a^{−1})′ = −a′/a^{2}.

Proposition 2.1 If A is an integral domain, a derivation d of A extends to the quotient field Qt(A) in a unique way.

Proof. For \frac{a}{b} ∈ Qt(A), we must have (\frac{a}{b})′ = \frac{a′b − ab′}{b^{2}} , so there is uniqueness.

We extend the derivation to Qt(A) by defining (\frac{a}{b})′ := \frac{a′b − ab′}{b^{2}} . If c ∈ A \ {0}, we have

(\frac{ac}{bc})′ = \frac{(ac)′bc − ac(bc)′}{b^{2}c^{2}} = \frac{(a′c + ac′)bc − ac(b′c + bc′)}{b^{2}c^{2}} = \frac{a′b − ab′}{b^{2}} ,

so the definition is independent of the choice of the representative. Now we have

\begin{matrix}(\frac{a}{b} + \frac{c}{d})′ = (\frac{ad + bc}{bd})′ = \frac{(ad + bc)′bd − (ad + bc)(bd)′}{b^{2}d^{2}} = \\ \frac{(a′d + ad′ + b′c + bc′)bd − (ad + bc)(b′d + bd′)}{b^{2}d^{2}} = \frac{a′b − ab′}{b^{2}} + \frac{c′d − cd′}{d^{2}} , \end{matrix}

\begin{matrix}(\frac{a}{b} · \frac{c}{d})′ = (\frac{ac}{bd})′ = \frac{(ac)′bd − ac(bd)′}{b^{2}d^{2}} = \frac{(a′c + ac′)bd − ac(b′d + bd′)}{b^{2}d^{2}} = \\ \frac{(a′b − ab′)c}{b^{2}d} + \frac{(c′d − cd′)a}{d^{2}b} = \frac{a′b − ab′}{b^{2}} · \frac{c}{d} + \frac{a}{b} · \frac{c′d − cd′}{d^{2}} . \end{matrix}

□

Remark 2.1 If A is a commutative ring with a derivation and S a multiplicative system of A, following the same steps as in the proof of proposition 2.1, we can prove that the derivation of A extends to the ring S^{−1}A in a unique way.

2.2 Differential rings

Definition 2.2 A differential ring is a commutative ring with identity endowed with a derivation. A differential field is a differential ring which is a field.

Examples.

1. Every commutative ring A with identity can be made into a differential ring with the trivial derivation defined by d(a) = 0, ∀a ∈ A.

Over ℤ and over ℚ, the trivial derivation is the only possible one, since d(1) = 0, and by induction, d(n) = d((n − 1) + 1) = 0 and so d(n/m) = 0.

2. The ring of all infinitely differentiable functions on the real line with the usual derivative is a differential ring.

3. The ring of analytic functions in the complex plane with the usual derivative is a differential ring. In this case, it is an integral domain and so the derivation extends to its quotient field which is the field of meromorphic functions.

4. Let A be a differential ring, let A[X] be the polynomial ring in one indeterminate over A. A derivation in A[X] extending that of A should satisfy (∑ a_{i}X^{i})′ = ∑(a′_{i}X^{i} + a_{i}iX^{i−1}X′). We can then extend the derivation of A to A[X] by assigning to X′ an arbitrary value in A[X]. Analogously, if A is a field, we can extend the derivation of A to the field A(X) of rational functions. By iteration, we can give a differential structure to A[X_{1}, …, X_{n}] for a differential ring A and to A(X_{1}, …, X_{n}) for a differential field A.

5. Let A be a differential ring. We consider the ring A[X_{i}] of polynomials in the indeterminates X_{i}, i ∈ ℕ ∪ {0}. By defining X′_{i} = X_{i+1}, a unique derivation of A[X_{i}] is determined. We change notation and write X = X_{0}, X^{(n)} = X_{n}. We call this procedure the adjunction of a differential indeterminate and we use the notation A{X} for the resulting differential ring. The elements of A{X} are called differential polynomials in X (they are ordinary polynomials in X and its derivatives).

If A is a differential field, then A{X} is a differential integral domain and its derivation extends uniquely to the quotient field. We denote this quotient field by A⟨X⟩, its elements are differential rational functions of X.

6. If A is a differential ring, we can define a derivation in the ring M_{n×n}(A) of square n × n matrices by defining the derivative of a matrix as the matrix obtained by applying the derivation of A to all its entries. Then for n ≥ 2, M_{n×n}(A) is a noncommutative ring with derivation.

In any differential ring A, the elements with derivative 0 form a subring C, called the ring of constants. If A is a field, so is C. The field of constants contains the image of the ring morphism ℤ → A, 1 ↦ 1. In the sequel, C_{K} will denote the constant field of a differential field K.

Definition 2.3 Let I be an ideal of a differential ring A. We say that I is a differential ideal if a ∈ I ⇒ a′ ∈ I, that is if d(I) ⊂ I.

If I is a differential ideal of the differential ring A, we can define a derivation in the quotient ring A/I by d(\overline{a}) = \overline{d(a)}. It is easy to check that this definition does not depend on the choice of the representative in the coset and indeed defines a derivation in A/I.

Definition 2.4 If A and B are differential rings, a map f : A → B is a differential morphism if it satisfies

1. f(a + b) = f(a) + f(b), f(ab) = f(a)f(b), ∀a, b ∈ A; f(1) = 1.

2. f(a)′ = f(a′), ∀a ∈ A.

If I is a differential ideal, the natural morphism A → A/I is a differential morphism. The meaning of differential isomorphism, differential automorphism is clear.

Proposition 2.2 If f : A → B is a differential morphism, then Ker f is a differential ideal and the isomorphism \overline{f} : A/ Ker f → Im f is a differential isomorphism.

Proof. For a ∈ Ker f, we have f(a′) = f(a)′ = 0, so a′ ∈ Ker f, hence Ker f is a differential ideal.

For any a ∈ A, we have (\overline{f}(\overline{a}))′ = (f(a))′ = f(a′) = \overline{f}(\overline{a′}) = \overline{f}(\overline{a}′), so \overline{f} is a differential isomorphism. □

2.3 Differential extensions

An inclusion A ⊂ B of differential rings is an extension of differential rings if the derivation of B restricts to the derivation of A. If S is a subset of B, we denote by A{S} the differential A-subalgebra of B generated by S over A, that is the smallest subring of B containing A, the elements of S and their derivatives. If K ⊂ L is an extension of differential fields, S a subset of L, we denote by K⟨S⟩ the differential subfield of L generated by S over K. If S is a finite set, we say that the extension K ⊂ K⟨S⟩ is differentially finitely generated.

Proposition 2.3 If K is a differential field, K ⊂ L a separable algebraic field extension, the derivation of K extends uniquely to L. Moreover, every K-automorphism of L is a differential one.

Proof. If K ⊂ L is a finite extension, we have L = K(α), for some α, by the primitive element theorem. If P (X) is the irreducible polynomial of α over K, by applying the derivation to P (α) = 0, we obtain P^{(d)}(α) + P′(α)α′ = 0, where P^{(d)} denotes the polynomial obtained from P by deriving its coefficients and P′ the derived polynomial. So, α′ = −P^{(d)}(α)/P′(α) and the derivation extends uniquely.

Let us look now at the existence. We have L ≃ K[X]/(P). We can extend the derivation of K to K[X] by defining X′ := −P^{(d)}(X)h(X) for h(X) ∈ K[X] such that h(X)P′(X) ≡ 1 (mod P). If h(X)P′(X) = 1+k(X)P (X), we have d(P (X)) = P^{(d)}(X)+P′(X)d(X) = P^{(d)}(X)+P′(X)(−P^{(d)}(X)h(X)) = P^{(d)}(X)(1 − P′(X)h(X)) = −P^{(d)}(X)k(X)P (X). Therefore (P) is a differential ideal and the quotient field K[X]/(P) is a differential ring.

The general case K ⊂ L algebraic is obtained from the finite case by applying Zorn lemma.

Now, if σ is a K-automorphism of L, σ^{−1}dσ is also a derivation of L extending that of K and by uniqueness, we obtain σ^{−1}dσ = d, and so dσ = σd, which gives that σ is a differential automorphism. □

Remark 2.2 Let K be a differential field with positive characteristic p (for example 𝔽_{p}(T) with derivation given by T′ = 1), let P (X) = X^{p} −a ∈ K[X], with a ∉ K^{p}, and let α be a root of P. If the element a ∈ K is not a constant, then it is not possible to extend the derivation of K to L := K(α). If the element a is a constant, we can extend the derivation of K to L by assigning to α′ any value in L.

Definition 2.5 If K ⊂ L is a differential field extension, α an element in L, we say that α is

- a primitive element over K if α′ ∈ K;

- an exponential element over K if α′/α ∈ K.

2.4 The ring of differential operators

Let K be a differential field with a nontrivial derivation d. A linear differential operator L with coefficients in K is a polynomial in d,

L = a_{n}d^{n} + a_{n−1}d^{n−1} + ⋯ + a_{1}d + a_{0}, with a_{i} ∈ K.

If a_{n} ≠ 0, we say that L has degree n. If a_{n} = 1, we say that L is monic. The ring of linear differential operators with coefficients in K is the noncommutative ring K[d] of polynomials in the variable d with coefficients in K where d satisfies the rule da = a′ + ad for a ∈ K. We have deg(L_{1}L_{2}) = deg(L_{1}) + deg(L_{2}) and then the only left or right invertible elements of K[d] are the elements of K \ {0}. A differential operator acts on K and on differential extensions of K with the interpretation d(y) = y′. To the differential operator L = a_{n}d^{n} + a_{n−1}d^{n−1} + ⋯ + a_{1}d + a_{0}, we associate the linear differential equation

L(Y) = a_{n}Y^{(n)} + a_{n−1}Y^{(n−1)} + ⋯ + a_{1}Y′ + a_{0}Y = 0.

As for the polynomial ring in one variable over the field K, we have a division algorithm on both left and right.

Lemma 2.1 For L_{1}, L_{2} ∈ K[d] with L_{2} ≠ 0, there exist unique differential operators Q_{l}, R_{l} (resp. Q_{r}, R_{r} ) in K[d] such that

L_{1} = Q_{l}L_{2} + R_{l} and deg R_{l} < deg L_{2}
(resp. L_{1} = L_{2}Q_{r} + R_{r} and deg R_{r} < deg L_{2}.)

The proof of this fact follows the same steps as in the polynomial case.

Corollary 2.1 For each left (resp. right) ideal I of K[d], there exists an element L ∈ K[d], unique up to a factor in K \ {0}, such that I = K[d]L (resp. I = LK[d]).

Taking into account this corollary, for two linear differential operators L_{1}, L_{2}, the left greatest common divisor will be the unique monic generator of K[d]L_{1} + K[d]L_{2} and the left least common multiple will be the unique monic generator of K[d]L_{1} ∩ K[d]L_{2}. Analogously, we can define right GCD and LCM. We can compute left and right GCD with a modified version of Euclides algorithm.

3 Picard-Vessiot extensions

3.1 Homogeneous linear differential equations

From now on, K will denote a field of characteristic zero.

We consider homogeneous linear differential equations over a differential field K, with field of constants C:

L(Y) := Y^{(n)} + a_{n−1}Y^{(n−1)} + ⋯ + a_{1}Y′ + a_{0}Y = 0, a_{i} ∈ K.

If K ⊂ L is a differential extension, the set of solutions of L(Y) = 0 in L is a C_{L}-vector space, where C_{L} denotes the constant field of L. We want to see that its dimension is at most equal to the order n of L.

Definition 3.1 Let y_{1}, y_{2}, …, y_{n} be elements in a differential field K. The determinant

W = W (y_{1}, y_{2}, …, y_{n}) := | \begin{matrix}y_{1} & y_{2} & ⋯ & y_{n} \\ y′_{1} & y′_{2} & ⋯ & y′_{n} \\ ⋮ & ⋮ & ... & ⋮ \\ y_{1}^{(n−1)} & y_{2}^{(n−1)} & ⋯ & y_{n}^{(n−1)} \end{matrix} |

is the wrońskian (determinant) of y_{1}, y_{2}, …, y_{n}.

Proposition 3.1 Let K be a differential field with field of constants C, and let y_{1}, …, y_{n} ∈ K. Then y_{1}, …, y_{n} are linearly independent over C if and only if W (y_{1}, …, y_{n}) ≠ 0.

Proof. Let us assume that y_{1}, …, y_{n} are linearly dependent over C, let ∑_{i=1}^{n} c_{i}y_{i} = 0, c_{i} ∈ C not all zero. By differentiating n−1 times this equality, we obtain ∑_{i=1}^{n} c_{i}y_{i}^{(k)} = 0, k = 0, …, n − 1. So the columns of the wrońskian are linearly dependent, hence W (y_{1}, …, y_{n}) = 0.

Reciprocally, let us assume W (y_{1}, …, y_{n}) = 0. We then have n equalities ∑_{i=1}^{n} c_{i}y_{i}^{(k)} = 0, k = 0, …, n − 1, with c_{i} ∈ K not all zero. We can assume c_{1} = 1 and W (y_{2}, …, y_{n}) ≠ 0. By differentiating equality k, we obtain ∑_{i=1}^{n} c_{i}y_{i}^{(k+1)} + ∑_{i=2}^{n} c′_{i}y_{i}^{(k)} = 0 and subtracting equality (k + 1), we get ∑_{i=2}^{n} c′_{i}y_{i}^{(k)} = 0, k = 0, …, n − 2. We then obtain a system of homogeneous linear equations in c′_{2}, …, c′_{n} with determinant W (y_{2}, …, y_{n}) ≠ 0, so c′_{2} = ⋯ = c′_{n} = 0, that is, the c_{i} are constants. □

Taking this proposition into account, we can say ”linearly (in)dependent” over constants without ambiguity, since the condition of (non)cancellation of the wrońskian is independent of the field.

Proposition 3.2 Let L(Y) = 0 be a homogeneous linear differential equation of order n over a differential field K. If y_{1}, …, y_{n+1} are solutions of L(Y) = 0 in a differential extension L of K, then W (y_{1}, …, y_{n+1}) = 0.

Proof. The last row in the wrońskian is (y_{1}^{(n)}, …, y_{n+1}^{(n)}), which is a linear combination of the preceding ones. □

Corollary 3.1 L(Y) = 0 has at most n solutions in L linearly independent over the field of constants. □

If L(Y) = 0 is a homogeneous linear differential equation of order n over a differential field K, y_{1}, …, y_{n} are n solutions of L(Y) = 0 in a differential extension L of K, linearly independent over the field of constants, we say that {y_{1}, …, y_{n}} is a fundamental set of solutions of L(Y) = 0 in L. Any other solution of L(Y) = 0 in L is a linear combination of y_{1}, …, y_{n} with constant coefficients. The next proposition can be proved straightforwardly.

Proposition 3.3 Let L(Y) = 0 be a homogeneous linear differential equation of order n over a differential field K and let {y_{1}, …, y_{n}} be a basis of the solution space of L(Y) = 0 in a differential extension L of K. Let z_{j} = ∑_{i=1}^{n} c_{ij}y_{i}, j = 1, …, n, with c_{ij} constants, then

W (z_{1}, …, z_{n}) = det(c_{ij}) · W (y_{1}, …, y_{n}).

3.2 Existence and uniqueness of the Picard-Vessiot extension

We define now the Picard-Vessiot extension of a homogeneous linear differential equation which is the analogue of the splitting field of a polynomial.

Definition 3.2 Given a homogeneous linear differential equation L(Y) = 0 of order n over a differential field K, a differential extension K ⊂ L is a Picard-Vessiot extension for L if 1. L = K⟨y_{1}, …, y_{n}⟩, where y_{1}, …, y_{n} is a fundamental set of solutions of

L(Y) = 0 in L.

2. Every constant of L lies in K, i.e. C_{K} = C_{L}.

Remark 3.1 Let k be a differential field, K = k⟨z⟩, with z′ = z, and consider the differential equation Y′ −Y = 0. As z is a solution to this equation, if we are looking for an analogue of the splitting field, it would be natural to expect that the Picard-Vessiot extension for this equation would be the trivial extension of K. Now, if we adjoin a second differential indeterminate and consider L = K⟨y⟩, with y′ = y, the extension K ⊂ L satisfies condition 1 in definition 3.2. Now, we have (y/z)′ = 0, so the extension K ⊂ L adds the new constant y/z. Hence condition 2 in the definition of the Picard-Vessiot extension guarantees its minimality.

In the case when K is a differential field with algebraically closed field of constants C, we shall prove that there exists a Picard-Vessiot extension L of K for a given homogeneous linear differential equation L defined over K and that it is unique up to differential K-isomorphism.

The idea for the existence proof is to construct a differential K-algebra containing a full set of solutions of the differential equation

L(Y) = Y^{(n)} + a_{n−1}Y^{(n−1)} + ⋯ + a_{1}Y′ + a_{0}Y = 0

and then to make the quotient by a maximal differential ideal to obtain an extension not adding constants.

We consider the polynomial ring in n^{2} indeterminates

K[Y_{ij}, 0 ≤ i ≤ n − 1, 1 ≤ j ≤ n]

and extend the derivation of K to K[Y_{ij}] by defining

(1)  \begin{matrix}Y′_{ij} = Y_{i+1,j}, 0 ≤ i ≤ n − 2, \\ Y′_{n−1,j} = −a_{n−1}Y_{n−1,j} − ⋯ − a_{1}Y_{1j} − a_{0}Y_{0j}. \end{matrix}

Note that this definition is correct, as we can obtain the preceding ring by defining the ring K{X_{1}, …, X_{n}} in n differential indeterminates and making the quotient by the differential ideal generated by the elements

X_{j}^{(n)} + a_{n−1}X_{j}^{(n−1)} + ⋯ + a_{1}X′_{j} + a_{0}X_{j}, 1 ≤ j ≤ n,

that is the ideal generated by these elements and their derivatives. Let R := K[Y_{ij}][W^{−1}] be the localization of K[Y_{ij}] in the multiplicative system of the powers of W = det(Y_{ij}). The derivation of K[Y_{ij}] extends to R in a unique way. The algebra R is called the full universal solution algebra for L.

From the next two propositions we shall obtain that a maximal differential ideal P of the full universal solution algebra R is a prime ideal, hence R/P is an integral domain and that the quotient field of R/P has the same field of constants as K.

Proposition 3.4 Let K be a differential field and K ⊂ R be an extension of differential rings. Let I be a maximal element in the set of proper differential ideals of R. Then I is a prime ideal.

Proof. By passing to the quotient R/I, we can assume that R has no proper differential ideals. Then we have to prove that R is an integral domain. Let us assume that a, b are nonzero elements in R with ab = 0. We claim that d^{k}(a)b^{k+1} = 0, ∀k ∈ ℕ. Indeed ab = 0 ⇒ 0 = d(ab) = ad(b) + d(a)b and, multiplying this equality by b, we obtain d(a)b^{2} = 0. Now, if it is true for k, 0 = d(d^{k}(a)b^{k+1}) = d^{k+1}(a)b^{k+1} + (k + 1)d^{k}(a)b^{k}d(b) and, multiplying by b, we obtain d^{k+1}(a)b^{k+2} = 0.

Let J now be the differential ideal generated by a, that is, the ideal generated by a and its derivatives. Let us assume that no power of b is zero. By the claim, all elements in J are then zero divisors. In particular J ≠ R and, as J contains the nonzero element a, J is a proper differential ideal of R, which contradicts the hypothesis. Therefore, some power of b must be zero.

As b was an arbitrary zero divisor, we have that every zero divisor in R is nilpotent, in particular a^{n} = 0, for some n. We choose n to be minimal. Then 0 = d(a^{n}) = na^{n−1}d(a). As K ⊂ R, we have na^{n−1} ≠ 0 and so d(a) is a zero divisor. We have then proved that the derivative of a zero divisor is also a zero divisor and so a and all its derivatives are zero divisors and hence nilpotent. In particular, J ≠ R, so J would be proper and we obtain a contradiction, proving that R is an integral domain. □

Proposition 3.5 Let K be a differential field, with field of constants C, and let K ⊂ R be an extension of differential rings, such that R is an integral domain, finitely generated as a K-algebra. Let L be the quotient field of R. We assume that C is algebraically closed and that R has no proper differential ideals. Then, L does not contain new constants, i.e. C_{L} = C.

Proof. 1. First we prove that the elements in C_{L} \ C cannot be algebraic over K. If α ∈ \overline{K} \ K, from the proof of proposition 2.3, we have α′ = −P^{(d)}(α)/P′(α), for P (X) = X^{k} + a_{k−1}X^{k−1} + ⋯ + a_{1}X + a_{0} the irreducible polynomial of α over K. Then α′ = 0 ⇒ P^{(d)}(X) = a′_{k−1}X^{k−1} + ⋯ + a′_{1}X + a′_{0} = 0, so P (X) ∈ C[X] and α ∈ C.

2. Next we have C_{L} ⊂ R. Indeed for any b ∈ C_{L}, we have b = f/g, with f, g ∈ R. We consider the ideal of denominators of b, J = {h ∈ R : hb ∈ R}. We have h ∈ J ⇒ hb ∈ R ⇒ (hb)′ = h′b ∈ R ⇒ h′ ∈ J. Then J is a differential ideal. By hypothesis, R does not contain proper differential ideals, so J = R, hence b = 1.b ∈ R.

3. Here we show that for any b ∈ C_{L}, there exists an element c ∈ C such that b − c is not invertible in R. Then the ideal (b − c)R is a differential ideal different from R, and is therefore zero. Thus b = c ∈ C.

We now use some results from algebraic geometry. Let \overline{K} be the algebraic closure of K, \overline{R} = R ⊗_{K} \overline{K}. If the element b ⊗ 1 − c ⊗ 1 = (b − c) ⊗ 1 is not a unit in \overline{R}, then the element b − c will be nonunit in R. So we can assume that K is algebraically closed. Let V be the affine algebraic variety with coordinate ring R. Then b defines a K-valued function f over V. By Chevalley’s theorem (theorem 7.2), its image f(V) is a constructible set in the affine line 𝔸^{1} and hence either a finite set of points or the complement of a finite set of points. In the second case, as C is infinite, there exists c ∈ C such that f(v) = c, for some v ∈ V so that f − c vanishes at v and so b − c belongs to the maximal ideal of v. Hence, b − c is a nonunit. If f(V) is finite, it consists of a single point, since R is a domain and therefore V is irreducible. So, f is constant and b lies in K, hence in C. □

Theorem 3.1 Let K be a differential field with algebraically closed constant field C. Let L(Y) = 0 be a homogeneous linear differential equation defined over K. Let R be the full universal solution algebra for L and let P be a maximal differential ideal of R. Then P is a prime ideal and the quotient field L of the integral domain R/P is a Picard-Vessiot extension of K for L.

Proof. R is differentially generated over K by the solutions of L(Y) = 0 and by the inverse of the wrońskian, so R/P as well. By proposition 3.4, P is prime. As P is a maximal differential ideal, R/P does not have proper differential ideals, so by proposition 3.5, C_{L} = C. Moreover, the wrońskian is invertible in R/P and so in particular is nonzero in L. We have then that L contains a fundamental set of solutions of L and is differentially generated by it over K. Hence L is a Picard-Vessiot extension of K for L. □

In order to obtain uniqueness of the Picard-Vessiot extension, we first prove a normality property.

Proposition 3.6 Let L_{1}, L_{2} be Picard-Vessiot extensions of K for a homogeneous linear differential equation L(Y) = 0 of order n and let K ⊂ L be a differential field extension with C_{L} = C_{K}. We assume that σ_{i} : L_{i} → L are differential K-morphisms, i = 1, 2. Then σ_{1}(L_{1}) = σ_{2}(L_{2}).

Proof. Let V_{i} := {y ∈ L_{i} : L(y) = 0}, i = 1, 2, V := {y ∈ L : L(y) = 0}. Then V_{i} is a C_{K}-vector space of dimension n and V is a C_{K}-vector space of dimension at most n. Since σ_{i} is a differential morphism, we have σ_{i}(V_{i}) ⊂ V, i = 1, 2 and so, σ_{1}(V_{1}) = σ_{2}(V_{2}) = V. From L_{i} = K⟨V_{i}⟩, i = 1, 2, we get σ_{1}(L_{1}) = σ_{2}(L_{2}). □

Corollary 3.2 Let K ⊂ L ⊂ M be differential fields. Assume that L is a Picard-Vessiot extension of K and that M has the same constant field as K. Then any differential K-automorphism of M sends L onto itself. □

Corollary 3.3 An algebraic Picard-Vessiot extension is a normal algebraic extension. □

In the next theorem we establish uniqueness up to K-isomorphism of the Picard-Vessiot extension.

Theorem 3.2 Let K be a differential field with algebraically closed field of constants C. Let L(Y) = 0 be a homogeneous linear differential equation defined over K. Let L_{1}, L_{2} be two Picard-Vessiot extensions of K for L(Y) = 0. Then there exists a differential K-isomorphism from L_{1} to L_{2}.

Proof. We can assume that L_{1} is the Picard-Vessiot extension constructed in theorem 3.1. The idea of proof is to construct a differential extension K ⊂ E with C_{E} = C and differential K-morphisms L_{1} → E, L_{2} → E and apply proposition 3.6. We consider the ring A := (R/P) ⊗_{K} L_{2}, which is a differential ring finitely generated as a L_{2}-algebra, with the derivation defined by d(x ⊗ y) = dx ⊗ y + x ⊗ dy. Let Q be a maximal proper differential ideal of A. Its preimage in R/P by the map R/P → A defined by a ↦ a ⊗ 1 is zero, as R/P does not contain proper differential ideal, and it cannot be equal to R/P, as, in this case, Q would be equal to A. So R/P injects in A/Q by a ↦ \overline{a ⊗ 1}, and the map L_{2} → A/Q given by b ↦ \overline{1 ⊗ b} is also injective. Now by proposition 3.4, Q is prime and so A/Q is an integral domain. Let E be its quotient field. Now we can apply proposition 3.5 to the L_{2}-algebra A/Q and obtain C_{E} = C_{L_{2}} = C_{K}. By applying proposition 3.6 to the maps L_{1} ↪ A/Q ↪ E and L_{2} ↪ A/Q ↪ E we obtain that there exists a differential K-isomorphism L_{1} → L_{2}. □

We now state together the results obtained in Theorems 3.1 and 3.2.

Theorem 3.3 Let K be a differential field with algebraically closed field of constants C, let L(Y) = Y^{(n)} + a_{n−1}Y^{(n−1)} + ⋯ + a_{1}Y′ + a_{0}Y = 0 be defined over K. Then there exists a Picard-Vessiot extension L of K for L and it is unique up to differential K-isomorphism.

We end this section with a proposition which will be used to obtain the Fundamental Theorem of Picard-Vessiot Theory. The reader can compare this result with the analogue property of Galois extensions in classical Galois Theory.

Proposition 3.7 a) If K ⊂ L is a Picard-Vessiot extension for L(Y) = 0 and x ∈ L \ K, then there exists a differential K-automorphism σ of L such that σ(x) ≠ x.

b) Let K ⊂ L ⊂ M be extensions of differential fields, where K ⊂ L and K ⊂ M are Picard-Vessiot. Then any σ ∈ G(L|K) can be extended to a differential automorphism of M.

Proof. a) We can assume that L is the quotient field of R/P with R the full universal solution algebra for L and P a maximal differential ideal of R. Let x = a/b, with a, b ∈ R/P. Then x ∈ A := (R/P)[b^{−1}] ⊂ K. We consider the differential K-algebra T = A ⊗_{K} A ⊂ L ⊗_{K} L. Let z = x ⊗ 1 − 1 ⊗ x ∈ T. Since x ∉ K, we have z ≠ 0, z′ ≠ 0 (if z was a constant, it would be in K) and z is no nilpotent (z^{n} = 0, for a minimal n would imply nz^{n−1}z′ = 0). We localize T at z and pass to the quotient T [1/z]/Q by a maximal differential ideal Q of T [1/z]. Since z is a unit, its image \overline{z} in T [1/z]/Q is nonzero. We have maps τ_{i} : A → T [1/z]/Q, i = 1, 2, induced by w ↦ w ⊗ 1, w ↦ 1 ⊗ w. The maximality of P implies that R/P has no nontrivial differential ideals, so neither has A, hence the τ_{i} are injective. Therefore they both extend to differential K-embeddings of L into the quotient field E of T [1/z]/Q. By proposition 3.5, E is a no new constants extension of K, so by proposition 3.6, τ_{1}(L) = τ_{2}(L). On the other hand, τ_{1}(x) − τ_{2}(x) = \overline{z} ≠ 0, so τ_{1}(x) ≠ τ_{2}(x). Thus τ = τ_{1}^{−1}τ_{2} is a K-differential automorphism of L with τ(x) ≠ x.

b) As L ⊂ M is Picard-Vessiot (for the same differential equation L as K ⊂ M, seen as defined over L), we can assume that M is the quotient field of R_{1}/P, where R_{1} = L⊗_{K} R with R the full universal solution algebra for L and P a maximal differential ideal of R_{1}. Then the extension of σ ∈ G(L|K) to M is induced by σ ⊗ Id_{R}. □

Corollary 3.4 If K ⊂ L is a Picard-Vessiot extension with differential Galois group G(L|K), we have L^{G(L|K)} = K, i.e. the subfield of L which is fixed by the action of G(L|K) is equal to K.

Proof. The inclusion K ⊂ L^{G(L|K)} is clear, the inclusion L^{G(L|K)} ⊂ K is given by Proposition 3.7 a). □

4 Differential Galois group

Definition 4.1 If K ⊂ L is a differential field extension, the group G(L|K) of differential K-automorphisms of L is called differential Galois group of the extension K ⊂ L. In the case when K ⊂ L is a Picard-Vessiot extension for L(Y) = 0, the group G(L|K) of differential K-automorphisms of L is also referred to as the Galois group of L(Y) = 0 over K. We shall use the notation Gal_{K}(L) or Gal(L) if the base field is clear from the context.

We want to see now that the differential Galois group of a Picard-Vessiot extension is a linear algebraic group. First we see that the Galois group of a homogeneous linear differential equation of order n defined over the differential field K is isomorphic to a subgroup of the general linear group GL(n, C) over the constant field C of K. Indeed, if y_{1}, y_{2}, …, y_{n} is a fundamental set of solutions of L(Y) = 0, for each σ ∈ Gal(L) and for each j ∈ {1, …, n}, σ(y_{j}) is again a solution of L(Y) = 0, and so σ(y_{j}) = ∑_{i=1}^{n} c_{ij}y_{i}, for some c_{ij} ∈ C_{K}. Thus we can associate to each σ ∈ Gal(L) the matrix (c_{ij}) ∈ GL(n, C). Moreover, as L = K⟨y_{1}, …, y_{n}⟩, a differential K-automorphism of L is determined by the images of the y_{j}. Hence, we obtain an injective morphism Gal(L) → GL(n, C) given by σ ↦ (c_{ij}). We shall see in proposition 4.1 below that Gal(L) is closed in GL(n, C) with respect to the Zariski topology (which is defined in section 7). First, we look at some examples.

4.1 Examples

Example 4.1 We consider the differential extension L = K⟨α⟩, with α′ = a ∈ K such that a is not a derivative in K. We say that L is obtained from K by adjunction of an integral. We shall prove that α is transcendent over K, K ⊂ K⟨α⟩ is a Picard-Vessiot extension and G(K⟨α⟩|K) is isomorphic to the additive group of C = C_{K}.

Let us assume that α is algebraic over K and write P (X) = X^{n} + ∑_{i=1}^{n} b_{i}X^{n−i} its irreducible polynomial over K. Then 0 = P (α) = α^{n} + ∑_{i=1}^{n} b_{i}α^{n−i} ⇒ 0 = nα^{n−1}a + b′_{1}α^{n−1} + terms of degree < n − 1 ⇒ na + b′_{1} = 0 ⇒ a = (−b_{1}/n)′ which gives a contradiction.

We prove now that K⟨α⟩ does not contain new constants. Let us assume that the polynomial ∑_{i=0}^{n} b_{i}α^{n−i}, with b_{i} ∈ K, is constant. Differentiating, we obtain 0 = b′_{0}α^{n}+(nb_{0}a+b′_{1})α^{n−1}+terms of degree < n−1 ⇒ b′_{0} = nb_{0}a+b′_{1} = 0 ⇒ a = −b′_{1}/nb_{0} = (−b_{1}/nb_{0})′, contradicting the hypothesis. Let us assume that the rational function f(α)/g(α) is constant, with g monic, of degree

≥ 1, minimal. Differentiating, we obtain 0 = \frac{f(α)′g(α)a − f(α)g(α)′a}{g(α)^{2}} ⇒

\frac{f(α)}{g(α)} = \frac{f(α)′}{g(α)′}, with g(α)′ a nonzero polynomial of lower degree that g, since

g(α) is not a constant and g is monic. This is a contradiction.

We observe that 1 and α are solutions of Y′′ − \frac{a′}{a} Y′ = 0, linearly independent over the constants, so K ⊂ K⟨α⟩ is a Picard-Vessiot extension.

A differential K-automorphism of K⟨α⟩ maps α to α + c, with c ∈ C and a mapping α ↦ α + c induces a differential K-automorphism of K⟨α⟩, for

each c ∈ C. So G(K⟨α⟩|K) ≃ C ≃ \{( \begin{matrix}1 & c \\ 0 & 1 \end{matrix} )\} ⊂ GL(2, C).

Example 4.2 We consider the differential extension L = K⟨α⟩, with α′/α = a ∈ K \ {0}. We say that L is obtained from K by adjunction of the exponential of an integral. It is clear that K⟨α⟩ = K(α) and α is a fundamental set of solutions of the differential equation Y′ − aY = 0. We assume that C_{L} = C_{K}. We shall prove that if α is algebraic over K, then α^{n} ∈ K for some n ∈ ℕ. The Galois group G(L|K) is isomorphic to the multiplicative group of C = C_{K} if α is transcendent over K and to a finite cyclic group if α is algebraic over K.

Let us assume that α is algebraic over K and let P (X) = X^{n}+a_{n−1}X^{n−1}+ ⋯ + a_{0} its irreducible polynomial. Differentiating, we get 0 = P (α)′ = P^{(d)}(α) + P′(α)α′ = P^{(d)}(α) + P′(α)aα = anα^{n} + ∑_{k=0}^{n−1}(a′_{k} + aka_{k})α^{k}. Then P divides this last polynomial and so a′_{k} + aka_{k} = ana_{k} ⇒ a′_{k} = a(n − k)a_{k}, 0 ≤ k ≤ n − 1. Hence (α^{n−k}/a_{k})′ = 0. In particular, α^{n} = ca_{0} for some c ∈ C_{L} = C_{K}, hence α^{n} = b ∈ K. Then P (X) divides X^{n} − b and so P (X) = X^{n} − b.

For σ ∈ G(L|K), we have σ(α)′ = σ(α′) = σ(aα) = aσ(α) ⇒ (σ(α)/α)′ = 0 ⇒ σ(α) = cα for some c ∈ C_{L} = C_{K}. If α is transcendent over K, for each c ∈ C_{K}, we can define a differential K-automorphism of L by α ↦ cα. If α^{n} = b ∈ K, then σ(α)^{n} = σ(α^{n}) = σ(b) = b ⇒ c^{n} = 1 ⇒ c must be an nth root of unity and Gal(L|K) is a finite cyclic group.

Example 4.3 We consider a differential field K, an irreducible polynomial P (X) ∈ K[X] of degree n and a splitting field L of P (X) over K. We shall see that K ⊂ L is a Picard-Vessiot extension. We know by proposition 2.3 that we can extend the derivation in K to L in a unique way by defining for each root x of P (X) in L, x′ = −P^{(d)}(x)h(x) for h(X) ∈ K[X] such that h(X)P′(X) ≡ 1(mod P). Moreover by reducing modulo P, we can obtain an expression of x′ as a polynomial in x of degree smaller than n. By deriving the expression obtained for x′, we obtain an expression for x′′ as a polynomial in x which again by reducing modulo P will have degree smaller than n. Iterating the process, we obtain expressions for the successive derivatives of x as polynomials in x of degree smaller than n. Therefore x, x′, …, x^{(n−1)} are linearly dependent over K. If we write down this dependence relation, we obtain a homogeneous linear differential equation with coefficients in K satisfied by all the roots of the polynomial P. Now, let us assume that, while computing the successive derivatives of a root x of P, the first dependence relation found gives the differential equation

(2)  Y^{(k)} + a_{k−1}Y^{(k−1)} + ⋯ + a_{1}Y′ + a_{0}Y = 0, a_{i} ∈ K, k ≤ n.

Then, there exist k roots x_{1}, …, x_{k} of P with W (x_{1}, …, x_{k}) ≠ 0 since otherwise we would have found a differential equation of order smaller than k satisfied by all the roots of P. Hence, L is a Picard-Vessiot extension of K for the equation (2) and by proposition 2.3 the differential Galois group of K ⊂ L coincides with its algebraic Galois group.

4.2 The differential Galois group as a linear algebraic group

Proposition 4.1 Let K be a differential field with field of constants C, L = K⟨y_{1}, …, y_{n}⟩ a Picard-Vessiot extension of K. There exists a set S of polynomials F (X_{ij}), 1 ≤ i, j ≤ n, with coefficients in C such that

1) If σ is a differential K-automorphism of L and σ(y_{j}) = ∑_{i=1}^{n} c_{ij}y_{i}, then F (c_{ij}) = 0, ∀F ∈ S.

2) Given a matrix (c_{ij}) ∈ GL(n, C) with F (c_{ij}) = 0, ∀F ∈ S, there exists a differential K-automorphism σ of L such that σ(y_{j}) = ∑_{i=1}^{n} c_{ij}y_{i}.

Proof. Let K{Z_{1}, …, Z_{n}} be the ring of differential polynomials in n indeterminates over K. We define a differential K-morphism from K{Z_{1}, …, Z_{n}} in L by Z_{j} ↦ y_{j}. The kernel Γ is a prime differential ideal of K{Z_{1}, …, Z_{n}}. Let L[X_{ij}], 1 ≤ i, j ≤ n be the ring of polynomials in the indeterminates X_{ij} with the derivation defined by X′_{ij} = 0. We define a differential K-morphism from K{Z_{1}, …, Z_{n}} to L[X_{ij}] such that Z_{j} ↦ ∑_{i=1}^{n} X_{ij}y_{i}. Let Δ be the image of Γ in this mapping. Let {w_{k}} be a basis of the C-vector space L. We write each polynomial in Δ as a linear combination of the w_{k} with coefficients polynomials in C[X_{ij}]. We take S to be the collection of all these coefficients.

1. Let σ be a differential K-automorphism of L and σ(y_{j}) = ∑_{i=1}^{n} c_{ij}y_{i}. We consider the diagram

Z_{j}  ↦ y_{j}
Z_{j}  K{Z_{1}, …, Z_{n}} ⟶ L
|  |  |
|  |  | σ
↓  ↓  ↓
∑ X_{ij}y_{i}  L[X_{ij}]  ⟶ L
X_{ij}  ↦ c_{ij}

It is clearly commutative. The image of Γ by the upper horizontal arrow followed by σ is 0. Its image by the left vertical arrow followed by the lower horizontal one is Δ evaluated in X_{ij} = c_{ij}. Therefore all polynomials of Δ vanish at c_{ij}. Writing this down in the basis {w_{k}}, we see that all polynomials of S vanish at c_{ij}.

2. Let us now be given a matrix (c_{ij}) ∈ GL(n, C) such that F (c_{ij}) = 0 for every F in S. We define a differential morphism

\begin{matrix}K\{Z_{1}, …, Z_{n}\} & → & K\{y_{1}, …, y_{n}\} \\ Z_{j} & ↦ & ∑_{i} c_{ij}y_{i} \end{matrix} .

This morphism is the composition of the left vertical arrow and the lower horizontal one in the diagram above. By the hypothesis on (c_{ij}), and the definition of the set S, we see that the kernel of this morphism contains Γ and so, we have a K-morphism

\begin{aligned}σ : & K\{y_{1}, …, y_{n}\} & → & K\{y_{1}, …, y_{n}\} \\ & y_{j} & ↦ & ∑_{i} c_{ij}y_{i} \end{aligned} .

It remains to prove that it is bijective. If u is a nonzero element in the kernel I, then u cannot be algebraic over K, since in this case, the constant term of the irreducible polynomial of u over K would be in I and then I would be the whole ring. But, if u is transcendent, we have

trdeg[K{y_{1}, …, y_{n}} : K] > trdeg[K{σy_{1}, …, σy_{n}} : K].

On the other hand,

trdeg[K{y_{j}, σy_{j}} : K] = trdeg[K{y_{j}, c_{ij}} : K] = trdeg[K{y_{j}} : K]

and analogously we obtain trdeg[K{y_{j}, σy_{j}} : K] = trdeg[K{σy_{j}} : K], which gives a contradiction. Since the matrix (c_{ij}) is invertible, the image contains y_{1}, …, y_{n} and so σ is surjective.

Therefore we have that σ is bijective and can be extended to an automorphism

σ : K⟨y_{1}, …, y_{n}⟩ → K⟨y_{1}, …, y_{n}⟩.
□

This proposition gives that G(L|K) is a closed (in the Zariski topology) subgroup of GL(n, C) and then a linear algebraic group (see section 8.1).

Remark 4.1 The proper closed subgroups of GL(1, C) ≃ C^{∗} are finite and hence cyclic groups. So for a homogeneous linear differential equation of order 1 the only possible Galois groups are C^{∗} or a finite cyclic group, as we saw directly in Example 4.2 above.

Remark 4.2 In Example 4.1 above, the element α is a solution of the nonhomogeneous linear equation Y′ − a = 0 and we saw that K ⊂ K⟨α⟩ is a

Picard-Vessiot extension for the equation Y′′ − \frac{a′}{a} Y′ = 0. More generally, we

can associate to the equation L(Y) = Y^{(n)}+a_{n−1}Y^{(n−1)}+⋯+a_{1}Y′+a_{0}Y = b,

the homogeneous equation \overline{L}(Y) = 0, where \overline{L} = (d − \frac{b′}{b})L. It is easy to check that if y_{1}, …, y_{n} is a fundamental set of solutions of L(Y) = 0 and y_{0} is a particular solution of L(Y) = b, then y_{0}, y_{1}, …, y_{n} is a fundamental set of solutions of \overline{L}(Y) = 0.

Remark 4.3 The full universal solution algebra K[Y_{ij}][W^{−1}] constructed before proposition 3.4 is clearly isomorphic, as a K-algebra, to K ⊗_{C} C[GL(n, C)], where C[GL(n, C)] = C[X_{11}, …, X_{nn}, 1/ det] denotes the coordinate ring of the algebraic group GL(n, C) (see section 8.1). If we let GL(n, C) act on itself by right translations, i.e.

\begin{matrix}GL(n, C) × GL(n, C) & → & GL(n, C) \\ (g, h) & ↦ & hg^{−1} \end{matrix} ,

the corresponding action of GL(n, C) on C[GL(n, C)] is

\begin{matrix}GL(n, C) × C[GL(n, C)] & → & C[GL(n, C)] \\ (g, f) & ↦ & ρ_{g}(f) : h ↦ f(hg) \end{matrix} .

(see section 8.4). If we take f to be the function X_{ij} sending a matrix in GL(n, C) to its entry ij, we have ρ_{g}(X_{ij})(h) = X_{ij}(hg) = (hg)_{ij} = ∑_{k=1}^{n} h_{ik}g_{kj}.

Now to an element σ ∈ G = G(L|K) such that σ(Y_{ij}) = ∑ g_{kj}Y_{ik}, we associate the matrix (g_{ij}) ∈ GL(n, C). So the isomorphism

\begin{matrix}K[Y_{ij}][W^{−1}] & → & K ⊗_{C} C[GL(n, C)] \\ Y_{ij} & ↦ & X_{i+1,j} \end{matrix}

is also an isomorphism of G-modules.

Moreover, via the K-algebra isomorphism between K[Y_{ij}][W^{−1}] and K ⊗_{C} C[GL(n, C)] we can make GL(n, C) act on the full universal solution algebra R = K[Y_{ij}][W^{−1}]. Then, if P is the maximal differential ideal of R considered in theorem 3.1, the Galois group G(L|K) can be defined as {σ ∈ GL(n, C) : σ(P) = P }. So the Galois group G(L|K) is the stabilizer of the C-vector subspace P of R. Using C-bases of P and Ann(P) ⊂ Hom(R, C), we can write down equations for G(L|K) in GL(n, C). This gives a second proof that G(L|K) is a closed subgroup of the algebraic group GL(n, C).

Proposition 4.2 Let K be a differential field with field of constants C. Let K ⊂ L be a Picard-Vessiot extension with differential Galois group G. Let T be the K-algebra R/P considered in theorem 3.1. We have an isomorphism of \overline{K}[G]-modules \overline{K} ⊗_{K} T ≃ \overline{K} ⊗_{C} C[G], where \overline{K} denotes the algebraic closure of the field K.

Proof. We shall use two lemmas. For any field F, we denote by F [Y_{ij}, 1/ det] the polynomial ring in the indeterminates Y_{ij}, 1 ≤ i, j ≤ n localized with respect to the determinant of the matrix (Y_{ij}).

Lemma 4.1 Let L be a differential field with field of constants C. We consider A := L[Y_{ij}, 1/ det] and extend the derivation on L to A by setting Y′_{ij} = 0. We consider B := C[Y_{ij}, 1/ det] as a subring of L[Y_{ij}, 1/ det]. The map I ↦ IA from the set of ideals of B to the set of differential ideals of A is a bijection. The inverse map is given by J ↦ J ∩ B.

Proof. Choose a basis {v_{s}}_{s∈S_{1}} of L over C, including 1. Then {v_{s}}_{s∈S_{1}} is also a free basis of the B-module A. The differential ideal IA consists of the finite sums ∑_{s} λ_{s}v_{s} with all λ_{s} ∈ I. Hence IA ∩ B = I.

We prove now that any differential ideal J of A is generated by I = J ∩B. Let {u_{s}}_{s∈S_{2}} be a basis of B over C. Any element b ∈ J can be written uniquely as a finite sum ∑_{s} μ_{s}u_{s}, with μ_{s} ∈ L. By the length l(b) we will mean the number of subindices s with μ_{s} ≠ 0. By induction on the length of b, we shall show that b ∈ IA. When l(b) = 0, 1, the result is clear. Assume l(b) > 1. We may suppose that μ_{s_{1}} = 1 for some s_{1} ∈ S_{2} and μ_{s_{2}} ∈ L \ C for some s_{2} ∈ S_{2}. Then b′ = ∑_{s} μ′_{s}u_{s} has a length smaller than l(b) and so b′ ∈ IA. Similarly (μ_{s_{2}}^{−1}b)′ ∈ IA. Therefore (μ_{s_{2}}^{−1})′b = (μ_{s_{2}}^{−1}b)′ − μ_{s_{2}}^{−1}b′ ∈ IA. Since C is the field of constants of L, one has (μ_{s_{2}}^{−1})′ ≠ 0 and so b ∈ IA. □

Lemma 4.2 Let K be a differential field with field of constants C. Let K ⊂ L be a Picard-Vessiot extension with differential Galois group G(L|K). We consider A := L[Y_{ij}, 1/ det], B := K[Y_{ij}, 1/ det]. The map I ↦ IA from the set of ideals of B to the set of G(L|K)-stable ideals of A is a bijection. The inverse map is given by J ↦ J ∩ B.

Proof. The proof is similar to that of lemma 4.1. We have to verify that any G(L|K)-stable ideal J of A is generated by I = J ∩ B. Let {u_{s}}_{s∈S} be a basis of B over K. Any element b ∈ J can be written uniquely as a finite sum ∑_{s} μ_{s}u_{s}, with μ_{s} ∈ L. By the length l(b) we will mean the number of subindices s with μ_{s} ≠ 0. By induction on the length of b, we shall show that b ∈ IA. When l(b) = 0, 1, the result is clear. Assume l(b) > 1. We may suppose that μ_{s_{1}} = 1 for some s_{1} ∈ S. If all μ_{s} ∈ K, then b ∈ IA. If not, there exists some s_{2} ∈ S with μ_{s_{2}} ∈ L \ K. For any σ ∈ G, the length of σb − b is less that l(b). Thus σb − b ∈ IA. By proposition 3.7 a), there exists a σ with σμ_{s_{2}} ≠ μ_{s_{2}}. As above, one finds σ(μ_{s_{2}}^{−1}b) − μ_{s_{2}}^{−1}b ∈ IA. Then (σμ_{s_{2}}^{−1} − μ_{s_{2}}^{−1})b = σ(μ_{s_{2}}^{−1}b) − μ_{s_{2}}^{−1}b − σ(μ_{s_{2}}^{−1})(σb − b) ∈ IA. As σμ_{s_{2}}^{−1} − μ_{s_{2}}^{−1} ∈ L^{∗}, it follows that b ∈ IA. □

Proof of Proposition 4.2.

We consider the K-algebra R = K[Y_{ij}, 1/ det] with derivation defined by

\begin{matrix}Y′_{ij} = Y_{i+1,j}, 0 ≤ i ≤ n − 2, \\ Y′_{n−1,j} = −a_{n−1}Y_{n−1,j} − ⋯ − a_{1}Y_{1j} − a_{0}Y_{0j}. \end{matrix}

as in section 3.2. We consider as well the L-algebra L[Y_{ij}, 1/ det] with derivation defined by the derivation in L and the preceding formulae. We consider now the C-algebra C[X_{st}, 1/ det] where X_{st}, 1 ≤ s, t ≤ n are indeterminates, det denotes the determinant of the matrix (X_{st}) and recall that C[X_{st}, 1/ det] is the coordinate algebra C[GL(n, C)] of the algebraic group GL(n, C). We consider the action of the group G on GL(n, C) by translation on the left, i.e.

\begin{matrix}G × GL(n, C) & → & GL(n, C) \\ (g, h) & ↦ & gh \end{matrix}

which gives the following action of G on C[GL(n, C)]

\begin{matrix}G × C[GL(n, C)] & → & C[GL(n, C)] \\ (g, f) & ↦ & λ_{g}(f) : h ↦ f(g^{−1}h) \end{matrix}

If we take f to be X_{st}, the action of an element σ of G on X_{st} is multiplication on the left by the inverse of the matrix of σ as an element in GL(n, C). We consider C[X_{st}, 1/ det] with this G-action and the inclusion C[X_{st}, 1/ det] ⊂ L[X_{st}, 1/ det]. Now we define the relation between the indeterminates Y_{ij} and X_{st} to be given by (Y_{ij}) = (r_{ab})(X_{st}), where r_{ab} are the images of the Y_{ab} in the quotient R/P of the ring R by the maximal differential ideal P. We observe that the G-action we have defined on the X_{st} is compatible with the G-action on L if we take the Y_{ij} to be G-invariant. Now, the definition of the derivation for the Y_{ij} and the r_{ab} gives X′_{st} = 0. We have then the following rings

K[Y_{ij}, \frac{1}{det}] ⊂ L[Y_{ij}, \frac{1}{det}] = L[X_{st}, \frac{1}{det}] ⊃ C[X_{st}, \frac{1}{det}]

each of them endowed with a derivation and a G-action which are compatible with each other. Combining lemmas 4.1 and 4.2, we obtain a bijection between the set of differential ideals of K[Y_{ij}, 1/ det] and the set of G(L|K)-stable ideals of C[X_{st}, 1/ det]. A maximal differential ideal of the first ring corresponds to a maximal G(L|K)-stable ideal of the second. So, Q = P L[Y_{ij}, 1/ det] ∩ C[X_{st}, 1/ det] is a maximal G(L|K)-stable ideal of the ring C[X_{st}, 1/ det]. By its maximality, Q is a radical ideal and defines a subvariety W of GL(n, C), which is minimal with respect to G(L|K)-invariance. Thus W is a left coset in GL(n, C) for the group G(L|K) seen as a subgroup of GL(n, C). Now, by going to the algebraic closure \overline{K} of K, we have an isomorphism from G\overline{K} to W\overline{K} and, correspondingly, an isomorphism \overline{K} ⊗_{C} C[G] ≃ \overline{K} ⊗_{C} C[W] between the coordinate rings.

On the other hand, we have ring isomorphisms

L ⊗_{K} T = L ⊗_{K} (K[Y_{ij}, \frac{1}{det}]/P)
≃ L[Y_{ij}, \frac{1}{det}]/(P L[Y_{ij}, \frac{1}{det}]) ≃ L ⊗_{C} (C[X_{st}, \frac{1}{det}]/Q)

and so L ⊗_{K} T ≃ L ⊗_{C} C[W]. We then have \overline{L} ⊗_{K} T ≃ \overline{L} ⊗_{C} C[W], for \overline{L} the algebraic closure of L. This corresponds to an isomorphism of affine varieties V\overline{L} ≃ W\overline{L}, where we denote by V the affine subvariety of GL(n, K) corresponding to the ideal P of K[Y_{ij}, 1/ det]. But both W and V are defined over K and so, by proposition 7.4, we obtain V\overline{K} ≃ W\overline{K}. Coming back to the corresponding coordinate rings, we obtain \overline{K} ⊗_{K} T ≃ \overline{K} ⊗_{C} C[W]. Composing with the isomorphism obtained above, we have \overline{K} ⊗_{K} T ≃ \overline{K} ⊗_{C} C[G], as desired. □

Corollary 4.1 Let K ⊂ L be a Picard-Vessiot extension with differential Galois group G(L|K). We have

dim G(L|K) = trdeg[L : K].

Proof. The dimension of the algebraic variety G is equal to the Krull dimension of its coordinate ring C[G] (see section 7). It can be checked that the Krull dimension of a C-algebra remains unchanged when tensoring by a field extension of C. Then proposition 4.2 gives that the Krull dimension of C[G] is equal to the Krull dimension of the algebra T (where T denotes as in proposition 4.2 the K-algebra R/P considered in theorem 3.1), which by Noether’s normalization Lemma (proposition 7.8) is equal to the transcendence degree of L over K. □

30

5 Fundamental theorem

The aim of this section is to establish the fundamental theorem of Picard-Vessiot theory, which is analogous to the fundamental theorem in classical Galois theory.

If K ⊂ L is a Picard-Vessiot extension and F an intermediate differential field, i.e. K ⊂ F ⊂ L, it is clear that F ⊂ L is a Picard-Vessiot extension (for the same differential equation as K ⊂ L, viewed as defined over F) with differential Galois group G(L|F) = {σ ∈ G(L|K) : σ_{|F} = Id_{F} }. If H is a subgroup of G(L|K), we denote by L^{H} the subfield of L fixed by the action of H, i.e. L^{H} = {x ∈ L : σ(x) = x , ∀σ ∈ H}. Note that L^{H} is stable under the derivation of L.

Proposition 5.1 Let K ⊂ L be a Picard-Vessiot extension, G(L|K) its differential Galois group. The correspondences

H ↦ L^{H} , F ↦ G(L|F)

define inclusion inverting mutually inverse bijective maps between the set of Zariski closed subgroups H of G(L|K) and the set of differential fields F with K ⊂ F ⊂ L.

Proof. It is clear that for H_{1}, H_{2} subgroups of G(L|K), we have H_{1} ⊂ H_{2} ⇒ L^{H_{1}} ⊃ L^{H_{2}} and that for F_{1}, F_{2} intermediate differential fields, F_{1} ⊂ F_{2} ⇒ G(L|F_{1}) ⊃ G(L|F_{2}).

It is also straightforward to see that, for a subgroup H of G, we have the equality L^{G(L|L^{H})} = L^{H}, and, for an intermediate field F, we have G(L|L^{G(L|F)}) = G(L|F).

We have to prove that L^{G(L|F)} = F for each intermediate differential field F of K ⊂ L and H = G(L|L^{H}) for each Zariski closed subgroup H of G(L|K). The first equality follows from the fact observed above that F ⊂ L is a Picard-Vessiot extension and corollary 3.4. For the second equality, it is clear that if H is a subgroup of G(L|K), the elements in H fix L^{H} elementwise. We shall prove now that, if H is a subgroup (not necessarily closed) of G = G(L|K), then H′ := G(L|L^{H}) is the Zariski closure of H in G. Assume the contrary, i.e. that there exists a polynomial f on GL(n, C) (where C = C_{K} and L|K is a Picard-Vessiot extension for an order n differential equation) such that f_{|H} = 0 and f_{|H′} ≠ 0. If L = K⟨y_{1}, …, y_{n}⟩, we consider the matrices A = (y_{j}^{(i)})_{0≤i≤n−1,1≤j≤n}, B = (u_{j}^{(i)})_{0≤i≤n−1,1≤j≤n}, where u_{1}, …, u_{n} are differential indeterminates. We let the Galois group act on the right, i.e we define the matrix M_{σ} of σ ∈ G(L|K) such that (σ(y_{1}), …, σ(y_{n})) = (y_{1}, …, y_{n})M_{σ}. We note that, as W (y_{1}, …, y_{n}) ≠ 0, the matrix A is invertible and we define the polynomial F (u_{1}, …, u_{n}) = f(A^{−1}B) ∈ L{u_{1}, …, u_{n}}. It has the property that F (σ(y_{1}), …, σ(y_{n})) = 0, for all σ ∈ H but not all σ ∈ H′. Assume we are taking F among all polynomials with the preceding property having the smallest number of nonzero monomials. We can assume that some coefficient of F is 1. For τ ∈ H, let τF be the polynomial obtained by applying τ to the coefficients of F. Then (τF)(σ(y_{1}), …, σ(y_{n})) = τ(F ((τ^{−1}σ(y_{1}), …, τ^{−1}σ(y_{n}))) = 0, for all σ ∈ H. So, F −τF is shorter than F and vanishes for (σ(y_{1}), …, σ(y_{n})) for all σ ∈ H. By the minimality assumption, it must vanish for (σ(y_{1}), …, σ(y_{n})), for all σ ∈ H′. If F − τF is not identically zero, we can find an element a ∈ L such that F − a(F − τF) is shorter than F and has the same property as F. So F − τF ≡ 0, for all τ ∈ H, which means that the coefficients of F are H-invariant. Therefore, F has coefficients in L^{H} = L^{H′}. Now, for σ ∈ H′, F (σ(y_{1}), …, σ(y_{n})) = (σF)(σ(y_{1}), …, σ(y_{n})) = σ(F (y_{1}, …, y_{n})) = 0. This contradiction completes the proof. □

Proposition 5.2 Let K ⊂ L be a differential field extension with differential Galois group G = G(L|K).

a) If H is a normal subgroup of G, then L^{H} is G-stable.

b) If F is an intermediate differential field of the extension, which is Gstable, then G(L|F) is a normal subgroup of G. Moreover the restriction morphism

\begin{matrix}G(L|K) & → & G(F |K) \\ σ & ↦ & σ_{|F} \end{matrix}

induces an isomorphism from the quotient G/G(L|F) into the group of all differential K-automorphisms of F which can be extended to L.

Proof. a) For σ ∈ G, a ∈ L^{H}, we want to see that σa ∈ L^{H}. If τ ∈ H, we have τσa = σa ⇔ σ^{−1}τσa = a and this last equality is true as a ∈ L^{H} and σ^{−1}τσ ∈ H, by the normality of H. b) To see that G(L|F) is normal in G, we must see that for σ ∈ G, τ ∈ G(L|F), σ^{−1}τσ belongs to G(L|F), i.e. it fixes every element a ∈ F. Now σ^{−1}τσa = a ⇔ τσa = σa and this last equality is true since σa ∈ F because F is G-stable. Now as F is G-stable, we can define a morphism φ : G(L|K) → G(F |K) by σ ↦ σ_{|F}. The kernel of φ is G(L|F) and its image consists of those differential K-automorphisms of F which can be extended to L. □

Definition 5.1 We shall call an extension of differential fields K ⊂ L normal if for each x ∈ F \K, there exists an element σ ∈ G(L|K) such that σ(x) ≠ x.

Proposition 5.3 Let K ⊂ L be a Picard-Vessiot extension, G := G(L|K) its differential Galois group.

a) Let H be a closed subgroup of G. If H is normal in G, then the differential field extension K ⊂ F := L^{H} is normal.

b) Let F be a differential field with K ⊂ F ⊂ L. If K ⊂ F is a Picard-Vessiot extension, then the subgroup H = G(L|F) is normal in G(L|K). In this case, the restriction morphism

\begin{matrix}G(L|K) & → & G(F |K) \\ σ & ↦ & σ_{|F} \end{matrix}

induces an isomorphism G(L|K)/G(L|F) ≃ G(F |K).

Proof. a) By proposition 3.7, for x ∈ F \ K, there exists σ ∈ G such that σx ≠ x. By proposition 5.2 a), we know that F is G-stable, hence σ_{|F} is an automorphism of F.

b) By corollary 3.2, F is G-stable. Then by proposition 5.2 b), H = G(L|F) is a normal subgroup of G = G(L|K).

For the last part, taking into account proposition 5.2 b), it only remains to prove that the image of the restriction morphism is the whole group G(F |K) which comes from proposition 3.7 b). □

The next proposition establishes the more difficult part of the Fundamental Theorem, namely that the intermediate field F corresponding to a normal subgroup of G is a Picard-Vessiot extension of K. This result is not proved in Kaplansky’s book [K], which refers to a paper by Kolchin [Ko1]. In fact, Kolchin establishes the fundamental theorem for strongly normal extensions and characterizes Picard-Vessiot extensions as strongly normal extensions with a linear algebraic group. Our proof is inspired in [P-S] and [Ż] but not all details of it can be found there. The proof given in [M] uses a different algebra T.

Proposition 5.4 Let K ⊂ L be a Picard-Vessiot extension, G(L|K) its differential Galois group. If H is a normal closed subgroup of G(L|K), then the extension K ⊂ L^{H} is a Picard-Vessiot extension.

Proof. Let us explain first the idea of the proof. Assume that we have a finitely generated K-subalgebra T of L satisfying the following conditions.

a) T is G-stable and its quotient field Qt(T) is equal to L,

b) for each t ∈ T, the C-vector space generated by {σt : σ ∈ G} is finite dimensional,

c) the subalgebra T^{H} = {t ∈ T : σt = t, ∀σ ∈ H} is a finitely generated K-algebra,

d) F := L^{H} is the quotient field Qt(T^{H}) of T^{H}.

With all these assumptions, let us prove that T^{H} is generated over K by the space of solutions of a homogeneous linear differential equation with coefficients in K. First let us observe that, as H ⊲ G, T^{H} is G-stable, i.e. τ(T^{H}) = T^{H}, for all τ ∈ G. Indeed, let t ∈ T^{H}, τ ∈ G. We want to see that τt ∈ T^{H}. For σ ∈ H, we have στt = τt ⇔ (τ^{−1}στ)t = t and the last equality is true as the normality of H implies τ^{−1}στ ∈ H. Thus T^{H} is a G-stable subalgebra of T and the restriction of the action of G to T^{H} gives an action of the quotient group G/H on T^{H}.

We now take a finite-dimensional subspace V_{1} ⊂ T^{H} over C which generates T^{H} as a K-algebra and which is G-stable. Note that such a V_{1} exists by conditions b) and c). Let z_{1}, …, z_{m} be a basis of V_{1}, then the wronskian W (z_{1}, …, z_{m}) is not zero. The differential equation in Z

\frac{W (Z, z_{1}, …, z_{m})}{W (z_{1}, …, z_{m})} = 0

is satisfied by any z ∈ V_{1}. Now, by expanding the determinant in the numerator with respect to the first column, we see that each coefficient of the equation is a quotient of two determinants and that all these determinants are multiplied by the same factor det σ_{|V_{1}} under the action of the element σ ∈ G. So these coefficients are fixed by the action of G and so, by using corollary 3.4, we see that they belong to K. Thus T^{H} = K⟨V_{1}⟩, where V_{1} is a space of solutions of a linear differential equation with solutions in K. Therefore F = L^{H} = Qt(T^{H}) is a Picard-Vessiot extension of K.

Let T now be the K-algebra R/P considered in the construction of the Picard-Vessiot extension (see theorem 3.1). We shall prove that T satisfies the conditions stated above.

a) By construction G acts on T and the quotient field Qt(T) of T is equal to L.

b) Taking into account remark 4.3, we can apply lemma 8.3a) and obtain that the orbit of an element t ∈ T by the action of G generates a finite dimensional C-vector space.

c) We consider the isomorphism of G-modules given by proposition 4.2 and restrict the action to the subgroup H. The group H acts on both \overline{K} ⊗_{K} T and \overline{K} ⊗_{C} C[G] by acting on the second factor. We then have \overline{K} ⊗_{K} T^{H} ≃ \overline{K} ⊗_{C} C[G]^{H}. By proposition 8.10, C[G]^{H} ≃ C[G/H] as C-algebras. Now C[G/H] is a finitely generated C-algebra and so \overline{K} ⊗_{K} T^{H} is a finitely generated \overline{K}-algebra. Now we apply the following two lemmas to obtain that T^{H} is a finitely generated K-algebra.

Lemma 5.1 Let K be a field, \overline{K} an algebraic closure of K, A a Kalgebra. If \overline{K} ⊗_{K} A is a finitely generated \overline{K}-algebra, then there exists a finite extension K̃ of K such that K̃⊗_{K}A is a finitely generated K̃-algebra.

Proof. Let \{v_{s}\}_{s∈S} be a K-basis of \overline{K} and let \{λ_{i} ⊗ a_{i}\}_{i=1,...,n} generate \overline{K} ⊗_{K} A as a \overline{K}-algebra. If we write down the elements λ_{i} in the K-basis of \overline{K}, only the v′_{s}s with s in some finite subset S′ of S are involved. We take K̃ = K({v_{s}}_{s∈S′}). Then the elements {v_{s} ⊗ a_{i}}_{s∈S′,i=1,...,n} generate K̃ ⊗_{K} A as a K̃-algebra. □

Lemma 5.2 Let K be a field, A a finitely generated K-algebra and let U be a finite group of automorphisms of A. Then the subalgebra A^{U} = {a ∈ A : σa = a , ∀σ ∈ U} of A is a finitely generated K-algebra.

Proof. For each element a ∈ A, let us define

S(a) = \frac{1}{N} ∑_{σ∈U} σa, where N = |U|,

and let us consider the polynomial

P_{a}(T) = ∏_{σ∈U}(T − σa) = T^{N} + ∑_{i=1}^{N}(−1)^{i}a_{i}T^{N−i}.

The coefficients a_{i} are the symmetric functions in the roots of P_{a}(T) and by the Newton formulae can be expressed in terms of the S(a^{i}), i = 1, …, N. Let u_{1}, …, u_{m} now generate A as a K-algebra. We consider the subalgebra B of A^{U} generated by the elements S(u_{i}^{j}), i = 1, …, m, j = 1, …, N. We have P_{u_{i}}(u_{i}) = 0 and so u_{i}^{N} can be written as a linear combination of 1, …, u_{i}^{N−1} with coefficients in B. Hence each monomial u_{1}^{a_{1}} … u_{m}^{a_{m}} can be written in terms of monomials u_{1}^{a_{1}} … u_{m}^{a_{m}}, with a_{i} < N and coefficients in B. Therefore each element a ∈ A can be written in the form

a = ∑_{a_{i}<N} φ_{a_{1}...a_{m}}u_{1}^{a_{1}} … u_{m}^{a_{m}}, with φ_{a_{1}...a_{m}} ∈ B.

Now, if a ∈ A^{U}, we have

a = S(a) = ∑_{a_{i}<N} φ_{a_{1}...a_{m}}S(u_{1}^{a_{1}} … u_{m}^{a_{m}}).

Thus A^{U} can be generated over K by the finite set

{S(u_{1}^{a_{1}} … u_{m}^{a_{m}})}_{a_{i}<N} ∪ {S(U_{i}^{N})}_{i=1,...,m}.
□

Now by applying lemma 5.1 to \overline{K} ⊗_{K} T^{H}, we obtain that K̃ ⊗_{K} T^{H} is a finitely generated K̃-algebra for some finite extension K ⊂ K̃ and then also a finitely generated K-algebra. Now we can assume that the extension K ⊂ K̃ is normal and consider the Galois group U = Gal(K̃|K) acting on K̃ ⊗_{K} A on the left factor. By applying lemma 5.2, we can conclude that T^{H} ≃ K ⊗_{K} T^{H} ≃ K̃^{U} ⊗_{K} T^{H} ≃ (K̃ ⊗_{K} T^{H})^{U} is a finitely generated K-algebra.

d) We prove now that L^{H} is the quotient field of T^{H}.

Let a ∈ L^{H} \ {0}. We want to write a as a quotient of elements in T^{H}. We consider the ideal J = {t ∈ T : ta ∈ T } of denominators of a. Since a is H-invariant, J is H-stable, i.e. HJ = J. Let s ∈ J \ {0}. Taking into account remark 4.3, we can apply lemma 8.3a) and obtain that the elements τs, τ ∈ H generate a finite dimensional vector space E over C. Let s_{1}, …, s_{p} be a basis of E and w = W (s_{1}, …, s_{p}) be the wronskian. By expanding the determinant with respect to the first row, we see that w ∈ J. We have τw = det(τ_{|E}) · w, for all τ ∈ H. We note that τ ↦ det(τ_{|E}) defines a character χ of H, i.e. an algebraic group morphism χ : H → 𝔾_{m}(C), where 𝔾_{m} denotes the multiplicative group. We say that w is a semi-invariant with weight χ (see section 8.8). Let t = wa. It belongs to T, because w ∈ J, and is a semi-invariant with the same weight as w, because a is H-invariant. So a can be written as t/w the quotient of two semi-invariants. If we find a semi-invariant u with weight 1/χ, then we would have a = (tu)/(wu) the quotient of two invariants as desired. We consider the subalgebra of T consisting of the semi-invariants of weight 1/χ, that is T_{1/χ} = {t ∈ T : τt = t/χ(τ) , ∀τ ∈ H}. We want to prove T_{1/χ} ≠ 0.

To this end, we first consider the action of H on the coordinate ring C[G] of the algebraic group G and prove C[G]_{η} ≠ 0, for each character η of H. Let us denote X(H) the character group of the group H. Let H_{0} be the intersection of the kernels of all characters of H. It is a normal subgroup of H and contains the commutator subgroup of H, so H/H_{0} is commutative. By theorem 8.2, H/H_{0} is isomorphic to the direct product of its closed subgroups (H/H_{0})_{s} = {h ∈ H/H_{0} : h is semisimple} and (H/H_{0})_{u} = {h ∈ H/H_{0} : h is unipotent}. We recall that an element x ∈ GL(n, C) is called nilpotent if x^{k} = 0 for some k ∈ ℕ, unipotent if it is the sum of the identity element and a nilpotent element, semisimple if it is diagonalizable over C. By lemma 8.8, (H/H_{0})_{u} is conjugate to a subgroup of the upper triangular unipotent group U(n, C). Hence a nontrivial character of (H/H_{0})_{u} would give a nontrivial character of the additive group 𝔾_{a}(C), but 𝔾_{a}(C) does not have nontrivial characters (see section 8.8), so (H/H_{0})_{u} does not have nontrivial characters either. We then have X(H) = X(H/H_{0}) = X((H/H_{0})_{s}). We write H′ for (H/H_{0})_{s}. If η is a character of H′, we have η ∈ C[H′] and moreover, for each x, y ∈ H′, we have (x.η)(y) = η(xy) = η(x)η(y) which gives x.η = η(x)η, so η is a semi-invariant of weight η and we get C[H′]_{η} ≠ 0. Now the inclusion H′ ↪ G/H_{0} corresponds to an epimorphism between the coordinate rings π : C[G/H_{0}] → C[H′]. We want to see that π_{|C[G/H_{0}]_{η}} : C[G/H_{0}]_{η} → C[H′]_{η} is also an epimorphism. Let a be a nonzero element in C[H′]_{η}. Let α ∈ C[G/H_{0}] such that π(α) = a. By lemma 8.3a), there exists a finite dimensional H′-stable subspace E_{1} of C[G/H_{0}] containing α. As H′ is semisimple and commutative, it is diagonalizable, i.e. conjugate in the general linear group to a subgroup of the group of diagonal matrices (cf. lemma 8.8). Therefore the representation of H′ on E_{1} diagonalizes in a certain basis α_{1}, ⋯, α_{p}. We can choose it such that α_{1}, ⋯, α_{l}, with l < p are a basis of E_{1} ∩ Ker π. We have α = ∑_{j=1}^{n} c_{j}α_{j} ⇒ τ(α) = ∑_{j=1}^{n} c_{j}η_{j}(τ)α_{j}, then π(τ(α)) = ∑_{j=1}^{n} c_{j}η_{j}(τ)π(α_{j}) and, on the other hand, π(τ(α)) = τ(π(α)) = τ(a) = η(τ) ∑_{j=1}^{n} c_{j}π(α_{j}). We have c_{j} ≠ 0 for some j > l and so η_{j}(τ) = η(τ) which gives that α_{j} is a semi-invariant with weight η. We then obtain 0 ≠ C[G/H_{0}]_{η} ⊂ C[G]_{η}.

Now we consider again the isomorphism of G-modules given by proposition 4.2 with action restricted to the subgroup H. As the group H acts on both \overline{K} ⊗_{K} T and \overline{K} ⊗_{C} C[G] by acting on the second factor, we have C[G]_{1/χ} ≠ 0 ⇒ (\overline{K} ⊗_{C} C[G])_{1/χ} ≠ 0 ⇒ (\overline{K} ⊗_{K} T)_{1/χ} ≠ 0 ⇒ T_{1/χ} ≠ 0. To obtain the last implication, we use the fact that if t ∈ \overline{K} ⊗_{K} T, we have t ∈ K̃ ⊗_{K} T, for some finite extension K̃ of K. We can assume that K ⊂ K̃ is a normal extension and take U = G(K̃|K). Then, if t ∈ (K̃ ⊗_{K} T)_{1/χ}, the element ∑_{σ∈U} σt is a semi-invariant with weight 1/χ (as H acts in K̃ ⊗_{K} T by acting on the right factor and U by acting on the left factor, both actions commute) and belongs to K ⊗_{K} T ≃ T.

□

Now, propositions 5.1, 5.3 and 5.4 together establish the fundamental theorem of Picard-Vessiot theory.

Theorem 5.1 (Fundamental Theorem) Let K ⊂ L be a Picard-Vessiot extension, G(L|K) its differential Galois group.

1. The correspondences

H ↦ L^{H} , F ↦ G(L|F)

define inclusion inverting mutually inverse bijective maps between the set of Zariski closed subgroups H of G(L|K) and the set of differential fields F with K ⊂ F ⊂ L.

2. The intermediate differential field F is a Picard-Vessiot extension of K if and only if the subgroup H = G(L|F) is normal in G(L|K). In this case, the restriction morphism

\begin{matrix}G(L|K) & → & G(F |K) \\ σ & ↦ & σ_{|F} \end{matrix}

induces an isomorphism G(L|K)/G(L|F) ≃ G(F |K).

40

6 Liouville extensions

The aim of this section is to characterize linear differential equations solvable by quadratures. This is the analogue of characterization of algebraic equations solvable by radicals.

6.1 Liouville extensions

Definition 6.1 A differential field extension K ⊂ L is called a Liouville extension if there exists a chain of intermediate differential fields K = F_{1} ⊂ F_{2} ⊂ ⋯ ⊂ F_{n} = L such that F_{i+1} = F_{i}⟨α_{i}⟩, where each α_{i} is either a primitive element over F_{i}, i.e. α′_{i} ∈ F_{i}, or an exponential element over F_{i}, i.e. α′_{i}/α_{i} ∈ F_{i}.

Proposition 6.1 Let L be a Liouville extension of the differential field K, having the same field of constants as K. Then the differential Galois group G(L|K) of L over K is solvable.

Proof. We assume that the extension K ⊂ L has a chain of intermediate differential fields as in definition 6.1. From examples 4.1 and 4.2, we obtain that K ⊂ F_{2} is a Picard-Vessiot extension with commutative differential Galois group. By corollary 3.2, every K-differential automorphism of L sends F_{2} onto itself. By proposition 5.2 b), G(L|F_{2}) is a normal subgroup of G(L|K) and G(L|K)/G(L|F_{2}) is a subgroup of G(F_{2}|K), hence commutative. By iteration, we obtain that G(L|K) is solvable. □

The next proposition is the first step for a converse of proposition 6.1. In fact we shall consider generalized Liouville extensions, admitting also algebraic extensions as constructing blocks.

Proposition 6.2 Let K ⊂ L be a normal extension of differential fields. Assume that there exist elements u_{1}, …, u_{n} ∈ L such that for every differential automorphism σ of L we have

(3)  σ u_{j} = a_{1j} u_{1} + ⋯ + a_{j−1,j} u_{j−1} + a_{jj} u_{j} , j = 1, …, n,

with a_{ij} constants in L (depending on σ). Then K⟨u_{1}, …, u_{n}⟩ is a Liouville extension of K.

Proof. The first of the equations (3) is σu_{1} = a_{11}u_{1}. Differentiating, we obtain σu′_{1} = a_{11}u′_{1} and so u′_{1}/u_{1} is invariant under each σ (we can assume u_{1} ≠ 0 for otherwise it could simply be suppressed). By the normality of K ⊂ L, we obtain u′_{1}/u_{1} ∈ K. Hence the adjunction of u_{1} to K is the adjunction of an exponential. Next we divide each of the next n − 1 equations by the equation σu_{1} = a_{11}u_{1} and differentiate. The result is

σ (\frac{u_{j}}{u_{1}} )′ = \frac{a_{2j}}{a_{11}} (\frac{u_{2}}{u_{1}} )′ + ⋯ + \frac{a_{j−1,j}}{a_{11}} (\frac{u_{j−1}}{u_{1}} )′ + \frac{a_{jj}}{a_{11}} (\frac{u_{j}}{u_{1}} )′ .

This is a set of equations of the same form as (3) in the elements (u_{j}/u_{1})′, with j = 2, …, n. By induction on n, the adjunction of (u_{j}/u_{1})′ to K yields a Liouville extension. Then adjoining u_{j}/u_{1} themselves means adjoining integrals. □

6.2 Generalized Liouville extensions

Definition 6.2 A differential field extension K ⊂ L is called a generalized Liouville extension if there exists a chain of intermediate differential fields K = F_{1} ⊂ F_{2} ⊂ ⋯ ⊂ F_{n} = L such that F_{i+1} = F_{i}⟨α_{i}⟩, where each α_{i} es either a primitive element over F_{i}, or an exponential element over F_{i}, or is algebraic over F_{i}.

Theorem 6.1 Let K be a differential field with algebraically closed field of constants C. Let L be a Picard-Vessiot extension of K. Assume that the identity component G_{0} of G = G(L|K) is solvable. Then L can be obtained from K by a finite normal extension, followed by a Liouville extension.

Proof. Let F = L^{G_{0}}. We know by proposition 8.1 that G_{0} is a normal subgroup of G of finite index. Then K ⊂ F is a finite normal extension and G(L|F) ≃ G_{0}. Then by theorem 8.3, we can apply proposition 6.2 and obtain that F ⊂ L is a Liouville extension. □

To prove an inverse to this theorem we shall use the following lemma.

Lemma 6.1 Let K be a differential field with algebraically closed field of constants C. Let L be a Picard-Vessiot extension of K. Let L_{1} = L⟨z⟩ be an extension of L with no new constants. Write K_{1} = K⟨z⟩. Then K_{1} ⊂ L_{1} is a Picard-Vessiot extension and its differential Galois group is isomorphic to G(L|L ∩ K_{1}).

Proof. It is clear that K_{1} ⊂ L_{1} is a Picard-Vessiot extension as both fields have the same field of constants and the extension is generated by the solutions of the differential equation associated to the Picard-Vessiot extension K ⊂ L. By corollary 3.2, any K-differential automorphism of L_{1} sends L onto itself. Thus restriction to L gives a morphism φ : G(L_{1}|K_{1}) → G(L|K). An automorphism of L_{1} in Ker φ fixes both K_{1} and L and so is the identity. Hence φ is injective and G(L_{1}|K_{1}) is isomorphic to a closed subgroup of G(L|K). The corresponding intermediate field of the extension K ⊂ L is L∩K_{1} and by the fundamental theorem 5.1 we get G(L_{1}|K_{1}) ≃ G(L|L∩K_{1}). □

Theorem 6.2 Let K be a differential field with algebraically closed field of constants C. Let L be a Picard-Vessiot extension of K. Assume that L can be embedded in a differential field M which is a generalized Liouville extension of K with no new constants. Then the identity component G_{0} of G = G(L|K) is solvable (whence by theorem 6.1, L can be obtained from K by a finite normal extension, followed by a Liouville extension).

Proof. We make an induction on the number of steps in the chain from K to M. Let K⟨z⟩ be the first step. Then, by induction, the differential Galois group of L⟨z⟩ over K⟨z⟩ has a solvable component of the identity. By lemma 6.1, this group is isomorphic to the subgroup H of G corresponding to L ∩ K⟨z⟩. Assume that z is algebraic over K. Then, H has finite index in G. In this case, by proposition 8.1, G^{0} = H^{0}, hence solvable. If z is either an integral or an exponential, by examples 4.1 and 4.2, K⟨z⟩ is a Picard-Vessiot extension of K with commutative Galois group. Thus all differential fields between K and K⟨z⟩ are normal over K. In particular, L ∩ K⟨z⟩ is normal over K with a commutative differential Galois group. Thus H is normal in G with G/H commutative. So by lemma 8.10, the identity component G^{0} of G is solvable. □

44

7 Appendix on algebraic varieties

In this appendix, we gather some topics on algebraic varieties which are used in the Picard-Vessiot theory, and develop them as far as possible using an elementary approach. For the proofs of the results and more details on algebraic geometry we refer the reader to [Hu], [Kl] and [Sp].

In this section C will denote an algebraically closed field.

7.1 Affine varieties

The set C^{n} = C × ⋯ × C will be called affine n-space and denoted by 𝔸^{n}. We define an affine variety as the set of common zeros in 𝔸^{n} of a finite collection of polynomials. To each ideal I of C[X_{1}, …, X_{n}] we associate the set V(I) of its common zeros in 𝔸^{n}. By Hilbert’s basis theorem, the C-algebra C[X_{1}, …, X_{n}] is Noetherian, hence each ideal of C[X_{1}, …, X_{n}] has a finite set of generators. Therefore the set V(I) is an affine variety. To each subset S ⊂ 𝔸^{n} we associate the collection I(S) of all polynomials vanishing on S. It is clear that I(S) is an ideal and that we have inclusions S ⊂ V(I(S)), I ⊂ I(V(I), which are not equalities in general. We define the radical \sqrt{I} of an ideal I by

\sqrt{I} := \{f(X) ∈ C[X_{1}, …, X_{n}] : f(X)^{r} ∈ I for some r ≥ 1\}.

It is an ideal containing I. A radical ideal is an ideal equal to its radical. We can easily see the inclusion \sqrt{I} ⊂ I(V(I)). Equality is given by the next theorem.

Theorem 7.1 (Hilbert’s Nullstellensatz) If I is any ideal in C[X_{1}, …, X_{n}], then

\sqrt{I} = I(V(I)).

As a consequence, we have that V and I set a bijective correspondence between the collection of all radical ideals of C[X_{1}, …, X_{n}] and the collection of all affine varieties of 𝔸^{n}.

The following proposition is easy to prove.

Proposition 7.1 The correspondence V satisfies the following equalities:

a) 𝔸^{n} = V(0), ∅ = V(C[X_{1}, …, X_{n}]), b) If I and J are two ideals of C[X_{1}, …, X_{n}], V(I) ∪ V(J) = V(I ∩ J),

c) If I_{α} is an arbitrary collection of ideals of C[X_{1}, …, X_{n}], ∩_{α}V(I_{α}) = V(∑_{α} I_{α}).

We have then that affine varieties in 𝔸^{n} satisfy the axioms of closed sets in a topology. This is called Zariski topology. Hilbert’s basis theorem implies the descending chain condition on closed sets and therefore the ascending chain condition on open sets. Hence 𝔸^{n} is a Noetherian topological space. This implies that it is quasicompact. However the Hausdorff condition fails.

Recall that a topological space X is said to be irreducible if it cannot be written as the union of two proper, nonempty, closed subsets. Recall as well that a Noetherian topological space X can be written as a union of its irreducible components, i.e. its finitely many maximal irreducible subspaces.

Proposition 7.2 A closed set V in 𝔸^{n} is irreducible if and only if its ideal I(V) is prime. In particular, 𝔸^{n} itself is irreducible.

Proof. Write I = I(V). Suppose that V is irreducible and let f_{1}, f_{2} ∈ C[X_{1}, …, X_{n}] such that f_{1}f_{2} ∈ I. Then each x ∈ V is a zero of f_{1} or f_{2}, hence V ⊂ V(I_{1}) ∪ V(I_{2}), for I_{i} the ideal generated by f_{i}, i = 1, 2. Since V is irreducible, it must be contained within one of these two sets, i.e. f_{1} ∈ I or f_{2} ∈ I, and I is prime.

Reciprocally, assume that I is prime but V = V_{1} ∪ V_{2}, with V_{1}, V_{2} closed in V. If none of the V_{i}’s covers V, we can find f_{i} ∈ I(V_{i}) but f_{i} ∉ I, i = 1, 2. But f_{1}f_{2} vanish on V, so f_{1}f_{2} ∈ I, contradicting that I is prime. □

A principal open set of 𝔸^{n} is the set of nonzeros of a single polynomial. We note that principal open sets are a basis of the Zariski topology. We recall that a subspace of a topological space is irreducible if and only if its closure is. The closure in the Zariski topology of a principal open set is the whole affine space. Hence, as 𝔸^{n} is irreducible, we obtain that principal open sets are irreducible.

If V is closed in 𝔸^{n}, each polynomial f(X) ∈ C[X_{1}, …, X_{n}] defines a C-valued function on V. But different polynomials may define the same function. It is clear that we have a 1-1 correspondence between the distinct polynomial functions on V and the residue class ring C[X_{1}, …, X_{n}]/I(V).

We denote this ring by C[V] and call it the coordinate ring of V. It is a finitely generated algebra over C and is reduced (i.e. without nonzero nilpotent elements) because I(V) is a radical ideal. If V is an affine variety, f ∈ C[V], we define V_{f} := {P ∈ V : f(P) ≠ 0} which is clearly an open subset of V.

If V is irreducible, equivalently if I(V) is a prime ideal, C[V] is an integral domain. We can then consider its field of fractions C(V), which is called function field of V. Elements f ∈ C(V) are called rational functions on V. Any rational function can be written f = g/h, with g, h ∈ C[V]. In general, this representation is not unique. We can only give f a well defined value at a point P if there is a representation f = g/h, with h(P) ≠ 0. In this case we say that the rational function f is regular at P. The domain of definition of f is defined to be the set

dom(f) = {P ∈ V : f is regular at P }.

Proposition 7.3 Let V be an irreducible variety. For a rational function f ∈ C(V), the following hold

a) dom(f) is open and dense in V.

b) dom(f) = V ⇔ f ∈ C[V].

c) If h ∈ C[V] and V_{h} := {P ∈ V : h(P) ≠ 0}, then dom(f) ⊃ V_{h} ⇔ f ∈ C[V][1/h].

Part b) of the above proposition says that the polynomial functions are precisely the rational functions that are ”everywhere regular”.

The local ring of V at a point P ∈ V is the ring

{f ∈ C(V) : f is regular at P }.

It is isomorphic to the ring C[V]_{𝔐_{P}} obtained by localizing the ring C[V] at the maximal ideal 𝔐_{P} = {f ∈ C[V] : f(P) = 0}. This is indeed a local ring, i.e. it has a unique maximal ideal, namely 𝔐_{P} C[V]_{𝔐_{P}}.

We shall see now that a principal open set can be seen as an affine variety. If V_{f} = {x ∈ 𝔸^{n} : f(x) ≠ 0}, for some f ∈ C[X_{1}, …, X_{n}], the points of V_{f} are in 1-1 correspondence with the points of the closed set of 𝔸^{n+1}:

{(x_{1}, …, x_{n}, x_{n+1}) : f(x_{1}, …, x_{n}) x_{n+1} − 1 = 0}, hence V_{f} has an affine variety structure and its coordinate ring is C[V_{f}] = C[X_{1}, …, X_{n}, 1/f], i.e. the ring C[X_{1}, …, X_{n}] localized in the multiplicative system of the powers of f(X).

More generally, for V an affine variety, f ∈ C[V], the algebra of regular functions on the principal open set V_{f} := {x ∈ V : f(x) ≠ 0} is the algebra C[V]_{f}, i.e. the algebra C[V] localized in the multiplicative system {f^{n}/n ≥ 0}.

Now let V ⊂ 𝔸^{n}, W ⊂ 𝔸^{m} be arbitrary affine varieties. A morphism φ : V → W is a mapping of the form φ(x_{1}, …, x_{n}) = (φ_{1}(x), …, φ_{m}(x)), where φ_{i} ∈ C[V]. A morphism φ : V → W is continuous for the Zariski topologies involved. Indeed if Z ⊂ W is the set of zeros of polynomial functions f_{i} on W, then φ^{−1}(Z) is the set of zeros of the functions f_{i} ∘ φ on V. With a morphism φ : V → W, an algebra morphism φ^{∗} : C[W] → C[V] is associated, defined by φ^{∗}(f) = f ∘ φ. If φ : V → W is a morphism for which φ(V) is dense in W, then φ^{∗} is injective. The morphism φ : V → W is an isomorphism if there exists a morphism ψ : W → V such that ψ ∘ φ = Id_{V} and φ ∘ ψ = Id_{W}, or equivalently φ^{∗} : C[W] → C[V] is an isomorphism of C-algebras (with its inverse being ψ^{∗}). We say that the varieties V, W defined over the same field C are isomorphic if there exists an isomorphism φ : V → W. If V is an algebraic variety defined over C and L is a field containing C, we shall denote by V_{L} the variety obtained from V by extending scalars to L. The coordinate ring of V_{L} is L[V] = L⊗C[V]. It is clear that if V, W are affine varieties defined over C, we have V ≃ W ⇒ V_{L} ≃ W_{L}. The next proposition gives the converse of this implication for algebraically closed fields.

Proposition 7.4 Let K, L be algebraically closed fields, K ⊂ L. Let V, W be affine algebraic varieties defined over K. Let V_{L}, W_{L} be the varieties obtained from V, W by extending scalars to L. If V_{L} and W_{L} are isomorphic, then V and W are isomorphic.

Proof. As the statement ”V and W are isomorphic” can be written in the first order language of the theory of fields, the proposition follows from the fact that the theory of algebraically closed field is model complete (see [F-J] Corollary 8.5). □

We will often need to consider maps on an affine variety V which are not everywhere defined, so we introduce the following concept.

Definition 7.1 a) A rational map φ : V → 𝔸^{n} is an n-tuple (φ_{1}, …, φ_{n}) of rational functions φ_{1}, …, φ_{n} ∈ C(V). The map φ is called regular at a point P of V if all φ_{i} are regular at P. The domain of definition dom(φ) is the set of all regular points of φ, i.e. dom(φ) = ∩_{i=1}^{n} dom(φ_{i}).

b) For an affine variety W ∈ 𝔸^{n}, a rational map φ : V → W is a rational map φ : V → 𝔸^{n} such that φ(P) ∈ W for all regular points P ∈ dom(φ).

Proposition 7.5 Let φ : V → W a morphism of varieties. Then φ(V) contains a nonempty open subset of its closure \overline{φ(V)}.

Given a rational map φ : V → W, it is not always possible to define a morphism φ^{∗} : C(W) → C(V) given by φ^{∗}(f) = f ∘ φ. In order to determine when this is possible, we introduce the following concept.

Definition 7.2 A rational map φ : V → W is called dominant if φ(dom(φ)) is a Zariski dense subset of W.

Proposition 7.6 For irreducible affine varieties V and W, the following hold.

a) Every dominant rational map φ : V → W induces a C-linear morphism φ^{∗} : C(W) → C(V).

b) If f : C(W) → C(V) is a C-linear morphism, then there exists a unique dominant rational map φ : V → W with f = φ^{∗}.

c) If φ : V → W and ψ : W → X are dominant, then ψ ∘ φ : V → X is also dominant and (ψ ∘ φ)^{∗} = φ^{∗} ∘ ψ^{∗}.

Definition 7.3 Let V, W be irreducible affine varieties. A rational map φ : V → W is called birational (or a birational equivalence) if there is a rational map ψ : W → V with φ ∘ ψ = Id_{W} and ψ ∘ φ = Id_{V}.

Definition 7.4 Two irreducible varieties V and W are said to be birationally equivalent if there is a birational equivalence φ : V → W.

Proposition 7.7 Let V, W be irreducible affine varieties. For a rational map φ : V → W, the following statements are equivalent. a) φ is birational.

b) φ is dominant and φ^{∗} : C(W) → C(V) is an isomorphism.

c) There are open sets V_{0} ⊂ V and W_{0} ⊂ W such that the restriction φ_{|V_{0}} : V_{0} → W_{0} is an isomorphism.

7.2 Abstract affine varieties

We have considered so far affine varieties as closed subsets of affine spaces. We shall see now that they can be defined in an intrinsic way (i.e. not depending on an embedding in an ambient space) as topological spaces with a sheaf of functions satisfying adequate conditions.

Definition 7.5 A sheaf of functions on a topological space X is a function F which assigns to every nonempty open subset U ⊂ X a C-algebra F(U) of C-valued functions on U such that the following two conditions hold:

a) If U ⊂ U′ are two nonempty open subsets of X and f ∈ F(U′), then the restriction f_{|U} belongs to F(U).

b) Given a family of open sets U_{i}, i ∈ I, covering U and functions f_{i} ∈ F(U_{i}) for each i ∈ I, such that f_{i} and f_{j} agree on U_{i} ∩U_{j}, for each pair of indices i, j, there exists a function f ∈ F(U) whose restriction to each U_{i} equals f_{i}.

Definition 7.6 A topological space X together with a sheaf of functions O_{X} is called a geometric space. We refer to O_{X} as the structure sheaf of the geometric space X.

Definition 7.7 Let (X, O_{X}) and (Y, O_{Y}) be geometric spaces. A morphism

φ : (X, O_{X}) → (Y, O_{Y})

is a continuous map φ : X → Y such that for every open subset U of Y and every f ∈ O_{Y} (U), the function φ^{∗}(f) = f ∘ φ belongs to O_{X}(φ^{−1}(U)).

Remark 7.1 We shall often denote the morphism φ : (X, O_{X}) → (Y, O_{Y}) by φ : X → Y.

Example 7.1 Let X be an affine variety. To each nonempty open set U ⊂ X we assign the ring O_{X}(U) of regular functions on U. Then (X, O_{X}) is a geometric space. Moreover the two notions of morphism agree.

Let (X, O_{X}) be a geometric space and Z be a subset of X with induced topology. We can make Z into a geometric space by defining O_{Z}(V) for an open set V ⊂ Z as follows: a function f : V → C is in O_{Z}(V) if and only if there exists an open covering V = ∪_{i}V_{i} in Z such that for each i we have f_{|V_{i}} = g_{i}_{|V_{i}} for some g_{i} ∈ O_{X}(U_{i}) where U_{i} is an open subset of X containing V_{i}. It is not difficult to check that O_{Z} is a sheaf of functions on Z. We will refer to it as the induced structure sheaf and denote it by O_{X}_{|Z}. Note that if Z is open in X then a subset V ⊂ Z is open in Z if and only if it is open in X, and O_{X}(V) = O_{Z}(V).

Let X be a topological space and X = ∪_{i}U_{i} be an open cover. Given sheaves of functions O_{U_{i}} on U_{i} for each i, which agree on each U_{i} ∩ U_{j}, we can define a natural sheaf of functions O_{X} on X by gluing the O_{U_{i}} . Let U be an open subset in X. Then O_{X}(U) consists of all functions on U, whose restriction to each U ∩ U_{i} belongs to O_{U_{i}}(U ∩ U_{i}).

Let (X, O_{X}) be a geometric space. If x ∈ X we denote by v_{x} the map from functions on X to C obtained by evaluation at x:

v_{x}(f) = f(x).

Definition 7.8 A geometric space (X, O_{X}) is called an abstract affine variety if the following three conditions hold.

a) O_{X}(X) is a finitely generated C-algebra, and the map from X to the set Hom_{C}(O_{X}(X), C) of C-algebra morphisms defined by x ↦ v_{x} is a bijection.

b) For each f ∈ O_{X}(X), f ≠ 0, the set

X_{f} := {x ∈ X : f(x) ≠ 0}

is open, and every nonempty open set in X is a union of some X_{f}’s .

c) O_{X}(X_{f}) = O_{X}(X)_{f}, where O_{X}(X)_{f} denotes the C-algebra O_{X}(X) localized at f.

Remark 7.2 It can be checked that affine varieties with sheaves of regular functions are abstract affine varieties. We claim that, conversely, every abstract affine variety is isomorphic (as a geometric space) to an affine variety with the sheaf of regular functions. Indeed, let (X, O_{X}) be an abstract affine variety. Since O_{X}(X) is a finitely generated algebra of functions, we can write O_{X}(X) = C[X_{1}, …, X_{n}]/I for some radical ideal I. By the property a) of abstract affine varieties and the Nullstellensatz (theorem 7.1), we can identify X with V(I) as a set, and O_{X}(X) with the ring of regular functions on V(I). The Zariski topology on V(I) has the principal open sets as its base, so it now follows from b) that the identification of X and V(I) is a homeomorphism. Finally, by c), O_{X}(X_{f}) and the ring of regular functions on the principal open set X_{f} are also identified. This is enough to identify O_{X}(U) with the ring of regular functions on U for any open set U, as regularity is a local condition.

The preceding argument shows that the affine variety can be recovered completely from its algebra O_{X}(X) of regular functions, and conversely.

Example 7.2 In view of remark 7.2, a closed subset of an abstract affine variety is an abstract affine variety (as usual, with the induced sheaf).

7.3 Auxiliary results

We shall now define the product of two affine varieties. If V ⊂ 𝔸^{n}, W ⊂ 𝔸^{m} are closed subsets, then V × W ⊂ 𝔸^{n} × 𝔸^{m} = 𝔸^{n+m} is clearly a closed set, hence the cartesian product of two affine varieties is an affine variety. We have an isomorphism C[V × W] ≃ C[V] ⊗ C[W].

We shall introduce now the notion of dimension of an affine variety. If X is a topological space, we define the dimension of X to be the supremum of all integers n such that there exists a chain Z_{0} ⊂ Z_{1} ⊂ ⋯ ⊂ Z_{n} of distinct irreducible closed subsets of X. We define the dimension of an affine variety to be its dimension as a topological space. For example dim 𝔸^{n} = n. Clearly the dimension of an affine variety is the maximum of the dimensions of its irreducible components. For a ring A, we define the Krull dimension of A to be the supremum of all integers n such that there exists a chain P_{0} ⊂ P_{1} ⊂ ⋯ ⊂ P_{n} of distinct prime ideals of A. If V ⊂ 𝔸^{n} is an affine variety, by proposition 7.2, irreducible closed subsets of V correspond to prime ideals of C[X_{1}, …, X_{n}] containing I(V) and these in turn correspond to prime ideals of C[V]. Hence the dimension of V is equal to the Krull dimension of its coordinate ring C[V]. Now by Noether’s normalization lemma below (proposition 7.8), if V is irreducible, the Krull dimension of C[V] is equal to the transcendence degree trdeg[C(V) : C] of the function field C(V) of V over C.

Proposition 7.8 (Noether’s normalization Lemma) Let C be an arbitrary field, R a finitely generated integral domain over C with quotient field F, d = trdeg[F : C]. Then there exist elements y_{1}, …, y_{d} ∈ R, algebraically independent over C such that R is integral over C[y_{1}, …, y_{d}].

A subset of a topological space X is called locally closed if it is the intersection of an open set with a closed set. A finite union of locally closed sets is called a constructible set.

Theorem 7.2 (Chevalley’s theorem) Let φ : V → W be a morphism of varieties. Then φ maps constructible sets to constructible sets. In particular, φ(V) is constructible in W.

We now define the tangent space of an affine variety at a point. If V is an affine variety in 𝔸^{n} defined by polynomials f(X_{1}, …, X_{n}), x = (x_{1}, …, x_{n}) a point in V, we define the tangent space to V at the point x as the linear variety Tan(V)_{x} ⊂ 𝔸^{n} defined by the vanishing of all d_{x}f = ∑_{i=1}^{n}(∂f/∂X_{i})(x)(X_{i} −x_{i}), for f ∈ I(V). If 𝔐_{x} is the maximal ideal of C[V] consisting of the functions vanishing at x, we have C[V]/𝔐_{x} ≃ C, hence 𝔐_{x}/𝔐_{x}^{2} is a C-vector space. It can be proved that Tan(V)_{x} ≃ (𝔐_{x}/𝔐_{x}^{2})^{∗}, where ^{∗} denotes the dual vector space, i.e. (𝔐_{x}/𝔐_{x}^{2})^{∗} = Hom(𝔐_{x}/𝔐_{x}^{2}, C). Note that the definition of the tangent space as (𝔐_{x}/𝔐_{x}^{2})^{∗} is intrinsic, i.e. does not depend on an embedding of the affine variety in an ambient space.

For any point x in an affine variety V we have dim Tan(V)_{x} ≥ dim V. We say that x is a simple point if we have equality. It can be proved that the subset of simple points of V is dense in V. A variety is called nonsingular if all its points are simple.

We now state a version of Zariski’s main theorem. For its proof, we refer the reader to [Sp].

Theorem 7.3 Let φ : X → Y be a morphism of irreducible varieties that is bijective and birational. Assume Y to be nonsingular. Then φ is an isomorphism.

We end this appendix with a proposition which will be used in the construction of the quotient of an algebraic group by a subgroup.

Proposition 7.9 Let X and Y be irreducible varieties and let φ : X → Y be a dominant morphism. Let r := dim X − dim Y. There is a nonempty open subset U of X with the following properties.

a) The restriction of φ to U is an open morphism U → Y;

b) If Y′ is an irreducible closed subvariety of Y and X′ an irreducible component of φ^{−1}(Y′) that intersects U, then dim X′ = dim Y′ + r. In particular, if y ∈ Y, any irreducible component of φ^{−1}y that intersects U has dimension r;

c) If C(X) is algebraic over C(Y), then for all x ∈ U the number of points of the fiber φ^{−1}(φx) equals [C(X) : C(Y)].

Remark 7.3 In proposition 7.9, a) can be replaced by the following stronger property: a’) For any variety Z, the restriction of φ to U defines an open morphism U × Z → Y × Z.

8 Appendix on algebraic groups

In this appendix, we introduce the notion of algebraic group and develop some important points in this theory, such as the concept of solvable algebraic group, the existence of quotients and Lie-Kolchin theorem. Throughout the appendix, C will denote an algebraically closed field of characteristic 0.

8.1 The notion of algebraic group

Definition 8.1 An algebraic group over C is an algebraic variety G defined over C, endowed with a structure of group and such that the two maps μ : G × G → G, where μ(x, y) = xy and ι : G → G, where ι(x) = x^{−1}, are morphisms of varieties.

Translation by an element y ∈ G, i.e. x ↦ xy is clearly a variety automorphism of G, and therefore all geometric properties at one point of G can be transferred to any other point, by suitable choice of y. For example, since G has simple points (see section 7), all points must be simple, hence G is nonsingular.

Examples.

The additive group 𝔾_{a} is the affine line 𝔸^{1} with group law μ(x, y) = x+y, so ι(x) = −x and e = 0. The multiplicative group 𝔾_{m} is the principal open set C^{∗} ⊂ 𝔸^{1} with group law μ(x, y) = xy, so ι(x) = x^{−1} and e = 1. Each of these two groups is irreducible, as a variety, and has dimension 1. It can be proven that they are the only algebraic groups (up to isomorphism) with these two properties.

The general linear group GL(n, C) is the group of all invertible n × n matrices with entries in C with matrix multiplication. The set M(n, C) of all n×n matrices over C may be identified with the affine space of dimension n^{2} and GL(n, C) with the principal open subset defined by the nonvanishing of the determinant. Viewed thus as an affine variety, GL(n, C) has a coordinate ring generated by the restriction of the n^{2} coordinate functions X_{ij}, together with 1/ det(X_{ij}). The formulas for matrix multiplication and inversion make it clear that GL(n, C) is an algebraic group. Notice that GL(1, C) = 𝔾_{m}.

Taking into account that a closed subgroup of an algebraic group is again an algebraic group, we can construct further examples. We consider the following subgroups of GL(n, C): the special linear group SL(n, C)

:= {A ∈ GL(n, C) : det A = 1}; the upper triangular group T(n, C) := {(a_{ij}) ∈ GL(n, C) : a_{ij} = 0, i > j}; the upper triangular unipotent group U(n, C) := {(a_{ij}) ∈ GL(n, C) : a_{ii} = 1, a_{ij} = 0, i > j}; the diagonal group D(n, C) := {(a_{ij}) ∈ GL(n, C) : a_{ij} = 0, i ≠ j}.

The direct product of two or more algebraic groups, i.e. the usual direct product of groups endowed with the Zariski topology, is again an algebraic group. For example D(n, C) may be viewed as the direct product of n copies of 𝔾_{m}, while affine n-space may be viewed as the direct product of n copies of 𝔾_{a}.

8.2 Connected algebraic groups

Let G be an algebraic group. We assert that only one irreducible component of G contains the unit element e. Indeed, let X_{1}, …, X_{m} be the distinct irreducible components containing e. The image of the irreducible variety X_{1}×⋯×X_{m} under the product morphism is an irreducible subset X_{1} ⋯ X_{m} of G which again contains e. So X_{1} ⋯ X_{m} lies in some X_{i}. On the other hand, each of the components X_{1}, …, X_{m} clearly lies in X_{1} ⋯ X_{m}. Then m must be 1.

Denote by G^{0} this unique irreducible component of e and call it the identity component of G.

Proposition 8.1 Let G be an algebraic group.

a) G^{0} is a normal subgroup of finite index in G, whose cosets are the connected as well as irreducible components of G.

b) Each closed subgroup of finite index in G contains G^{0}.

c) Every finite conjugacy class of G has at most as many elements as [G : G^{0}].

Proof. a) For each x ∈ G^{0}, x^{−1}G^{0} is an irreducible component of G containing e, so x^{−1}G^{0} = G^{0}. Therefore G^{0} = (G^{0})^{−1}, and further G^{0}G^{0} = G^{0}, i.e. G^{0} is a (closed) subgroup of G. For any x ∈ G, xG^{0}x^{−1} is also an irreducible component of G containing e, so xG^{0}x^{−1} = G^{0} and G^{0} is normal. Its (left or right) cosets are translates of G^{0}, and so must also be irreducible components of G; as G is a Noetherian space there can only be finitely many of them. Since they are disjoint, they are also the connected components of G. b) If H is a closed subgroup of finite index in G, then each of its finitely many cosets is also closed. The union of those cosets distinct from H is also closed and then, H is open. Therefore the left cosets of H give a partition of G^{0} into a finite union of open sets. Since G^{0} is connected and meets H, we get G^{0} ⊂ H.

c) Write n = [G : G^{0}] and assume that there exists an element x ∈ G with a finite conjugacy class having a number of elements exceeding n. The mapping from G to G defined by a ↦ axa^{−1} is continuous. The inverse image of each conjugate of x is closed and, as there are finitely many of them, also open. This yields a decomposition of G into more than n open and closed sets, a contradiction. □

We shall call an algebraic group G connected when G = G^{0}. As is usual in the theory of linear algebraic groups, we shall reserve the word ”irreducible” for group representations.

The additive group 𝔾_{a}(C) and the multiplicative group 𝔾_{m}(C) are connected groups. The group GL(n, C) is connected as it is a principal open set in the affine space of dimension n^{2}. The next proposition will allow us to deduce the connectedness of some other algebraic groups. We first establish the following lemma.

Lemma 8.1 Let U, V be two dense open subsets of an algebraic group G. Then G = U · V.

Proof. Since inversion is a homeomorphism, V^{−1} is again a dense open set. So is its translate xV^{−1}, for any given x ∈ G. Therefore, U must meet xV^{−1}, forcing x ∈ U · V. □

For an arbitrary subset M of an algebraic group G, we define the group closure GC(M) of M as the intersection of all closed subgroups of G containing M.

Proposition 8.2 Let G be an algebraic group, f_{i} : X_{i} → G, i ∈ I, a family of morphisms from irreducible varieties X_{i} to G, such that e ∈ Y_{i} = f_{i}(X_{i}) for each i ∈ I. Set M = ∪_{i∈I}Y_{i}. Then

a) GC(M) is a connected subgroup of G.

b) For some finite sequence a = (a_{1}, …, a_{n}) in I, GC(M) = Y_{a_{1}}^{e_{1}} … Y_{a_{n}}^{e_{n}}, e_{i} = ±1.

Proof. We can if necessary enlarge I to include the morphisms x ↦ f_{i}(x)^{−1} from X_{i} to G. For each finite sequence a = (a_{1}, …, a_{n}) in I, set Y_{a} := Y_{a_{1}} … Y_{a_{n}}. The set Y_{a} is constructible, as it is the image of the irreducible variety X_{a_{1}}×⋯×X_{a_{n}} under the morphism f_{a_{1}}×⋯×f_{a_{n}} composed with multiplication in G, and moreover \overline{Y}_{a} is an irreducible variety passing through e. Given two finite sequences b, c in I, we have \overline{Y}_{b}\overline{Y}_{c} ⊂ \overline{Y}_{(b,c)}, where (b, c) is the sequence obtained from b and c by juxtaposition. Indeed, for x ∈ Y_{c}, the map y ↦ yx sends Y_{b} into Y_{(b,c)}, hence by continuity \overline{Y}_{b} into \overline{Y}_{(b,c)}, i.e. \overline{Y}_{b}Y_{c} ⊂ \overline{Y}_{(b,c)}. In turn, x ∈ \overline{Y}_{b} send Y_{c} into \overline{Y}_{(b,c)}, hence \overline{Y}_{c} as well. Let us now take a sequence a for which \overline{Y}_{a} is maximal. For each finite sequence b, we have \overline{Y}_{a} ⊂ \overline{Y}_{a}\overline{Y}_{b} ⊂ \overline{Y}_{(a,b)} = \overline{Y}_{a}. Setting b = a, we have \overline{Y}_{a} stable under multiplication. Choosing b such that Y_{b} = Y_{a}^{−1}, we also have \overline{Y}_{a} stable under inversion. We have then that \overline{Y}_{a} is a closed subgroup of G containing all Y_{i} so \overline{Y}_{a} = GC(M), proving a).

Since Y_{a} is constructible, lemma 8.1 shows that \overline{Y}_{a} = Y_{a} · Y_{a} = Y_{(a,a)}, so the sequence (a, a) satisfies b). □

Corollary 8.1 Let G be an algebraic group, Y_{i}, i ∈ I, a family of closed connected subgroups of G which generate G as an abstract group. Then G is connected. □

Corollary 8.2 The algebraic groups SL(n, C), U(n, C), D(n, C), T(n, C) (see section 8.1) are connected.

Proof. Let U_{ij} be the group of all matrices with 1’s on the diagonal, arbitrary entry in the (i, j) position and 0’s elsewhere, for 1 ≤ i, j ≤ n, i ≠ j. Then the U_{ij} are isomorphic to 𝔾_{a}(C), and so connected, and generate SL(n, C). Hence by corollary 8.1, SL(n, C) is connected. The U_{ij} with i < j generate U(n, C), whence U(n, C) is connected.

The group D(n, C) is the direct product of n copies of 𝔾_{m}(C), whence connected. Finally, T(n, C) is generated by U(n, C) and D(n, C), whence is also connected. □

8.3 Subgroups and morphisms

Proposition 8.3 Let H be a subgroup of an algebraic group G, \overline{H} its closure.

a) \overline{H} is a subgroup of G. b) If H is constructible, then H = \overline{H}.

Proof. a) Inversion being a homeomorphism, it is clear that \overline{H}^{−1} = \overline{H^{−1}} = \overline{H}. Similarly, translation by x ∈ H is a homeomorphism, so x \overline{H} = \overline{xH} = \overline{H}, i.e. H\overline{H} ⊂ \overline{H}. In turn, if x ∈ \overline{H}, Hx ⊂ \overline{H}, so \overline{H} x = \overline{Hx} ⊂ \overline{H}. This says that \overline{H} is a group.

b) If H is constructible, it contains a dense open subset U of \overline{H}. Since \overline{H} is a group, by part a), lemma 8.1 shows that \overline{H} = U · U ⊂ H · H = H. □

For a subgroup H of a group G we define the normalizer N_{G}(H) of H in G as

N_{G}(H) = {x ∈ G : xHx^{−1} = H}.

If a subgroup H′ of G is contained in N_{G}(H), we say that H′ normalizes H.

Proposition 8.4 Let A, B be closed subgroups of an algebraic group G. If B normalizes A, then AB is a closed subgroup of G.

Proof. Since B ⊂ N_{G}(A), AB is a subgroup of G. Now AB is the image of A × B under the product morphism G × G → G; hence it is constructible, and therefore closed by proposition 8.3 b). □

By definition a morphism of algebraic groups is a group homomorphism which is also a morphism of algebraic varieties.

Proposition 8.5 Let φ : G → G′ be a morphism of algebraic groups. Then

a) Ker φ is a closed subgroup of G.

b) Im φ is a closed subgroup of G′.

c) φ(G^{0}) = φ(G)^{0}

Proof. a) φ is continuous and Ker φ is the inverse image of the closed set {e}.

b) φ(G) is a subgroup of G′. It is also a constructible subset of G′, by theorem 7.2 , so it is closed by proposition 8.3 b).

c) φ(G^{0}) is closed by b) and connected; hence it lies in φ(G)^{0}. As it has finite index in φ(G), it must be equal to φ(G)^{0}, by proposition 8.1b). □

8.4 Linearization of affine algebraic groups

We have seen that any closed subgroup of GL(n, C) is an affine algebraic group. We shall see now that the converse is also true.

Let G be an algebraic group, V an affine variety. We say that V is a G-variety if the algebraic group G acts on the affine variety V, i.e. we have a morphism of algebraic varieties

\begin{matrix}G × V & → & V \\ (g, v) & ↦ & g · v \end{matrix}

satisfying g_{1}.(g_{2}.v) = (g_{1}g_{2}).v, for any g_{1}, g_{2} in G, v in V, and e.v = v, for any v ∈ V.

Let V, W be G-varieties. A morphism φ : V → W is a G-morphism, or is said to be equivariant if φ(g.v) = g.φ(v), for g ∈ G, v ∈ V.

The action of G over V induces an action of G on the coordinate ring C[V] of V defined by

\begin{matrix}G × C[V] & → & C[V] \\ (g, f) & ↦ & g.f : v ↦ f(g^{−1} · v) \end{matrix} .

In particular, we can consider two different actions of G on its coordinate ring C[G] associated to the action of G on itself by left or right translations. To the action of G on itself by left translations defined by

\begin{matrix}G × G & → & G \\ (g, h) & ↦ & gh \end{matrix}

corresponds the action

\begin{matrix}G × C[G] & → & C[G] \\ (g, f) & ↦ & λ_{g}(f) : h ↦ f(g^{−1}h) \end{matrix} .

To the action of G on itself by right translations defined by

\begin{matrix}G × G & → & G \\ (g, h) & ↦ & hg^{−1} \end{matrix}

corresponds the action

\begin{matrix}G × C[G] & → & C[G] \\ (g, f) & ↦ & ρ_{g}(f) : h ↦ f(hg) \end{matrix} .

We can use right translations to characterize membership in a closed subgroup:

Lemma 8.2 Let H be a closed subgroup of an algebraic group G, I the ideal of C[G] vanishing on H. Then H = {g ∈ G : ρ_{g}(I) ⊂ I}.

Proof. Let g ∈ H. If f ∈ I, ρ_{g}(f)(h) = f(hg) = 0 for all h ∈ H, hence ρ_{g}(f) ∈ I, i.e. ρ_{g}(I) ⊂ I. Assume now ρ_{g}(I) ⊂ I. In particular, if f ∈ I, then ρ_{g}(f) vanishes at e ∈ H, then f(g) = f(eg) = ρ_{g}(f)(e) = 0, so g ∈ H. □

Lemma 8.3 Let G be an algebraic group and V an affine variety both defined over an algebraically closed field C. Assume that G acts on V and let F be a finite dimensional subspace of the coordinate ring C[V] of V.

a) There exists a finite dimensional subspace E of C[V] including F which is stable under the action of G.

b) F itself is stable under the action of G if and only if φ^{∗}F ⊂ C[G] ⊗_{C} F, where φ : G × V → V is given by φ(g, x) = g^{−1}.x

Proof. a) If we prove the result in the case in which F has dimension 1, we can obtain it for a finite dimensional F by summing up the subspaces E corresponding to the subspaces of F generated by one vector of a chosen basis of F. So we may assume that F =< f > for some f ∈ C[V]. Let φ : G × V → V be the morphism giving the action of G on V, φ^{∗} : C[V] → C[G × V] = C[G] ⊗ C[V] the corresponding morphism between coordinate rings. Let us write φ^{∗}f = ∑ g_{i} ⊗ f_{i} ∈ C[G] ⊗ C[V] (note that this expression is not unique). For g ∈ G, x ∈ V, we have (g.f)(x) = f(g^{−1}.x) = f(φ(g^{−1}, x)) = (φ^{∗}f)(g^{−1}, x) = ∑ g_{i}(g^{−1})f_{i}(x) and then g.f = ∑ g_{i}(g^{−1})f_{i}. So every translate g.f of f is contained in the finite dimensional C-vector space of C[V] generated by the functions f_{i}. So E = ⟨g.f | g ∈ G⟩ is a finite-dimensional G-stable vector space containing f.

b) If φ^{∗}F ⊂ C[G] ⊗_{C} F, then the proof of a) shows that the functions f_{i} can be taken to lie in F, i.e. F is stable under the action of G. Conversely, let F be stable under the action of G and extend a vector space basis {f_{i}} of F to a basis {f_{i}} ∪ {h_{j}} of C[V]. If φ^{∗}f = ∑ r_{i} ⊗ f_{i} + ∑ s_{j} ⊗ h_{j}, for g ∈ G, we have g.f = ∑ r_{i}(g^{−1})f_{i} + ∑ s_{j}(g^{−1})h_{j}. Since this element belongs to F, the functions s_{j} must vanish identically on G, hence must be 0. We then have φ^{∗}F ⊂ C[G] ⊗_{C} F. □

Theorem 8.1 Let G be an affine algebraic group. Then G is isomorphic to a closed subgroup of some GL(n, C).

Proof. Choose generators f_{1}, …, f_{n} for the coordinate algebra C[G]. By applying lemma 8.3 a), we can assume that the f_{i} are a C-basis of a Cvector space F which is G-stable when considering the action of G by right translations. If φ : G × G → G is given by (g, h) ↦ hg, by lemma 8.3 b), we can write φ^{∗}f_{i} = ∑_{j} m_{ij} ⊗ f_{j}, where m_{ij} ∈ C[G]. Then ρ_{g}(f_{i})(h) = f_{i}(hg) = ∑_{j} m_{ij}(g) ⊗ f_{j}(h), whence ρ_{g}(f_{i}) = ∑_{j} m_{ij}(g) ⊗ f_{j}. In other words, the matrix of ρ_{g}|F in the basis {f_{i}} is (m_{ij}(g)). This shows that the map ψ : G → GL(n, C) defined by g ↦ (m_{ij}(g)) is a morphism of algebraic groups.

Notice that f_{i}(g) = f_{i}(eg) = ∑ m_{ij}(g)f_{j}(e), i.e. f_{i} = ∑ f_{j}(e)m_{ij}. This shows that the m_{ij} also generate C[G]; in particular, ψ is injective. Moreover the image group G′ = ψ(G) is closed in GL(n, C) by proposition 8.5 b). To complete the proof we therefore only need to show that ψ : G → G′ is an isomorphism of varieties. But the restriction to G′ of the coordinate functions X_{ij} are sent by ψ∗ to the respective m_{ij}, which were just shown to generate C[G]. So ψ^{∗} is surjective, and thus identifies C[G′] with C[G]. □

8.5 Homogeneous spaces

Let G be an algebraic group. A homogeneous space for G is a G-variety V on which G acts transitively. An example of homogeneous space for G is V = G with the action given by left or right translations introduced in section 8.4.

Lemma 8.4 Let V be a G-variety.

a) For v ∈ V, the orbit G.v is open in its closure.

b) There exist closed orbits.

Proof. By applying proposition 7.5 to the morphism G → V, g ↦ g.v, we obtain that G.v contains a nonempty open subset U of its closure. Since G.v is the union of the open sets g.U, g ∈ G, assertion a) follows. It implies that for v ∈ V, the set S_{v} = \overline{G.v} \ G.v is closed. It is also G-stable, hence a union of orbits. As the descending chain condition on closed sets is satisfied, there is a minimal set S_{v}. By a), it must be empty. Hence the orbit G.v is closed, proving b). □

Lemma 8.5 Let G be an algebraic group and G^{0} its identity component. Let V be a homogeneous space for G.

a) Each irreducible component of V is a homogeneous space for G^{0}.

b) The components of V are open and closed and V is their disjoint union.

Proof. Let V′ be the orbit of G^{0} in V. Since G acts transitively on V, it follows from proposition 8.1 that V is the disjoint union of finitely many translates g.V′. Each of them is a G^{0}-orbit and is irreducible. It follows from lemma 8.4 that all G^{0}-orbits are closed. Now a) and b) readily follow. □

Proposition 8.6 Let G be an algebraic group and let φ : V → W be an equivariant morphism of homogeneous spaces for G. Put r = dim V −dim W.

a) For any variety Z the morphism (φ, Id) : V × Z → W × Z is open.

b) If W′ is an irreducible closed subvariety of W and V′ an irreducible component of φ^{−1}W′, then dim V′ = dim W′ + r. In particular, if y ∈ W, then all irreducible components of φ^{−1}y have dimension r.

Proof. Using lemma 8.5, we reduce the proof to the case when G is connected and V, W are irreducible. Then φ is surjective, hence dominant. Let U ∈ V be an open subset with the properties of proposition 7.9 and remark 7.3. Then all translates g.U enjoy the same properties. Since these cover V, we have a) and b). □

8.6 Decomposition of algebraic groups

Let x ∈ End V, for V a finite dimensional vector space over C. Then x is nilpotent if x^{n} = 0 for some n (equivalently if 0 is the only eigenvalue of x). At the other extreme, x is called semisimple if the minimal polynomial of x has distinct roots (equivalently if x is diagonalizable over C). If x ∈ End V, by Jordan decomposition, we obtain

Lemma 8.6 Let x ∈ End V.

a) There exist unique x_{s}, x_{n} ∈ End V such that x_{s} is semisimple, x_{n} is nilpotent and x = x_{s} + x_{n}. b) There exist polynomials P (T), Q(T) ∈ C[T], without constant term such

that x_{s} = P (x), x_{n} = Q(x). In particular x_{s} and x_{n} commute with any endomorphism of V which commutes with x.

c) If W_{1} ⊂ W_{2} are subspaces of V, and x maps W_{2} into W_{1}, then so do x_{s} and x_{n}.

d) Let y ∈ End V. If xy = yx, then (x+y)_{s} = x_{s} +y_{s} and (x+y)_{n} = x_{n} +y_{n}. □

If x ∈ GL(V), its eigenvalues are nonzero, and so x_{s} is also invertible. We can write x_{u} = 1 + x_{s}^{−1}x_{n} and then we obtain x = x_{s} + x_{n} = x_{s}(1 + x_{s}^{−1}x_{n}) = x_{s} · x_{u}. We call an element in GL(V) unipotent it it is the sum of the identity and a nilpotent endomorphism or, equivalently, if 1 is its unique eigenvalue. For x ∈ GL(V), the decomposition x = x_{s} · x_{u}, with x_{s} semisimple, x_{u} unipotent, is unique. Clearly the only element in GL(V) which is both semisimple and unipotent is identity. From lemma 8.6, we obtain

Lemma 8.7 Let x ∈ GL(V).

a) There exist unique x_{s}, x_{u} ∈ GL(V) such that x_{s} is semisimple, x_{u} is unipotent, x = x_{s}x_{u} and x_{s}x_{u} = x_{u}x_{s}.

b) x_{s} and x_{u} commute with any endomorphism of V which commutes with x.

c) If W is a subspace of V stable under x, then W is stable under x_{s} and x_{u}.

d) Let y ∈ GL(V). If xy = yx, then (xy)_{s} = x_{s}y_{s} and (xy)_{u} = x_{u}y_{u}. □

If G is a linear algebraic group, we consider the subsets

G_{s} = {x ∈ G : x = x_{s}} and G_{u} = {x ∈ G : x = x_{u}}.

Let us denote by T (n, C) (resp. D(n, C)) the ring of all upper triangular (resp. all diagonal) matrices in M(n, C). A subset M of M(n, C) is said to be triangularizable (resp. diagonalizable) if there exists x ∈ GL(n, C) such that xMx^{−1} ⊂ T (n, C) (resp. D(n, C)).

Lemma 8.8 If M ⊂ M(n, C) is a commuting set of matrices, then M is triangularizable. If M has a subset N consisting of diagonalizable matrices, N can be diagonalized at the same time.

Proof. Let V = C^{n} and proceed by induction on n. If x ∈ M, λ ∈ C, the subspace W = Ker(x − λI) is evidently stable under the endomorphisms of V which commute with x, hence it is stable under M. Unless M consists of scalar matrices (then we are done), it is possible to choose x and λ such that 0 ≠ W ≠ V. By induction, there exists a nonzero v_{1} ∈ W such that Cv_{1} is M-stable. Applying the induction hypothesis next to the induced action of M on V/Cv_{1}, we obtain v_{2}, … v_{n} ∈ V completing the basis for V, such that M stabilizes each subspace Cv_{1} + ⋯ + Cv_{i} (1 ≤ i ≤ n). The transition from the canonical basis of V to (v_{1}, … v_{n}) therefore triangularizes M.

Now if N does not already consist of scalar matrices, we can choose x above to lie in N. Since x is diagonalizable, V = W ⊕ W′, where the sum W′ of remaining eigenspaces of x is nonzero. As before, both W and W′ are M-stable. The induction hypothesis allows us to choose basis of W and W′ which triangularize M while simultaneously diagonalizing N. □

Theorem 8.2 Let G be a commutative linear algebraic group. Then G_{s}, G_{u} are closed subgroups, connected if G is connected, and the product map φ : G_{s} × G_{u} → G is an isomorphism of algebraic groups.

Proof. As G is commutative, by lemma 8.7 d), G_{s} and G_{u} are subgroups of G. The subset G_{u} is closed since the subset of all unipotent matrices x in GL(V) can be defined as the zero set of the polynomials implied by (x−1)^{n} = 0. As G is commutative, by lemma 8.7 a), φ is a group isomorphism. By lemma 8.8, we may assume that G ⊂ T(n, C) and G_{s} ⊂ D(n, C). This forces G_{s} = G ∩ D(n, C), so G_{s} is also closed. Moreover, φ is a morphism of algebraic groups.

It has to be shown that the inverse map is a morphism of algebraic groups. To this end, it suffices to show that x ↦ x_{s} and x ↦ x_{u} are morphisms. Since, x_{u} = x_{s}^{−1}x, if the first map is a morphism, the second will also be. Now, if x ∈ G, x_{s} is the diagonal part of x, hence x ↦ x_{s} is a morphism. Furthermore, if G is connected, so are G_{s} and G_{u} since there are morphic images of G. □

8.7 Solvable algebraic groups

For a group G, we denote by [x, y] the commutator xyx^{−1}y^{−1} of two elements x, y ∈ G. If A and B are two subgroups of G we denote by [A, B] the subgroup generated by all commutators [a, b] with a ∈ A, b ∈ B. The identity

(4)  z[x, y]z^{−1} = [zxz^{−1}, zyz^{−1}]

shows that [A, B] is normal in G if both A and B are normal in G. We denote by Z(G) the center of a group G, i.e.

Z(G) = {x ∈ G : xy = yx , ∀y ∈ G}.

Lemma 8.9 a) If the index [G : Z(G)] is finite, then [G, G] is finite.

b) Let A, B be normal subgroups of G, and suppose the set S = {[x, y] : x ∈ A, y ∈ B} is finite. Then [A, B] is finite.

Proof. a) Let n = [G : Z(G)] and let S be the set of all commutators in G. Then S generates [G, G]. For x, y ∈ G, it is clear that [x, y] depends only on the cosets of x, y modulo Z(G); in particular, Card S ≤ n^{2}. Given a product of commutators, any two of them can be made adjacent by suitable conjugation, e.g. [x_{1}, y_{1}][x_{2}, y_{2}][x_{3}, y_{3}] = [x_{1}, y_{1}][x_{3}, y_{3}][z^{−1}x_{2}z, z^{−1}y_{2}z], where z = [x_{3}, y_{3}]. Therefore, it is enough to show that the (n + 1)th power of an element of S is the product of n elements of S, in order to conclude that each element of [G, G] is the product of at most n^{3} factors from S. This in turn will force [G, G] to be finite. Now [x, y]^{n} ∈ Z(G) and so we can write [x, y]^{n+1} = y^{−1}[x, y]^{n}y[x, y] = y^{−1}[x, y]^{n−1}[x, y^{2}]y, and the last expression can be written as a product of n commutators by using identity (4).

b) We can assume that G = AB. Taking into account identity (4), we see that G acts on S by inner automorphisms. If H is the kernel of the resulting morphism from G in the group Sym(S) of permutations of S, then clearly, H is a normal subgroup of finite index in G. Moreover, H centralizes C = [A, B]. It follows that H ∩ C is central in C and of finite index. By a), [C, C] is finite (as well as normal in G, since C ⊲ G). So we can replace G by G/[C, C], i.e. we can assume that C is abelian.

Now the commutators [x, y], x ∈ A, y ∈ C, lie in S and commute with each other. As C is abelian and normal in G, [x, y]^{2} = (xyx^{−1})^{2}y^{−2} = [x, y^{2}] is another such commutator. This clearly forces [A, C] to be finite (as well as normal in G). Replacing G by G/[A, C], we may further assume that A centralizes C. This implies that the square of an arbitrary commutator is again a commutator. It follows that [A, B] is finite. □

Proposition 8.7 Let A, B be closed subgroups of an algebraic group G.

a) If A is connected, then [A, B] is closed and connected. In particular, [G, G] is connected if G is.

b) If A and B are normal in G, then [A, B] is closed (and normal) in G. In particular, [G, G] is always closed.

Proof. a) For each b ∈ B, we can define the morphism φ_{b} : A → G, a ↦ [a, b]. Since A is connected and φ_{b}(e) = e, by proposition 8.2, the group generated by all φ_{b}(A), b ∈ B is closed and connected and this is exactly [A, B].

b) It follows from part a) that [A^{0}, B] and [A, B^{0}] are closed, connected (as well as normal) subgroups of G, so by proposition 8.4 their product C is a closed normal subgroup of G. To show that [A, B] is closed, it therefore suffices to show that C has finite index in [A, B], which is a purely grouptheoretic question. In the abstract group G/C, the image of A^{0} (resp. B^{0}) centralizes the image of B (resp. A). Since the indices [A : A^{0}] and [B : B^{0}] are finite, this implies that there are only finitely many commutators in G/C constructible from the images of A and B. Lemma 8.9 b) then guarantees that [A, B]/C is finite. □

For an abstract group G, we define the derived series D^{i}G inductively by

D^{0}G = G , D^{i+1}G = [D^{i}G, D^{i}G], i ≥ 0.

We say that G is solvable if its derived series terminates in e.

If G is an algebraic group, D^{1}G = [G, G] is a closed normal subgroup of G, connected if G is, by proposition 8.7. By induction the same holds true for all D^{i}G. If G is a connected solvable algebraic group of positive dimension, we have dim[G, G] < dim G.

It is easy to see that an algebraic group G is solvable if and only if there exists a chain of closed subgroups G = G_{0} ⊃ G_{1} ⊃ ⋯ ⊃ G_{n} = e such that G_{i} ⊲ G_{i−1} and G_{i−1}/G_{i} is abelian, for i = 1, …, n.

The following results from group theory are well known.

Proposition 8.8 a) Subgroups and homomorphic images of a solvable group are solvable.

b) If N is a normal solvable group of G for which G/N is solvable, then G itself is solvable.

c) If A and B are normal solvable subgroups of G, so is AB. □

The following lemma is used in the characterization of Liouville extensions.

Lemma 8.10 Let G be an algebraic group, H a closed subgroup of G. Suppose that H is normal in G and G/H is abelian. Suppose further that the identity component H^{0} of H is solvable. Then the identity component G^{0} of G is solvable.

Proof. We have [G, G] ⊂ H, whence [G^{0}, G^{0}] ⊂ H. By proposition 8.7, [G^{0}, G^{0}] is connected. Hence [G^{0}, G^{0}] ⊂ H^{0}. By hypothesis H^{0} is solvable, whence [G^{0}, G^{0}] is solvable and then G^{0} is solvable. □

Example 8.1 We consider the groups T(n, C) and U(n, C). We know by corollary 8.2 that they are connected. We shall now see that they are solvable. Write T = T(n, C), U = U(n, C). First, since the diagonal entries in the product of two upper triangular matrices are just the respective products of diagonal entries it is clear that [T, T] ⊂ U. Now we know that U is generated by the subgroups U_{ij} with i < j, each of them isomorphic to 𝔾_{a} (see the proof of corollary 8.2). By proposition 8.7, we have that [D, U_{ij}] ⊂ U_{ij} is closed and connected, and clearly this group is nontrivial. Then U_{ij} ⊂ [D, U_{ij}] ⊂ [T, T]. We have then proved [T, T] = U.

Now we want to prove that U is solvable. This will imply that T is solvable as well. Let us denote by T the full set of upper triangular matrices viewed as a ring. The subset N of matrices with 0 diagonal is a 2-sided ideal of T . So each ideal power N^{h} is again a two-sided ideal. For an element u ∈ U, such that u = 1 + a, with a ∈ N, we have (1 + a)^{−1} = 1 − a + a^{2} − a^{3} + ⋯ + (−1)^{n−1}a^{n−1}. If we set U_{h} = 1 + N^{h}, we obtain [U_{h}, U_{l}] ⊂ U_{h+l}. In particular, U is solvable.

The next theorem establishes that the connected solvable subgroups of GL(n, C) are exactly the conjugate subgroups of T(n, C).

Theorem 8.3 (Lie-Kolchin) Let G be a connected solvable subgroup of GL(n, C), n ≥ 1. Then G is triangularizable.

Proof. Let V = C^{n}. Let us assume first that G is reducible, i.e. that V admits a nontrivial invariant subspace W. If a basis of W is extended to a basis of V, the matrices representing G have the form

( \begin{matrix}φ(x) & ∗ \\ 0 & ψ(x) \end{matrix} ) .

The morphism x ↦ φ(x) is a morphism of algebraic groups. As G is connected, φ(G) ⊂ GL(W) is also connected as well as solvable (proposition 8.8 a)). By induction on n, φ(G) can be triangularized. Analogously, we obtain that ψ(G) can be triangularized as well. We then obtain the triangularization for G itself. We may then assume that G is irreducible.

By proposition 8.7, the commutator subgroup [G, G] of G is connected, so by induction on the length of the derived series, we can assume that [G, G] is in triangular form.

Let V_{1} be the subspace of V generated by all common eigenvectors of [G, G]. We have V_{1} ≠ 0, since the triangular form of [G, G] yields at least one common eigenvector. Now, for each x ∈ G, y ∈ [G, G], we have x^{−1}yx ∈ [G, G], hence for each v ∈ V_{1}, (x^{−1}yx)(v) = λv, for some λ ∈ C, equivalently y(xv) = λxv. So, V_{1} is G-stable. Since G is irreducible, V_{1} = W, which means that [G, G] is in diagonal form.

Now, any element in [G, G] is a diagonal matrix. Its conjugates in G are again in [G, G], hence also diagonal. The only possible conjugates are then obtained by permuting the eigenvalues. Hence each element in [G, G] has a finite conjugacy class. By proposition 8.1c), [G, G] lies in the center of G.

Assume that there is a matrix y ∈ [G, G] which is not a scalar. Let λ be an eigenvalue of y, and W the corresponding eigenspace. Since y commute with all elements in G, W is G-invariant, hence W = V, y = λ · 1.

Since [G, G] is the commutator subgroup of G, its elements have determinant 1. Hence the diagonal entries must be n-th roots of unity. There are only a finite number of these, so [G, G] is finite. But by proposition 8.7, [G, G] is connected, then [G, G] = 1, which means that G is commutative. The result then follows from lemma 8.8. □

8.8 Characters and semi-invariants

Definition 8.2 A character of an algebraic group G is a morphism of algebraic groups G → 𝔾_{m}.

For example, the determinant defines a character of GL(n, C). If χ_{1}, χ_{2} are characters of an algebraic group G, so is their product defined by (χ_{1}χ_{2})(g) = χ_{1}(g)χ_{2}(g). This product gives the set X(G) of all characters of G the structure of a commutative group.

Examples.

1. A morphism χ : 𝔾_{a} → 𝔾_{m} would be given by a polynomial χ(x) satisfying χ(x + y) = χ(x)χ(y). We obtain then X(𝔾_{a}) = 1.

2. Given a character χ of SL(n, C), by composition with the morphism 𝔾_{a} → SL(n, C), x ↦ I + xe_{ij}, where we denote by e_{ij} the matrix with entry 1 in the position (i, j) and 0’s elsewhere, we obtain a character of 𝔾_{a}. As the subgroups U_{ij} = {I + xe_{ij} : x ∈ C} generate SL(n, C), we obtain X(SL(n, C)) = 1.

3. A character of 𝔾_{m} is defined by x ↦ x^{n}, for some n ∈ ℤ, hence X(𝔾_{m}) ≃ ℤ. As D(n, C) ≃ 𝔾_{m} × ⋯ × 𝔾_{m}, we obtain X(D(n, C)) ≃ ℤ × ⋯ × ℤ.

If G is a closed subgroup of GL(V), for each χ ∈ X(G), we define V_{χ} = {v ∈ V : g.v = χ(g)v for all g ∈ G}. Evidently V_{χ} is a G-stable subspace of V. Any nonzero element of V_{χ} is called a semi-invariant of G of weight χ. Conversely if v is any nonzero vector which spans a G-stable line in V, then it is clear that g.v = χ(g)v defines a character χ of G.

More generally, if φ : G → GL(V) is a rational representation, then the semi-invariants of G are by definition those of φ(G).

Lemma 8.11 Let φ : G → GL(V) be a rational representation. Then the subspaces V_{χ}, χ ∈ X(G), are in direct sum; in particular, only finitely many of them are nonzero.

Proof. Otherwise, we could choose minimal n ≥ 2 and nonzero vectors v_{i} ∈ V_{χ_{i}}, for distinct χ_{i}, 1 ≤ i ≤ n, such that v_{1} + ⋯ + v_{n} = 0. Since the χ_{i} are distinct, χ_{1}(g) ≠ χ_{2}(g) for some g ∈ G. But 0 = φ(g)(∑ v_{i}) = ∑ χ_{i}(g)v_{i}, so ∑ χ_{1}(g)^{−1}χ_{i}(g)v_{i} = 0. The coefficient of v_{2} is different from 1; so we can subtract this equation from the equation ∑ v_{i} = 0 to obtain a nontrivial dependence involving ≤ n − 1 characters, contradicting the choice of n. □

We assume now that H is a closed normal subgroup of G and consider the spaces V_{χ} for χ ∈ X(H). We claim that each element of φ(G) maps V_{χ} in some V_{χ′}. To prove this claim, we can assume that G ⊂ GL(V). If g ∈ G, h ∈ H, v ∈ V_{χ}, then h.(g.v) = (hg).v = g(g^{−1}hg).v = g.(χ(g^{−1}hg).v) = χ(g^{−1}hg)g.v and the function h ↦ χ(g^{−1}hg) is clearly a character χ′ of H, so g maps V_{χ} into V_{χ′}.

8.9 Quotients

The aim of this section is to prove that if G is an algebraic group and H a closed normal subgroup of G, then the quotient G/H has the natural structure of an algebraic group, with coordinate ring C[G/H] ≃ C[G]^{H}.

If V is a finite dimensional C-vector space, then GL(V) acts on exterior powers of V by g.(v_{1} ∧ ⋯ ∧ v_{k}) = g.v_{1} ∧ ⋯ ∧ g.v_{k}. If M is a d-dimensional subspace of V, it is especially useful to look at the action on L = ∧^{d}M, which is a 1-dimensional subspace of ∧^{d}V.

Lemma 8.12 For g ∈ GL(V), we have (∧^{d}g)(L) = L if and only if gM = M.

Proof. The ”if” part is clear. For the other implication, we can choose a basis v_{1}, …, v_{n} in V such that v_{1}, …, v_{d} is a basis of M, and, for some l ≥ 0, v_{l+1}, …, v_{l+d} is a basis of gM. By hypothesis (∧^{d}g)(v_{1} ∧⋯∧v_{d}) is a multiple of v_{1} ∧ ⋯ ∧ v_{d} but, on the other hand, it is a multiple of v_{l+1} ∧ ⋯ ∧ v_{l+d} forcing l = 0. □

Proposition 8.9 Let G be an algebraic group, H a closed subgroup of G. Then there is a rational representation φ : G → GL(V) and a 1-dimensional subspace L of V such that H = {g ∈ G : φ(g)L = L}

Proof. Let I be the ideal in C[G] vanishing on H. It is a finitely generated ideal. By lemma 8.3, there exists a finite dimensional subspace W of C[G], stable under all ρ_{g}, g ∈ G, which contains a given finite generating set of I. Set M = W ∩ I, so M generates I. Notice that M is stable under all ρ_{g}, g ∈ H, since by lemma 8.2, H = {g ∈ G : ρ_{g}I = I}. We claim that H = {g ∈ G : ρ_{g}M = M}. Assume that we have ρ_{g}M = M. As M generates I, we have ρ_{g}I = I, hence g ∈ H.

Now take V = ∧^{d}W, L = ∧^{d}M, for d = dim M. By lemma 8.12, we have the desired characterization of H. □

Theorem 8.4 Let G be an algebraic group, H a closed normal subgroup of G. Then there is a rational representation ψ : G → GL(W) such that H = Ker ψ.

Proof. By proposition 8.9, there exists a morphism φ : G → GL(V) and a line L such that H = {g ∈ G : φ(g)L = L}. Since each element in H acts on L by scalar multiplication, this action has an associated character χ_{0} : H → C. Consider the sum in V of all nonzero V_{χ} for all characters χ of H. By lemma 8.11, this sum is direct and of course includes L. Moreover, we saw in the last paragraph in section 8.8 that φ(G) permutes the various V_{χ} since H is normal in G. So we can assume that V itself is the sum of the V_{χ}.

Now let W be the subspace of End V consisting of those endomorphisms which stabilize each V_{χ}, χ ∈ X(H). There is a natural isomorphism W ≃ ⊕ End V_{χ}. Now GL(V) acts on End V by conjugation. Notice that the subgroup φ(G) stabilizes W, since φ(G) permutes the V_{χ} and W stabilizes each of them. We then obtain a group morphism ψ : G → GL(W) given by ψ(g)(h) = φ(g)_{|W} h φ(g)_{|W}^{−1}; so ψ is a rational representation. Let us check now H = Ker ψ. If g ∈ H, then φ(g) acts as a scalar on each V_{χ}, so conjugating by φ(g) has no effect on W, hence g ∈ Ker ψ. Conversely, let g ∈ G, ψ(g) = I. This means that φ(g) stabilizes each V_{χ} and commutes with End V_{χ}. But the center of End V_{χ} is the set of scalars, so φ(g) acts on each V_{χ} as a scalar. In particular, φ(g) stabilizes L ⊂ V_{χ_{0}}, forcing g ∈ H. □

Corollary 8.3 The quotient G/H can be given a structure of linear algebraic group endowed with an epimorphism π : G → G/H.

Proof. We consider the representation ψ : G → GL(W) with kernel H given by theorem 8.4 and its image Y = Im ψ. By theorem 7.2, Y is a constructible set and, as it is a subgroup of GL(W), by proposition 8.3, it is a closed subgroup of GL(W). We have a group isomorphism G/H ≃ Y, hence we can translate the linear algebraic group structure of Y to G/H. Moreover ψ induces an epimorphism of algebraic groups π : G → G/H. □

Definition 8.3 Let G be an algebraic group, H a closed subgroup of G. A Chevalley quotient of G by H is a variety X together with a surjective morphism π : G → X such that the fibers of π are exactly the cosets of H in G.

In corollary 8.3, we have established that there exists a Chevalley quotient of an algebraic group G by a closed normal subgroup H. However it is not clear if Chevalley quotients are unique up to isomorphism.

Definition 8.4 Let G be an algebraic group, H a closed subgroup of G. A categorical quotient of G by H is a variety X together with a morphism π : G → X that is constant on all cosets of H in G with the following universal property: given any other variety Y and a morphism φ : G → Y that is constant on all cosets of H in G there is a unique morphism \overline{φ} : X → Y such that φ = \overline{φ} ∘ π.

It is clear that categorical quotients are unique up to unique isomorphism. Our aim is to prove that Chevalley quotients are categorical quotients. We then will have a quotient of G by H defined uniquely up to isomorphism and satisfying the universal property.

Theorem 8.5 Chevalley quotients are categorical quotients.

Proof. First we construct a categorical quotient in the category of geometric spaces. Define G/H to be the set of cosets of H in G. Let π : G → G/H be the map defined by x ↦ xH. Give G/H the structure of topological space by defining U ⊂ G/H to be open if and only if π^{−1}(U) is open in G. Next define a sheaf O = O_{G/H} of C-valued functions on G/H as follows: if U ⊂ G/H is open, then O(U) is the ring of functions f on U such that f ∘ π is regular on π^{−1}(U) (this defines indeed a sheaf of functions). In order to check the universal property, let ψ : G → Y be a morphism of geometric spaces constant on the cosets of H in G. We get the induced map of sets \overline{ψ} : G/H → Y, xH ↦ ψ(x), satisfying clearly ψ = \overline{ψ} ∘ π. We prove that \overline{ψ} is a morphism of geometric spaces. To check continuity, take an open subset V ⊂ Y and note that U := \overline{ψ}^{−1}(V) is open in G/H, by the definition of the topology in G/H and the continuity of ψ. Finally, for f ∈ O_{Y} (U), \overline{ψ}^{∗}(f) ∈ O_{G/H}, because π^{∗}(\overline{ψ}^{∗}(f)) ∈ O_{G}(ψ^{−1}(V)).

Now we take (G/H, π) as above and let (X, ψ) be a Chevalley quotient. Using the universal property established above, we get a unique Gequivariant morphism \overline{ψ} : G/H → X such that ψ = \overline{ψ} ∘ π. We will prove that \overline{ψ} is an isomorphism of geometric spaces, which will imply that G/H is a variety and that X is a categorical quotient.

By lemma 8.5, we can assume that G is a connected algebraic group. First of all, it is clear that \overline{ψ} is a continuous bijection. If U ⊂ G/H is open, then \overline{ψ}(U) = ψ(π^{−1}(U)) and by proposition 8.6 a), it follows that \overline{ψ}(U) is open, which implies that \overline{ψ} is a homeomorphism.

In order to prove that \overline{ψ} is an isomorphism, the following has to be established: If U is a principal open set in X, the homomorphism of C-algebras O_{X}(U) → O_{G/H}(\overline{ψ}^{−1}(U)) defined by \overline{ψ}^{∗} is an isomorphism. By definition of O_{G/H} this means that, for any regular function f on V = ψ^{−1}(U) such that f(gh) = f(g), ∀g ∈ V, h ∈ H, there is a unique regular function F on U such that F (ψ(g)) = f(g). Let Γ = {(g, f(g)) : g ∈ V } ⊂ V × 𝔸^{1} be the graph of f and put Γ′ = (ψ, Id)(Γ), so Γ′ ⊂ U × 𝔸^{1}. Since Γ is closed in V × 𝔸^{1}, proposition 8.6 a) shows that (ψ, Id)(V × 𝔸^{1} \ Γ) = U × 𝔸^{1} \ Γ is open in U × 𝔸^{1}. Hence Γ′ is closed in U × 𝔸^{1}. Let λ : Γ′ → U be the morphism induced by the projection on the first component. It follows from the definition that λ is bijective and birational. By Zariski’s main theorem 7.3, λ is an isomorphism. This implies that there exists a regular function F on U such that Γ′ = {(u, F (u)) : u ∈ U}, which is what we wanted to prove. This finishes the proof of the theorem. □

We recall that the action of G on itself by translation on the left gives an action of G on its coordinate ring C[G] defined by λ_{g}(f)(g′) = f(g^{−1}g′) for f ∈ C[G], g, g′ ∈ G (see section 8.4).

Proposition 8.10 Let G be an algebraic group, H a closed normal subgroup of G. We have C[G/H] ≃ C[G]^{H}.

Proof. We consider the epimorphism π given by corollary 8.3. If f ∈ C[G/H], then f̃ = f ∘ π ∈ C[G]. Moreover, for h ∈ H, g ∈ G, we have λ_{h}(f̃)(g) = f̃(h^{−1}g) = (f ∘ π)(h^{−1}g) = f(π(h^{−1}g)) = f(π(g)) = f̃(g), so λ_{h}(f̃) = f̃ and f̃ ∈ C[G]^{H}.

If f ∈ C[G]^{H}, then f is a morphism G → 𝔸^{1} which is constant on the cosets of H in G. Hence, by the universal property of the quotient G/H established in theorem 8.5, there exists F ∈ C[G/H] such that f = F ∘ π. □

9 Suggestions for further reading

1. In section 2.4, we introduced the ring K[d] of differential operators. To a linear differential equation, we can associate a differential module, i.e. a K[d]-module. The concept of differential module allows to study differential equations in a more intrinsic way. The reader can look at [P-S] chap. 2 for a detailed exposition and at [Mo] and [Ż] for more advanced applications.

2. In his lecture at the 1966 International Congress of Mathematicians [Ko2], E. Kolchin raised two important problems in the Picard-Vessiot theory.

1. Given a linear differential equation L(Y) = 0 over a differential field K, determine its Galois group (direct problem).

2. Given a differential field K, with field of constants C, and a linear algebraic group G defined over C, find a linear differential equation defined over K with Galois group G (inverse problem).

The paper [S] is a very good survey on direct and inverse problems in differential Galois theory.

3. Linear differential equations defined over the field ℂ(T) of rational functions over the field ℂ of complex numbers can be given a more analytic treatment. In this context we can define the singularities of the differential equation as the poles of its coefficients. By considering analytic prolongation of the solutions along paths avoiding singular points, one can define the monodromy group of the equation, which is a subgroup of the Galois group. In the case of equations of Fuchsian type, the Galois group is equal to the Zariski closure of the monodromy group. Some interesting topics in the analytic theory of differential equations are the Riemann-Hilbert problem, Stokes phenomena, hypergeometric equations and their generalizations. The interested reader can consult [Ż] , [Mo] and [P-S], as well as the bibliography given there.

4. In the recent years, Morales and Ramis have used differential Galois theory to obtain non-integrability criteria for Hamiltonian systems, which generalize classical results of Poincaré and Liapunov as well as more recent results of Ziglin. This theory is presented in the monograph [Mo] and is also discussed in [Ż].

5. Some interesting contributions to the theory of differential fields have been made by model theorists. The proof of the existence of a differential closure for a differential field uses methods of model theory in an essential way. The first proof of the existence of an algorithm to determine the Galois group of a linear differential equation is as well model theoretical (see [Hr]). The paper [P] is an interesting survey on the relation between differential algebra and model theory.

10 Bibliography

[Ba] H. Bass et al. eds., Selected works of Ellis Kolchin with commentary, American Mathematical Society, 1999.

[B] A. Borel, Linear Algebraic Groups, Graduate Texts in Mathematics 126, Springer, 1991. [F-J] M.D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete 11, Springer, 1986.

[Hu] K. Hulek, Elementary algebraic geometry, Student Mathematical Library vol. 20, American Mathematical Society, 2003.

[H] J.E. Humphreys, Linear Algebraic Groups, Graduate Texts in Mathematics 21, Springer, 1981.

[Hr] E. Hrushovski, Computing the Galois group of a linear differential equation, in: T.Crespo, Z. Hajto (eds.), Proceedings of the Differential Galois Theory workshop, Banach Center Publications 58, Warszawa 2002, pp. 97-138.

[K] I. Kaplansky, An introduction to differential algebra, Hermann, 1976.

[Kl] A. Kleshchev, Lectures on Algebraic Groups, http://darkwing.uoregon.edu/ klesh/teaching/AGLN.pdf

[Ko1] E.R. Kolchin, On the Galois theory of differential fields, Amer. J. Math. 77 (1955), 868-894; [Ba] pp. 261-287.

[Ko2] E.R. Kolchin, Some problems in differential algebra, Proceedings of the International Congress of Mathematicians, Moscow, 1968, pp.269-276: [Ba] pp. 357-364.

[M] A. R. Magid, Lectures on Differential Galois Theory, University Lecture Series 7, American Mathematical Society, 1997.

[Mo] J.J. Morales Ruiz, Differential Galois Theory and non-integrability of Hamiltonian systems, Progress in Mathematics 179, Birkhäuser, 1999.

[P] B. Poizat, Les corps différentiellement clos, compagnons de route de la théorie del modèles, [Ba] pp. 555-565. [P-S] M. van der Put, M.F. Singer, Galois Theory of Linear Differential Equations, Grundlehren der mathematischen Wissenschaften 328, Springer, 2003.

[S] M.F. Singer, Direct and inverse problems in differential Galois theory,

[Ba] pp. 527-554.

[Sp] T.A. Springer, Linear Algebraic Groups, Progress in Mathematics 9, Birkhäuser, 1998.

[Ż] H. Żoła̧dek, The Monodromy Group, Monografie Matematyczne Instytut Matematyczny PAN 67, Birkhäuser, 2006.