ON A GENERIC INVERSE DIFFERENTIAL GALOIS PROBLEM FOR GL_{n}

LOURDES JUAN

Abstract. In this paper we construct a generic Picard-Vessiot extension for the general linear groups. In the case when the differential base field has finite transcendence degree over its field of constants we provide necessary and sufficient conditions for solving the inverse differential Galois problem for this groups via specialization from our generic extension.

Introduction

Le F be a differential field of characteristic zero with algebraically closed field of constants C. In this paper we give an affirmative answer, for the group GL_{n}(C), to the following

Generic Inverse Differential Galois Problem: For a connected algebraic group G over C find a generic Picard-Vessiot extension of F with differential Galois group G.

By generic extension we mean a Picard-Vessiot extension of a generic field that contains F and such that every Picard-Vessiot extension of F for G in the usual sense can be obtained from the generic one by specialization.

We point out that any such specialization will provide a solution to the inverse differential Galois problem in the usual sense, namely, to determine, given F and C as above, and a linear algebraic group G over C, what differential field extensions E ⊃ F are Picard-Vessiot extensions with differential Galois group G and, in particular, whether there are any.

We use the terminology of A. Magid’s book [13]. In [13] the reader may also find definitions and proofs of some results from differential Galois theory that will be recalled.

Our construction of a generic Picard-Vessiot extension of F with group GL_{n}(C) may be summarized as follows: Let F {Y_{ij}} be the ring of differential polynomials over F in the differential indeterminates Y_{ij}, 1 ≤ i, j ≤ n. Our generic base field will be the differential quotient field F ⟨Y_{ij}⟩ of F {Y_{ij}}. Let X_{ij}, 1 ≤ i, j ≤ n, be algebraically independent over F ⟨Y_{ij}⟩. Extend the derivation on F ⟨Y_{ij}⟩ to the polynomial ring F ⟨Y_{ij}⟩[X_{ij}] by letting D(X_{ij}) = ∑_{ℓ=1}^{n} Y_{iℓ}X_{ℓj}. This derivation extends in a natural way to the quotient field F ⟨Y_{ij}⟩(X_{ij}) (note that F ⟨Y_{ij}⟩(X_{ij}) is also the function field of the group obtained from GL_{n}(C) by extending scalars from C to F ⟨Y_{ij}⟩). Then

Date: April 12, 2001.

1991 Mathematics Subject Classification. Primary 12H05; Secondary 12F12, 12Y05, 20G15.

Key words and phrases. Constants, differential field, differential Galois group, differential specialization, inverse differential Galois problem, generic, multivariable division algorithm, new constant, Picard-Vessiot extension, wronskian.

This paper was written while the author held an NSF-funded postoctoral fellowship at the MSRI.

The author wishes to thank the MSRI for its hospitality.

Theorem 1. The differential field extension F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is a generic Picard-Vessiot extension of F with differential Galois group GL_{n}(C).

The main step in proving Theorem 1 is to show that F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is a nonew-constant extension. A new constant in F ⟨Y_{ij}⟩(X_{ij}) must be the quotient of two relatively prime Darboux polynomials, that is, two polynomials p_{1}, p_{2} ∈ F {Y_{ij}}[X_{ij}] that satisfy D(p_{i}) = qp_{i} for some q ∈ F {Y_{ij}}[X_{ij}]. We use Gröbner basis techniques to show that the only Darboux polynomials in F {Y_{ij}}[X_{ij}] are those of the form ℓ det[X_{ij}]^{a}, where ℓ ∈ F and a ∈ ℕ. This implies that there are no two such relatively prime Darboux polynomials and, therefore, no new constants.

To show that the extension is generic we let E ⊇ F be any Picard-Vessiot extension of F with group GL_{n}(C). Then E is isomorphic to F (X_{ij}), the function field of the group obtained from GL_{n}(C) by extending scalars from C to F, as a GL_{n}(C)- module and as an F -module, and there are f_{ij} ∈ F such that the derivation on E is given by D_{E} = ∑_{ℓ=1}^{n} f_{iℓ}X_{ℓj}. In this situation, the Picard-Vessiot extension F (X_{ij}) ⊇ F is obtained from the Picard-Vessiot extension C⟨f_{ij}⟩(X_{ij}) ⊇ C⟨f_{ij}⟩ by extension of scalars from C to F. Therefore, any Picard-Vessiot extension E ⊃ F can be obtained from F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ via the specialization Y_{ij} ↦ f_{ij}. That is, F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is a generic Picard-Vessiot extension of F for GL_{n}(C).

Now, suppose that F has finite transcendence degree over C say,

F = C(t_{1}, …, t_{m})[z_{1}, …, z_{k}],

where the t_{i} are algebraically independent over C and the z_{i} are algebraic over C(t_{1}, …, t_{m}). Consider the differential field F (X_{ij}) with derivation D_{E} as above. Let C denote its field of constants. Let F {Y_{ij}}[X_{ij}] be the differential ring with the derivation defined above. For k ≥ 1 let 𝕋_{k} denote the set of monomials in the both the t_{i} and X_{ij} of total degree less than or equal to k. Fix a term order on the set 𝕋 of monomials in the t_{i} and the X_{ij} and let W_{k}(Y_{ij}) denote the wronskian of 𝕋_{k} relative to that order (the order will only affect the wronskian by a sign). The following theorem summarizes our specialization results:

Theorem 2. F (X_{ij}) ⊃ F is a Picard-Vessiot extension for GL_{n}(C) if and only if all the wronskians W_{k}(Y_{ij}) map to nonzero elements in F (X_{ij}) via the specialization Y_{ij} ↦ f_{ij} ∈ F.

The above condition on the wronskians means that all the sets 𝕋_{k}, for k ≥ 1, are linearly independent over C. This is in turn equivalent to the fact that the set of all the t_{i} and all the X_{ij} are algebraically independent over C. Unfortunately, Theorem 2 gives infinitely many conditions. We do not know at present how to use these conditions to effectively construct solutions to the inverse problem, and this constitutes an interesting open problem.

A specialization as in Theorem 2, however, is known to exist by a result of C. Mitschi and M. Singer [17]. They give a constructive algebraic solution to the inverse problem for all connected linear algebraic groups (and, in particular, for GL_{n}(C)) when F has finite transcendence degree over C. An interesting direction of research in connection with the previous open problem is to try to fully describe all possible solutions (isomorphic and non-isomorphic) that may arise in this situation.

The work of Mitschi and Singer in [17] makes use of the logarithmic derivative and an inductive technique developed by Kovacic [11], [12], to lift a solution to the inverse problem from G/R_{u}, where R_{u} is the unipotent radical of G, to the full group G. Using this machinery Kovacic proved that it is enough to find a solution to the inverse problem for reductive groups (observe that G/R_{u} is reductive). In [19], van der Put explains and partly proves the results in [17].

In the introduction of [17] the authors briefly review previous work on the inverse problem such as results of Bialynicki-Birula in [3], Kovacic [11], [12], Ramis [20], [21], Singer [24], Tretkoff and Tretkoff [26], Beukers and Heckman [2], Katz [10], Duval and Mitschi [7], Mitschi [15], [16], Duval and Loday-Richaud [6], Ulmer and Weil [27] and Singer and Ulmer [25]. A more extensive survey on the inverse problem can be found in M. Singer’s [23].

The constructive algebraic solutions to the inverse differential Galois problem for connected algebraic groups that are currently available are based on Kolchin’s Main Structure Theorem for Picard-Vessiot extensions (see Theorem 1.1.1 below). In particular, a corollary to this theorem (see Theorem 1.1.2) establishes that if E ⊃ F is Picard-Vessiot and G is, for example, unipotent or solvable or G = GL_{n} or G = SL_{n}, then E is isomorphic as an F -module and as a G-module to the function field of the group G_{F} obtained from G by extension of scalars from C to F. Therefore, to get a Picard-Vessiot extension E ⊃ F with group G (if it exists) one can begin by taking E to be the function field of G_{F} and then the problem reduces to extending the derivation from F to E in such a way that E ⊃ F is Picard-Vessiot for that derivation. In this paper we use this approach for our construction.

The idea of tackling the inverse problem by constructing generic extensions is inpired by the works of E. Noether [18] for the Galois theory of algebraic equations. Following her approach, L. Goldman in [8] introduced the notion of a generic differential equation with group G. Goldman explicitly constructed a generic equation with group G for some groups. However, after specializing Goldman’s equation the group of the new equation obtained is a subgroup of the original group. In order to solve the inverse problem by this means, we need to keep the original group as the group of the equation after specialization. Goldman’s generic equation for GL_{n} is equivalent to Magid’s general equation of order n (Example 5.26 in [13]).

More work in the spirit of Goldman’s generic equation came some years later in J. Miller’s dissertation [14]. He defined the notion of hilbertian differential field and gave a sufficient condition for the generic equation with group G to specialize to an equation over such a field with group G as well. However, as pointed out by Mitschi and Singer in [17], his condition was stronger than the analogous one for algebraic equations and this made the theory especially difficult to apply for those groups that were not already known to be Galois groups.

This paper contains the results of the author’s Ph.D. dissertation [9]. I wish to thank my Ph.D. advisor Andy Magid for the many valuable research meetings that we had. I am also grateful to Michael Singer for many enlightening conversations on the inverse problem.

1. Preliminaries

1.1. Notation and some basic results from Differential Galois Theory. We fix a differential field F with algebraically closed field of constants C. If E ⊇ F is a differential field extension then the group of differential automorphisms of E over F is denoted by G(E/F).

If G is a linear algebraic group over C and K is an overfield of C we denote by G_{K} the group obtained from G by extending scalars from C to K.

We will show that the differential field F ⟨Y_{ij}⟩(X_{ij}) to be defined in 1.2 is a Picard-Vessiot extension of F ⟨Y_{ij}⟩ with differential Galois group GL_{n}(C). Note that F ⟨Y_{ij}⟩(X_{ij}) is the function field of G_{K} with G = GL_{n}(C) and K = F ⟨Y_{ij}⟩. The following two results provide the rationale for choosing such function field for the generic Picard-Vessiot extension. The proofs can be found in [13] (Theorem 5.12 and Corollary 5.29 respectively).

Theorem 1.1.1 (Kolchin Structure Theorem). Let E ⊇ F be a Picard-Vessiot extension, let G ≤ G(E/F) be a Zariski closed subgroup and let T be the set of all f in E that satisfy a linear homogeneous differential equation over K = E^{G}. Then T is a finitely generated G-stable differential K-algebra with function field E, and if \overline{K} denotes the algebraic closure of K, then there is a G-algebra isomorphism

\overline{K} ⊗_{K} T → \overline{K} ⊗_{C} C[G].

Note that C[G] denotes the affine coordinate ring of G and that the target of the above isomorphism is the affine coordinate ring of the group G\overline{K} obtained from G by extension of scalars from C to \overline{K}.

Theorem 1.1.2. Let E ⊇ F be a Picard-Vessiot extension, let G ≤ G(E/F) be a Zariski closed subgroup with E^{G} = F. Let \overline{F} be an algebraic closure of F, and suppose the Galois cohomology H^{1}(\overline{F} /F, G(\overline{F})) is a singleton. Let T (E/F) be the set of all f in E that satisfy a linear homogeneous differential equation over F. Then there are F - and G-isomorphisms T (E/F) → F [G_{F}] and E → F (G_{F}). In particular, this holds if G is unipotent or solvable, or if G = GL_{n}(C) or if G = SL_{n}.

We will use the following characterization of Picard-Vessiot extension given in [13] (Proposition 3.9):

Theorem 1.1.3. Let E ⊇ F be a differential field extension. Then E is a Picard-Vessiot extension if and only if:

1. E = F ⟨V ⟩, where V ⊂ E is a finite-dimensional vector space over C;

2. There is a group G of differential automorphisms of E with G(V) ⊇ V and E^{G} = F;

3. E ⊃ F has no new constants.

In particular, if the above conditions hold and if {y_{1}, …, y_{n}} is a C-basis of V, then E is a Picard-Vessiot extension of F for the linear homogeneous differential operator

L(Y) = \frac{w(Y, y_{1}, …, y_{n})}{w(y_{1}, …, y_{n})}

where w(−) denotes the wronskian determinant and L^{−1}(0) = V.

In our case the base field is F ⟨Y_{ij}⟩ and G = GL_{n}(C). We want to show that F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is Picard-Vessiot. We first show (Corollary 1.3.6) that the field of constants of F ⟨Y_{ij}⟩ is C. Then, conditions 1. and 2. in Theorem 1.1.3 are easily verified with V the C-span of the X_{ij} and G = GL_{n}(C). Therefore, our main goal henceforth will be to show that F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is a no-new-constant extesion.

1.2. The differential fields F⟨Y_{ij}⟩(X_{ij}) and F(X_{ij}). Let Y_{ij}, 1 ≤ i, j ≤ n, be differential indeterminates over F. For convenience, denote the k-th derivative D^{(k)}(Y_{ij}) by Y_{ij,k}, for k ≥ 0, so that D(Y_{ij,k}) = Y_{ij,k+1}, k ≥ 0. As usual, D^{(0)}(Y_{ij}) = Y_{ij,0} represents the original element Y_{ij}. In this situation we will omit the k-subindex and write Y_{ij} instead of Y_{ij,0}.

Let F {Y_{ij}} be the ring of differential polynomials in the Y_{ij} and F ⟨Y_{ij}⟩ its differential quotient field. By that we mean the usual quotient field endowed with the natural derivation:

D(\frac{p}{q}) = \frac{D(p)q − pD(q)}{q^{2}}.

for p, q ∈ F {Y_{ij}}, where D is the derivation on F {Y_{ij}}.

Next let X_{ij}, 1 ≤ i, j ≤ n, be algebraically independent over F ⟨Y_{ij}⟩. We consider the differential ring R = F {Y_{ij}}[X_{ij}] with derivation extending the derivation on F {Y_{ij}} by a formula

D(X_{ij}) = ∑_{ℓ=1}^{n} Y_{iℓ}X_{ℓj}.

As above, this derivation extends to the quotient field

Q = F ⟨Y_{ij}⟩(X_{ij})

in a natural way.

Henceforth we fix the differential field Q as defined above.

Likewise we will regard the polynomial ring F [X_{ij}] as a differential ring with derivation extending the derivation on F by a formula

D(X_{ij}) = ∑_{ℓ=1}^{n} f_{iℓ}X_{ℓj}

with f_{ij} ∈ F. Then, as before, we extend this derivation to the quotient field F (X_{ij}) in a natural way.

The multinomial notation a_{α}X^{α} will be used to denote a term of the form a_{α_{11}···α_{kℓ}}X_{11}^{α_{11}} ⋯ X_{kℓ}^{α_{kℓ}}.

The ring F [X_{ij}] is assumed to be ordered with the degree reverse lexicographical order (degrevlex). That is, the set

𝕋^{n^{2}} = {X^{β} | X = (X_{ij}), β = (β_{ij}) ∈ ℕ^{n^{2}}}

of the power products in the X_{ij} is ordered by X_{11} > ⋯ > X_{1n} > ⋯ > X_{n1} > ⋯ > X_{nn}, and

X^{α} < X^{β} ⟺ \begin{cases}∑_{j=1}^{n} ∑_{i=1}^{n} α_{ij} < ∑_{j=1}^{n} ∑_{i=1}^{n} β_{ij} \\ or \\ ∑_{j=1}^{n} ∑_{i=1}^{n} α_{ij} = ∑_{j=1}^{n} ∑_{i=1}^{n} β_{ij}, and the first coordinates \\ α_{ij}, β_{ij} from the right which are different satisfy α_{ij} > β_{ij}. \end{cases}

Henceforth the leading power product of a polynomial in F [X_{ij}] is assumed to be with respect to this order.

The above definitions as well as the Multivariable Division Algorithm that will be used in Remark 1.4.14, can be found in [1].

1.3. Darboux polynomials and the constants of F⟨Y_{ij}⟩(X_{ij}). We need to show that the field of constants C of Q = F ⟨Y_{ij}⟩(X_{ij}) coincides with the field of constants C of F. In this section we will show (Corollary 1.3.4) that this can be reduced to proving that the only Darboux polynomials in R are, up to a scalar multiple in F, powers of det[X_{ij}].

Definition 1.3.1. Let D be a derivation on the polynomial ring A = k[Y_{1}, …, Y_{s}]. A polyomial p ∈ A is called a Darboux polynomial if there is a polynomial q ∈ A such that D(p) = qp. That is, p divides D(p).

Darboux polynomials correspond to generators of principal differential ideals in A. Chapter I of J.A. Weil’s Thesis [28] is devoted to constants and Darboux polynomials in Differential Algebra.

The following basic proposition (proven in [28] for A as in Definition 1.3.1) characterizes new constants for the extension Q ⊃ F in terms of Darboux polynomials:

Proposition 1.3.2. Let p_{1}, p_{2} ∈ R = F {Y_{ij}}[X_{ij}], p_{1}, p_{2} ≠ 0, be relatively prime. Then D(\frac{p_{1}}{p_{2}} ) = 0, if and only if p_{1} and p_{2} are Darboux polynomials. Moreover, if q_{1}, q_{2} ∈ R are such that D(p_{1}) = q_{1}p_{1} and D(p_{2}) = q_{2}p_{2}, then q_{1} = q_{2}.

Proof. For the necessity of the condition we have

D (\frac{p_{1}}{p_{2}} ) = \frac{D(p_{1})p_{2} − p_{1}D(p_{2})}{p_{2}^{2}} = 0,

thus D(p_{1})p_{2} − p_{1}D(p_{2}) = 0, that is

D(p_{1})p_{2} = p_{1}D(p_{2}). (1)

Since p_{1} and p_{2} are relatively prime, the last equation implies that p_{1} divides D(p_{1}) and p_{2} divides D(p_{2}).

Now, let q_{1}, q_{2} ∈ R be such that D(p_{1}) = q_{1}p_{1} and D(p_{2}) = q_{2}p_{2}, respectively. Then it follows from (1) that

q_{1}p_{1}p_{2} = q_{2}p_{1}p_{2}.

Hence, q_{1} = q_{2}.

The proof of the converse is obvious.

Proposition 1.3.3. Let p = ℓ det[X_{ij}]^{a}, with ℓ ∈ F. Then p is a Darboux polynomial in R with q = \frac{ℓ′}{ℓ} + a ∑_{i=1}^{n} Y_{ii}.

Proof. We have

D(p) = ℓ′ det[X_{ij}]^{a} + ℓ(a det[X_{ij}]^{a−1}D(det[X_{ij}])).

And,

D(det[X_{ij}]) = ( ∑_{i=1}^{n} Y_{ii}) det[X_{ij}].

Thus,

D(p) = ℓ′ det[X_{ij}]^{a} + ℓ(a det[X_{ij}]^{a−1}( ∑_{i=1}^{n} Y_{ii}) det[X_{ij}]

= (\frac{ℓ′}{ℓ} + a ∑_{i=1}^{n} Y_{ii}) ℓ det[X_{ij}]^{a}.

That is,

p = ℓ det[X_{ij}]^{a}

is a Darboux polynomial in R with

q = \frac{ℓ′}{ℓ} + a ∑_{i=1}^{n} Y_{ii}.

Corollary 1.3.4. Suppose that if p ∈ R is a Darboux polynomial then p = ℓ det[X_{ij}]^{a}, with ℓ ∈ F, a ∈ ℕ and q = \frac{ℓ′}{ℓ} +a ∑_{i=1}^{n} Y_{ii}. Then F ⟨Y_{ij}⟩(X_{ij}) ⊃ F is a no-new-constant extension.

Proof. F ⟨Y_{ij}⟩(X_{ij}) is the fraction field of F {Y_{ij}}[X_{ij}]. Thus, if an element f ∈ F ⟨Y_{ij}⟩(X_{ij}) satisfies D(f) = 0 and f ∉ F then, by proposition 1.3.2, f = p_{1}/p_{2} with p_{1}, p_{2} ∈ R relatively prime Darboux polynomials. If the hypothesis is true, this contradicts the fact that p_{1} and p_{2} are relatively prime.

Next we show that the constants of F ⟨Y_{ij}⟩ coincide with the constants of F. For simplicity, if h(Y) ∈ F {Y_{ij}}, we will use the notation h′(Y) for D(h(Y)). Notice that this is not the usual meaning h′(Y) = ∑ h′_{α}Y^{α}.

Proposition 1.3.5. If h(Y) ∈ F {Y_{ij}} satisfies h′(Y) = g(Y)h(Y) for some g(Y) ∈ F {Y_{ij}} then h(Y) ∈ F. That is, there are no non-trivial Darboux polynomials in F {Y_{ij}}.

Proof. Suppose that the hypothesis of the proposition is true. According to our notation D^{(k)}(Y_{ij}) = Y_{ij,k}. Consider the set of subindices {ij, k}, i, j, k ∈ ℕ, ordered with the lexicographical order. That is, {i_{1}j_{1}, k_{1}} > {i_{2}j_{2}, k_{2}} if and only if the first coordinates s_{1} and s_{2} from the left, for s = i, j, k above, which are different satisfy s_{1} > s_{2}.

Let {mn, t} be the largest subindex such that Y_{mn,t} occurs in h(Y).

Let h(Y) = ∑_{α} a_{α}Y_{11}^{α_{11}} ⋯ Y_{mn,t}^{α_{mn,t}}. Then

h′(Y) = ∑_{α} a′_{α}Y_{11}^{α_{11}} ⋯ Y_{mn,t}^{α_{mn,t}} + ∑_{α} a_{α}α_{11}Y_{11}^{α_{11}−1}Y_{11,1}^{α_{11,1}+1} ⋯ Y_{mn,t}^{α_{mn,t}}

+ ⋯ + ∑_{α} a_{α}α_{mn,t}Y_{11}^{α_{11}} ⋯ Y_{mn,t}^{α_{mn,t}−1}Y_{mn,t+1}

= h_{1}(Y_{11}, ⋯, Y_{mn,t}) + ( ∑_{α} a_{α}α_{mn,t}Y_{11}^{α_{11}} ⋯ Y_{mn,t}^{α_{mn,t}−1})Y_{mn,t+1}
= g(Y)h(Y).

For Y_{mn,t+1} = D(Y_{mn,t}) we have {mn, t + 1} > {mn, t}, thus it does not occur in h(Y) by the choice of {mn, t}. Also, it does not occur in h_{1}(Y_{11}, ⋯, Y_{mn,t}). Thus the above equation implies that Y_{mn,t+1} must occur in g(Y). Let g_{t+1}(Y) be its coefficient in g(Y) and let

h_{2}(Y) = ∑_{α} a_{α}α_{mn,t}Y_{11}^{α_{11}} ⋯ Y_{mn,t}^{α_{mn,t}−1}.

Then we have

h(Y)g_{t+1}(Y)Y_{mn,t+1} = h_{2}(Y)Y_{mn,t+1}

or

h(Y)g_{t+1}(Y) = h_{2}(Y).

But the total degree of h_{2}(Y) is strictly less than the total degree of h(Y). This forces h(Y) ∈ F.

Corollary 1.3.6. The field of constants of F ⟨Y_{ij}⟩ coincides with C, the field of constants of F.

Proof. This is a consequence of Propositions 1.3.2 and 1.3.5.

1.4. Darboux polynomials in R = F{Y_{ij}}[X_{ij}]. Proposition 1.3.3 shows that scalar multiples of det[X_{ij}] and its powers are Darboux polynomials in R. Corollary 1.3.4 implies that if these are the only Darboux polynomials in R then we are done since consequently there will be no new constants in Q. In this section we will show that that is the case, namely, the only Darboux polynomials in R are those of the form p = ℓ det[X_{ij}]^{a}, with ℓ ∈ F, a ∈ ℕ and q = \frac{ℓ′}{ℓ} + a ∑_{i=1}^{n} Y_{ii}.

Remarks. 1.4.1 (Derivative of a power product in the X_{ij}). Let

X^{α} = X_{11}^{α_{11}} ⋯ X_{1n}^{α_{1n}} ⋯ X_{n1}^{α_{n1}} ⋯ X_{nn}^{α_{nn}},

then

D(X^{α}) = ( ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii}) X^{α}

+ ∑_{i=1}^{n} ∑_{j=1}^{n} ( ∑_{ℓ>i} α_{ij}Y_{iℓ}X_{11}^{a_{11}} ⋯ X_{ij}^{α_{ij}−1} ⋯ X_{ℓj}^{α_{ℓj}+1} ⋯ X_{nn}^{α_{nn}}

+ ∑_{ℓ<i} α_{ij}Y_{iℓ}X_{11}^{a_{11}} ⋯ X_{ℓj}^{α_{ℓj}+1} ⋯ X_{ij}^{α_{ij}−1} ⋯ X_{nn}^{α_{nn}}).

1.4.2. Given X^{α} as in 1.4.1, we need to know which power products in the X_{ij} contain in their derivatives a Y -multiple of X^{α}. That is, we need to find the power products X^{β} such that D(X^{β}) contains an expression of the form Y_{rt}X^{α}. By Remark 1.4.1 such power products are

X^{α_{rs,t}} = \begin{cases}X_{11}^{α_{11}} ⋯ X_{rs}^{α_{rs}+1} ⋯ X_{ts}^{α_{ts}−1} ⋯ X_{nn}^{α_{nn}} & if r < t \\ X_{11}^{α_{11}} ⋯ X_{ts}^{α_{ts}−1} ⋯ X_{rs}^{α_{rs}+1} ⋯ X_{nn}^{α_{nn}} & if r > t \end{cases}

for 1 ≤ r, s ≤ n, t ≠ r, and X^{α} itself.

1.4.3. Let p ∈ R. Since D(X_{ij}) = ∑_{ℓ=1}^{n} Y_{iℓ}X_{ℓj}, then the total degree of p with respect to the X_{ij} does not change after differentiation. Therefore, if D(p) = qp then q ∈ F {Y_{ij}}.

Proposition 1.4.4. Let p ∈ R. Write it as p = ∑_{α} p_{α}(Y)X^{α}, with p_{α}(Y) ∈ F {Y_{ij}}. Then for any α with p_{α}(Y) ≠ 0, the coefficient of X^{α} in D(p) is

p′_{α}(Y) + p_{α}(Y) ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii} + ∑_{i=1}^{n} ∑_{j=1}^{n}(α_{ij} + 1) ∑_{ℓ≠i} p_{α_{ij,ℓ}}(Y)Y_{iℓ},

where α_{ij,ℓ} is the exponent vector of the power product

X^{α_{ij,ℓ}} = \begin{cases}X_{11}^{α_{11}} ⋯ X_{ij}^{α_{ij}+1} ⋯ X_{ℓj}^{α_{ℓj}−1} ⋯ X_{nn}^{α_{nn}} & if i < ℓ \\ X_{11}^{α_{11}} ⋯ X_{ℓj}^{α_{ℓj}−1} ⋯ X_{ij}^{α_{ij}+1} ⋯ X_{nn}^{α_{nn}} & if ℓ > i \end{cases}

as in Remark 1.4.2.

Proof. This is a direct consequence of Remarks 1.4.1 and 1.4.2.

Proposition 1.4.5. Let p ∈ R and suppose that D(p) = qp, for some q ∈ F {Y_{ij}}. Then p ∈ F [X_{ij}].

Proof. Let p = ∑_{α} p_{α}(Y)X^{α}. Then

D(p) = ∑_{α} p′_{α}(Y)X^{α} + p_{α}(Y)D(X^{α})
= qp
= ∑_{α} q(Y)p_{α}(Y)X^{α}.

By Proposition 1.4.4, for each α with p_{α}(Y) ≠ 0 the corresponding coefficient of X^{α} in D(p) is

D(p)_{α} = p′_{α}(Y) + p_{α}(Y) ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii}

+ ∑_{i=1}^{n} ∑_{j=1}^{n}(α_{ij} + 1) ∑_{ℓ≠i} p_{α_{ij,ℓ}}(Y)Y_{iℓ}.

Since D(p) = qp, it must be D(p)_{α} = q(Y)p_{α}(Y) or, equivalently,

q(Y)p_{α}(Y) = p′_{α}(Y) + p_{α}(Y) ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii}

+ ∑_{i=1}^{n} ∑_{j=1}^{n}(α_{ij} + 1) ∑_{ℓ≠i} p_{α_{ij,ℓ}}(Y)Y_{iℓ}.

This means that for each α, the coefficient p_{α}(Y) of X^{α} in p divides the expression

p′_{α}(Y) + ∑_{i=1}^{n} ∑_{j=1}^{n}(α_{ij} + 1) ∑_{ℓ≠i} p_{α_{ij,ℓ}}(Y)Y_{iℓ}.

Thus, for each α, there is u_{α}(Y) such that

p_{α}(Y)u_{α}(Y) = p′_{α}(Y) + ∑_{i=1}^{n} ∑_{j=1}^{n}(α_{ij} + 1) ∑_{ℓ≠i} p_{α_{ij,ℓ}}(Y)Y_{iℓ}.

As in the proof of Proposition 1.3.5, order the triples {ij, k}, i, j, k ∈ ℕ, with the lexicographical order. Let {mn, t} be the largest subindex such that Y_{mn,t} occurs in p. We have D(Y_{mn,t}) = Y_{mn,t+1} and {mn, t + 1} > {mn, t}.

Now, for each α such that Y_{mn,t} occurs in p_{α}(Y) we have that Y_{mn,t+1} will occur in p′_{α}(Y) but not in p_{α}(Y) or in

∑_{i=1}^{n} ∑_{j=1}^{n}(α_{ij} + 1) ∑_{ℓ≠i} p_{α_{ij,ℓ}}(Y)Y_{iℓ}

by the choice of {mn, t}. Therefore, it must occur in p_{α}(Y)u_{α}(Y). Let

p_{α}(Y) = ∑ a_{β}Y_{11}^{β_{11}}Y_{12}^{β_{12}} ⋯ Y_{mn,t}^{β_{mn,t}}

then

p′_{α}(Y) = ∑ a′_{β}Y_{11}^{β_{11}} ⋯ Y_{mn,t}^{β_{mn,t}}
+ ∑ a_{β} β_{11} Y_{11}^{β_{11}−1}Y_{11,1}^{β_{11,1}+1} ⋯ Y_{mn,t}^{β_{mn,t}} + …
+ ∑ a_{β} β_{mn,t} Y_{11}^{β_{11}} ⋯ Y_{mn,t}^{β_{mn,t}−1}Y_{mn,t+1}.

So Y_{mn,t+1} occurs in p′_{α}(Y) only in

∑ a_{β} β_{mn,t} Y_{11}^{β_{11}} ⋯ Y_{mn,t}^{β_{mn,t}−1}Y_{mn,t+1}
= ( ∑ a_{β} β_{mn,t} Y_{11}^{β_{11}} ⋯ Y_{mn,t}^{β_{mn,t}−1})Y_{mn,t+1}
= v(Y)Y_{mn,t+1}.

Since Y_{mn,t+1} occurs in p_{α}(Y)u_{α}(Y) and not in p_{α}(Y) it must occur in u_{α}(Y). Let u_{α,t+1}(Y) be the coefficient of Y_{mn,t+1} in u_{α}(Y). Then it has to be

p_{α}(Y)u_{α,t+1}(Y)Y_{mn,t+1} = v(Y)Y_{mn,t+1}.

The above equation implies that p_{α}(Y) divides v(Y). But this is impossible since the total degree of v(Y) is strictly less than the total degree of p_{α}(Y). This contradiction yields the result.

Lemma 1.4.6. Let p ∈ F [X_{ij}] and suppose that there is q ∈ F {Y_{ij}} such that D(p) = qp. Then q is a linear polynomial in the Y_{ij}. If β = (β_{ij}) is sucht that X^{β} occurs in p, then for 1 ≤ i ≤ n the coefficient of Y_{ii} in q is ∑_{j=1}^{n} β_{ij}. In particular, the sums ∑_{j=1}^{n} β_{ij}, for 1 ≤ i ≤ n, are independent of the choice of X^{β}.

Proof. We have p = ∑ a_{β}X^{β}, with a_{β} ∈ F.

Thus,

D(p) = ∑ a′_{β}X^{β} + a_{β}D(X^{β})
= qp
= ∑ q(Y)a_{β}X^{β}.

By Proposition 1.4.4, the coefficient of X^{β} in D(p) is

a′_{β} + a_{β} ∑_{i=1}^{n} ∑_{j=1}^{n} β_{ij}Y_{ii} + ∑_{i=1}^{n} ∑_{j=1}^{n}(β_{ij} + 1) ∑_{ℓ≠i} a_{β_{ij,ℓ}}Y_{iℓ}.

Hence, it must be

q(Y)a_{β} = a′_{β} + a_{β}( ∑_{i=1}^{n} ∑_{j=1}^{n} β_{ij}Y_{ii} + ∑_{i=1}^{n} ∑_{j=1}^{n}(β_{ij} + 1) ∑_{ℓ≠i} a_{β_{ij,ℓ}}Y_{iℓ}).

From this,

q(Y) = \frac{a′_{β}}{a_{β}} + ∑_{i=1}^{n} ∑_{j=1}^{n} β_{ij}Y_{ii} + ∑_{i=1}^{n} ∑_{j=1}^{n}(β_{ij} + 1) ∑_{ℓ≠i} \frac{a_{β_{ij,ℓ}}}{a_{β}} Y_{iℓ}.

The coefficient of Y_{ii} in the above expression is ∑_{j=1}^{n} β_{ij}, for 1 ≤ i ≤ n. Since this expression for q is valid for any index β, the “in particular” part follows immediately.

Corollary 1.4.7. Let p be as in Lemma 1.4.6. Let X^{α} be the leading power product of p. Let X^{β} be any power product with non-zero coefficient in p. Then ∑_{j=1}^{n} β_{ij} = ∑_{j=1}^{n} α_{ij}, for 1 ≤ i ≤ n. Thus p is homogeneous of degree ∑_{j=1}^{n} ∑_{i=1}^{n} α_{ij}.

Proof. This is an immediate consequence of the “in particular” part in Lemma 1.4.6.

Corollary 1.4.8. Let p ∈ F [X_{ij}] and suppose that D(p) = qp, for some q ∈ F {Y_{ij}}. Let X^{α} be the leading power product of p, and let ℓ ∈ F be its coefficient. Then

q = \frac{ℓ′}{ℓ} + ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii}.

Proof. By Proposition 1.4.4 and since D(p) = qp, the coefficient of X^{α} in D(p) is

ℓq = ℓ′ + ℓ ( ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii} + ∑_{i=1}^{n} ∑_{j=1}^{n}(α_{ij} + 1) ∑_{ℓ≠i} p_{α_{ij,ℓ}}Y_{iℓ}). (1)

The p_{α_{ij,k}} are the coefficients of the power products X^{α_{ij,k}} inp, with α_{ij,k} ≠ α, such that D(X^{α_{ij,k}}) contains an expression of the form Y_{st}X^{α}. By Remark 1.4.2, these power products are

X^{α_{rs,t}} = \begin{cases}X_{11}^{α_{11}} ⋯ X_{rs}^{α_{rs}+1} ⋯ X_{ts}^{α_{ts}−1} ⋯ X_{nn}^{α_{nn}} & if r < t \\ X_{11}^{α_{11}} ⋯ X_{ts}^{α_{ts}−1} ⋯ X_{rs}^{α_{rs}+1} ⋯ X_{nn}^{α_{nn}} & if r > t \end{cases} ,

all of which violate Corollary 1.4.7 for i = r and i = t. Therefore it must be p_{α_{ij,k}} = 0, for all 1 ≤ i, j ≤ n; k ≠ i. But now, substituting back in (1), we see that

ℓq = ℓ′ + ℓ ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii}.

Hence,

q = \frac{ℓ′}{ℓ} + ∑_{i=1}^{n} ∑_{j=1}^{n} α_{ij}Y_{ii}.

Our next step in order to show that the Darboux polynomials p ∈ R have the desired form will be to show that such a p is not reduced with respect to det[X_{ij}]. For that we will show that the leading power product of p is a power of the leading power product of det[X_{ij}]. First, we have

Lemma 1.4.9. Let p ∈ F [X_{ij}] be such that D(p) = qp, q ∈ F {Y_{ij}}. Let X^{α} be its leading power product. Then α_{ij} = 0 for j ≠ n − i + 1 and α_{i,n−i+1} > 0, 1 ≤ i ≤ n. That is, X^{α} = X_{1n}^{α_{1n}}X_{2,n−1}^{α_{2,n−1}} ⋯ X_{n1}^{α_{n1}}.

Proof. To prove that α_{ij} = 0 for j ≠ n − i + 1 we first show that α_{ij} = 0 for j > n − k + 1, i ≥ k, 2 ≤ k ≤ n. Indeed, for k = 2 we have j > n − 1, so j = n and

D(X^{α}) = α_{nn} ∑_{k=1}^{n−1} Y_{nk}X_{11}^{α_{11}} ⋯ X_{kn}^{α_{kn}+1} ⋯ X_{nn}^{α_{nn}−1} + …

Since q has no Y_{ij} with i ≠ j, each term in D(X^{α}) containing such a Y_{ij} must be cancelled. In particular we need to cancel the terms containing

Y_{nj}X_{11}^{α_{11}} ⋯ X_{jn}^{α_{jn}+1} ⋯ X_{nn}^{α_{nn}−1}

for 1 ≤ j ≤ n − 1 above. For that we can only use the derivatives of power products of the form

X^{α_{nl,j}} =
X_{11}^{α_{11}} ⋯ X_{jℓ}^{α_{jℓ}−1} ⋯ X_{jn}^{α_{jn}+1} ⋯ X_{n1}^{α_{n1}} ⋯ X_{nℓ}^{α_{nℓ}+1} ⋯ X_{nn}^{α_{nn}−1}, ℓ < n.

But these are all strictly greater than X^{α} (the leading power product of p), and they may not occur in p. As a consequence, it has to be α_{nn} = 0. Now let k > 2 be such that α_{in} = 0 for i ≥ k. Then

X^{α} =
X_{11}^{α_{11}} ⋯ X_{k−1,n}^{α_{k−1,n}} ⋯ X_{k,n−1}^{α_{k,n−1}}X_{k+1,1}^{α_{k+1,1}} ⋯ X_{k+1,n−1}^{α_{k+1,n−1}} ⋯ X_{n,n−1}^{α_{n,n−1}}

and

D(X^{α}) = α_{k−1,n}( ∑_{i<k−1} Y_{k−1,i}X_{11}^{α_{11}} ⋯ X_{in}^{α_{in}+1} ⋯ X_{k−1,n}^{α_{k−1,n}−1} ⋯ X_{n,n−1}^{α_{n,n−1}}

+ ∑_{i>k−1} Y_{k−1,i}X_{11}^{α_{11}} ⋯ X_{k−1,n}^{α_{k−1,n}−1} ⋯ X_{in}^{α_{in}+1} ⋯ X_{n,n−1}^{α_{n,n−1}}) + …

Likewise, we need to cancel all the terms in D(X^{α}) that contain Y_{k−1,i}, for i ≠ k−1. In particular, we need to cancel

Y_{k−1,i}X_{11}^{α_{11}} ⋯ X_{in}^{α_{in}+1} ⋯ X_{k−1,n}^{α_{k−1,n}−1} ⋯ X_{n,n−1}^{α_{n,n−1}},

for i < k − 1. For that we can only use the power products of the form

X^{α_{k−1,ℓ,i}} =
X_{11}^{α_{11}} ⋯ X_{iℓ}^{α_{iℓ}−1} ⋯ X_{in}^{α_{in}+1} ⋯ X_{k−1,ℓ}^{α_{k−1,ℓ}+1} ⋯ X_{k−1,n}^{α_{k−1,n}−1} ⋯ X_{n,n−1}^{α_{n,n−1}},

for i < k − 1.

But all of them are strictly greater than X^{α} and cannot occur in p. Thus, it has to be α_{k−1,n} = 0. Since this argument is valid for any k > 2, it follows that α_{kn} = 0, for 2 ≤ k ≤ n. This makes the statement that α_{ij} = 0 for j > n − k + 1, i ≥ k, true for k = 2.

Now assume that k is such that α_{ij} = 0 for j > n − k + 1, i ≥ k. So

X^{α} =
X_{11}^{α_{11}} ⋯ X_{1n}^{α_{1n}} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}}X_{k+1,1}^{α_{k+1,1}} ⋯ X_{k+1,n−k+1}^{α_{k+1,n−k+1}} ⋯ X_{n,n−k+1}^{α_{n,n−k+1}}

and for i > k

α_{i,n−k+1}Y_{ij}X_{11}^{α_{11}} ⋯ X_{1n}^{α_{1n}} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}+1} ⋯ X_{i,n−k+1}^{α_{i,n−k+1}−1} ⋯ X_{n,n−k+1}^{α_{n,n−k+1}}

occurs in D(X^{α}). Thus we need to cancel it. For that we can only use the derivatives of power products of the form

X^{α_{ij,k}} =
X_{11}^{α_{11}} ⋯ X_{kj}^{α_{kj}−1} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}+1} ⋯ X_{ij}^{α_{ij}+1} ⋯ X_{i,n−k+1}^{α_{i,n−k+1}−1} ⋯ X_{n,n−k+1}^{α_{n,n−k+1}}

with j < n − k + 1 since α_{kj} = 0 for all j > n − k + 1 by hypothesis. But all such power products are strictly greater than X^{α} and therefore they cannot occur in p. This forces α_{i,n−k+1} = 0 for i > k. We can repeat this process until k = n and get α_{ij} = 0 for all j > n − k + 1, i ≥ k, 2 ≤ k ≤ n, that is,

X^{α} =
X_{11}^{α_{11}} ⋯ X_{1n}^{α_{1n}}X_{21}^{α_{21}} ⋯ X_{2,n−1}^{α_{2,n−1}}X_{31}^{α_{31}} ⋯ X_{n−1,2}^{α_{n−1,2}}X_{n1}^{α_{n1}}.

Now we show that α_{ij} = 0 for j < n − k + 1, 1 ≤ k ≤ n − 1, i ≤ k. The process is analogous to what we just did. First we show that α_{i1} = 0 for i < n. Indeed, for each i we have for ℓ > i that

α_{i1}Y_{iℓ}X_{11}^{α_{11}} ⋯ X_{i1}^{α_{i1}−1} ⋯ X_{ℓ1}^{α_{ℓ1}+1} ⋯ X_{n1}^{α_{n1}}

occurs in D(X^{α}). So, in order to cancel it, we need to use the derivatives of power products of the form

X^{α_{ij,ℓ}} =
X_{11}^{α_{11}} ⋯ X_{i1}^{α_{i1}−1} ⋯ X_{ij}^{α_{ij}+1} ⋯ X_{ℓ1}^{α_{ℓ1}+1} ⋯ X_{ℓj}^{α_{ℓj}−1} ⋯ X_{n1}^{α_{n1}}

with j > 1, all of which are strictly greater than X^{α} if ℓ < n, and for ℓ = n we cannot simply have one of those since α_{nj} = 0 for j ≠ 1. Thus such power products cannot occur in p and it has to be α_{i1} = 0 for i < n.

Let k ≤ n − 1 be such that α_{ij} = 0 for j < n − k + 1, i ≤ k. We have

X^{α} =
X_{1,n−k+1}^{α_{1,n−k+1}} ⋯ X_{1n}^{α_{1n}} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}} ⋯ X_{n1}^{α_{n1}}

and for all i < k, ℓ > i, we have that

α_{i,n−k+1}Y_{iℓ}X_{1,n−k+1}^{α_{1,n−k+1}} ⋯ X_{i,n−k+1}^{α_{i,n−k+1}−1} ⋯ X_{ℓ,n−k+1}^{α_{ℓ,n−k+1}+1} ⋯ X_{n1}^{α_{n1}}

occurs in D(X^{α}) and in order to cancel it we only have the derivatives of power products of the form

X^{α_{ij,ℓ}} =
X_{1,n−k+1}^{α_{1,n−k+1}} ⋯ X_{i,n−k+1}^{α_{i,n−k+1}−1} ⋯ X_{ij}^{α_{ij}+1} ⋯ X_{ℓ,n−k+1}^{α_{ℓ,n−k+1}+1} ⋯ X_{ℓj}^{α_{ℓj}−1} ⋯ X_{n1}^{α_{n1}}

with j > n − k + 1 since α_{ij} = 0 for i ≤ k, j < n − k + 1.

For ℓ < k, all these power products are strictly greater than X^{α} and therefore they cannot occur in p. For ℓ ≥ k we cannot simply have such power products since for ℓ ≥ k, α_{ℓj} = 0 if j > n − k + 1. Thus it has to be α_{i,n−k+1} = 0 for i ≤ k − 1.

We can repeat this process until k = n − 1 and get α_{ij} = 0, j < n − k + 1, i ≤ k, 1 ≤ k ≤ n − 1. This completes the proof of the first part of the lemma.

To prove that α_{i,n−i+1} ≠ 0, for all 1 ≤ i ≤ n, suppose that there is i such that α_{i,n−i+1} = 0 and let j ≠ i be such that α_{j,n−j+1} ≠ 0. Then D(X^{α}) will contain

α_{j,n−j+1}Y_{ji}X_{1n}^{α_{1n}} ⋯ X_{j,n−j+1}^{α_{j,n−j+1}−1} ⋯ X_{i,n−j+1} ⋯ X_{n1}^{α_{n1}} + … if i > j

or

α_{j,n−j+1}Y_{ji}X_{1n}^{α_{1n}} ⋯ X_{i,n−j+1} ⋯ X_{j,n−j+1}^{α_{j,n−j+1}−1} ⋯ X_{n1}^{α_{n1}} + … if i < j.

As noted above, since q does not contain any Y_{ij} with i ≠ j, we need to cancel the terms in D(p) involving either of the above. But that is impossible since α_{ij} = 0 for all j and by Corollary 1.4.7 all the power products

X_{11}^{β_{11}} ⋯ X_{ij}^{β_{ij}} ⋯ X_{nn}^{β_{nn}}

in p must have β_{ij} = 0 for j = 1, …, n. In particular, we cannot have in p power products of the form X^{α_{j,n−j+1,i}} as in Remark 1.4.2

Next we show that the exponents α_{st} of the X_{st} in X^{α}, the leading power product of p, are all equal:

Lemma 1.4.10. Let p ∈ F [X_{ij}] be such that D(p) = qp, q ∈ F {Y_{ij}}. Let

X^{α} = X_{1n}^{α_{1n}}X_{2,n−1}^{α_{2,n−1}} ⋯ X_{n1}^{α_{n1}}

be its leading power product. Then α_{i,n−i+1} = α_{1n}, for i > 1, that is, if a = α_{1n}, then

X^{α} = (X_{1n}X_{2,n−1} ⋯ X_{n1})^{a}.

Proof. Let ℓ be the coefficient of X^{α} in p. We have

D(ℓX_{1n}^{α_{1n}}X_{2,n−1}^{α_{2,n−1}} ⋯ X_{n1}^{α_{n1}}) =

( ∑_{i=1}^{n} α_{i,n−i+1}ℓY_{ii})X_{1n}^{α_{1n}}X_{2,n−1}^{α_{2,n−1}} ⋯ X_{n1}^{α_{n1}}

+ α_{1n}ℓ ∑_{k≠1} Y_{1k}X_{1n}^{α_{1n}−1} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}} ⋯ X_{kn} ⋯ X_{n1}^{α_{n1}}

+ ℓ ∑_{1<i} α_{i,n−i+1} ∑_{k>i} Y_{ij}X_{1n}^{α_{1n}} ⋯ X_{i,n−i+1}^{α_{i,n−i+1}−1} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}} ⋯ X_{k,n−i+1} ⋯ X_{n1}^{α_{n1}}

+ ℓ ∑_{1<i} α_{i,n−i+1} ∑_{k>i} Y_{ij}X_{1n}^{α_{1n}} ⋯ X_{k,n−i+1} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}} ⋯ X_{i,n−i+1}^{α_{i,n−i+1}−1} ⋯ X_{n1}^{α_{n1}}

+ ℓ′X_{1n}^{α_{1n}}X_{2,n−1}^{α_{2,n−1}} ⋯ X_{n1}^{α_{n1}}.

In order to cancel

α_{1n}ℓY_{1k}X_{1n}^{α_{1n}−1} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}} ⋯ X_{kn} ⋯ X_{n1}^{α_{n1}}, k ≠ 1,

above, we can only use the derivatives of the power product

X^{α_{1,n−k+1,k}} =
X_{1,n−k+1} ⋯ X_{1n}^{α_{1n}−1} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}−1} ⋯ X_{kn} ⋯ X_{n1}^{α_{n1}},

since for j ≠ n − k + 1 we have α_{kj} = 0.

Let a_{α_{1,n−k+1,k}} be the coefficient of X^{α_{1,n−k+1,k}} in p. Then

a_{α_{1,n−k+1,k}} = −ℓα_{1n} (2)

On the other hand, in order to cancel

α_{k,n−k+1}ℓY_{k1}X_{1,n−k+1} ⋯ X_{1n}^{α_{1n}} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}−1} ⋯ X_{n1}^{α_{n1}}, k ≠ 1

above, the only power product that we can use is, again,

X^{α_{kn,1}} = X_{1,n−k+1} ⋯ X_{1n}^{α_{1n}−1} ⋯ X_{k,n−k+1}^{α_{k,n−k+1}−1} ⋯ X_{kn} ⋯ X_{n1}^{α_{n1}}
= X^{α_{1,n−k+1,k}},

since α_{1j} = 0 for j ≠ n. Thus it must be

a_{α_{1,n−k+1,k}} = −ℓα_{k,n−k+1} (3)

as well.

From (2) and (3) it follows that, for k ≠ 1, α_{1n} = α_{k,n−k+1}.

As a consequence of the above results we obtain the following expression for q:

Corollary 1.4.11. Let p ∈ F [X_{ij}] and suppose that D(p) = qp, q ∈ F {Y_{ij}}. Let X^{α} be the leading power product of p. Let a ∈ ℕ be such that

X^{α} = (X_{1n}X_{2,n−1} ⋯ X_{n1})^{a}

and let ℓ ∈ F be the coefficient of X^{α} in p. Then

q = \frac{ℓ′}{ℓ} + a ∑_{i=1}^{n} Y_{ii}.

Proof. This is a consequence of Corollary 1.4.8 and Lemma 1.4.10.

Corollary 1.4.12. Let p be as in Corollary 1.4.11. Then p is homogeneous of degree na.

Proof. This is a consequence of Corollary 1.4.6 and Lemma 1.4.10

Lemma 1.4.10 implies that p is not reduced with respect to det[X_{ij}]. Since this is a key point in the proof of our main result we restate it as the following

Theorem 1.4.13. Let p ∈ F [X_{ij}] be such that D(p) = qp, q ∈ F {Y_{ij}}. Let X^{α} be its leading power product. Then

X^{α} = (X_{1n}X_{2,n−1} ⋯ X_{n1})^{a} = lp(det[X_{ij}])^{a}.

Thus p is not reduced with respect to det[X_{ij}].

Note. If f is a polynomial, lp(f) denotes its leading power product with respect to a given order.

Proof. This is just a restatement of Lemma 1.4.10.

Remark 1.4.14. Let p_{1}, p_{2} ∈ F [X_{ij}] be two polynomials such that lp(p_{1}) = X^{α} = lp(p_{2}). Then we can write p_{1} = f p_{2} + r where f ∈ F and r is reduced with respect to p_{2}. Indeed, since lp(p_{1}) = lp(p_{2}), we have that lp(p_{2}) divides lp(p_{1}). So p_{1} is not reduced with respect to p_{2} . We may apply the Multivariable Division Algorithm to p_{1} and p_{2}, to get f, r ∈ F [X_{ij}], such that p_{1} = f p_{2} + r, with r reduced with respect to p_{2} and lp(p_{1}) = lp(f)lp(p_{2}). The last equation implies that lp(f) = 1. Hence, f ∈ F.

We are now ready to prove our main result on the form of the Darboux polynomials in R:

Theorem 1.4.15. Let p ∈ F [X_{ij}] and q ∈ F {Y_{ij}} be polynomials in R that satisfy the Darboux condition D(p) = qp. Then there is a ∈ ℕ and ℓ ∈ F such that

p = ℓ det[X_{ij}]^{a}

and

q = \frac{ℓ′}{ℓ} + a ∑_{i=1}^{n} Y_{ii}.

Proof. Let q_{1} = ∑_{i=1}^{n} a Y_{ii}, so that,

D(det[X_{ij}]^{a}) = q_{1} det[X_{ij}]^{a} = (q − \frac{ℓ′}{ℓ}) det[X_{ij}]^{a}.

By Remark 1.4.14 we can write p = ℓ det[X_{ij}]^{a} + r, with r reduced with respect to det[X_{ij}]^{a}. Now,

D(p) = D(ℓ det[X_{ij}]^{a}) + D(r)
= ℓ′ det[X_{ij}]^{a} + ℓ(q − \frac{ℓ′}{ℓ}) det[X_{ij}]^{a} + D(r)
= ℓ′ det[X_{ij}]^{a} + qℓ det[X_{ij}]^{a} − ℓ′ det[X_{ij}]^{a} + D(r)
= qℓ det[X_{ij}]^{a} + D(r).

On the other hand, we have

D(p) = qp
= qℓ det[X_{ij}]^{a} + qr.

Therefore, it has to be D(r) = qr. But r is reduced with respect to det[X_{ij}]^{a}. It follows, by Theorem 1.4.13, that r = 0. The statement about the form of q is just the content of Corollary 1.4.11.

Now, for 1 ≤ i, j ≤ n, let D_{E(ij)} ∈ Lie(GL_{n}(C)) be the derivation given by multiplication by the matrix E(ij), with 1 in position (i,j) and zero elsewhere. The set {D_{E(ij)}| 1 ≤ i, j ≤ n} is a basis for Lie(GL_{n}(C)) with respect to which the derivation in Theorem 1.4.15 is expressed. We show that the result does not depend on the basis chosen on Lie(GL_{n}(C)):

Theorem 1.4.16. Let D_{st}, 1 ≤ s, t ≤ n, be any basis of Lie(GL_{n}(C)). Define a derivation in the ring R = F {Y_{ij}}[X_{ij}] by D = ∑ Y_{st}D_{st}. Let p and q be polynomials in R that satisfy the Darboux condition D(p) = qp. Then there is a ∈ ℕ and ℓ ∈ F such that p = ℓ det[X_{ij}]^{a} and q = \frac{ℓ′}{ℓ} + a ∑_{i=1}^{n} Y_{ii}.

Proof. Since {D_{E(ij)}| 1 ≤ i, j ≤ n} is a basis of Lie(GL_{n}(C)) we have

D_{st} = ∑ c_{st,ij}D_{E(ij)},

with c_{st,ij} ∈ C. Thus,

D = ∑_{s,t} Y_{st}D_{st}
= ∑_{s,t} Y_{st} ∑_{i,j} c_{st,ij}D_{E(ij)}
= ∑_{i,j} ∑_{s,t} c_{st,ij}Y_{st}D_{E(ij)}
= ∑_{i,j} Z_{ij}D_{E(ij)},

where Z_{ij} = ∑_{s,t} c_{st,ij}Y_{st}. Now, [c_{st,ij}] is a matrix of change of basis so it is invertible. Also the c_{st,ij} are contants for D, thus the map Z_{ij,k} → Y_{ij,k} is a differential bijection. In other words, the differential rings

R = F {Y_{ij}}[X_{ij}], D

and

R′ = F {Z_{ij}}[X_{ij}], D

are isomorphic and therefore we can apply Theorem 1.4.15 to R′.

2. A Generic Picard-Vessiot Extension for GL_{n}(C)

In this section we prove the statement about the generic Picard-Vessiot extension for F with differential Galois group GL_{n}(C). We provide some specialization properties in the case when F has finite transcendence degree over C.

2.1. The generic extension.

Theorem 2.1.1. F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is a generic Picard-Vessiot extension with differential Galois group GL_{n}(C).

Proof. First we need to show that F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is a Picard-Vessiot extension with differential Galois group GL_{n}(C). We will use the characterization of Theorem 1.1.3. We have

1. F ⟨Y_{ij}⟩(X_{ij}) = F ⟨Y_{ij}⟩⟨V ⟩, where V ⊂ F ⟨Y_{ij}⟩(X_{ij}) is the finite dimensional vector space over C spanned by the X_{ij}.

2. The group G = GL_{n}(C) acts as a group of differential automorphisms of F ⟨Y_{ij}⟩(X_{ij}) with G(V) ⊆ V and F ⟨Y_{ij}⟩(X_{ij})^{G} = F ⟨Y_{ij}⟩. This follows from the fact that F ⟨Y_{ij}⟩(X_{ij}) is the function field of GL_{n}(C)_{F ⟨Y_{ij}⟩}.

3. F ⟨Y_{ij}⟩(X_{ij}) ⊇ F ⟨Y_{ij}⟩ has no new constants. This is a consequence of Proposition 1.3.2, Corollary 1.3.4 and Theorem 1.4.15.

Now, suppose that E ⊇ F is a Picard-Vessiot extension of F with differential Galois group GL_{n}(C). By Theorems 1.1.1 and 1.1.2, we have that in this situation E is isomorphic to F (X_{ij}) (the function field of GL_{n}(C)_{F}) as a GL_{n}(C)-module and as an F -module. Any GL_{n}(C) equivariant derivation D_{E} on F (X_{ij}) extends the derivation on F in such a way that

D_{E}(X_{ij}) = ∑_{ℓ=1}^{n} f_{iℓ}X_{ℓj}

with f_{ij} ∈ F. Since E ⊃ F is a Picard-Vessiot extension for GL_{n}(C), then so is C⟨f_{ij}⟩(X_{ij}) ⊃ C⟨f_{ij}⟩, the derivation on C⟨f_{ij}⟩(X_{ij}) being the corresponding restriction of D_{E}. From this Picard-Vessiot extension one can retrieve F (X_{ij}) ⊃ F by extension of scalars from C to F. In this way, any Picard-Vessiot extension E ⊃ F with differential Galois group GL_{n}(C) can be obtained from F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ via the specialization Y_{ij} ↦ f_{ij}. This means that F ⟨Y_{ij}⟩(X_{ij}) ⊃ F ⟨Y_{ij}⟩ is a generic Picard-Vessiot extension of F for GL_{n}(C).

2.2. Specializing to a Picard-Vessiot extension of F. In this section we give necessary and sufficient conditions for a specialization Y_{ij} ↦ f_{ij}, f_{ij} ∈ F, with C⟨f_{ij}⟩(X_{ij}) ⊃ C⟨f_{ij}⟩ a Picard-Vessiot extension, to exist. We restrict ourselves to the case when F has finite transcendence degree over C.

Our goal is to find f_{ij} ∈ F such that the specialization (homomorphism) from C{Y_{ij}} to F given by Y_{ij} ↦ f_{ij} is such that C⟨f_{ij}⟩(X_{ij}) ⊃ C⟨f_{ij}⟩, with derivation given by D(X_{ij}) = ∑_{ℓ=1}^{n} f_{iℓ}X_{ℓj}, has no new constants. We have:

Theorem 2.2.1. Let F = C(t_{1}, …, t_{m})[z_{1}, …, z_{k}] where the t_{i} are algebraically independent over C and the z_{i} are algebraic over C(t_{1}, …, t_{m}). Assume that the derivation on F has field of constants C and that it extends to F (X_{ij}) so that D(f ⊗ X_{ij}) = D(f) ⊗ X_{ij} + f ⊗ ∑_{ℓ=1}^{n} f_{iℓ}X_{ℓj} on F ⊗ C[X_{ij}]. Let C be the field of constants of F (X_{ij}). Then C = C if and only if the set of all the t_{i} and all the X_{ij} are algebraically independent over C.

Proof. (Sufficiency) Suppose that C properly contains C. Let r be the transcendence degree of C over C. Since C is algebraically closed, r has to be at least one.

We have the tower of fields

C ⊂ C ⊂ C(X_{ij}) ⊂ F (X_{ij})

where the transcendence degree of C ⊂ C(X_{ij}) is n^{2} and the transcendence degree of C ⊂ F (X_{ij}) is n^{2} + m. Since r ≥ 1 the transcendence degree ℓ of C(X_{ij}) ⊂ F (X_{ij}) has to be ℓ < m and therefore there is an algebraic relation among the t_{i} over C(X_{ij}). Let g(X_{ij}), f_{i}(X_{ij}) ∈ C[X_{ij}], g(X_{ij}) ≢ 0, be such that

t^{δ_{s}} + \frac{f_{s−1}(X_{ij})}{g(X_{ij})}t^{δ_{s−1}} + ⋯ + \frac{f_{0}(X_{ij})}{g(X_{ij})} = 0.

Then

g(X_{ij})t^{δ_{s}} + f_{s−1}(X_{ij})t^{δ_{s−1}} + ⋯ + f_{0}(X_{ij}) = 0.

Since the f_{i}(X_{ij}) and g(X_{ij}) are polynomials in the X_{ij} with coefficients in C, the last equation gives an algebraic relation among the t_{i} and the X_{ij} over C.

For the necessity we only need to point out that by construction the set of all the t_{i} and all the X_{ij} are algebraically independent over C.

Now to check whether the set of all the t_{i} and all the X_{ij} are algebraically independent over C, we let 𝕋_{k}, k ≥ 1, denote the set of monomials in both the t_{i} and the X_{ij} of total degree less than or equal to k. Then the set of all the t_{i} and all the X_{ij} are algebraically independent over C if and only if, for each k, the set 𝕋_{k} is linearly independent over C.

Fix a term order on the set 𝕋 of all monomials in both the t_{i} and the X_{ij} and let W_{k} denote the wronskian of the set 𝕋_{k} relative to that order. Then the above condition is equivalent to the fact that W_{k} ≠ 0 for k ≥ 1. Now go back to C{Y_{ij}}[X_{ij}] and extend scalars from C to F. Let W_{k}(Y_{ij}) be the Wronskian of 𝕋_{k} in F ⊗ C{Y_{ij}}[X_{ij}].

Then, the condition of Theorem 2.2.1 for finding a specialization Y_{ij} ↦ f_{ij} so that C⟨f_{ij}⟩(X_{ij}) ⊃ C⟨f_{ij}⟩ has no new constants can be expressed as follows:

Theorem 2.2.2. There is a specialization of the Y_{ij} with no new constants if and only if there are f_{ij} ∈ F such that all the wronskians W_{k}(Y_{ij}), k ≥ 1, map to non-zero elements under Y_{ij} ↦ f_{ij}.

3. Results for connected linear algebraic groups

We do not know at present whether generic Picard-Vessiot extensions exist for arbitrary connected linear algebraic groups. However, the proofs of the specialization theorems in 2.2 can be easily generalized for such groups.

3.1. Specialization results. We point out that the proofs of Theorems 2.2.1 and 2.2.2 do not make any special use of the fact that G = GL_{n}(C).

Let F = C(t_{1}, …, t_{m})[z_{1}, …, z_{k}] where the t_{i} are algebraically independent over C and the z_{i} are algebraic over C(t_{1}, …, t_{m}).

Let Y_{1}, …, Y_{n} be differential indeterminates over F and X_{1}, …, X_{n} algebraically independent over F ⟨Y_{i}⟩.

We consider the group G to be a connected linear algebraic group over C with function field C(G) = C(X_{i}).

If {D_{1}, …, D_{n}} is a basis for Lie(G), let D_{Y} = ∑_{i=1}^{n} Y_{i} D_{i} be a G-equivariant derivation on F ⟨Y_{i}⟩(X_{i}) and D = ∑_{i=1}^{n} f_{i} D_{i}, f_{i} ∈ F, a specialization of D_{Y} to a G-equivariant derivation on F (X_{i}). Let C be the field of constants of F (X_{i}) for this derivation. We have,

Theorem 3.1.1. Let F, C and C be as above. Then C = C if and only if the set of all the t_{i} and the X_{i} are algebraically independent over C.

Now, fix an order in the set 𝕋 of monomials in both the t_{i} and the X_{i} and let W_{k}(Y_{i}) be the wronskian (with respect to this order) of the monomials in both the t_{i} and the X_{i} of degree less than or equal to k computed in F ⊗ C{Y_{i}}[X_{i}]. Then, Theorem 3.1.2. There is a specialization of the Y_{i} with no new constants if and only if there are f_{i} ∈ F such that all the wronskians W_{k}(Y_{i}), k ≥ 1, map to non-zero elements under Y_{i} ↦ f_{i}.

For the proofs of Theorems 3.1.1 and 3.1.2 we only need to replace the X_{ij} with X_{i} , the Y_{ij} with Y_{i} and n^{2} with n in the proofs of Theorems 2.2.1 and 2.2.2.

Observe that the proofs of Theorems 3.1.1 and 3.1.2 do not use the fact that C(X_{i}) is the function field of G. However, this hypothesis is used in the following theorem to show that F (X_{i}) ⊃ F is a Picard-Vessiot extension with group G.

Under the hypothesis and notation of Theorems 3.1.1 and 3.1.2 we have:

Theorem 3.1.3. F (X_{i}) ⊃ F is a Picard-Vessiot extension with Galois group G if and only if the set of all the t_{i} and all the X_{i} are algebraically independent over the field of constants C of F (X_{i}).

Proof. First assume that F (X_{i}) ⊃ F is a Picard-Vessiot extension. Then the field of constants C of F (X_{i}) coincides with C. So we can apply Theorem 3.1.1 and get the result.

Conversely, if the set of all the t_{i} and all the X_{i} are algebraically independent over C, by Theorem 3.1.1, F (X_{i}) ⊃ F is a no new constant extension. On the other hand, F (X_{i}) is obtained from C(X_{i}) by the extension of scalars:

F (X_{i}) = q.f.(F ⊗_{C} C(X_{i}))
= q.f.(F ⊗_{C} C[G])

where C[G] is the coordinate ring of G and G acts on F ⊗_{C} C[G] fixing F. So, G ⊆ G(F (X_{i})/F). Counting dimensions we get that G = G(F (X_{i})/F) since C(X_{i}) = C(G), the function field fo G. Finally, F (X_{i}) = F ⟨V ⟩, where V is the finitedimensional vector space over C spanned by the X_{i}. By Theorem 1.1.3, F (X_{i}) ⊃ F is a Picard-Vessiot extension.

Applying Theorems 3.1.2 and 3.1.3 we also obtain:

Theorem 3.1.4. There is a specialization of the Y_{i} such that F (X_{i}) ⊃ F is a Picard-Vessiot extension if and only if there are f_{i} ∈ F such that all the W_{k}(Y_{i}), k ≥ 1, map to non-zero elements via Y_{i} ↦ f_{i}.

4. Computing new constants

Let C be an algebraically closed field with trivial derivation. Let F = C(t_{1}, …, t_{m})[z_{1}, …, z_{k}] where the t_{i} are algebraically independent over C and the z_{i} are algebraic over C(t_{1}, …, t_{m}). Assume that the derivation on F has field of constants C and that it extends to F (X_{ij}) so that

D(f ⊗ X_{ij}) = D(f) ⊗ X_{ij} + f ⊗ ∑_{ℓ=1}^{n} f_{iℓ}X_{ℓj}

on F ⊗ C[X_{ij}], for certain f_{ij} ∈ F. By Theorem 2.2.1, if there is an algebraic relation among the set of all the t_{i} and all the X_{ij} over the field of constants C of F (X_{ij}) then C properly contains C.

In this section we will produce a new constant from such an algebraic relation. In order to simplify the computations we will assume F = C. So, in particular, the coefficients f_{ij} in the derivation above are constant. In this situation, since the transcendence degree of F over C is zero, if C ⫌ C, the condition of Theorem 2.2.1 means that the X_{ij} are algebraically dependent over C.

We will restrict ourselves to the case n = 2 and use a particular linear dependence relation.

Extend the derivation on F to F (X_{11}, X_{12}, X_{21}, X_{22}) by letting

D(X_{ij}) = ∑_{ℓ=1}^{2} f_{iℓ}X_{ℓj},

where the f_{ij} are such that the wronskian W_{1} = w(X_{11}, X_{12}, X_{21}, X_{22}) = 0. That is, the X_{ij} are linearly dependent over C. Furthermore, assume that the linear relation among the X_{ij} is such that there are β_{12}, β_{21}, β_{22} ∈ C with

X_{11} = β_{12}X_{12} + β_{21}X_{21} + β_{22}X_{22} (1)

and that X_{12}, X_{21} and X_{22} are linearly independent. In order to simplify the computations we will also assume that det[f_{ij}] = 0.

We want to find a, b, c ∈ F such that p = aX_{12} + bX_{21} + cX_{22} is a Darboux polynomial in F [X_{ij}], that is D(aX_{12} + bX_{21} + cX_{22}) = q(aX_{12} + bX_{21} + cX_{22}) for certain q ∈ F.

We have,

D(a X_{12} + b X_{21} + c X_{22})
= a(f_{11}X_{12} + f_{12}X_{22}) + b(f_{21}X_{11} + f_{22}X_{21}) + c(f_{21}X_{12} + f_{22}X_{22})
= bf_{21}X_{11} + (af_{11} + cf_{21})X_{12} + bf_{22}X_{21} + (af_{12} + cf_{22})X_{22}
= bf_{21}(β_{12}X_{12} + β_{21}X_{21} + β_{22}X_{22}) + (af_{11} + cf_{21})X_{12} + bf_{22}X_{21}
+(af_{12} + cf_{22})X_{22}
= (af_{11} + bf_{21}β_{12} + cf_{21})X_{12} + b(f_{22} + f_{21}β_{12})X_{21}
+(af_{12} + bf_{21}β_{22} + cf_{22})X_{22}
= qaX_{12} + qbX_{21} + qcX_{22}.

Therefore,

\begin{aligned}[a(f_{11} − q) + bf_{21}β_{12} + cf_{21} & ]X_{12} + b(f_{22} + f_{21}β_{12} − q)X_{21} \\ & + (af_{12} + bf_{21}β_{22} + c(f_{22} − q)X_{22} = 0. \end{aligned} (2)

Since we are assuming that X_{12}, X_{21} and X_{22} are linearly independent their coefficients in (2) must be equal to zero. So we have the following homogeneous linear system in a, b, c:

(f_{11} − q) a +  f_{21}β_{12} b +  f_{21} c = 0
(f_{22} + f_{21}β_{12} − q) b  = 0
f_{12} a +  f_{21}β_{22} b + (f_{22} − q) c = 0

In order for the above system to have non-trivial solutions we need that

det [ \begin{matrix}f_{11} − q & f_{21}β_{12} & f_{21} \\ 0 & f_{22} + f_{21}β_{12} − q & 0 \\ f_{12} & f_{21}β_{22} & f_{22} − q \end{matrix} ] = 0.

But,

det [ \begin{matrix}f_{11} − q & f_{21}β_{12} & f_{21} \\ 0 & f_{22} + f_{21}β_{12} − q & 0 \\ f_{12} & f_{21}β_{22} & f_{22} − q \end{matrix} ]

= (f_{22} + f_{21}β_{12} − q) det [ \begin{matrix}f_{11} − q & f_{21} \\ f_{12} & f_{22} − q \end{matrix} ]

= (f_{22} + f_{21}β_{12} − q)(det[f_{ij}] − ( ∑_{i=1}^{2} f_{ii})q + q^{2})
= 0.

This gives either f_{22} + f_{21}β_{12} − q = 0 (3) or

det[f_{ij}] − ( ∑_{i=1}^{2} f_{ii})q + q^{2} = 0. (4)

From (3)-(4) we get q = f_{22} + f_{21}β_{12} (5) or

q = \frac{∑_{i=1}^{2} f_{ii} ± \sqrt{( ∑_{i=1}^{2} f_{ii})^{2} − 4 det[f_{ij}]}}{2} (6)

Since we are assuming that det[f_{ij}] = 0, (6) becomes:

q = \begin{cases}∑_{i=1}^{2} f_{ii}, or \\ 0 \end{cases} (7)

Choose q = ∑_{i=1}^{2} f_{ii} and assume that q ≠ 0, q ≠ f_{22} + f_{21}β_{12}. Then the second equation in the system implies that b = 0 and the system becomes:

\begin{matrix}−f_{22} & a & + & f_{21} c = & 0 \\ f_{12} & a & − & f_{11} c = & 0 \end{matrix}

If f_{22} ≠ 0 then the above system has the general solution

a = \frac{f_{21}}{f_{22}} c, where c ∈ C .

In particular, if we take c = 1 then p = \frac{f_{21}}{f_{22}} X_{12} + X_{22} satisfies

D(\frac{f_{21}}{f_{22}} X_{12} + X_{22}) = ( ∑_{i=1}^{2} f_{ii})(\frac{f_{21}}{f_{22}} X_{12} + X_{22}).

On the other hand we also have that

D(det[X_{ij}]) = ( ∑_{i=1}^{2} f_{ii}) det[X_{ij}].

Let

θ = \frac{\frac{f_{21}}{f_{22}} X_{12} + X_{22}}{det[X_{ij}]} .

We have,

D(θ)
= D( \frac{\frac{f_{21}}{f_{22}} X_{12} + X_{22}}{det[X_{ij}]} )

= \frac{D(\frac{f_{21}}{f_{22}} X_{12} + X_{22}) det[X_{ij}] − (\frac{f_{21}}{f_{22}} X_{12} + X_{22})D(det[X_{ij}])}{det[X_{ij}]^{2}}

= \frac{( ∑_{i=1}^{2} f_{ii})(\frac{f_{21}}{f_{22}} X_{12} + X_{22}) det[X_{ij}] − (\frac{f_{21}}{f_{22}} X_{12} + X_{22})( ∑_{i=1}^{2} f_{ii}) det[X_{ij}]}{det[X_{ij}]^{2}}
= 0.

That is, θ is a new constant in F (X_{ij}).

Now we show that under the restrictions that we imposed on the f_{ij} it is possible to find a non-zero f_{22}.

Since we have a linear dependence relation among the X_{ij}, the wronskian W_{1} must be equal to zero. This Wronskian can be expressed, up to a sign, as the following product of determinants:

W_{1} = | \begin{matrix}1 & 0 & 0 & 1 \\ f_{11} & f_{12} & f_{21} & f_{22} \\ A & B & E & F \\ C & D & G & H \end{matrix} | | \begin{matrix}X_{11} & X_{12} & 0 & 0 \\ X_{21} & X_{22} & 0 & 0 \\ 0 & 0 & X_{11} & X_{12} \\ 0 & 0 & X_{21} & X_{22} \end{matrix} | = M(f_{ij}) det[X_{ij}]^{2},

where

A = f′_{11} + f_{11}^{2} + f_{12}f_{21},
B = f′_{12} + f_{11}f_{12} + f_{12}f_{22}
C = f_{11}A + f_{21}B + A′
= 3f_{11}f′_{11} + 2f_{11}f_{12}f_{21} + 2f′_{12}f_{21} + f′′_{11} + f_{12}f′_{21} + f_{11}^{3},
D = f_{12}A + f_{22}B + B′
= 2f′_{11}f_{12} + f_{11}^{2}f_{12} + f_{12}^{2}f_{21} + f_{21}f_{22}^{2} + 2f′_{12}f_{22} + f_{11}f′_{12}
+f_{12}f′_{22} + f′′_{12} + f_{11}f_{12}f_{22},
E = f′_{21} + f_{21}f_{11} + f_{22}f_{21},
F = f′_{22} + f_{12}f_{21} + f_{22}^{2},
G = f_{11}E + f_{21}F + E′
= 2f′_{21}f_{11} + f_{21}f_{11}^{2} + f_{22}f_{21}f_{11} + 2f′_{22}f_{21} + f_{12}f_{21}^{2}
+f_{22}^{2}f_{21} + f′′_{21} + f_{21}f′_{11} + f_{22}f′_{21},
H = f_{22}F + f_{12}E + F′
= f_{21}f_{11}f_{12} + 2f_{22}f_{21}f_{12} + +3f_{22}f′_{22} + 2f_{12}f′_{21} + f′_{12}f_{21}
+f_{22}^{3} + f′′_{22}.

and

M(f_{ij}) = | \begin{matrix}1 & 0 & 0 & 1 \\ f_{11} & f_{12} & f_{21} & f_{22} \\ A & B & E & F \\ C & D & G & H \end{matrix} | .

We have after simplifying using the hypothesis that det[f_{ij}] = 0,

M(f_{ij}) = (f_{22} − f_{11})(f′_{12}f′′_{21} − f′_{21}f′′_{12}) + (f′_{22} − f′_{11})(f′′_{12}f_{21} − f_{12}f′′_{21})
− f′_{12}f′_{21}(f_{11} − f_{22})^{2} − f_{12}f_{21}(f′_{11} − f′_{22})
+ f_{12}f′_{21}(f_{11}f′_{11} + f_{22}f′_{22} − f′_{11}f_{22} − f_{11}f′_{22} + f′′_{22} − f′′_{11} + f_{12}f′_{21} − f′_{12}f_{21})
+ f′_{12}f_{21}(f_{11}f′_{11} + f_{22}f′_{22} − f′_{11}f_{22} − f_{11}f′_{22} + f′′_{11} − f′′_{22} + f′_{12}f_{21} − f_{12}f′_{21}).

Getting the above expression for M(f_{ij}) took long and involved computations. We first computed the determinant directly and then we checked the result using Dogson’s method [5], [22].

The wronskian W_{1} = 0 if and only if M(f_{ij}) = 0. Now, observe that if f_{12} = 0 then f′_{12} = 0 which implies that B = 0 and D = 0 as well. Therefore M(f_{ij}) = 0. So, if we let M(Y_{ij}) be the differential polynomial in the Y_{ij} whose specialization to the f_{ij} is M(f_{ij}) then M(Y_{ij}) is in the differential ideal

I = {det[Y_{ij}], Y_{12}}
= {Y_{11}Y_{22} − Y_{12}Y_{21}, Y_{12}}
= {Y_{11}Y_{22}, Y_{12}}

of C{Y_{11}, Y_{12}, Y_{21}, Y_{22}}. It is easy to see that Y_{22} is not in I. Indeed, suppose that

Y_{22} = p Y_{11}Y_{22} + q Y_{12} + r, (8)

where p, q ∈ C{Y_{11}, Y_{12}, Y_{21}, Y_{22}},

r = ∑_{i,j} [p_{i} (Y_{11}Y_{22})^{(i)} + q_{j} Y_{12}^{(j)}]

with p_{i}, q_{j} ∈ C{Y_{11}, Y_{12}, Y_{21}, Y_{22}}.

Now, consider the map

ψ : C{Y_{11}, Y_{21}, Y_{22}} ⟶ C[Y_{11}, Y_{21}, Y_{22}]

given by ψ(Y_{22}) = Y_{22} and ψ(Y_{ij}) = 0 for i, j ≠ 2. Let \overline{p} = ψ(p), \overline{q} = ψ(q), \overline{r} = ψ(r). We have that \overline{r} = 0 and (8) becomes

Y_{22} = 0.

which is impossible.

References

1. W. W. Adams and Ph. Loustaunau, An Introduction to Gröbner Bases, Graduate Studies in Mathematics, American Mathematical Society (1994).

2. F. Beukers, G. Heckmann, Monodromy for the hypergeometric function _{n}F_{n−1}, Invent. Math. 95 (1989), 325–354.

3. A. Bialynicki-Birula, On the inverse problem of Galois theory of differential fields, Bull. Amer. Math. Soc. 16 (1963), 960–964.

4. K. Boussel, Groupes de Galois des équations hypergéométriques, C. R. Acad. Sci. Paris. 309, I (1989), 587–589.

5. C. L. Dodgson, Condensation of determinants, Proc. Royal Soc. London 15 (1866), 150–155.

6. A. Duval, M. Loday-Richaud, Kovacic’s algorithm and its applications to some families of special functions, AAECC Journal 3 (1992).

7. A. Duval, C. Mitschi, Matrices de Stokes et groupe de Galois des équations hypergéométriques confluentes généralisées, Pacific J. Math. 138 (1989), 25–56.

8. L. Goldman, Specializations and the Picard–Vessiot theory, Trans. Amer. Math. Soc. 85 (1957), 327–356.

9. L. Juan, A Generic Picard–Vessiot extension with group GL_{n} and the inverse differential Galois problem, Ph. D. Thesis, University of Oklahoma (2000).

10. N. Katz, On the calculation of some differential Galois groups, Invent. Math. 87 (1987), 13–61.

11. J. Kovacic, The inverse problem in the Galois theory of differential fields, Ann. of Math. 89 (1969), 583–608.

12. J. Kovacic, On the inverse problem in the Galois theory of differential fields, Ann. of Math. 93 (1971), 269–284.

13. A. Magid, Lectures in Differential Galois Theory, University Lecture Series, American Mathematical Society (1994).

14. J. Miller On Differentially Hilbertian Differential Fields, Ph.D. thesis, Columbia University (1970).

15. C. Mitschi, Groupe de Galois des équations hypergéométriques confluentes généralisées, C. R. Acad. Sci. Paris. 309, I (1989), 217–220.

16. C. Mitschi, Differential Galois Groups of Generalized Hypergeometric Equations: An Approach Using Stokes Multipliers, Pacific J. Math. 176 2, (1996), 365–405.

17. C. Mitschi and M. F. Singer, Connected Linear Groups as Differential Galois Groups, Journal of Algebra. 184 (1996), 333–361.

18. E. Noether, Gleichungen mit vorgeschriebener Gruppen, Math. Ann. 78 (1918), 221–229.

19. M. van der Put, Recent Work on Differential Galois Theory, Séminaire BOURBAKI, 50ème année, (1997-1998), n^{o} 849.

20. J. -P. Ramis, About the solution of some inverse problems in differential Galois theory by Hamburger equations, in Differential Equations, Dynamical Systems, and Control Science, Elworthy, Everett, and Lee, eds., Lecture Notes in Pure and Applied Mathematics. 157 (1994), 277–300.

21. J. -P. Ramis, About the Inverse Problem in Differential Galois Theory: The Differential Abhyankar Conjecture, The Stokes Phenomenon and Hilbert’s 16th Problem; B. I. J. Braaksma, et al., eds. World Scientific, Singapore (1996)

22. D. P. Robins and H. Rumsey, Jr., Determinants and Alternating Sign Matrices, Advances in Mathematics. 62 (1986), 169–184.

23. M. F. Singer, Direct and Inverse Problems in Differential Galois Theory, Selected Works of Ellis Kolchin with Commentary, Bass, Buium, Cassidy eds., American Mathematical Society (1999), 527–554.

24. M. F. Singer, Moduli of linear differential equations on the Riemann sphere with fixed Galois groups, Pacific J. Math. 106, 2 (1993), 343–395.

25. M. F. Singer, F. Ulmer, Liouvillian and algebraic solutions of second and third order linear differential equations, Journal of Symbolic Computation. 16 (1993), pp. 37–74.

26. C. Tretkoff and M. Tretkoff, Solution of the inverse problem of differential Galois theory in the classical case, Amer. J. Math. 101 (1979), 1327–1332.

27. F. Ulmer, J. -A. Weil, Note on Kovacic’s algorithm, J. Symbolic Comput. 28, 2, (1996) 179–200.

28. J. -A. Weil, Constantes et polynômes de Darboux en algèbre différentielle: applications aux systèmes différentiels linéaires, Ph.D. Thesis, École Polytechnique de France (1995).

Mathematical Sciences Research Institute, 1000 Centennial Dr., Berkeley, California 94720, USA

  E-mail address: ljuan@msri.org
  URL: www.msri.org/people/members/ljuan/