Differential subfields of liouvillian extensions

Journal Pre-proof Differential subfields of Liouvillian extensions

Varadharaj R. Srinivasan

PII:

S0021-8693(20)30032-6

DOI:

https://doi.org/10.1016/j.jalgebra.2019.12.023

Reference:

YJABR 17516

To appear in:

Journal of Algebra

Received date:

27 September 2019

Please cite this article as: V.R. Srinivasan, Differential subfields of Liouvillian extensions, J. Algebra (2020), doi: https://doi.org/10.1016/j.jalgebra.2019.12.023.

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Differential Subfields of Liouvillian Extensions Varadharaj R. Srinivasan∗ Department of Mathematical Sciences, IISER Mohali, SAS Nagar, Punjab 140306

January 22, 2020

Abstract For an ordinary differential field k of characteristic zero, we will prove structure theorems for intermediate differential subfields of liouvillian extensions of k. We apply our results to produce large classes of non linear differential equations having no liouvillian solutions that are transcendental over k.

1

Introduction

Let k be a differential field of characteristic zero with a derivation denoted by 0 and Ck := {c ∈ k | c0 = 0} be its field of constants. A differential field extension E = k(t1 , · · · , tn ) of k will be called a generalised liouvillian extension of k if either (i) ti is algebraic over k(t1 , . . . , ti−1 ) or (ii) t0i ∈ k(t1 , . . . , ti−1 ) or (iii) t0i /ti ∈ k(t1 , . . . , ti−1 ) or (iv) t0i = αi ti + βi , where αi and βi are nonzero elements of k(t1 , . . . , ti−1 ). A generalised liouvillian extension E = k(t1 , · · · , tn ) of k, where the elements ti satisfy either (i) or (ii) or (iii) (respectively either (i) or (ii), respectively either (i) or (iii)) is called a liouvillian extension ([5], p.409) of k (respectively a primitive extension extension of k, respectively an exponential extension of k). Let f be a polynomial in n + 1 variables over the differential field k. We say that the differential equation f (y, y (1) , y (2) , · · · , y (n) ) = 0, where y (i) denote the i−th derivative of y, admits a liouvillian solution (or a “closed form” solution) if there is a liouvillian extension E of k such that CE = Ck and that E contains a solution of the differential equation. In order to determine whether or not a given differential equation admits such a solution, it is useful to have a structure theorem for differential ∗

The author is supported by the DST-SERB grant: EMR/2016/001516.

1

subfields of liouvillian extensions. In the main theorem of [10], Singer proved a structure theorem for a special kind of liouvillian field extension called generalized elementary field extensions. Using this structure theorem, Singer generalised and provided purely algebraic proofs for results of Mordukhay-Boltovsky (Corollary 1 of [10]) on the existence of elementary solutions of non linear differential equations. In a subsequent paper [9], Rosenlicht and Singer proved that if E is a generalised elementary extension field of k and K is a differential field intermediate to E and k and that K is algebraically closed in E then K is again a generalised elementary extension of k. Moreover, they also noted that relatively algebraically closed differential subfields of liouvillian extension fields need not be liouvillian1 . In this article we will prove the following structure theorems for differential subfields of liouvillian extensions. Theorem (A). Let k ⊆ K ⊆ E be differential fields, Ck be an algebraically closed field, CE = Ck and K be algebraically closed in E. If E is a liouvillian extension of k then K is a generalised liouvillian extension of k. If E is a primitive extension of k (respectively an exponential extension of k) then K is also a primitive extension of k (respectively an exponential extension of k). Theorem (B). Let k ⊆ K be differential fields, K be finitely generated over k as fields and tr.d(K|k) = 1. Let k(x) be the algebraic closure of k in K and suppose that there is an element in K − k(x) satisfying a linear differential equation of order ≥ 1 over k(x). Then there are elements y, z ∈ K such that (i) K = khx, y, zi, the field generated by elements x, y, z and their derivatives of all order over the field k. (ii) y ∈ K − k(x) is a solution of a linear homogeneous differential equation of order ≤ 2 over k(x) and z is algebraic over the differential field khx, yi. Furthermore, if there is no element in K − k(x) satisfying a first order linear differential equation over k(x) then we can choose y ∈ K − k so that there are a quadratic extension field L ⊆ K of k(x), where K is an algebraic closure of K, and an element y1 ∈ K(L) − L satisfying the following properties: 1. y and y1 are Ck −linearly independent solutions of an (irreducible) differential polynomial L (Y ) = Y 00 + αY 0 + βY, where α, β ∈ k(x). 2. y10 /y1 ∈ L − k(x), (y − y1 )0 /(y − y1 ) ∈ L − k(x) and k(x)hyi = k(x, y, y 0 ) is a quadratic extension of k(x, y). Theorem (A) is an analogue of Singer’s structure theorem on generalised elementary extensions2 , whereas, Theorem (B) not only strengthens Corollary 4 of Singer [11] (see Proposition 3.5) but also serves as an analogue of certain results of Kolchin3,4 and also applies to all intermediate differential 2 R 2 In Remark 2 of [9], the authors considered the differential field E := C(z)(ez , ez ), where C(z) is the field of rational functions one variable over complex numbers with the derivation d/dz. They observed that the equation y 0 = −2xy + 1 R 2 2 has a solution in E, namely, y = ( ez )/ez and proved that the differential field C(z)(y) is not a liouvillian extension of C(z). 2 Compare Theorem (A) with the Theorem of [10] . Also see Remarks 1 and 2 of [9] and Proposition 2.5 of Section 2.2. 3 See [2], Theorem 7 and [3], Theorem 3 on weakly and strongly normal differential field extensions of transcendence degree one. 4 There are differential fields satisfying the hypothesis of Theorem (B) without being weakly normal over k or without having an infinite group of differential automorphisms fixing k element-wise. Therefore Theorem (B) cannot be derived from these results of Kolchin. For further details, see Remark 2.

1

2

subfields, having transcendence degree one over k, of liouvillian extensions of k (see Remark 1). The condition in Theorem (A) that K is algebraically closed in E cannot be relaxed in general (see Example 3.1). We will apply our results and study non linear differential equations. We prove that if f is a rational function in one variable over k, E = k(t1 , · · · , tn ) is a liouvillian extension of k with CE = Ck , tn is transcendental over kn−1 and y 0 = f (y) for some element y ∈ E − kn−1 then f must be a polynomial of degree ≤ 2. We also prove that if C(z) denotes the field of rational functions in one complex variable z with the usual derivation d/dz then every liouvillian solution of the differential equation y 0 = αn y n + · · · + α2 y 2 , where αi ∈ C[z] and αn 6= 0 must be algebraic over C(z). We assume that the reader is familiar with the basic concepts in differential algebra5 . The rest of the article is organised as follows. Necessary preliminaries are developed in Section 2 and we prove our Theorems in Sections 2 and 3. In Section 4, applications of our results are discussed.

2

A Structure Theorem for Liouvillian Extensions

2.1

Notations and Preliminaries

All fields considered in this article are of characteristic zero and all differential fields are ordinary differential fields whose derivation maps are denoted by 0 . When we write “k ⊆ E be differential fields”, we mean that E is a differential field extension of k, that is, E is a field extension of k and the derivation of E restricts to the derivation of k. Elements of Ck are called constants and it is easy to see that Ck is a differential subfield of k. The Wronskian ω(x1 , . . . , xn ) of elements x1 , . . . , xn ∈ k is the determinant of the matrix   x1 x2 ... xn  x01 x02 ... x0n    .. .. ..   ..  . . . .  (n−1)

x1

(n−1)

x2

...

(n−1)

xn

and it can be easily shown that ω(x1 , . . . , xn ) = 0 if and only if x1 , . . . , xn are Ck −linearly independent. Let k ⊆ E be differential fields. For any subset S of E, the smallest differential subfield of E containing S and k will be denoted by khSi and when S = {x1 , · · · , xn } is a finite set, we use the notation khx1 , · · · , xn i instead of khSi. If E = k(t1 , · · · , tn ), the subfield k(t1 , · · · , ti−1 ) will be often denoted by ki , where it is understood that k0 is the field k. Note that when E = k(t1 , · · · , tn ) is a liouvillian extension of k, for each i, ki = ki−1 (ti ) = ki−1 hti i. The group of all field automorphisms of E fixing elements of k will be denoted by Gal(E|k). The subset of all automorphisms σ ∈ Gal(E|k) such that σ(x)0 = σ(x0 ) for all x ∈ E forms a subgroup of Gal(E|k), which will be denoted by DGal(E|k). The symbols [E : k] and tr.d(E|k) will represent the dimension of the k−vector space E and the transcendence degree of field E over k, respectively. Let k[∂] denotes the ring of differential operators over k and L ∈ k[∂] be of order m. If V := {v ∈ k | L (v) = 0} has dimension m as a Ck −vector space then we say that k contains a full set of solutions V of L (Y ) = 0. The differential field E is a Picard-Vessiot extension of k for a monic L ∈ k[∂] if CE = Ck and E = khV i, where V ⊆ E is a full set of all solutions of L (Y ) = 0. We say L ∈ k[∂] of order m is reducible ([7], p.37) if for 5

Standard references are [1], [6] and [7].

3

some p, 1 ≤ p  m there are operators Lm−p , Lp ∈ k[∂] of order m − p and p, respectively, such that L = Lm−p Lp . In this case, Lp is called a right-hand factor of L . In the next theorem and proposition, we gather several standard results from differential algebra and put it in a form convenient for later use. Theorem 2.1. Let k ⊆ M ⊆ E be differential fields with CE = Ck and denote DGal(E|k) by G. Suppose that there are elements w, z ∈ E − k such that w0 ∈ k and z 0 /z ∈ k. Then (i) if there is an element x ∈ E such that x is algebraic over k(w) and that either x0 /x ∈ k or x0 ∈ k then either xs ∈ k for some positive integer s or x + cw ∈ k, for some constant c, respectively. Similarly, if there is an element x ∈ E such that x is algebraic over k(z) and that either x0 ∈ k or x0 /x ∈ k then either x ∈ k or there are integers r, s, not both zero, such that xr z s ∈ k, respectively. (ii) If M ⊆ k(w) then M = k(w) and if M ⊆ k(z) then there is a positive integer s such that M = k(z s ). (iii) w is transcendental over k and for each σ ∈ G we have σ(w) = w + cσ for some cσ ∈ Ck , and thus G stabilises k(w). (iv) The map φ : G → (Ck , +) defined by φ(σ) = cσ is a group homomorphism, and when E = k(w), the field fixed by G equals k and φ is an isomorphism. (v) Either z n ∈ k for some integer n > 0 or z is transcendental over k, and for each σ ∈ G, σ(z) = cσ z for some cσ ∈ Ck − {0} and thus G stabilises k(z). (vi) The map φ : G → (Ck − {0}, ×) defined by φ(σ) = cσ is a group homomorphism. Moreover, if E = k(z) then the field fixed by G equals k, φ is an injective group homomorphism. Furthermore, φ(G) = (Ck − {0}, ×) if z is transcendental over k, and φ(G) is the subgroup of nth roots of unity of (Ck − {0}, ×) if n is the smallest positive integer such that z n ∈ k. Proof. Since these are standard results6 from differential algebra, we shall provide only a sketch of the proof with appropriate references. For a proof of (i) and (ii), we refer the reader to Lemma A1 and A2 of [8]. In fact, from Lemma A1 of [8], it follows that w is transcendental and that z is either transcendental over k or z n ∈ k. Note that for any σ ∈ G, σ(y 0 ) = σ(y)0 for all y ∈ E and therefore (σ(w) − w)0 = 0 and (σ(z)/z)0 = 0. Then σ(w) = w + cσ and σ(z) = dσ z for cσ , dσ ∈ Ck where dσ 6= 0. Thus, from these facts, (iii) and (v) follow. It is easy to see that φ : G → (Ck , +), as defined in (iv), is a group homomorphism. When E = k(w), we observe that the assignment w 7→ w + c for any constant c extends uniquely to a differential automorphism of E. This proves (iv). On the other hand, when z is transcendental, it is easy to see that the assignment z 7→ cz for any nonzero constant c extends uniquely to a differential automorphism of k(z). However, if z n ∈ k then such an assignment extends to a differential automorphism of k(z) only when c is an n−th root of unity and this proves (vi). Proposition 2.2. Let k(t) be a liouvillian extension of k with Ck(t) = Ck , L ∈ k[∂] and L (u) = β for some u ∈ k(t) − k and β ∈ k. (a) If t0 ∈ k then k contains a nonzero solution of L (Y ) = 0. 6

Compare (1)–(4) with [1], Lemmas 3.9 and 3.10.

4

(b) If t0 /t ∈ k then k(t) contains a nonzero solution of L (Y ) = 0 of the form atn for 0 6= a ∈ k and integer n 6= 0. Furthermore, if β 6= 0 then k contains a nonzero solution of L (Y ) = β. (c) If t is algebraic over k and β 6= 0 then k contains a solution of L (Y ) = β and k(t) − k contains a solution of L (Y ) = 0. Proof. Let t0 ∈ k or t0 /t ∈ k. Then k(t) is a Picard-Vessiot extension of k and from Examples 1.18 and 1.19, p.14 of [7], we know that the Picard-Vessiot ring of k(t) is k[t] when t0 ∈ k and k[t, t−1 ] when t0 /t ∈ k. Since L (u) = β implies that u is a solution of the monic homogeneous differential equation β( β1 L (Y ))0 = 0, we apply Corollary 1.38, p.30 of [7], and obtain that u belongs to the Picard-Vessiot ring of k(t). P i 0 0 m Let t0 ∈ k. Then u = m i=0 αi t ∈ k[t], where αm 6= 0 and therefore, u = αm t + terms involving m lower powers of t. Thus β = L (u) = L (am )t + lower powers of t and we obtain L (am ) = 0. Pm αi ti ∈ k[t, t−1 ] and note that since u ∈ k(t) − k, P αj 6= 0 for some Let t0 /t ∈ k. Then u = i=−l P m m i 0 0 i 0 j 6= 0. Now we have u = i=−l βi t , where i=−l (αi + iαi (t /t))t and therefore, β = L (u) = i i βi t = L (αi t ) for all i 6= 0 and β0 = L (α0 ). If t is transcendental then the distinct powers ti are linearly independent over k and it follows that βi = 0 for all i 6= 0. Thus L (αi ti ) P = 0 for all i 6= 0 i and β = L (α0 ). If t is algebraic then we shall choose m  [k[t] : k] and rewrite u = m i=0 αi t , where i αi ∈ k. Now, again from the linear independence of the distinct powers t , we obtain that L (αi ti ) = 0 for all i 6= 0 and β = L (α0 ). Let t be algebraic over k and E be a splitting field of t over k with Galois group G. Then T r(u) := P σ∈G σ(u) belongs to k. Since L (u) = β, we have L (T r(u)) = T r (L (u)) = |G|β. Thus L (α) = β for α := T r(u)/|G| ∈ k and clearly, L (u − α) = 0.

2.2

Structure Theorem

Our approach to the proof of Theorem (A) requires an understanding of irreducible differential equations that admit a liouvillian solution. Let E be a Picard-Vessiot extension of k of an irreducible monic L ∈ k[∂] of order m, V be the full set of all solutions of L (Y ) = 0 in E and suppose that L (Y ) = 0 admits a liouvillian solution. We shall now state various facts, concerning the Picard-Vessiot extension E, whose proofs can be found in [11], pp.70-71. All the solutions of L (Y ) = 0 are liouvillian and therefore E is also a liouvillian extension of k. Moreover, if a nonzero solution of L (Y ) = 0 is algebraic over k then all its solutions are algebraic over k and consequently, E is a finite (algebraic) Galois extension of k. Let G := DGal(E | k) and G0 be the identity component of G. Then the differential field fixed by all the elements of G0 is the algebraic closure k(x) of k in E and it follows that k(x) is a finite algebraic Galois extension of k, whose Galois group is isomorphic to the finite group G/G0 . For every nonzero y ∈ V there are differential automorphisms σ2 , . . . , σm ∈ G such that y, σ2 (y), . . . , σm (y) forms a Ck −basis of V . There is a nonzero solution y1 ∈ V such that y10 /y1 ∈ k(x). Since σ(y1 )0 /σ(y1 ) ∈ k(x) for all σ ∈ G, the Picard-Vessiot extension E has the following structure: E = k(x)(y1 , . . . , ym ), where k(x) is the algebraic closure of k in E and y1 , . . . , ym is a Ck −basis of V such that yi0 /yi ∈ k(x) for each i. The algebraic group G0 is isomorphic to ×r Gm (Ck ), where r = tr.d(E | k(x)). Thus in particular, one can find elements x1 , . . . , xr ∈ E, algebraically independent over k(x), such that E = k(x)(x1 , . . . , xr ) and x0i /xi ∈ k(x) ([6], p.72, Example 5.25). Assume that V has an element transcendental over k. Then, as noted earlier, all nonzero elements of V are transcendental over k. Therefore, L ∈ k[∂] is of order 1 if and only if y10 /y1 ∈ k and y1 is transcendental 5

over k. In this case, k(y1 ) = E and k(x) = k. Since y10 /y1 ∈ k(x), it follows that if V has an element transcendental over k then the order of L ∈ k[∂] is ≥ 2 if and only if k(x) % k. Outline of the proof of Theorem (A). Denote the algebraic closure of k in E by e k. Since K is algebraically closed in E, e k ⊆ K. Since E = k(t1 , · · · , tn ) remains a liouvillian extension over any of its intermediate differential fields, Theorem (A) follows, via an induction on the transcendence degree tr.d(E | k), once we produce an element w ∈ K − k˜ satisfying a first order differential equation over e k. e Now, to find such an element w ∈ K − k, we proceed as follows. First, among all pairs (L , w), where L ∈e k[∂] is monic, w ∈ K − e k and L (w) ∈ e k, we choose a pair such that the order of L is minimal. Existence of one such pair is guaranteed by [12], Theorem 2.2. Secondly, if D is a monic irreducible right-hand factor of L in e k[∂] then since D(w) ∈ K, by minimality of L , we have D(w) ∈ e k and thus e we obtain that L is irreducible over k. Finally, we apply Lemmas 2.3 and 2.4, where we analyse the structure of certain differential subfields of Picard-Vessiot extensions of irreducible operators, and show that the order of L is 1. The hypothesis that K is algebraically closed in E is also used to show that the case where L (w) = 0, w ∈ kn−1 (tn ) − kn−1 and tn is algebraic over kn−1 does not occur. Lemma 2.3. Let k ⊆ K ⊆ E be differential fields, E be a liouvillian extension of k, CE = Ck and Ck be an algebraically closed field. Let k be algebraically closed in K and k(x) be the algebraic closure of k in E. Suppose that E = khV i is also a Picard-Vessiot extension of k for a monic irreducible operator L ∈ k[∂], where V is the full set of solutions of L (Y ) = 0 in E. (i) If 0 6= y ∈ V ∩ K(x) and y 0 /y ∈ k(x) then SpanCk {σ(y) | σ ∈ Gal(K(x) | K)} = V and in particular, E = K(x). (ii) If V ∩ K 6= {0} then E = K(x). Proof. Let G := Gal(K(x) | K), W = SpanCk {σ(y) | σ ∈ G} and y1 , . . . , ym ∈ {σ(y) | σ ∈ G} be a Ck − basis of W and Y be a differential indeterminate. It is clear that W ⊆ K(x) and that W ⊆ V . A standard result in differential algebra ([6], Proposition 3.9) tells us that the coefficients of the differential operator ω(Y, y1 , . . . , ym ) (2.1) D(Y ) = ω(y1 , . . . , ym ) belong to the field fixed by G. Since k is algebraically closed in K, the fields K(x) and K are linearly disjoint over k. Then the group G is isomorphic to the finite Galois group of k(x) over k and the set of all elements fixed by G is the field K. Thus D ∈ K[∂]. On the other hand, since G stabilizes k(x), σ(y)0 /σ(y) ∈ k(x) for all σ ∈ G. Therefore, we can directly compute the Wronskians: ! m Y ω(y1 , . . . , ym ) = b yi , b ∈ k(x) and (2.2) i=1

ω(Y, y1 , . . . , ym ) =

m Y i=1

! yi

m X

! bi Y (i)

,

bi ∈ k(x), bm = b

(2.3)

i=1

and obtain that the coefficients of D belong to k(x) as well. Since K ∩ k(x) = k, D ∈ K[∂] ∩ k(x)[∂] = k[∂]. Now since W ⊆ V , D must be a right-hand factor of the irreducible operator L . This implies D = L and therefore V = W ⊆ K(x). Pm Let w = claim that y1 ∈ i=1 ci yi ∈ K ∩ V be a nonzero element and assume that c1 6= 0. We Pl K(x). To see this, consider the space Spank(x) {y1 , . . . , ym } and a nonzero element j=1 αi zi ∈ K(x)∩ 6

Spank(x) {y1 , . . . , ym } of shortest length l with z1 := y1 and distinct zi ∈ {y2 , . . . , ym }. Since w ∈ Spank(x) {y1 , . . . , ym }, such an l exists and it is clear that z1 , . . . , zl are k(x)−linearly independent. If l = 1 then since α1 ∈ k(x), we have y1 ∈ K(x). Now suppose that l ≥ 2. Then we have the derivative Pl 0 j=1 (αj + βj αj )zj ∈ K(x) and thus we obtain l X


α2 αj0 − α20 αj + (βj − β2 )α1 α2 zj ∈ K(x),

j=1,j6=2

which is of potentially shorter length, where βi = zi0 /zi . Therefore the coefficients of zj must be zero and in particular, we obtain (α1 /α2 )0 = (β2 − β1 )(α1 /α2 ). Then z2 /z1 = cα1 /α2 for some nonzero c ∈ Ck . This contradicts the fact that z2 and z1 are k(x)− linearly independent. Hence the claim. Now since y1 ∈ K(x) ∩ V and y10 /y1 ∈ k(x), (ii) follows from (i). e ⊆ U be differential fields, CU = Ck , Ck be an algebraically closed field and Lemma 2.4. Let k ⊆ k ∗ $ K ⊆ K e be a finite algebraic extension of K as well as a generalised liouvillian U be a liouvillian extension of k. Let K ∗ e Suppose that U contains a full set of extension of k such that both k and k ∗ are algebraically closed in K. solutions V of a monic irreducible operator L ∈ k[∂] and that K(x) = k ∗ hV i, where k(x) is the algebraic closure of k in the Picard-Vessiot extension khV i of k. Then the following statements hold. (i) There is an element y ∈ K − k ∗ such that y 0 /y ∈ k ∗ and that for any Q y1 ∈ V with y10 /y1 ∈ k(x), there p q ∗ are nonzero integers p, q such that y N r(y1 ) ∈ k , where Nr(y1 ) := σ∈ Gal(K(x) | K) σ(y1 ). (ii) K is an algebraic extension of k ∗ (y). e then w0 /w ∈ k ∗ . (iii) If 0 6= w ∈ V ∩ K Proof. Let y1 ∈ V be such that y10 /y1 ∈ k(x) ⊆ k ∗ (x). Since K(x) = k ∗ hV i, using Lemma 2.3 (i), we choose a Ck −basis y1 , . . . , ym ∈ {σ(y1 ) | σ ∈ G := Gal(K(x) | K)} of V . Then K(x) = e is a generalized liouvillian extension of k ∗ k ∗ (x)(y1 , . . . , ym ), where yi0 /yi ∈ k(x) ⊆ k ∗ (x). Since K e there is an element t ∈ K e transcendental over k ∗ , such and that k ∗ is algebraically closed in K, that t0 = αt + β for some α, β ∈ k ∗ . If β = 0 then y := tp ∈ K for some nonzero integer p and e y 0 /y = pα ∈ k ∗ . On the other hand if β 6= 0 then since K(x) = K(x)(z) = k ∗ (x)(y1 , . . . , ym )(z) for e algebraic over K, from Proposition 2.2, we obtain γ ∈ k ∗ such that γ 0 = αγ + β. Now some z ∈ K since (t − γ)0 /(t − γ) = α ∈ k ∗ ⊆ K and t − γ is algebraic over K, there is a nonzero integer p such that y := (t − γ)p ∈ K. Thus, in any event, there is an element y ∈ K − k ∗ such that y 0 /y = pα ∈ k ∗ . From Kolchin- Ostrowski [8], p. 369), we can find elements γ ∈ k ∗ (x) and Q integers s, ni Qm ni Theorem (see ni s s such that y = γ i=1 yi . Replacing y by y and now, abusing notation, we have y = γ m i=1 yi . Let P (X) = X n + · · · + a0 ∈ K[X] be the irreducible polynomial over K of y1 . Then each yi is a root of P and therefore N r(yi ) = N r(yj ) = (−1)|G| al0 ∈ K for all i, j, where l = |G|/n is an integer. Since y ∈ K, we obtain N r(y) = y |G| = N r(γ)N r(y1 )p , where N r(γ) ∈ k ∗ and 1  p ≤ m. This proves i. We now claim that K must be an algebraic extension of k ∗ (y). Let M denote the algebraic closure of e and observe that it is enough to show that K ⊆ M . the differential field k ∗ (y) in K

7

Note that N r(y1 )p = y |G| N r(γ)−1 ∈ k ∗ (y) ⊆ M . Thus K(x) e N r(y1 ) = (−1)|G| al0 ∈ M and it follows that a0 ∈ M . e are linearly disjoint over k ∗ and K e Since k ∗ (x) and K ∗ K(x) = k hV i, the fields M K and K(x) are linearly disjoint over K and thus P remains irreducible over e reM K. Furthermore, M K(x) = M hV i and that K mains a generalised liouvillian extension of M . Suppose that K 6⊆ M . Then M $ M K and therefore, from i, we obtain an element u ∈ M K − M such that e 0 p q K(x) u Q/u ∈ M and that u N r(y1 ) ∈ M, where N r(y1 ) = σ∈ Gal(M K(x) | M K) σ(y1 ) and p, q are nonzero integers. e Since P remains irreducible over M K and that |G| = K [K(x) : K] = [M K(x) : M K], we have N r(y1 ) = (−1)|G| al0 ∈ M . This implies up ∈ M . Since M is algee we obtain u ∈ M , a contradiction. braically closed in K, Thus we have shown that K ⊆ M , which implies K is a finite algebraic extension of k ∗ (y).

M K(x) MK

K(x) K

k ∗ (x) k∗

M K(x) MK

M (x) M

k ∗ (x) k∗

Let K = k ∗ (y, z), where z is algebraic over k ∗ . Then k ∗ hV i = K(x) = k ∗ (x, y, z) and since k ∗ hV i is Picard-Vessiot over k ∗ (x) with differential Galois group isomorphic to Gm (Ck ), we have k ∗ hV i = k ∗ (x, t), where t0 /t ∈ k ∗ (x) and t is transcendental over k ∗ (x). Note that G stabilizes k ∗ (x) and therefore for any σ ∈ G, we have σ(t)0 /σ(t) =: bσ ∈ k ∗ (x). Thus σ(t) ∈ k ∗ (x, t) − k ∗ (x) is a solution of the first order operator L := ∂ − bσ ∈ k ∗ (x)[∂]. Now, from Proposition 2.2, it follows that σ(t) = aσ tmσ for some aσ ∈ k ∗ (x) and integer mσ 6= 0. For similar reasons, y = atr and y1 = a1 tn for some a, a1 ∈ k ∗ (x) and nonzero integers r and n. Now, for any σ ∈ G, atr = y = σ(y) = σ(a)arσ trmσ . Since t is transcendental over k ∗ (x), mσ = 1 for all σ ∈ G. Then σ(y1 ) = σ(a1 )anσ tn for each σ ∈ G. Since the orbit of y1 under G spans V as a CP k − vector space ∗ hV i = k ∗ (x, t), we obtain that n = 1. Now, for any 0 6= v ∈ V , we have v = and k σ∈G cσ σ(y1 ) = P 0 ∗ 0 0 σ∈G (cσ σ(a1 )aσ ) t. Thus there is an element dv ∈ k (x) such that v = dv t. Then v /v = dv /dv + t /t ∈ e ∩ V then w0 /w ∈ k ∗ (x) ∩ K e = k∗ . k ∗ (x) for all nonzero v ∈ V . Thus, if 0 6= w ∈ K Proof of Theorem (A). Since E remains a liouvillian extension over any of its intermediate subfields, we first replace k with its algebraic closure in K. Now since K is algebraically closed in E, k is algebraically closed in E as well. Choose the smallest n ≥ 1 so that k $ K ∗ := K ∩ kn . Then K ∗ ∩ kn−1 = k and K ∗ is algebraically closed in kn . To prove the theorem, we shall first show that there is an element w ∈ K ∗ − k such that w0 = aw + b for some a, b ∈ k. Then since tr.d(E|k(w))  tr.d(E|k) and k(w) ⊆ K ⊆ E, the rest of the proof follows by an induction on tr.d(E | k). From [12], Theorem 2.2, we know that there is an element w ∈ K ∗ − k and L ∈ k[∂] such that L (w) = 0. Therefore, we shall choose a w ∈ K ∗ − k and L ∈ k[∂] of smallest possible order m such that L (w) = b ∈ k. Now we only need to show that m = 1. We accomplish this by dividing the rest of the proof into two cases depending on whether or not b = 0. Before we proceed with these cases, we observe that if Lm−p , Lp ∈ k[∂] are operators of orders m − p ≥ 1 and p ≥ 1 respectively, Lp is irreducible and that L = Lm−p (Lp (Y )) then since Lp (w) ∈ K ∗ , from our choice of w and m, we obtain Lp (w) ∈ k. This in turn implies that p = m and thus we shall further assume that L is irreducible. Suppose that b = 0 and let V be the set of all its solutions in a Picard-Vessiot extension of kn and 8

y1 , . . . , ym be a Ck −basis of V such that yi0 /yi ∈ k(x). Then for each i, ki hV i is a Picard-Vessiot extension of ki for L . Observe from Lemma 2.3 (ii) that khwi(x) = khV i. Suppose that tn is algebraic P over kn−1 and P (X) = pi=0 ai X i be the monic irreducible polynomial of w over kn−1 . Note that the algebraic closure e kn−1 of kn−1 in kn−1 hV i is a finite Galois extension of kn−1 . Let L ⊆ e kn−1 be the smallest Galois extension of kn−1 containing w. Then σ(w) ∈ V for any σ ∈ Gal(L | kn−1 ). Since the coefficients ai ∈ kn−1 of P are symmetric polynomial in {σ(w) | σ ∈ Gal(L | kn−1 )} over kn−1 , each ai ∈ khV i = khwi(x) and in particular, ai ∈ kn−1 is algebraic over khwi. Now since K ∗ ⊇ khwi and K ∗ is algebraically closed in kn , each ai ∈ K ∗ ∩ kn−1 = k. This makes w ∈ K ∗ − k algebraic over k, a contradiction. Thus tn must be transcendental over kn−1 . Next we suppose that t0n ∈ kn−1 . Then tn is transcendental over kn−1 and kn−1 (tn ) is a Picard-Vessiot extension of kn−1 having kn−1 [tn ] as its Picard-Vessiot ring. Therefore, w = αr trn + · · · + α0 ∈ kn−1 [tn ] with αr 6= 0 and r ≥ 1. Then 0 = L (w) = L (ar )trn + lower powers of tn and therefore L (ar ) = 0. Now we apply Lemma 2.3 (ii) and obtain that V ⊆ khar i(x) and thus w ∈ V must be algebraic over khar i ⊆ kn−1 . That is, w = αr trn + · · · + α0 is algebraic over kn−1 , which contradicts the fact that tn is transcendental over kn−1 . Finally, we suppose that tn is transcendental and t0n /tn ∈ kn−1 . We apply Lemma 2.4 (iii), with kn−1 in place of k ∗ and kn in place of both K and e and obtain that w0 /w ∈ kn−1 . Since K ∗ ∩ kn−1 = k, we have w0 /w ∈ k and thus m = 1. K Suppose that b 6= 0. From Proposition 2.2, it follows that there is a nonzero solution u ∈ kl of Lp (Y ) = 0 such that tr.d(kl | k)  tr.d(kn | k). We claim that u0 /u ∈ k and this indeed implies m = 1. g the algebraic closure of khui in kn , is a generalised liouvillian Applying induction we obtain that khui, g and khui(x) = khV i, we shall take U = kn hV i extension of k. Since k is algebraically closed in khui and apply Lemma 2.4 (iii) to obtain that u0 /u ∈ k. This implies that m = 1 and thus we have proved that K is a generalised liouvillian extension of k. Now we shall prove the second part of Theorem (A). We know that K must be a generalised liouvillian extension of k. Let e k be the algebraic closure of k in K and that e k $ K. Then there must be 0 e e an element t ∈ K − k such that t = αt + β for some α, β ∈ k. We shall now produce an element y ∈ K −e k such that y 0 ∈ k when E is a primitive extension of k and y 0 /y ∈ e k when E is an exponential extension of k. Then since e k(y) ⊆ K ⊆ E and tr.d(E | e k(y))  tr.d(E | e k), the rest of the proof follows from an induction on tr.d(E | k). Let E = k(t1 , · · · , tn ) be a primitive extension of k. Now From Proposition 2.2, we obtain a nonzero element z ∈ E such that z 0 /z = α. Since E = e k(t1 , . . . , tn ) is r e ki for some a primitive extension of k, we choose the smallest non negative integer i such that z ∈ e r 0 r e e nonzero integer r. Note that (z ) /z = rα ∈ k ⊆ ki−1 . Then from Theorem 2.1 (i) and (v), we observe that z r ∈ e ki−1 when t0i ∈ e ki−1 and z rs ∈ e ki−1 for some nonzero integer s when ti is algebraic over e ki−1 , m e respectively. Thus, z ∈ ki−1 for some nonzero integer m. Therefore, by minimality of i, z m ∈ e k. 0 e e e But k is algebraically closed in E as well and thus z ∈ k. Then y := t/z ∈ K − k and y = β/z ∈ k. Now let E = k(t1 , · · · , tn ) be an exponential extension k and to avoid triviality, assume β 6= 0. Then a repeated application of Proposition 2.2 will produce a nonzero element γ ∈ e k such that γ 0 = αγ + β. Observe that y := t − γ ∈ K − e k and that y 0 /y = α ∈ e k.  Corollary (A). Let E be a liouvillian extension of k, k be algebraically closed in E, CE = Ck and Ck be algebraically closed. (i) If there is a monic irreducible L ∈ k[∂] and b ∈ k such that L (Y ) = b has a nonzero solution in E then the order of L is 1. 9

(ii) If K is a differential field such that k $ K ⊆ E then there is an element in w ∈ K − k satisfying a first order differential equation over k. Proof. If E has a nonzero solution of L (y) = b then by Proposition 2.2, there is a nonzero y ∈ E such e of K in E is that L(y) = 0. Let K = khyi and observe from Theorem (A) that the algebraic closure K a generalised liouvillian extension of k. Let U := EhV i be a Picard-Vessiot extension of E for L and k(x) be the algebraic closure of k in khV i. Then K(x) = khyi(x) = khV i and therefore, we shall apply Lemma 2.4 (iii), with k in place of k ∗ , and obtain that y 0 /y ∈ k. Thus order of L is 1. As in the proof of Theorem (A), we choose w ∈ K − k and a monic L ∈ k[∂] of smallest order such that L (w) = b ∈ k. But such an L must irreducible over k and therefore by (i), the order of L must be 1. Converse of Theorem (A). Let k ⊆ k(t) be differential fields, where t0 = at + b for some a, b ∈ k. Consider the field k(t, w), where w is an indeterminate and define a derivation on k(t, w) such that w0 = aw. Then k(t, w) = k(w, t/w) and since (t/w)0 = b/w ∈ k(w), we observe that k(t) is a differential subfield of the liouvillian extension k(w, t/w). When Ck(t) = Ck and Ck is an algebraically closed field, one can construct a liouvillian extension E containing k(t) as a differential subfield and having field of constants as Ck as follows. Consider the differential subring k(t)[w] of k(t, w) and choose a maximal differential ideal m with w 6∈ m and form the quotient ring R := k(t)[w]/m. Let π : k(t)[w] → R denote the canonical surjection and make R a differential ring and π a differential ring homomorphism by defining π(x)0 := π(x0 ). Extend the derivation of R to the field of fractions K of R. It can be shown that CK = Ck (See Lemma 1.17 of [7]). Since π(t/w)0 = b/t and K = k(t, π(w)) = k(t, π(t/w)), we obtain that K is a liouvillian extension of k with CK = Ck . Now if k ⊆ k(t) is an algebraic extension then as noted earlier, Ck(t) is an algebraic extension of Ck . Since Ck is algebraically closed, we obtain Ck(t) = Ck . One can easily extend these arguments and prove the following Proposition 2.5. If K is a generalised liouvillian extension of k, CK = Ck and that Ck is an algebraically closed field then there is a liouvillian extension E of k such that k ⊆ K ⊆ E having CE = Ck .

3

Transcendence degree one extensions and Galois descent

Field of Constants. Let K be a differential field extension of k, K be an algebraic closure of K and let M be a differential field intermediate to k and K. If y ∈ M (K)−K be a constant such that y∈ / CK and Pn−1 Pn−1 n i n P (X) =P X + i=0 ai X be the monic irreducible polynomial of y over K then y + i=1 ai y i = 0 0 i 0 implies n−1 i=0 ai y = 0. Therefore ai = 0 for each i and we have ai ∈ CK . Thus y is algebraic over CK . Conversely if y ∈ M (K) is algebraic over CK then it is easy to see that y 0 = 0 and thus CM (K) is an algebraic extension of CK . In particular for M = K, we obtain CK = CK , which is an algebraically closed field. Now suppose that CK = Ck and that k is algebraically closed in K. Then CM (K) is an algebraic extension of Ck and the fields M and K are linearly disjoint over k. Therefore M becomes algebraically closed in M (K). Thus CM (K) = CM , and when M = k, we have Ck(K) = Ck . We record all these observations in the following Proposition 3.1. Let k ⊆ K be differential fields, K be an algebraic closure of K and M be a differential field intermediate to k and K. Then CM (K) is an algebraic extension of CK and in particular, CK equals the algebraically closed field CK . Furthermore, if CK = Ck then we have the following: 10

1. CM (K) is an algebraic extension of Ck and if k is algebraically closed in K then CM (K) = CM . 2. Ck(K) = Ck and CK = Ck = Ck . Outline of the proof of Theorem (B): Let k(x) be the algebraic closure of k in K. We first apply Corollary 4 of [11] to obtain an element v ∈ k(x)(K) − k(x) satisfying a first order linear differential equation over k(x). We then divide the proof into two cases depending on whether or not k(x)(K) − k(x) contains an element u such that u0 /u ∈ k(x). In Lemma 3.4, we show that when there is no such element u, there is an element in K − k(x) satisfying a first order differential equation over k(x) and when there is such an element u, there is an element in K − k(x) satisfying a linear homogeneous differential equation over k(x) of order at most 2. Our approach uses Galois descent of (differential) vector spaces and in particular, the following standard result. Theorem 3.2. Let M be an algebraic Galois extension of k with Galois group G and V be a M −vector space of dimension n. If V is a G− module such that σ(ry) = σ(r)σ(y) where r ∈ M and y ∈ V then there exist a M −basis {u1 , · · · , un } of V such that σ(ui ) = ui for all i and for all σ ∈ G. Proof. Follows from Propositions 1 and 2 of [4]. Proposition 3.3. Let k ⊆ E be differential fields such that tr.d(E|k) = 1 and CE = Ck . Suppose that there is no element w ∈ E − k such that w0 /w ∈ k and that there are elements u, v ∈ E − k such that v is transcendental over k, v 0 = av + b for some a, b ∈ k and u0 = αu + β for some α, β ∈ k. Then u = f v + g for some f, g ∈ k and f 6= 0. Proof. Since E is an algebraic extension of k(v), we shall apply Proposition 2.2 (c) (with k := k(v)) and obtain a nonzero x ∈ E such that x0 = αx. Then u/x ∈ E and (u/x)0 = β/x. From our hypothesis, we must have x ∈ k and since u/x is algebraic over k(v), we have u/x ∈ k(v). Thus we obtain u ∈ k(v). Let u = P/Q, where P, Q ∈ k[v], P, Q are relatively prime polynomials and that Q be monic. Then αP Q + βQ2 = P 0 Q − P Q0 and therefore Q must divide Q0 . If Q 6= 1 then Q0 = f Q, for some f ∈ k and this contradicts ) = 1. Suppose Pn thei hypothesis of our theorem. Therefore u = P . We claim that deg(P n that P = i=0 ai v with ai ∈ k, an 6= 0 and n ≥ 2. Comparing the coefficients of v and v n−1 of the equation P 0 = αP + β, we obtain a0n = (α − na)an a0n−1

− aan−1 + nban = (α − na)an−1 .

(3.1) (3.2)

We use Equation 3.1 and eliminate α−na from Equation 3.2 and obtain (an−1 /an )0 −a(an−1 /an )+nb = 0. Then we have ((an−1 /an ) + nv)0 = a((an−1 /an ) + nv) and since v is transcendental over k, we must have (an−1 /an ) + nv ∈ / k and this contradicts a hypothesis of our proposition. Thus deg(P ) = 1 and we obtain u = f v + g for some f, g ∈ k and f 6= 0. Lemma 3.4. Let k ⊆ K be differential fields, CK = Ck , tr.d(K|k) = 1 and k is algebraically closed in K. Let K be an algebraic closure of K and k be the algebraic closure of k in K. Suppose that there is an element v ∈ k(K) − k satisfying a first order linear differential equation over k and that there is no w ∈ k(K) − k such that w0 /w ∈ k. Then there are elements r, s ∈ k such that r 6= 0, rv + s ∈ K − k and that rv + s satisfies a first order linear differential equation over k. Suppose that there is an element v ∈ k(K) − k such that v 0 = av for some a ∈ k. Then either (a) there is an element y ∈ K − k such that y 0 /y ∈ k or 11

(b) there is a quadratic extension L of k and elements y1 ∈ K(L) − L, y ∈ K − k satisfying the following properties: (b1) y and y1 are Ck −linearly independent solutions of the irreducible differential polynomial L (Y ) = Y 00 + αY 0 + βY for some α, β ∈ k. (b2) y10 /y1 ∈ K(L) − L, (y − y1 )0 /(y − y1 ) ∈ K(L) − L, (y 0 )2 = γ1 yy 0 − γ2 y 2 + (4γ2 − γ12 )γ3 for some γ1 , γ2 , γ3 ∈ k and khyi = k(y, y 0 ) is a quadratic extension of k(y). Proof. Suppose that there is no w ∈ k(K) − k such that w0 /w ∈ k. Let v 0 = av + b, a, b ∈ k and G = Gal(K(k)|K). Since k is algebraically closed in K, k and K are linearly disjoint over k and the group G is naturally isomorphic to Gal(k|k). Observe that G stabilises the field k and for any σ ∈ G, we have σ(v)0 = σ(a)σ(v) + σ(b). Apply Proposition 3.3 and obtain (3.3)

σ(v) = fσ v + gσ

for some fσ , gσ ∈ k and fσ 6= 0. The set V :=Spank {1, v} is a (differential) k−vector space and from Equation 3.3, we know that G stabilises V .

K(k) K

V V ∩K

k k

Apply Theorem 3.2, and use the fact that K is the field fixed by G, to obtain a k−basis {y1 , y2 } ⊆ K of V . Since V 6⊆ k, we shall assume y1 ∈ / k and thus rv + s = y1 , where r, s ∈ k and r 6= 0. Observe that (rv + s)0 = α(rv + s) + β, (3.4) where α = (r0 /r) + a ∈ k and β = s0 + rb − s(r0 /r) − sa ∈ k. Since both rv + s and (rv + s)0 belong to K, we have (rv + s)0 = σ(α)(rv + s) + σ(β) for all σ ∈ G. Comparing this equation with 3.4, we obtain both α, β ∈ K ∩ k = k. Suppose that there is an element v ∈ k(K) − k such that v 0 = av for some a ∈ k and that (a) does not hold. Let M ⊆ k be a finite Galois extension of k such that v ∈ K(M ) and a ∈ M . Let G = Gal(K(M )|K) and observe that since K and M are linearly disjoint over k, M (K) is a finite Galois extension of K such that G ∼ = Gal(M |k). Since tr.d(K|k) = 1, M (K) is a finite algebraic extension of M (v) and therefore for any σ ∈ G, we have σ(v) is algebraic over M (v). Moreover, σ(v)0 /σ(v) = σ(a) ∈ M and from Proposition 3.1, we have CM (K) = CM . Now it follows from Theorem 2.1 (i) that for every σ ∈ G, there are integers lσ ≥ 1 and mσ such that σ(v)lσ = fσ v mσ for some fσ ∈ M . Since τ σ(v)lσ lτ = τ (fσlτ )τ (v mσ lτ ) = τ (fσlτ )fτmσ v mσ mτ and τ σ(v)lτ σ = fτ σ v mτ σ , we compute τ σ(v)lτ σ lσ lτ from
l these two equations and obtain τ (fσlτ )fτmσ τ σ v mσ mτ lτ σ = fτlσσlτ v mτ σ lσ lτ . Since v is transcendental over M , we must have (mτ σ /lτ σ ) = (mτ /lτ )(mσ /lσ ). Thus the map φ : G → Q − {0} defined by φ(σ) = mσ /lσ is a group homomorphism. However, G is a finite group and therefore φ(G) ⊆ {±1}. Then mσ ∈ {±lσ } and since σ(v)lσ = fσ v mσ , we must have either (σ(v)v)lσ ∈ M or (σ(v)v −1 )lσ ∈ M . But M is algebraically closed in M (K) forces lσ = 1 and thus for any σ ∈ G, σ(v) = fσ v mσ (3.5) for some nonzero fσ ∈ M and mσ ∈ {±1}. Note that if φ(G) = {1} then V = SpanM {v} is a G−module and therefore by Theorem 3.2 we obtain a nonzero element f ∈ M such that f v ∈ K.

12

But (f v)0 /(f v) = f 0 /f + v 0 /v ∈ k and this contradicts our assumption that (a) does not hold. Thus φ(G) = {±1}. Observe that ker(φ) is a normal subgroup of G of index 2 and let ker(φ)∗ be the image of the subgroup ker(φ) under the restriction isomorphism from G to Gal(M |k). Let L ⊆ M be the field fixed by subgroup ker(φ)∗ and note that [L : k] = 2. From the linear disjointness of L and K over k, it follows that the field K(L) is the field fixed by the group ker(φ), K(M ) is a Galois extension of K(L) with Galois group ker(φ) and [K(L) : K] = 2 as well. Note that for any σ ∈ ker(φ), we have σ(v) = fσ v for some fσ ∈ M and thus the M −vector space spanned by v is a ker(φ)−module. Applying Theorem 3.2, we obtain an element y1 = f v ∈ W , where W := K(L) ∩ V for some nonzero element f ∈ M . Let r1 := y10 /y1 and observe that r1 ∈ K(L) ∩ M = L. Since tr.d(K(L)|k) = 1, we observe that if r1 ∈ k ⊆ K then y1m ∈ K for some positive integer m and thus (y1m )0 /y1m = mr1 ∈ k. This again contradicts the assumption that (a) does not hold. Thus r1 ∈ L − k.

K(M ) K(L)

V

K

W

M

W ∩K

L k

Let λ be the non trivial automorphism of K(L) over K. Then replacing M with L and v with y1 in the above calculations, we obtain λ(y) = gλ y −1 , (3.6) for some gλ ∈ L. Let y2 = λ(y1 ) and r2 = λ(r1 ). Since r1 ∈ / k we have r1 6= r2 and it follows that y1 and y2 are linearly independent over CL (and therefore over Ck as well). Clearly λ fixes y := y2 + y1 and γ3 := y1 y2 and thus y, γ3 ∈ K. Since γ30 = (r1 + r2 )γ3 and r1 + r2 ∈ k, from our assumption that (a) does not hold, we obtain γ3 ∈ k. Let  0    r2 + r22 − r10 − r12 r2 (r10 + r12 ) − r1 (r20 + r22 ) 00 0 L (Y ) := ω(Y, y1 , y2 )/ω(y1 , y2 ) = Y + Y + Y (3.7) r1 − r2 r1 − r2 The coefficients of L are fixed by λ and therefore they belong to the field K ∩ L = k. Clearly, L (y) = 0 and since K is algebraic over k(y), it is also clear that khyi = k(y, y 0 ). If L ∈ k[∂] were reducible over k then there must be first order monic operators D and H in k[∂] such that L = H D. Now either D(y) ∈ k or D(y) ∈ K − k and H (D(y)) = 0. In any event, we obtain a contradiction to our assumption that K − k has no elements satisfying a first order differential equation over k. A straight forward computation shows that (y 0 )2 = γ1 yy 0 − γ2 y 2 + (4γ2 − γ12 )γ3 , where γ1 = r1 +r2 ∈ k, γ2 = r1 r2 ∈ k and γ3 = y1 y2 ∈ k. Therefore khyi must be a quadratic extension of k(y). Proof of Theorem (B). It is clear that any y ∈ K − k(x) is transcendental and for each such y there is an element z ∈ K − khx, yi such that z is algebraic over khx, yi and khx, y, zi = K. Thus, if there is an element y ∈ K − k(x) such that y 0 = 0 then K = k(x, y, z) becomes a liouvillian extension of k and therefore we shall assume that CK = Ck(x) . Let K be an algebraic closure of K. Suppose that w ∈ K − k(x) and L (w) = 0 for some linear differential polynomial over k(x) of degree ≥ 1. Then by Corollary 4 of [11], we have an element 13

v ∈ k(x)(K) − k(x) such that v 0 = av + b for a, b ∈ k(x). Moreover, from Proposition 3.1 we have Ck(x)(K) = Ck(x) . We now apply Lemma 3.4, with k(x) in place of k, to prove the theorem. Remark 1. Let k ⊆ E be a liouvillian extension, K be a differential field intermediate to E and k and tr.d(K|k) = 1. Then from Theorem 2.2 of [12], there is an element in K − k satisfying a linear differential polynomial of degree ≥ 1 over the algebraic closure of k in K. Since E is finitely generated over k, so is K over k. Thus, any such differential field K satisfies the hypothesis of Theorem (B) and therefore there are elements x, y and z in K such that K = khx, y, zi, k(x) is the algebraic closure of k in K, y satisfies a linear homogeneous differential polynomial of order at most 2 over k(x) and z is algebraic over khx, yi. Proposition 3.5. Let k ⊆ K be differential fields, u ∈ K be a solution of some linear differential equation over k and tr.d(khui|k) = 1. Then z belongs to some liouvillian extension of k and there exists an element y ∈ khui such that y satisfies a linear homogeneous differential polynomial of order ≤ 2 over the algebraic closure k1 of k in K and u is algebraic over k1 hyi. Proof. This proposition strengthens Corollary 4 of [11]. We will first show that u belongs to some liouvillian extension of k. From Corollary 4 of [11], we obtain elements a, b ∈ k and v ∈ khui such that v 0 = av + b and u is algebraic over the differential field k(v). Choose a finite algebraic extension M ⊆ k of k containing a, b such that u is algebraic over M (v). Clearly, M, M (v) and M (v, u) are all differential fields. As described in Section 2.2, we attach an indeterminate w to the field M (v, u) and define w0 = aw. Then M (v, u)(w) becomes a differential field and we have the following tower of differential fields k ⊆ M ⊆ M (w) ⊆ M (w, v/w) ⊆ M (w, v/w)(u) = M (v, u, w). Since u is algebraic over M (v) ⊆ M (w, v/w), (v/w)0 = b/w ∈ M (w) and w0 /w = a ∈ M , we see that M (v, u, w) is a liouvillian extension of k. Since khui is a differential subfield of the liouvillian extension M (v, u, w) of k, the second part of this proposition follows from Remark 1. Through the following example, we will show that K must be relatively algebraically closed in order to have the structure described in Theorem (A) and that we cannot, in general, require the element y from Remark 1 to satisfy a first order linear homogeneous differential equation over k(x). In particular, the situation described in Lemma 3.4 (b) does occur. Example 3.1. Let k = C(x) be the ordinary differential field with the derivation 0 := d/dx,√where √ C 6= {0} is any subfield of the complex numbers. The exponential extension E := k( x, e x ) is a Picard-Vessiot extension of k for the differential polynomial L (Y ) = Y 00 + √

1 0 1 Y − Y. 2x 4x

√

The set V :=spanC {e √x , e− x }√ forms a full set of solutions of L (Y ) = 0. Consider the field K = k(w, w0 ), where w = e x + e− x . Since L (w) = 0, it follows that k(w, w0 ) is a differential field and clearly tr.d(k(w, w0 )|k) = 1. Moreover (w0 )2 = (1/4x)(w2 − 4). Now we shall show that neither is √ there an element in E − k( x) (and therefore neither is there an element in k(w, w0 ) − k) satisfying a first order linear differential equation over k nor is there an element in k(w, w0 ) − k algebraic over k. √ We first show that there is no y ∈ E − k( x) such that y 0 /y ∈ k. Suppose not. Then, as done in the proof of Lemma 2.4, we shall apply Proposition 2.2 to the first order equation Y 0 − (y 0 /y)Y and 14

√ √ obtain y = ren x for some nonzero integer n and r ∈ k( x). √ Let G := √ DGal(E|k) √ be the differential x x − Galois group and note that G stabilises V and therefore σ(e ) = cσ e + dσ e x , where σ ∈ G and cσ , dσ ∈ C, not both zero. Since y 0 /y ∈ k, for any σ ∈ G, we have σ(y) = aσ y for some aσ ∈ C − {0}. Thus we obtain √ √
√ n σ(r) cσ e x + dσ e− x = aσ ren x . √ √ From this equation and from the fact that e x is transcendental over k( x), we conclude that dσ = 0. √ Thus σ(r) = bσ r for all σ ∈ G, where aσ /cσ = bσ ∈ C. Write r = α + β x and observe that the √ √ √ non-trivial automorphism of k( x) which maps x 7→ − x extends to a field automorphism τ of √ √ √ √ E where τ (e x ) = e− x . A simple computation shows that τ ((e x )0 ) = (τ (e x ))0 and it follows that √ √ τ ∈ G. Then τ (r) = bτ r only if α = 0 and thus r = β x. Since y 0 /y = r0 /r + n/(2 x) = c for some √ √ c ∈ k, we then have n/ x = c − (β 0 /β) − (1/2x) ∈ k. Since n 6= 0, we obtain x ∈ k, which is absurd! √ Thus E − k( x) has no elements y such that y 0 /y ∈ k. √ Next, we suppose that y ∈ E − k( x) such that y 0 = ay + b, where a, b ∈ k and b 6= 0. Applying Proposition 2.2, we obtain an element γ ∈ k such that γ 0 = aγ + β and thus (y − γ)0 /(y − γ) ∈ k. √ This contradicts our earlier observation that no such elements exist in E − k( x). Thus we only need to show that k(w, w0 ) − k has no elements algebraic over k. Since the algebraic closure of k √ √ √ in E is k( x), it is enough to show that x ∈ / k(w, w0 ). Suppose that x ∈ k(w, w0 ). Then since √ (w0 )2 = (1/4x)(w2 − 4), we have k(w, w0 ) = k(w, x). Therefore √ w0 = a + b x

for a, b ∈ k(w) with b 6= 0. Squaring both sides of the above equation and substituting (1/4x)(w2 − 4) for (w0 )2 , we obtain √ a2 + b2 x + ab x = (1/4x)(w + 2)(w − 2) ∈ k[w]. Since w +2 and w −2 are irreducible in k[w] and a, b ∈ k(w), we write a and b as quotients of relatively prime polynomials in w with coefficients in k and see that neither a nor b is zero. On the other hand, √ if ab 6= 0 then x ∈ k(w), which contradicts the fact that w is transcendental over k.  √

√

Thus we have found a differential field K = k(w, w0 ), where w = e x + e− z such that tr.d(K|k) = 1, w is a solution of a second order linear homogeneous equation L (Y ) = 0 which has two √ √ differential √ linearly independent solutions, namely, e x and e− x over k( x), k is algebraically closed in K and K − k does not contain any solution of a first order linear differential equation over k. Moreover √ K( x) = E has the structure described in Theorem (A). This shows that Lemma 3.4 (b) occurs and that Theorem (A) holds only for relatively algebraically closed intermediate differential fields. √ √ Remark 2. From Remark 1, it follows that if K is a differential field intermediate to E = C(x)( x, e x ) and k and tr.d(K|k) = 1 then Theorem (B) becomes applicable and we obtain that K = khx, y, zi, where x, y, z ∈ K are as described in the statement of Theorem (B). In [2], Theorem 7, p.809, Kolchin showed that if K is a differential field extension of k, k is algebraically closed in K, tr.d(K/k) = 1 and DGal(K|k) is infinite then K admits a structure similar to the one described in Theorem (B). However, Theorem (B) does not follow readily from this result of Kolchin since differential fields satisfying the hypothesis of Theorem (B) need not have infinite differential automorphism group. For instance, DGal(K|k) is not infinite if we take K = k(w, w0 ) as in Example 3.1. To see this, we first adopt notations from Example 3.1 and observe that if σ(w) and w are C−linearly independent for some σ ∈ DGal(K|k) then since V is of dimension 2 over C, we must have V ⊆ K and we obtain √ E = K. This contradicts the fact that x ∈ / K. Therefore for any σ ∈ DGal(K|k), σ(w) = cσ
w for some cσ ∈ C. Now since (w0 )2 = (1/4x)(w2 − 4), applying σ, we get (w0 )2 = (1/4x) w2 − (4/c2σ ) and subtracting these two equations, we get c2σ = 1. Thus DGal(K|k) is of order at most 2. It is also easy 15

to see that DGal(K|k) is the group of order 2 where the non trivial differential automorphism maps w to −w.

4

Applications to nonlinear differential equations

Let C be an algebraically closed field with the trivial derivation and k = C(z) be the ordinary differential field of rational functions in one variable z over C with the usual derivation d/dz which we denote by 0 . Consider the differential equation y 0 = αn y n + · · · + α1 y + α0 ,

(4.1)

where αi ∈ C(z) and αn 6= 0. When n = 3, such differential equations are called Abel equations of the first kind. Suppose that E is a liouvillian extension of k such that CE = Ck , y ∈ E satisfies the differential Equation 4.1 and y is transcendental over k. Let E be an algebraic closure of E and let k be the algebraic closure of k in E. Then we apply Theorem (A) to the liouvillian extension k(E) of k and obtain an element η ∈ k(y) − k such that η 0 = aη + b

(4.2)

for some a, b ∈ k. Consider the laurent series (y−adic) expansion of η, given by η = ar y r + ar+1 y r+1 + · · · ,

(4.3)

where r ∈ Z, ai ∈ k and ar 6= 0. Then we have
a ar y r + ar+1 y r+1 + · · · + b = f (y),

(4.4)

where f (y) = a0r y r + rar y r−1 (α0 + α1 y + · · · + αt y n )

(4.5)

+ a0r+1 y r+1 + (r + 1)ar+1 y r (α0 + α1 y + · · · + αt y n ) + · · · Proposition 4.1. The liouvillian solutions of the differential equation y 0 = αn y n + · · · + α2 y 2 , where αi ∈ C[z] and αn 6= 0, are algebraic over C(z). Proof. We first claim that r 6= 0 in Equation 4.5. Otherwise, we have a00 = aa0 + b and thus (η − a0 )0 = a(η − a0 ). Therefore we shall write η − a0 = as y s + as+1 y s+1 + · · · , where aj ∈ k for all j ≥ s and as 6= 0 and obtain that a0s = aas . Since (η − a0 )/as ∈ k(y) − k and ((η − a0 )/as )0 = 0, we arrive at a contradiction to our assumption that k(y) and k have the same field of constants C. Thus r 6= 0. Since we have 0 = α1 = α2 , f (y) = a0r y r + higher powers of y. Now from Equation 4.5, we have a0r = aar and thus (η/ar )0 = b/ar . This implies that there exist an element, namely η/ar , in k(y) − k whose derivative belongs to k. Thus, we shall assume that a = 0 in Equation 4.2 and consequently obtain b = f (y). 16

Let us assume for a moment that the coefficients of the powers of y in Equation 4.5 lie in C[z]. Then there are polynomials ar , br+1 , br+2 , · · · ∈ C[z] such that f (y) = a0r y r + b0r+1 y r+1 + · · · and therefore b = b0s for some s ∈ Z. This contradicts the assumption that k(y) and k have the same field of constants since (η − bs )0 = 0. Thus we have shown that there are no liouvillian solutions of the Equation 4.1 which are transcendental over C(z). Now we shall show that the coefficients of f (y) indeed lie in C[z]. It is clear that exactly one of the coefficients of f (y) must equal b and all the others must equal zero. Since r 6= 0, a0r = 0 and we have ar ∈ C ⊆ C[z]. For some integer s ≥ 1, assume that the elements ar , ar+1 , · · · , ar+s−1 belong to C[z]. Then the coefficient of y r+s of f (y) is a0r+s +

j X

(r + s − i)ar+s−i αi+1

(4.6)

i=1

Pj

for some j, 1 ≤ j ≤ n − 1. Since i=1 (r + s − i)ar+s−i αi+1 ∈ C[z] it has an antiderivative br+s ∈ C[z]. As noted earlier in the proof, (ar+s +br+s )0 = a0r+s +b0r+s 6= b and therefore we obtain (ar+s +br+s )0 = 0. Thus ar+s ∈ C[z].

References [1] I. Kaplansky, Introduction to Differential Algebra, Hermann, Paris 1957. 3, 4 [2] E.R. Kolchin, Galois Theory of Differential fields, Amer J. Math, Vol. 75, No. 4 (1953), pp. 753-824. 2, 15 [3] E.R. Kolchin, On Galois Theory of Differential fields, Amer J. Math, Vol. 77, No. 4 (1955), pp. 868-894. 2 [4] E. R. Kolchin and S. Lang, Existence of invariant bases, Proc. AMS, Vol. 11 (1960), pp. 140-148. 11 [5] E.R. Kolchin, Differential Algebra and Algebraic Groups, Academic Press INC (1973). 1 [6] A. Magid, Lectures on Differential Galois Theory, University Lecture Series. American Mathematical society 1994, 2nd edn. 3, 5, 6 [7] M. van der Put, M. F. Singer, Galois Theory of Linear Differential Equations, 328, Grundlehren der mathematischen Wissenshaften, Springer, Heidelberg, 2003. 3, 5, 10 [8] L. Rubel, M. Singer, Autonomous Functions, Jour. Diff eqns, Vol. 75, (1988), pp. 354-370. 4, 7 [9] M. Rosenlicht, M. Singer, On Elementary, Generalized Elementary, and Liouvillian Extension Fields, Contributions to Algebra, (H. Bass et.al., ed.), Academic Press (1977), pp. 329-342. 2 [10] M. Singer, Elementary Solutions of Differential Equations, Pacific J. Math, Vol. 59, no. 2, (1975), pp. 535-547. 2 [11] M. Singer, Solutions of Linear Differential Equations in Function Fields of One Variable, Proc. AMS, Vol. 54, no 1, (1976), pp. 69-72. 2, 5, 11, 13, 14 [12] Varadharaj R. Srinivasan, Liouvillian solutions of First Order Non Linear Differential Equations, Jour. Pure. Appl Algebra, Vol. 221, no. 2, (2017), pp. 411-421. 6, 8, 14

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