This is a preprint of: “On planar polynomial vector fields with elementary first integrals”, Colin J. Christopher, Jaume Llibre, Chara Pantazi, Sebastian Walcher, J. Differential Equations, vol. 267, 4572–4588, 2019. DOI: [10.1016/j.jde.2019.05.007] ON PLANAR POLYNOMIAL VECTOR FIELDS WITH ELEMENTARY FIRST INTEGRALS C. CHRISTOPHER1 , JAUME LLIBRE2 , CHARA PANTAZI3 AND SEBASTIAN WALCHER4 Abstract. We show that under rather general conditions a polynomial differential system having an elementary first integral already must ad- mit a Darboux first integral, and we explicitly characterize the vector fields in this class. We also investigate some exceptional cases, i.e. equa- tions admitting an elementary first integral but not a Darboux first inte- gral. In particular we provide a rather detailed discussion of exceptional elementary first integrals built from algebraic functions of prime degree. 1. Introduction and survey of results In the present note we discuss a polynomial vector field ( ) ∂ ∂ P X=P + Q , briefly X = (1) ∂x ∂y Q in C2 , and the associated ordinary differential equation ẋ = P (x, y), ẏ = Q(x, y). Our focus is on systems which admit a first integral that is elementary over the differential field K = C(x, y). (The pertinent notions will be introduced below; in particular we collect some basic definitions and facts in an appen- dix.) Our approach is based on the classical work [10] by Prelle and Singer. We recall some of their main results, specialized to the rational function field. Theorem 1. (Prelle and Singer [10]) (a) If the polynomial vector field X admits an elementary first integral then there exist an integer m ≥ 0, algebraic functions v, u1 , . . . , um over K and nonzero constants c1 , . . . , cm ∈ C such that m ( m ) ∑ X(ui ) ∑ X(v) + ci =X v+ ci log(ui ) = 0, (2) ui i=1 i=1 ∑ but v + m i=1 ci log(ui ) is not constant. The ci may be chosen linearly independent over the rational numbers Q. 1 http://www.gsd.uab.cat 2 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER (b) If the vector field X admits an elementary first integral then it admits an integrating factor of the special form µ = f1−d1 · · · fr−dr , (3) with irreducible and pairwise relatively prime polynomials f1 , . . . , fr , and exponents d1 , . . . , dr ∈ Q. The second statement includes the case when a rational first integral exists; in the particular case of a polynomial first integral there may be a constant ∑m integrating factor. Note that the existence of a first integral v + i=1 ci log(ui ) for (1) is equivalent to the existence of a first integral ∏ c exp(v) · ui i . (4) The aim of the present paper is to gain more insight into the nature of the algebraic functions v and ui , and to obtain a structural characterization of the polynomial vector fields which admit such a first integral. Thus we prove: Whenever the trace of v is not constant or the norm of some ui is not constant then the existence of a first integral (4) already implies the existence of a Darboux first integral (i.e., a first integral of this type with rational functions v and ui ). This yields a precise characterization of the vector fields which admit this first integral. With regard to the exceptional cases we prove some general results about the simplest one (the case of a first integral (2) with m = 1), and discuss some examples. Our results indicate that the name “exceptional” is justified. Before we state the results in detail, we briefly recall the general setting, with a particular focus on integrating factors of type (3). By Singer [14] and Christopher [3], the vector field (1) admits a Liouvillian first integral if it admits an integrating factor of type (3) with complex exponents di . Moreover, there exists a Liouvillian first integral if and only if there exists a Darboux integrating factor. (The existence of a Liouvillian first integral is not always associated to the existence of some invariant algebraic curve, see [7].) The special structure of polynomial vector fields admitting an integrating factor of type (3) (with exponents di ∈ C) has been discussed in a number of recent papers, e.g. [2, 9, 4, 5, 6]. We will use some of these results below. Returning to polynomial vector fields with elementary first integrals, the particular vector fields which even admit a rational first integral form a special class. On the one hand such vector fields are easy to construct by forming the Hamiltonian of a rational function and multiplying it by a suitable factor to obtain a polynomial (with relatively prime entries, if desired). On the other hand it may be problematic to decide for a given polynomial vector field whether it admits a rational first integral. There are several algorithms to compute rational first integrals of a given polynomial vector field when degree bounds are given; see e.g. Bostan et al. [1]. But determining such degree bounds a priori is an unsolved problem. ELEMENTARY FIRST INTEGRALS 3 We will frequently require that a vector field under consideration does not admit a rational first integral. This implies in particular (as will be shown in Section 3 below) that we may assume that v is not constant, m > 0 and the ui are not constant in (2). Our principal result states that under rather weak assumptions the ex- istence of an elementary first integral will already imply the existence of a Darboux first integral. Theorem 2. ∑mLet the vector field X given in (1) admit the elementary first integral v + i=1 ci log(ui ), where m > 0 and v, u1 , . . . , um are nonconstant algebraic functions over K = C(x, y), and furthermore c1 , . . . , cm are com- plex constants linearly independent over Q. If X does not admit a rational first integral then the following hold. (a) Whenever some uj has non-constant norm, or whenever v has non- constant trace, then X admits a Darboux first integral. (b) Moreover, if (3) is an integrating factor for X, with some uj having non- constant norm, then it is uniquely determined (up to a nonzero complex factor) and all di are nonnegative integers. We will prove this in Section 3. Moreover, given statement (a) of Theo- rem 2, the corresponding polynomial vector fields can be determined explic- itly, up to multiplication by a polynomial function, by invoking a result of Chavarriga et al. [2]; see Theorem 5 and Corollary 1. Theorem 2 leaves the problem to investigate “exceptional” cases admitting an elementary first integral (2) with all ui of constant norm, v of constant trace. In the present work we restrict our attention to the (presumably) simplest case m = 1. One may then take c1 = 1 with no further loss of generality. Thus we assume a relation X(u) X(v) + = 0, (5) u with u and v non-constant algebraic functions over K (but not both in K), with u of constant norm and v of constant trace. In Section 4 we will prove: Proposition 1. Let relation (5) be given, and assume that X admits no rational first integral. (a) The intersection K[u] ∩ K[v] is nontrivial, i.e. ̸= K. (b) If the degree of K[v] over K is a prime number, then K[v] ⊆ K[u]. If both degrees are prime numbers then K[v] = K[u]. We proceed to discuss the cases when both u and v are contained in an extension F of K whose degree is a prime number. For quadratic extensions we give a complete characterization as follows. Theorem 3. Assume relation (5) with u of constant norm and v of constant trace, both contained in a degree two extension of K, but neither contained in K. Then, with no loss of generality, one may take u and v to satisfy u2 + 2g · u + 1 = 0 and v = b(g + u) 4 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER with a non-constant rational function g and a nonzero rational function b. (a) There is a rational function s such that ( ) P = −s(· (g 2 − 1)by + (bg − 1)g)y , Q = s · (g 2 − 1)bx + (bg − 1)gx , defines a polynomial vector field. If one requires P and Q to have rela- tively prime entries then s is unique up to a factor in C∗ . (b) This vector field admits √ the elementary first integral v + log u and the integrating factor ( g − 1 · s)−1 . 2 (c) If this vector field admits a Darboux first integral then it admits a ratio- nal first integral. For the proof see Section 4; we also give some examples which do not admit a rational first integral. Turning to extensions of prime degree p > 2, we will show that the hy- pothesis of Proposition 1 is never satisfied. We have: Theorem 4. Let F = K[u] = K[v] be an extension of prime degree p, with v of constant trace and u of constant norm, and consider a polynomial vector field X admitting the elementary first integral (5). If X admits no rational first integral then p = 2. The proof will also be given in Section 4. While this result does not provide complete information on the possible elementary first integrals of type (5) (and the corresponding polynomial vector fields), it indicates that such cases are indeed exceptional. 2. Notions and known facts We start by recalling some notions and facts (see the appendix for more). We sketch some proofs, for easy reference and the reader’s convenience. Definition 1. We call a function ∏ exp(R) · Sici , with rational functions R and Si and complex constants ci , a Darboux func- tion. The notions of Darboux first integral and Darboux integrating factor are then self-explaining. Lemma 1. Let (1) be given, and H an analytic first integral of this vector field on an open set U . If the nonzero polynomial vector field e = Pe ∂ + Q X e ∂ ∂x ∂y also admits the first integral H on U then there exists a rational function R e = R · X. such that X ELEMENTARY FIRST INTEGRALS 5 e such that Proof. There exist analytic µ and µ µX = XH = −Hy ∂/∂x + Hx ∂/∂y = µ e eX, e = R · X on a nonempty open subset of C2 , see Section 5. Hence one has X and (e.g.) R = Pe/P shows that R is rational.  The vector fields with non-constant Darboux integrating factor (3) (with fixed polynomials and exponents) form a linear space F. The so-called trivial vector fields admitting (3) form a subspace F 0 ⊆ F. To construct these, start with an arbitrary polynomial g and define ( ) Zg = Zg(d1 ,··· ,dr ) = Hamiltonian vector field of g/ f1d1 −1 · · · frdr −1 . Then, according to [5], the trivial vector field r ∑ f f1d1 · · · frdr · Z g = f · Xg − (di − 1)g · Xfi (6) fi i=1 (with Xh denoting the Hamiltonian vector field of a function h) is polynomial and admits (3) by construction. Proposition 2. If d1 , . . . , dr are rational numbers then every element of F 0 admits a rational (hence elementary) first integral. ( ) Proof. Some integer power of g/ f1d1 −1 · · · frdr −1 is a rational function.  We remark that one has F = F 0 whenever the geometry of the underlying curves (including the line at infinity) is nondegenerate; see [4, 5]. In such cases the above Proposition applies. Next we recall a precise description of polynomial vector fields which admit a Darboux first integral. Theorem 5. (See Chavarriga et al. [2], and [9].) Let H(x, y) = f1λ1 · · · frλr exp(g/(f1n1 · · · frnr )) be a Darboux function with λ1 , · · · , λr ∈ C, n1 , · · · , nr ∈ N ∪ {0} and g ∈ C[x, y] coprime with fi whenever ni ̸= 0. Then H is a first integral of the polynomial vector field r ( r ) ∏ ∑ b= X f nk +1 · λk Xf /fk + Z (n1 +1,...,nr +1) k k g k=1 k=1 ∏ −(n +1) which, in turn, admits the integrating factor rk=1 fk k . Moreover, any polynomial vector field admitting the first integral H admits a rational inte- grating factor. 6 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER Sketch of proof. One verifies (as in [9]) directly that     ( r ) r   ∏ n ∑   ∏r   ∑r  ∏  r   r ∏ b= X fl l λi  fj  Xfi −g ni  fj  Xfi + fj  Xg     l=1 i=1 j=1 i=1 j=1 j=1 j ̸= i j ̸= i admits the first integral H. Comparison with equation (6) shows that X b admits a representation as stated. The remaining assertions are obvious, resp. follow from Lemma 1.  Corollary 1. If the vector field (1) admits a Darboux first integral but not a rational first integral then it admits (up to constant multiples) a unique integrating factor, and this integrating factor is rational. As for a converse, we have the following result which easily follows from a theorem of Rosenlicht. Proposition 3. If the vector field (1) admits a rational integrating factor then it admits a Darboux first integral. Proof. There exists a rational function µ such that µ · X is the Hamiltonian vector field of some analytic function H, hence Hx and Hy are rational. By Rosenlicht [13], Theorem 3, letting αD = Hx for D = ∂/∂x and αD = Hy ∑ = ∂/∂y, there exist ui and v in K, and ci ∈ C such that H = for D v + ci log ui ; thus exp H is a Darboux first integral.  3. Elementary and Darboux first integrals In this section we will prove Theorem 2. By Theorem 1 we may assume there exists an elementary first integral for X of the form (2), which is built from algebraic functions over K = C(x, y), and their logarithms. Given an algebraic function w, we denote its minimal polynomial by d ∑ Mw (T ) := T d + gi T d−i ∈ K[T ]. i=1 Here −g1 is the trace of w and (−1)d gd is the norm of w. (Trace and norm are also been used in Prelle and Singer [10]). The other zeros of Mw (in a suitable extension field F of K) are called the conjugates of w. (For algebraic notions and facts the reader may consult e.g. Lang [8].) The following observation is well-known. Lemma 2. If an algebraic function w is a first integral of (1) then all nonconstant coefficients of its minimal polynomial are first integrals of (1). Hence there exist rational first integrals. ELEMENTARY FIRST INTEGRALS 7 Proof. Indeed, X(w) = 0 implies 0 = X(Mw (w)) ( ) ∑ ∑ = X(w) · dwd−1 + d−1 i=1 (d − i)g i w d−i−1 + d i=1 X(gi )w d−i ∑d d−i , = i=1 X(gi )w which forces all X(gi ) = 0 by uniqueness of the minimal polynomial.  We note a few more useful properties. Lemma 3. (a) If the polynomial vector field X admits an elementary first integral (2) then it admits an algebraic integrating factor µ in F := K [v, u1 , . . . , um ]. (b) Moreover, if X admits no rational first integral and F = K[µ] with [F : K] = d then µd ∈ K. Proof. (i) We first show that the partial derivatives of v and all ui are contained in K[v, u1 , . . . , um ]. Indeed, let e ∑ e v + hi v e−i = 0; h1 , . . . , he ∈ K i=1 be a relation of minimal degree. Differentiate to obtain e−1 ∑ e ∑ 0 = vx · (ev e−1 + hi (e − i)v e−i−1 ) + hi,x v e−i i=1 i=1 and hence vx ∈ K(v) = K[v], see for more details Proposition 1.4 in Ch. V of [8]. The same proof applies to vy , and by the same arguments we obtain that ui,x and ui,y lie in K[ui ] for 1 ≤ i ≤ m. (ii) Thus, setting ∑ H := v + ci log(ui ) we see that Hx and Hy are contained in F. Since H is a first integral for X, we have Hy = µ · P, −Hx = µ · Q with some analytic function µ. Since Hx , Hy , P, Q are all contained in F, the same must hold for µ. Thus part (a) is proven. (iii) We turn to part (b). From the proof of Proposition 2 in Prelle and b is a normal extension of F with degree Singer [10] one sees : If F b [F : K] = n then µ ∈ K. But since K contains all nth roots of unity, n the polynomial T n − g splits over F, and therefore F = K[µ] itself is normal over K (see e.g. Lang [8], Ch. V, Thm. 3.3). Now one can invoke the argument from Prelle and Singer again.  8 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER Since we are interested in vector fields which do not admit a rational first integral, we may assume that the vector field X given in (1) admits an elementary first integral (2) with m > 0; equivalently we may require ( m ) ∏ ci X exp(v) · ui = 0 i=1 with m > 0. We may and will assume that the ci are linearly independent over the rational number field Q. Proof of Theorem 2. (i) Let F be the smallest normal extension of K which contains v and all ui , and denote its Galois group by G. Since K- automorphisms of F commute with derivations (see e.g. Rosenlicht [11], Proposition and p. 156), we also have ∑m X (σ(ui )) X (σ(v)) + ci = 0, σ(ui ) i=1 for all σ ∈ G, and summation yields ( ) m (∏ ) ∑ ∑ X σ∈G σ(ui ) X σ(v) + ci ∏ = 0. σ∈G i=1 σ∈G σ(ui ) ∑ Since R := ∏ σ∈G σ(v) is a positive integer multiple of the trace of v, and Si := σ∈G σ(ui ) is a power of the norm of ui with a positive integer exponent, 1 ≤ i ≤ m, this identity ∑ involves only rational func- tions.1 In∏particular, whenever R + ci log Si ̸= const., equivalently exp(R) · Sici ̸= const., we have a Darboux first integral. Note that whenever all ui have constant norm then the trace of v must also be constant since there exists no rational∏ first integral by assumption. (ii) We show in detail that exp(R) · Sici ̸= const. whenever some uk has non-constant norm. Thus let T be an irreducible polynomial and mk a nonzero integer such that Sk = T mk · Sek , with T dividing neither numerator nor denominator of Sek . Likewise we have Si = T mi · Sei , with an integer mi and T dividing neither numerator nor denominator ∏ of Sei , for all i ̸= k. As a preliminary step we show that Sici ̸= const. Due to the linear independence of the ci over Q and mk ̸= 0 we have ∏ c ∑ ∏ c ∑ Si i = T ci mi · Sei i , ci mi ̸= 0, and it suffices to show that ∏ e ∑ T · S ∗ := T · Se i ̸= const., i with ei := ci / cj mj . There exists (x0 , y0 ) ∈ C2 be such that T (x0 , y0 ) = 0 and S ∗ is analytic and does not vanish in this point. Thus T · S ∗ has a zero in (x0 , y0 ) but is not identically zero. [ ] 1The positive integer in question is the degree F : F b , with F b ⊂ F denoting the splitting field of v, respectively of ui . ELEMENTARY FIRST INTEGRALS 9 (iii) We next show the full assertion. Let R = T ℓ · R,e with T not dividing numerator or denominator of R. e Whenever ℓ ≥ 0 the argument from above still works to prove non-constancy. Thus assume ℓ < 0 and consider ∑ R∗ := T −ℓ (R + ci log Si ) = R e + (∑ ci mi ) T −ℓ log T + T −ℓ ∑ ci log Sei . There exists (x0 , y0 ) ∈ C2 at which Re and all Sei are analytic and nonzero but T (x0 , y0 ) = 0. Let γ(t) = (x0 , y0 ) + t · a, with a ∈ C2 parameterize a transversal line segment to the curve defined by T = 0. Then standard growth estimates yield ∑ e lim R∗ (t) = R(0) ̸= 0, thus |R + ci log Si | → ∞ as t → 0, t→0 and non-constancy follows as asserted. Thus part (a) is proven. (iv) The assertion of part (b) follows by Theorem 5 and Corollary 1.  4. The simplest exceptional setting In this section we will discuss exceptional vector fields which, by definition, admit an elementary first integral but not a Darboux first integral. We restrict attention to the simplest setting, starting from a first integral v + log u of X as in relation (5), with u and v nonconstant and algebraic over K (but not both in K), with u of constant norm and v of constant trace. Since u may be changed to α · u with some α ∈ C∗ and v may be changed to v + β with some β ∈ C, we may assume that u has norm one and v has trace zero. 4.1. Some generalities. In this subsection we prove that K[u] ∩ K[v] ̸= K under the assumptions of Proposition 1. We first remark that neither u nor v are elements of K, since this would imply u ∈ C (resp. v ∈ C) and the existence of a rational first integral follows from Lemma 2. The assertion of Proposition 1(a) is then a direct consequence of the following general fact; note that X(u)/u and X(v) are contained in the intersection due to (5). Lemma 4. Let s ̸= 0 be algebraic over K with constant norm, Y a derivation of K[s] extending a derivation of K, and Y (s)/s ∈ K. Then Y (s) = 0, hence s is constant unless Y admits a rational first integral. Proof. The minimal polynomial of s yields a relation sm + g1 sm−1 + · · · + gm−1 s + gm = 0 with all gi ∈ K (g0 := 1) and constant gm . Applying the derivation yields m ∑ m−1 ∑ 0 = Y (s) · gi−1 (m − i + 1)sm−i + Y (gi )sm−i , i=1 i=1 10 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER in view of Y (gm ) = 0. Letting Y (s) = h · s with h ∈ K, one finds m−1 ∑ mhsm + ((m − i)hgi + Y (gi )) sm−i = 0. i=1 On the other hand m ∑ mhsm + mhgi sm−i = 0 i=1 with the same leading coefficient. By uniqueness of the minimal polynomial, all coefficients must be equal, which implies mhgm = 0, and thus h = 0. So Y (s) = 0 and consequently s ∈ C unless Y admits a first integral in K.  Part (b) of Proposition 1 is a direct consequence of part (a) and multi- plicativity of the degree for field extensions. Proposition 4. Let F be of prime degree p, and let X admit an elementary first integral (5) which is not Darboux. Additionally, assume that X admits no rational first integral. Then the integrating factor µ of X (which is unique up to a nonzero scalar) satisfies F = K[µ] and there exists g ∈ K which is not a pth power in K such that µp − g = 0. (7) In particular F : K is a cyclic Galois extension of degree p. Proof. This is a direct consequence of Lemma 3. Note that µ ̸∈ K by Proposition 3.  4.2. Quadratic extensions. In this subsection we consider the smallest possible degrees of K[u] and K[v]. Thus in view of Proposition 1 we assume relation (5) with u and v contained in a degree two extension F = K[u] = K[v] of K. With u of norm one, there exists g ∈ K such that u2 + 2g · u + 1 = 0. (8) This relation may be rewritten in the form (u + g)2 = g 2 − 1 (9) which in turn implies u u ux = − gx , uy = − gy . (10) u+g u+g Moreover v = a + bu, with a, b ∈ K and b ̸= 0 (11) by Proposition 1. Using (8) and (11) we obtain v 2 = a2 + 2abu + b2 u2 2 2 ( 2 − b 2) + (2a − 2gb)bu = (a ) = a − b − 2a(a − gb) + 2(a − gb)v ELEMENTARY FIRST INTEGRALS 11 and trace zero forces a = gb, therefore v = b(g + u). Proof of Theorem 3. The assertions regarding u and v have already been shown. We determine the Hamiltonian vector field of H := b(u + g) + log u. With (10) and (9) one finds Hx = bx (u + g) + b(ux + gx ) + ux /u = bx (u + g) +(bggx − gx /(u + g) ) = (u √+ g)−1 · bx (g 2 − 1) + gx (bg − 1) ( ) = 1/ g 2 − 1 · bx (g 2 − 1) + gx (bg − 1) and a similar expression for Hy . For the rational vector field ( ) b −by (g 2 − 1) − gy (bg − 1) X= bx (g 2 − 1) + gx (bg − 1) there obviously exists a unique (up to a scalar factor) rational function s b is polynomial with relatively prime entries. Thus part (a) such that s · X holds, and (b) is satisfied by construction. To prove part (c), assume that the vector field admits a Darboux first integral. Then Corollary 1 yields a contradiction since the integrating factor from part (b) is not rational.  Theorem 3 describes all polynomial vector fields which admit an excep- tional elementary first integral of type (5) with u and v in a quadratic extension of K. One may start with arbitrary rational functions g (subject to u ̸∈ K) and b ̸= 0. But one has to ascertain that the vector field admits no rational first integral. To verify that there actually exist such vector fields, we discuss two examples. Example 1. This is known from Example 2 of Prelle and Singer [10]: Let g = x and b = y. The vector field ( ) b 1 − x2 X= xy − 1 √ admits the integrating factor 1/ 1 − x2 and the elementary first integral √ ( √ ) H(x, y) = y · x2 − 1 + log −x + x2 − 1 . Prelle and Singer [10] showed by differential-algebraic arguments that no rational first integral exists. Example 2. The polynomial differential system ẋ = 1 − x2 , ẏ = 1 − x2 − xy, admits the integrating factor (x2 − 1)−3/2 and the elementary first integral √ y ( √ ) H = e x2 −1 / x + x2 − 1 12 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER √ (or equivalently log H = √xy2 −1 − log(x + x2 − 1)). In order to show that this system has no Darboux first integral, it is again sufficient to prove there is no rational first integral. We proceed here using an argument different from Prelle and Singer. Obviously the only stationary points of the system are (1, 0) and (−1, 0), and the lines defined by x ± 1 = 0 are invariant algebraic curves with cofac- tors −x ± 1, respectively. We show that no other∑ invarianti algebraic curves exist. Thus assume that f (x, y) = 0, with f = N i=0 ϕi (x)y ∈ C[x, y], defines an invariant algebraic curve C and let (x0 , y0 ) ∈ C with x0 ̸∈ {−1, 1} and ϕN (x0 ) ̸= 0. Then going to the splitting field of f over C(x), one gets f (x, y) = ϕN (x0 )(y − ψ1 (x)) · · · (y − ψN (x)) with ψ1 , · · · , ψN algebraic functions. We may assume that C is defined by y = ψ1 (x) in a neighborhood of (x0 , y0 ). On the other hand the first integral H implies that ( ( √ )) √ y = c + log x + x2 − 1 x2 − 1 near (x0 , y0 ), with some c ∈ C. Since this is a transcendental function, we have a contradiction. To summarize: The only possible rational first integrals have the form (x − 1)m (x + 1)n , with integers m and n. But any integer linear combination m · (−x + 1) + n · (−x − 1) of the cofactors is nonzero unless m and n are both zero. Therefore, no rational first integral exists. 4.3. Extensions of odd prime degree. Finally, we will prove Theorem 4. We start with an auxiliary result. Lemma 5. Let p be a prime number, g ∈ K not a pth power in K, and F = K[µ] with µp = g. Then C(x) is algebraically closed in F; thus every ρ ∈ F that is algebraic over C(x) already lies in C(x). Proof. Since g is not constant, we may assume that gy ̸= 0. Moreover we may assume that there is a prime polynomial q such that g = q ℓ ge, qy ̸= 0, q divides neither numerator nor denominator of ge and ℓ ̸= 0 is not an integer multiple of p. (i) We note the identity 1 gy µy = ·µ p g which follows from logarithmic differentiation of µp = g. (ii) We first show that ρy = 0. Indeed, from the minimal polynomial of ρ over C(x) one obtains a relation m−1 ∑ ρm + ai ρm−i = 0; a1 , . . . , am−1 ∈ C(x), i=0 ELEMENTARY FIRST INTEGRALS 13 of smallest degree m, and differentiation yields ( m−1 ) ∑ 0 = ρy mρm−1 + (m − i)ai ρm−i−1 i=0 with the term in brackets nonzero. (iii) Now consider the unique representation p−1 ∑ ρ= bi µi with bi ∈ K. i=0 Using part (ii) we find p−1 ∑ p−1 ∑ 0 = ρy = bi,y µi + ibi µi−1 µy i=0 i=1 p−1 ( ∑ ) i gy = b0,y + bi,y + bi µi p g i=1 Since all coefficients in the last sum must be zero, we find that ∂ ( p i) b g = 0, 0 ≤ i < p. ∂y i Thus b0 ∈ C(x), and moreover there exist ki ∈ C(x) such that bi (x, y)p = ki (x)/g(x, y)i , 1 ≤ i < p. (iv) Now assume ki ̸= 0 and let q ∈ C[x, y]\C[x] be as above. Let bi = q mi ·ebi with q dividing neither numerator nor denominator of ebi . Comparing exponents of q yields p · |mi | = i · |ℓ|. This is a contradiction, since p divides neither factor on the right-hand side. Therefore all ki = 0, and ρ ∈ K.  Proof of Theorem 4. Assume that the prime number p satisfies p > 2, and that v + log u is an elementary first integral, with F = K[u] = K[v] of degree p over K. Let µ be an integrating factor for system (1). Then µp = g ∈ K by Lemma 3, and we have, by the integrating factor condition, ux vx + = −µQ. u Denote by ζ a primitive pth root of unity. Applying the automorphism σ defined by µ 7→ ζµ to this identity (and recalling that σ commutes with all derivations) we get σ(u)x σ(v)x + = −ζµ Q σ(u) σ(u)x ⇒ ζ −1 σ(v)x + ζ −1 = −µQ. σ(u) 14 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER Combining these relations, we obtain ( ) σ(u)x ux v − ζ −1 σ(v) x − ζ −1 + = 0. σ(u) u To this relation one can apply the Theorem from Rosenlicht [12], with ∂ the identifications k := C(x), K = F, t = y and .′ = ∂x (hence t′ = 0 ∈ k). The hypotheses of this theorem are satisfied, since by Lemma 5, k is algebraically closed in K, and due to p > 2 we have that 1 and ζ −1 are linearly independent over the rational number field. Rosenlicht’s theorem now implies that u ∈ k = C(x). This in turn implies F = K[u] = K; a contradiction.  In the proof of Theorem 4 we did not use all the available information. We collect some facts to conclude the paper, and to point towards further research. Remark 1. (a) In the situation of Lemma 3 we have K ⊆ K[µ] ⊆ F ⊆ Fb and K[µ] : K is Galois as well as F b : K. Therefore, with the Galois group G of F, b K[µ] is the fixed point field of a normal subgroup H of G, and G/H is cyclic of order d. This shows that strong restrictions exist with regard to the possible field extensions F related to elementary first integrals of the form (2) which are not logarithms of Darboux first integrals. (b) In particular, Theorem 4 characterizes every field extension F : K related to an elementary first integral which has the form (5), but is not a logarithm of a Darboux first integral, with the property that its smallest normal extension F b has a simple Galois group. 5. Appendix In this section we recall some familiar notions, terminology and facts. Definition 2. Let U be a nonempty open subset of C2 . We say that a nonconstant (possibly multivalued) function H : U −→ C is a first integral of the vector field X on U if and only if the relation X(H) = 0 holds on U. Given some nonconstant analytic function g on U , the analytic vector field ∂g ∂ ∂g ∂ Xg := − + ∂y ∂x ∂x ∂y is called the Hamiltonian vector field of g. By construction it admits the first integral g. We say that a nonconstant (multivalued) function µ : U −→ C is an integrating factor of the vector field X on U if and only if X(µ) = −(Px + Qy )µ on U ; in other words div(µ · X) = 0. ELEMENTARY FIRST INTEGRALS 15 If H is a first integral of the vector field X, then there is a unique inte- grating factor µ satisfying ∂H ∂H µP = and µQ = − . ∂y ∂x Next we recall some properties of invariant algebraic curves. Definition 3. Let f ∈ C[x, y] be an irreducible polynomial. The algebraic curve defined by f (x, y) = 0 is an invariant algebraic curve of the polynomial vector field (1) if there exists a polynomial K ∈ C[x, y] (called the cofactor of f ) such that Xf = Kf. Such a condition is necessary and sufficient for the curve to be an invariant set, thus a union of trajectories of the vector field (1). If the vector field (1) admits an integrating factor (3) then the algebraic curves defined by fi = 0 are invariant. Moreover, if Ki is the cofactor of fi , respectively, then ∑ di Ki + div X = 0. We are furthermore concerned with elementary first integrals of (1). Thus consider the differential field K = C(x, y) with derivations ∂/∂x and ∂/∂y. ∂ ∂ Every derivation of K has the form Y = R ∂x + S ∂y with rational R and S. Definition 4. (a) An extension field L of K is called elementary if there is a finite tower of extension fields K = L0 ⊂ L1 ⊂ · · · ⊂ Ln = L such that Li+1 is obtained from Li by adjoining an algebraic element, an exponential (i.e. an element w such that Y (w)/w ∈ Li for some derivation Y ) or a logarithm (i.e. an element w such that Z(w) = Z(a)/a for some a ∈ Li and some derivation Z). (b) An extension field L is called Liouvillian over K if there is a finite tower of extension fields K = L0 ⊂ L1 ⊂ · · · ⊂ Ln = L such that Li+1 is obtained from Li by adjoining an algebraic element, an integral element (i.e. an element w such that Z(w) = a for some a ∈ Li ), or an ex- ponential. Every element of L is called a Liouvillian function of two variables. One may regard such field extensions as abstract objects, but in the present work we realize them (locally, on suitable open subsets of C2 ) as analytic functions of two variables. The class of Liouvillian functions obvi- ously contains the class of elementary functions, and the inclusion is proper. Acknowledgements The second author is partially supported by an AGAUR grant 2014SGR- 568. The second and third author are also supported by MINECO-FEDER grant MTM2016-77278-P, and a MINECO grant MTM2013-40998-P. The third author is partially supported by an AGAUR grant 2014SGR-504, and 16 C. CHRISTOPHER, J. LLIBRE, C. PANTAZI AND S. WALCHER the MECC-FEDER grant MTM2015-65715-P. The fourth author acknowl- edges the hospitality of the Mathematics Department at UAB when this manuscript was prepared. References [1] A. Bostan, G. Chéze, T. Cluzeau, J.-A. Weil: Efficient algorithms for computing rational first integrals and Darboux polynomials of planar polynomial vector fields. Math. Comp. 85 (2016), 1393–1425. [2] J. Chavarriga, H. Giacomini, J. Giné, J. Llibre: Darboux integrability and the inverse integrating factor. J. Differential Eqs. 194 (2003), 116–139. [3] C. 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Singer: Elementary first integrals of differential equations. Trans. Amer. Math. Soc. 279 (1983), 613–636. [11] M. Rosenlicht: Liouville’s theorem on functions with elementary integrals. Pacific J. Math. 24 (1968), 153–161. [12] M. Rosenlicht: On the explicit solvability of certain transcendental equations. Publ. Math. IHES 36 (1969), 15–22. [13] M. Rosenlicht: On Liouville’s theory of elementary functions. Pacific J. Math. 65 (1976), 485–492. [14] M.F. Singer: Liouvillian first integrals of differential equations. Trans. Amer. Math. Soc. 333 (1992), 673–688. 1 School of Computing, Electronics and Mathematics, Plymouth, Devon PL4 8AA United Kingdom E-mail address: C.Christopher@plymouth.ac.uk 2 Departament de Matemàtiques, Universitat Autònoma de Barcelona, Edi- fici C, 08193 Bellaterra, Barcelona, Catalonia, Spain E-mail address: jllibre@mat.uab.cat 3 Departament de Matemàtica Aplicada I, Universitat Politècnica de Cata- lunya, (EPSEB), Av. Doctor Marañón, 44–50, 08028 Barcelona, Catalonia, Spain E-mail address: chara.pantazi@upc.edu 4 Lehrstuhl A für Mathematik, RWTH Aachen, 52056 Aachen, Germany E-mail address: walcher@mathA.rwth.aachen.de