Simple Differential Field Extensions and Effective Bounds

Abstract

We establish several variations on Kolchin’s differential primitive element theorem, and conjecture a generalization of Pogudin’s primitive element theorem. These results are then applied to improve the bounds for the effective Differential Lüroth theorem.

J. Freitag—Thanks Dave Marker, Omar León Sanchez, and Gabriela Jeronimo for useful conversations related to this work. JF was partially supported by NSF MSPRF 1204510.

W. Li—Partially supported by a National Key Basic Research Project of China (2011CB302400) and by grants from NSFC (60821002, 11301519) and thanks the University of California, Berkeley for providing a good research environment during her appointment as a Visiting Scholar.

1 Notation

Throughout this paper, $\mathcal {U}$ will be a fixed sufficiently large saturated differentially closed field of characteristic zero with a derivation operator $\delta$. An element $c\in \mathcal {U}$ such that $\delta (c)=0$ is called a constant. In this paper, all the differential fields under discussion are subfields of $\mathcal {U}$ and subscripts denote differentiation.

Let F be a differential subfield of $\mathcal {U}$ and $S\subset \mathcal {U}$. We denote respectively by F[S], F(S), ${F}\{S\}$, and $F\langle S\rangle$ the smallest subring, the smallest subfield, the smallest differential subring, and the smallest differential subfield of $\mathcal {U}$ containing F and S.

The set S is said to be differentially dependent over F if the set $(\delta ^k a)_{a\in S,k\ge 0}$ is algebraically dependent over F, and otherwise, S is said to be differentially independent over F. In the case $S=\{a\}$, we also say that a is differentially algebraic or differentially transcendental over F respectively. A maximal subset $\varOmega$ of S which is differentially independent over F is said to be a differential transcendence basis of $F\langle S\rangle$ over F. We use $\mathrm{{d.tr.deg}}\,F\langle S\rangle /{F}$ to denote the differential transcendence degree of $F\langle S\rangle$ over F, which is the cardinality of $\varOmega$. Considering F and $F\langle S\rangle$ as algebraic fields, we denote the algebraic transcendence degree of $F\langle S\rangle$ over F by $\mathrm{{tr.deg}}\,F\langle S\rangle /{F}$.

We use $F\{y_1,\ldots ,y_n\}$ to denote the differential polynomial ring over F. Given a differential polynomial $f\in F\{y_1,\ldots ,y_n\}$, the order of f w.r.t. $y_i$ is the greatest number k such that $y_{i}^{(k)}$ appears effectively in f, which is denoted by ${\mathrm{ord}}(f,y_{i})$. And if $y_{i}$ does not appear in f, then we set ${\mathrm{ord}}(f,y_{i})=-\infty$. The order of f is defined to be $\max _{i}\,\mathrm{ord}(f,y_{i})$. A (resp. radical, prime) differential ideal is a (resp. radical, prime) algebraic ideal $\mathcal {I}$ of $F\{y_1,\ldots ,y_n\}$ satisfying $\delta (\mathcal {I})\subset \mathcal {I}$.

By affine space, we mean $\mathbb A^n=\mathcal {U}^n$. An element $\eta =(\eta _1,\ldots ,\eta _n)\in \mathbb A^n$ is called a differential zero of $f\in F\{y_1,\ldots ,y_n\}$ if $f(\eta )=0$. The set of all differential zeros of $\varSigma \subset F\{y_1,\ldots ,y_n\}$, is called a differential variety defined over F, denoted by $\mathbb V(\varSigma )$. All the differential varieties in this paper are assumed to be subsets of $\mathbb A^n$. For a differential variety V which is defined over F, we denote $\mathbb {I}(V)$ to be the set of all differential polynomials in $F\{y_1,\ldots ,y_n\}$ that vanish at every point of V.

A differential ideal $\mathcal {I}\subset F\{y_1,\ldots ,y_n\}$ is prime if and only if it has a generic point, that is, a point $\eta \in \mathbb V(\mathcal {I})$ such that for any $f\in F\{y_1,\ldots ,y_n\}$, $f(\eta )=0 \Leftrightarrow f\in \mathcal {I}$. Let $\mathcal {I}$ be a prime differential ideal with a generic point $(\eta _1,\ldots ,\eta _n).$ Then there exist d and h such that for sufficiently large t,

$${\mathrm{{tr.deg}}}\,F(\eta _i^{(k)}:1\le i\le n;k\le t)/F=d(t+1)+h.$$

The polynomial $\omega _{\mathcal {I}}(t)=d(t+1)+h$ is called the Kolchin polynomial of $\mathcal {I}$ and the corresponding d, h are called the differential dimension and order of $\mathcal {I}$. When $\bar{a}$ is a tuple in a differential field extension of a differential field K, then we write $\omega _{\bar{a} /K} (t)$ for $\omega _{ I (\bar{a} /K) } (t)$ where $I ( \bar{a} /K)$ is the differential ideal of all differential polynomials over K which vanish at $\bar{a}$.

2 Introduction

In this note, we discuss various primitive element theorems for ordinary differential field extensions. The oldest such result we consider goes back to Kolchin [7, p. 728], where in fact he proved the primitive element theorem in the more general partial differential settings. Here, we restrict to consider the ordinary differential field extensions, so for convenience, we state Kolchin’s primitive element theorem in the ordinary differential case as follows:

Theorem 2.1

Let F be a differential field containing at least one nonconstant. Let $E= F \langle a_1, \ldots , a_n \rangle$ and suppose that $\mathrm{{d.tr.deg}}\,E/F=0$. Then there is some $b \in E$ such that $E = F \langle b \rangle .$

Pogudin [12] generalized Kolchin’s theorem to the case that F is a constant field, under the assumption that E contains a nonconstant. In this note, we give a mild generalization of Kolchin’s theorem, and conjecture a generalization of Pogudin’s theorem. We also illustrate how these generalizations are useful for improving the bounds on a problem of effective differential algebra.

Let $F \langle u \rangle$ denote the fraction field of the differential polynomial ring $F \{ u \}$ in one variable. Ritt [14] proved the analog of Lüroth’s theorem:

Theorem 2.2

Let K be a differential field such that $F \subset K \subset F \langle u \rangle$. Then there is some element $g \in K$ such that $K = F \langle g \rangle$.

Ritt’s original formulation is for fields of meromorphic functions, but the general theorem follows from this case via Seidenberg’s embedding theorem [18] (Kolchin first proved the general theorem in [8, 9]). More recent work has focused on computational aspects of the Lüroth’s theorem in the case that K is finitely generated over F. To be more precise, suppose $K=F\langle P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u)\rangle$, then computing a Lüroth generator of K/F and giving order and degree bounds for a Lüroth generator are problems of effective differential algebra. Following Kolchin’s idea, if $A(y)\in K\{y\}$ is the minimal differential polynomial of x over K w.r.t. the canonical ranking, then for any pair $(a, b)\in K^2$ of coefficients of the polynomial A satisfying that $a/b\notin F$, this a/b can serve as a Lüroth generator [9]. Thus, using the language of modern differential characteristic sets, a Lüroth generator can be computed in the following way: Given a characteristic set $Q_1(u)y_1-P_1(u), \ldots ,Q_n(u)y_n-P_n(u)$ of a prime differential ideal $\mathcal {I}\subset F\{u,y_1,\ldots ,y_n\}$ w.r.t. the elimination ranking $u<y_1<\cdots <y_n$, compute a characteristic set $B_1(y_1,y_2),\ldots ,B_{n-1}(y_1,\ldots ,y_n), B_0(y_1,\ldots ,y_n,u)$ of $\mathcal {I}$ w.r.t. the elimination ranking $y_1<\cdots<y_n<u$. Rewrite $B_0(y_1,\ldots ,y_n,u)=\sum _{i}f_i(y_1,\ldots ,y_n)\theta _i(u)$, if $\zeta =\frac{f_{i_1}(P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u))}{f_{i_2}(P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u))}\notin F$, then $K=F\langle \zeta \rangle$. Based on this idea, Gao and Xu [6] gave an algorithmic proof of the differential Lüroth theorem, but did not consider bounds for the degrees or order of the generator. D’Alfonso et al. [3] proved the following effective version of the theorem:

Theorem 2.3

Let F be an ordinary differential field of characteristic 0, u differentially transcendental over F and $K= F\langle P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u)\rangle$, where $P_j, Q_j\in F\{ u \}$ are relatively prime differential polynomials of order at most $e\ge 1$ (i.e. at least one derivative of u occurs in $P_j$ or $Q_j$ for some j) and degree bounded by d such that each $P_j/Q_j\notin F$. Then, any Lüroth generator v of K / F can be written as the quotient of two relatively prime differential polynomials $P(u), Q(u)\in F\{u\}$ with order bounded by $\min \{\mathrm{ord}(P_j/Q_j): 1\le j \le n\}$ and total degree bounded by $\min \{(d+1)^{(e+1)n}, (nd(e+1)+1)^{2e+1}\}$.

The connection between the differential Lüroth theorem and the primitive element theorem is related to improving the degree bounds. We should note that our manipulations are not designed to attack the problem of bounding the order, but note that as Kolchin proved in [8], any two Lüroth generators $\omega _1$ and $\omega _2$ are related by the formula $\omega _2=(a\omega _1+b)/(c\omega _1+d)$ for some $a,b,c,d\in F$, so any two Lüroth generators should have the same order [3, see the remarks at the end of Subsect. 3.1]. Our technique is most easily employed in the case that the field F posseses a nonconstant element. In this case, the ideas of our techniques essentially derive from a mild generalization of Theorem 2.1. In the case that F is a constant differential field, our ongoing work is related to an attempt to generalize Pogudin’s recent primitive element theorem [12].

In the case that F contains a nonconstant element, we improve the degree bounds as follows:

Theorem 2.4

Let F be an ordinary differential field of characteristic 0 containing a nonconstant element. Let u be differentially transcendental over F and $K= F\langle P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u)\rangle$, where $P_j, Q_j\in F\{ u \}$ are relatively prime differential polynomials of order at most $e\ge 1$ (i.e. at least one derivative of u occurs in $P_j$ or $Q_j$ for some j) and degree bounded by d such that each $P_j/Q_j\notin F$ In Theorem 2.3. Then, any Lüroth generator v of K over F can be written as the quotient of two coprime differential polynomials $P(u), Q(u)\in F\{u\}$ with degree bounded by

$$\min \{(\lceil {n/2}{\rceil } \cdot d+1)^{2(e+1)}, (d+1)^{n(e+1)} , (nd(e+1)+1)^{2e+1} \}.$$

In the case that the base field consists of constants, we improve the bound as follows:

Theorem 2.5

Suppose F is a field of constants, u be differentially transcendental over F and $K= F\langle P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u)\rangle$ with $P_i,Q_i$ satisfying the same conditions as in Theorem 2.4. Then the total degree of a Lüroth generator of K over F is bounded by

$$\min \{(d(n+e-1)+1)^{2(e+1)}, (nd(e+1)+1)^{2e+1} \}.$$

The paper is organized as follows. In Sect. 3, we will prove various primitive element theorems for differential fields. In Sect. 4, we will utilize our embedding results to establish the improved bounds for the differential Lüroth’s Theorem.

3 Variations of the Primitive Element Theorem for Differential Field Extensions

Throughout this section, all differential fields which appear will be assumed to be subfields of $\mathcal U$, the fixed sufficiently large saturated differentially closed field of characteristic zero given in Sect. 1.

Lemma 3.1

[15, p. 35]. Suppose F contains at least one nonconstant element. If $f \in F\{u\}$ is a nonzero differential polynomial with order r, then for any nonconstant $\eta \in {F}$, there exists an element $c_{0}+c_{1}\eta +c_{2}\eta ^2+\cdots +c_{r}\eta ^{r}$ which does not annul f, where $c_{0}, \ldots , c_{r}$ are constants in F.

Remark 3.2

Note that in Lemma 3.1, we can always select the $c_i$ from the rational number field $\mathbb Q$. Indeed, let $x_0,\ldots ,x_r$ be arbitrary constants, i.e., the $x_i$ are algebraically independent over F and $x_i'=0$. Since $f(\sum _{i=0}^rc_{i}\eta ^{r})\ne 0$, $g(x_0,\ldots ,x_n)=f(x_{0}+x_{1}\eta +x_{2}\eta ^2+\cdots +x_{r}\eta ^{r})$ is a nonzero polynomial in $F\langle \eta \rangle [x_0,\ldots ,x_r]$. Since $\mathbb Q$ is an infinite field, by induction on r, it is easy to show that there exists $(d_0,\ldots ,d_r)\in \mathbb Q^{r+1}$ such that $f(d_{0}+d_{1}\eta +d_{2}\eta ^2+\cdots +d_{r}\eta ^{r})\ne 0$.

Also, if $f \in F\{u_1,\ldots ,u_n\}$ is a nonzero differential polynomial with order bounded by r, then for any nonconstant $\eta \in {F}$, there exist $c_{ij}\in \mathbb Q\,(1\le i\le n;0\le j\le r)$ such that $f(\sum _{i=0}^rc_{0i}\eta ^{i},\ldots ,\sum _{i=0}^rc_{ni}\eta ^{i})\ne 0$. We justify this by induction on n. The above paragraph shows that it is valid for $n=1$. Suppose it holds for $n-1$. Regard $f(u_1,\ldots ,u_n)$ as a polynomial in $u_1,\ldots ,u_{n-1}$ with coefficients in $F\{u_n\}$, then by the induction hypothesis, there exist $c_{ij}\in \mathbb Q\,(1\le i\le n-1)$ such that $g(u_n)=f(\sum _{j=0}^rc_{1j}\eta ^j,\ldots ,\sum _{j=0}^rc_{n-1,j}\eta ^j,u_n)\ne 0$. Thus, from the case $n=1$, there exist $c_{nj}\in \mathbb Q$ such that $g(\sum _{j=0}^rc_{nj}\eta ^j)=f(\sum _{j=0}^rc_{1j}\eta ^j,\ldots ,\sum _{j=0}^rc_{nj}\eta ^j)\ne 0$.

Lemma 3.3

Let $F_1 \subset F$ be differential fields. Suppose that $F_1$ is not a field of constants. Then the Kolchin closure of $F_1^n$ over F is $\mathbb A^n$.

Proof

It suffices to show that $\mathbb I(F_1^n)$, the set of all differential polynomials over F which vanish at $F_1^n$, is the zero differential ideal. Since F contains at least a nonconstant, say $\eta$, for any nonzero $f\in F\{y_1,\ldots ,y_n\}$, by Remark 3.2, there exist $c_{ij}\in \mathbb Q\,(0\le j\le {\mathrm{ord}}(f))$ such that $f(\sum _{j}c_{1j}\eta ^j,\ldots ,\sum _{j}c_{nj}\eta ^j)\ne 0$. Note that for each i, $\sum _{j}c_{ij}\eta ^j\in F_1$. Thus, $\mathbb I(F_1^n)=[0]$. Hence, the Kolchin closure of $F_1^n$ is $\mathbb V(0)=\mathbb A^n$.    $\square$

Note that Kolchin’s proof [7, p. 728] for Theorem 2.1 as well as Seidenberg’s proof for [17, Theorem 1] implies the following result:

Proposition 3.4

Let $L = F\langle \alpha _1,\ldots ,\alpha _n \rangle$ and $\mathrm{{d.tr.deg}}\, L/F=0$. Let F be a subfield of F such that F contains a nonconstant. Then, there exist $c_1,\ldots ,c_n\in F$ such that $L = F\langle c_1\alpha _1+\cdots +c_n\alpha _n \rangle$.

The following result is a straightforward implication of Proposition 3.4 and here we will give a new proof from the geometric point of view.

Proposition 3.5

Let $K= F \langle \alpha _1 , \ldots , \alpha _n \rangle$ with F containing at least one nonconstant and $\mathrm{{d.tr.deg}}\, K/F=d>0$. Assume without loss of generality that $\alpha _1, \ldots , \alpha _d$ is a differential transcendence basis of K over F. Then for any nonconstant subfield $F_1$ of F, there exist $c_{d+1},\ldots ,c_{n}\in F_1$ such that $K= F \langle \alpha _1, \ldots , \alpha _d ,\sum _{i=d+1}^nc_i\alpha _{i} \rangle$.

Proof

Consider the affine differential variety $V \subset \mathbb A^{n-d}$ given by the locus of $\alpha _ {d+1} , \ldots , \alpha _n$ over $F\langle \alpha _1 , \ldots , \alpha _d \rangle$. The variety V is $F\langle \alpha _1 , \ldots , \alpha _d \rangle$-definable each of whose points are differentially algebraic over $F\langle \alpha _1 , \ldots , \alpha _d \rangle$. Let $\bar{u} = (u_{d+1}, \ldots , u_n )$ be a tuple which are differentially independent over $F\langle \alpha _1 , \ldots , \alpha _d \rangle$. Then we claim that the map $\phi _{\bar{u}} : V \rightarrow \mathbb A^1$ given by $\bar{x} = (x_{d+1} , \ldots , x_n) \rightarrow \sum _{i=d+1} ^n u_i x_i$ is injective. For if not, then there are two points $\bar{a}=(a_{d+1}, \ldots , a_ n )$ and $\bar{b}=(b_{d+1}, \ldots , b_ n )$ such that $\sum _{i=d+1} ^n u_i a_i = \sum _{i=d+1} ^n u_i b_i$. But $\bar{a}$ and $\bar{b}$ are differentially algebraic over $F\langle \alpha _1 , \ldots , \alpha _d \rangle$, and so $\bar{u}$ is $\delta$-transcendental over $F\langle \alpha _1 , \ldots , \alpha _d , \bar{a}, \bar{b} \rangle$. But now we have a contradiction, because $\sum _{i=d+1} ^n u_i (a_i-b_i)=0$, and not all of the $a_i-b_i$ can be zero, since $\bar{a} \ne \bar{b}$.

The injectivity of the map $\phi _{\bar{u}}$ on the $F\langle \alpha _1 , \ldots , \alpha _d \rangle$-definable set V is a first order property (over $F\langle \alpha _1 , \ldots , \alpha _d \rangle$) of the tuple $\bar{u}$. Since $\bar{u}$ is generic over $F\langle \alpha _1 , \ldots , \alpha _d \rangle$, it follows by quantifier elimination that for all $\bar{v}$ in an $F\langle \alpha _1 , \ldots , \alpha _d \rangle$-open subset $U \subset \mathbb A^{n-d}$, the map $\phi _{\bar{v}}$ is injective. There is a point $\bar{\gamma }= (\gamma _{d+1}, \ldots , \gamma _n )$ of $F^{n-d}$ in U by Lemma 3.3 applied to F relative to $F \langle \alpha _1, \ldots , \alpha _d \rangle$.

Now let $W \subset \mathbb A^n$ be the locus of $\alpha _1, \ldots , \alpha _n$ over F. Then the map $\pi _{\gamma }: \mathbb A^n \rightarrow \mathbb A^{d+1}$ given by $(x_1, \ldots , x_n) \mapsto (x_1, \ldots , x_d, \sum _{i=d+1} ^n \gamma _i x_i )$ is injective on the fiber above $\alpha _1, \ldots , \alpha _d$ in W (the proper subvariety of W with $x_1=\alpha _1, \ldots , x_d =\alpha _d$). By the genericity of $\bar{\alpha }\in W$ over F, it follows that $\pi _{\gamma }$ is injective on a Kolchin open subset of $\mathbb A^n$. So $F \langle \bar{\alpha }\rangle \cong F \langle W \rangle \cong F \langle \pi _{ \gamma } (W) \rangle \cong F \langle \alpha _1, \ldots , \alpha _d , \sum _{i=d+1} ^n \gamma _i \alpha _i \rangle$ completing the proof.    $\square$

Next, we will establish Proposition 3.5 through the use of differential Chow forms, which enables us to compute $\beta _i$ effectively. Without assuming a transcendence basis beforehand, we restate the proposition as follows:

Proposition 3.6

Assume F is a differential field with at least one nonconstant. Suppose that $K= F \langle \alpha _1 , \ldots , \alpha _n \rangle$ is a finitely generated differential field extension of F such that the differential transcendence degree of K over F is d. Then for any subfield $F\subset F$ containing a nonconstant, there are $\beta _0,\ldots ,\beta _{d} \in K$ which are F-linear combinations of $\alpha _{1} , \ldots , \alpha _n$ such that

$$K= F \langle \beta _0, \ldots , \beta _d \rangle .$$

Proof

Let $\mathcal {I}=\mathbb {I}((\alpha _1,\ldots ,\alpha _n))\subset F\{y_1,\ldots ,y_n\}$. Then $\mathcal {I}$ is of differential dimension d. Let

$$L_i=u_{i0}+u_{i1}y_1+\cdots +u_{in}y_n\,(i=0,1,\ldots ,d)$$

be a system of $d+1$ generic differential hyperplanes where all the $u_{ij}$ are differentially independent over F. Denote

$$u_i=(u_{i0},u_{i1},\ldots ,u_{in})\,(i=0,\ldots ,d)\,\,\text {and}\,\,u=\{u_{ij}:i=0,\ldots ,d; j\ne 0\}.$$

Let $\mathcal {P}=[\mathcal {I},L_0,\ldots ,L_d]\subset F\{y_1,\ldots ,y_n,u_0,\ldots ,u_d\}$. Assume $G(u_0,\ldots ,u_d)$ is the differential Chow form of $\mathcal {I}$ and $\mathrm{ord}(G)=h$. Then by the property of the differential Chow form [5, Lemma 4.10],

$$\frac{\partial G}{\partial u_{00}^{(h)}}y_j-\frac{\partial G}{\partial u_{0j}^{(h)}}\in \mathcal {P}\,\,(j=1,\ldots ,n).$$

Since $\xi =(\alpha _1,\ldots ,\alpha _n;-\sum \limits _{j=1}^nu_{0j}\alpha _j,u_{01},\ldots ,u_{0n};\ldots ;-\sum \limits _{j=1}^nu_{dj}\alpha _j,u_{d1},\ldots ,u_{dn})$ is a generic point of $\mathcal {P}$ and $\frac{\partial G}{\partial u_{00}^{(h)}}\notin \mathcal {P}$, $\frac{\partial G}{\partial u_{00}^{(h)}}(\xi )\ne 0$. Now regard $\frac{\partial G}{\partial u_{00}^{(h)}}(\xi )$ as a differential polynomial in u with coefficients in K, which is nonzero. By Lemma 3.1, there exists $a_{ij}\in F$ (for any $u_{ij}\in u$) such that

$$\frac{\partial G}{\partial u_{00}^{(h)}}(\alpha _1,\ldots ,\alpha _n;-\sum _{j=1}^na_{0j}\alpha _j,a_{01},\ldots ,a_{0n};\ldots ;-\sum _{j=1}^na_{dj}\alpha _j,a_{d1},\ldots ,a_{dn})\ne 0.$$

For each $k=0,1,\ldots ,d$, let $g_k$ be the differential polynomial in $F\{u_{00},\ldots ,u_{d0}\}$ obtained from $\frac{\partial G}{\partial u_{0k}^{(h)}}$ by replacing $u_{ij}\in u$ by $a_{ij}$. Then $g_{0}$ is a nonzero differential polynomial which satisfies $g_0(-\sum _{j=1}^na_{0j}\alpha _j,\ldots ,-\sum _{j=1}^na_{dj}\alpha _j)\ne 0$.

Let $\beta _i=-\sum _{j=1}^na_{ij}\alpha _j$ for $i=0,\ldots ,d$. We claim that $K=F\langle \beta _0,\ldots ,\beta _d\rangle$. Let $\bar{L}_i=u_{i0}+a_{i1}y_1+\cdots +a_{in}y_n\,(i=0,\ldots ,d)$ and

$$\mathcal {P}_1=[\mathcal {I},\bar{L}_0,\ldots ,\bar{L}_d]\subset F\{y_1,\ldots ,y_n,u_{00},\ldots ,u_{d0}\} .$$

Clearly, $\mathcal {P}_1$ is a prime differential ideal with a generic point $(\alpha _1,\ldots ,\alpha _n, \beta _0,\ldots ,\beta _d)$. Since $\frac{\partial G}{\partial u_{00}^{(h)}}y_j-\frac{\partial G}{\partial u_{0j}^{(h)}}\in \mathcal {P}$, it is clear that $g_0y_j-g_j\in \mathcal {P}_1$ for each $j=1,\ldots ,n$. Thus, $g_0y_j-g_j$ vanishes at $(\alpha _1,\ldots ,\alpha _n,\beta _0,\ldots ,\beta _d)$, which implies

$$\alpha _j=g_j(\beta _0,\ldots ,\beta _d)/g_0(\beta _0,\ldots ,\beta _d).$$

Hence, $K=F\langle \beta _0,\ldots ,\beta _d\rangle$.    $\square$

We use the following two examples to illustrate the method given in the proof of Proposition 3.6 to compute the generators of the required forms. In the both examples, differential Chow forms can be computed using either the characteristic set method as described in [5, Remark 4.4] or the algorithms for computing differential Chow forms given in [10].

Example 3.7

Let $F=\mathbb {Q}(x)$ with derivation $\delta =\frac{d}{d x}$. Let $K=\mathbb {Q}(x)\langle \alpha _1,\alpha _2\rangle$ where $\alpha _1, \alpha _2$ are the generic solutions of $y'+1, z'$ respectively. Let $\mathcal {I}=\mathbb {I}((\alpha _1,\alpha _2))$ $\subset F\{y,z\}$. Then $\mathcal {I}$ is of differential dimension 0. Take $L_0=u_0+u_1y+u_2z$. Then the differential Chow form of $\mathcal {I}$ is

$$G=(u_1u_2'-u_2u_1')u_0''-2u_1'(u_1u_2'-u_1'u_2)-u_1''(u_0u_2'-u_2u_0')+u_2''(u_0u_1'-u_0'u_1+u_1^2).$$

So the separant of G is $S_G=u_1u_2'-u_2u_1'$. We can take $u_1=-1$ and $u_2=-x$ which does not annul $S_G$. Hence, $\alpha _1+x\alpha _2$ is a primitive element of K / F, that is, $K=F\langle \alpha _1+x\alpha _2\rangle$.

Example 3.8

Let $F=\mathbb {Q}(x)$ with $\delta =\frac{d }{d x}$. Let $K=\mathbb {Q}(x)\langle u+x, u'+x, u''\rangle$ where u is differentially transcendental over F. Clearly, the differential transcendence degree of K over F is 1. Let $u_i=(u_{i0},u_{i1},u_{i2},u_{i3})\,(i=0,1)$. Then we can compute the differential Chow form $G({u}_0, {u}_1)$ of $\mathbb {I}((u+x, u'+x, u''))\subset F\{y_1, y_2, y_3\}$, which is a differential polynomial of order 2 and differential degree 6. The separant of G is

$$\begin{aligned} S_G= & {} u_{13}(u_{11}u_{12}u_{03}^2-u_{11}u_{13}u_{02}u_{03}-u_{12}u_{11}u_{03}^2+u'_{12}u_{13}u_{01}u_{03}-u'_{03}u_{11}u_{13}u_{02} \\&+\; u'_{03}u_{12}u_{13}u_{01}+u'_{13}u_{11}u_{02}u_{03}-u'_{13}u_{12}u_{01}u_{03}+u'_{02}u_{11}u_{13}u_{03} -u'_{02}u_{13}^2u_{01} \\&-\; u'_{01}u_{12}u_{13}u_{03}+u'_{01}u_{13}^2u_{02}-u_{11}^2u_{03}^2+u_{11}u_{12}u_{02}u_{03}+2u_{11}u_{13}u_{01}u_{03} \\&-\; u_{11}u_{13}u_{02}^2-u_{12}^2u_{01}u_{03}+u_{12}u_{13}u_{01}u_{02}-u_{13}^2u_{01}^2). \end{aligned}$$

We can take $u_{01}= 1, u_{02} = 0, u_{03} = x, u_{11} = x, u_{12} = 1, u_{13} = 1$ which does not annul $S_G$. Hence, $\beta _1=(u+x)+tu'', \beta _2=x(u+x)+(u'+x)+u''$ is a set of generators of K / F, that is, $K=F\langle \beta _1, \beta _2\rangle$.

The proofs of Lemma 3.3 and Proposition 3.5 can be generalized to the case of a differential field with finitely many commuting derivations $\delta _1,\ldots ,\delta _m$, under the assumption that F contains m elements $\beta _1,\ldots ,\beta _n$ whose Jacobian, det$(\delta _i(\beta _j))$, is nonzero [7], but the proof of Proposition 3.6 is not suited for the partial case.

Proposition 3.5 does not hold in the case that the field F is the constant field:

Example 3.9

Let F be the rational number field with the trivial derivation. Let x, y be two constants in a differential extension field of F, which are algebraically independent over F. Consider $K=F\langle x, y \rangle =F(x,y)$. Then K is of differential transcendence degree 0, but the transcendence degree of K over F is 2. Clearly, there is no $a,b\in F$ such that $K=F(ax+by).$

When F is a constant differential field, although Proposition 3.5 is not valid, we have the following similar result, which can be regarded as a consequence of Seidenberg’s proof [17].

Proposition 3.10

Assume F is a differential field of constants. Suppose that $K= F \langle \alpha _1 , \ldots , \alpha _n \rangle$ is a finitely generated differential field extension of F such that the differential transcendence degree of K over F is $d>0$. Suppose $\alpha _1$ is differentially transcendental over F. Then there exist $\beta _1, \ldots , \beta _{d} \in K$ such that $K= F \langle \alpha _1, \beta _1, \ldots , \beta _{d} \rangle$ and each $\beta _i$ is an F-linear combination of $\alpha _2 , \ldots , \alpha _n$ and powers of $\alpha _1$ bounded by h where h is the order of the differential Chow form of $\alpha _2, \ldots , \alpha _n$ over $F \langle \alpha _1 \rangle$. In particular, there exist $c_{ijk}\in \mathbb Q$ such that $\beta _i=\sum _{j=2}^n(\sum _{k=0}^{h}c_{ijk}\alpha _1^k)\alpha _j, i=1,\ldots ,d$ and $K= F \langle \alpha _1, \beta _1, \ldots , \beta _{d} \rangle$.

Proof

Let $K_1=F\langle \alpha _1\rangle$. Then the differential transcendence degree of K over $K_1$ is $d-1$. Consider the differential ideal $\mathcal {I}=\mathbb {I}((\alpha _2,\ldots ,\alpha _n))\subset K_1\{y_2,\ldots ,y_n\}$. Suppose the order of $\mathcal {I}$ is equal to h. Suppose $G(u_0,\ldots ,u_{d-1})$ is the differential Chow form of $\mathcal {I}$, then $\mathrm{ord}(G)= h$. Applying the similar method as in the proof of Proposition 3.6 to $K/K_1$, by Remark 3.2, we can find $c_{ijk}\in \mathbb Q$ such that for $\beta _i=\sum _{j=2}^n(\sum _{k=0}^{h}c_{ijk}\alpha _1^k)\alpha _j$, $K=F\langle \alpha _1,\beta _1,\ldots ,\beta _{d}\rangle$.    $\square$

Also, in the case that F is a constant differential field, one can establish Proposition 3.5 when making an additional assumption on the elements $\alpha _1, \ldots , \alpha _n$. The additional assumption uses terminology from model theory, which we will now introduce. Our conventions are designed to deliver the model theoretic notions in the differential algebraic setting, where some of the notions can be given significantly simpler definitions than in the general setting.

We remind the reader that $\mathcal U$ is a universal differential field. Let $X \mathbb A^n$ be a constructible set in the Kolchin topology over F; that is, X is a boolean combination of affine differential varieties over F. Then we say X is orthogonal to the constants if for any differential field extension K of F, any element c of the constant field $\mathcal C$ of $\mathcal U$, and any $\bar{a} \in X$, we have the equality of the Kolchin polynomials:

$$\omega _{\bar{a} / K \langle c \rangle } (t) = \omega _{ \bar{a} / K } (t).$$

This implies that if c is transcendental over K, then c is transcendental over $K \langle a \rangle$.

The notion defined in the previous paragraph is a special case of the general notion defined in [1, see Ziegler’s article, page 40 for additional details].

Proposition 3.11

Suppose that $K= F \langle \alpha _1 , \ldots , \alpha _n \rangle$ is a finitely generated differential field extension of F such that the differential transcendence degree of K over F is d. Assume without loss of generality that $\alpha _1, \ldots , \alpha _d$ are differentially independent over F. Suppose that $loc((\alpha _{d+1} , \ldots , \alpha _n) /F \langle \alpha _1, \ldots , \alpha _d \rangle )$ is orthogonal to the constants. Then there is $\beta _{d+1} \in K$ such that $K= F \langle \alpha _1, \ldots , \alpha _d ,\beta _{d+1} \rangle$ and $\beta _{d+1}$ is an $\mathbb Q$-linear combination of $\alpha _{d+1} , \ldots , \alpha _n$.

Proof

First, we claim that the general result follows from the case in which $n=d+2$. This follows inductively, noting that $\mathbb Q$-linear combinations of $\mathbb Q$-linear combinations of $\alpha _{d+1} , \ldots , \alpha _n$ are again $\mathbb Q$-linear combinations of $\alpha _{d+1} , \ldots , \alpha _n$ and $\mathbb Q$-linear combinations preserve orthogonality to the constants. So, without loss of generality, assume that $n=d+2$.

Let $X = loc _{F \langle \alpha _1, \ldots , \alpha _d \rangle } (a,b),$ the Kolchin closure of (a, b) over the ground field $F \langle \alpha _1, \ldots , \alpha _d \rangle$. Let $c_1, c_2 \in \mathcal C$ be independent transcendental constants over $F \langle \alpha _1, \ldots , \alpha _d \rangle$. We claim the map

$$\phi _{ \bar{c}}: X \rightarrow \mathbb A^1$$

given by $(x,y) \mapsto c_1 x + c_2 y$ is injective on a Kolchin open subset of X. If this is not the case, there are $(x_1,y_1), (x_2,y_2) \in X$ such that $(x_i, y_i)$ is generic on X over $F \langle \alpha _1, \ldots , \alpha _d , c_1, c_2 \rangle$ (which implies that $c_1,c_2$ are independent transcendentals over $F \langle \alpha _1, \ldots , \alpha _d , x_i , y_i \rangle$ for $i=1,2$) such that

$$ c_1 x_1 + c_2 y _1 = c_1 x_2 + c_2 y _2. $$

But now taking $K= F \langle \alpha _1, \ldots , \alpha _d , c_1, x_1,y_1 \rangle$, we can see that $c_2$ is not transcendental over $K \langle x_2, y_2 \rangle$ as $c_2 = c_1 \cdot \frac{x_2 - x_1 }{y_1 -y_2}$ is in $K \langle x_2, y_2 \rangle$. But this implies:

$$\omega _{(x_2,y_2 ) / K \langle c_2 \rangle } (t) \ne \omega _{ (x_2,y_2 ) / K } (t),$$

and this contradicts the assumption that X is orthogonal to the constants.

So, there is a Kolchin open subset $U \subset X$ such that the map $\phi _{\bar{c} } |_U$ is an injective map. Injectivity is a definable property of the map $\phi _{\bar{c} } |_U$, and it holds for the generic point in $\mathcal C^2$, so for some Zariski open (the Kolchin open subsets of $\mathcal C$ are Zariski open) subset $U_1 \subset \mathcal C$, for all $\bar{c}' \in U_1$, the map $\phi _{\bar{c} '} |_U$ is injective. By the density of $\mathbb Q^2$ in $\mathcal C^2$, there are $\bar{q} =q_1,q_2$ for which $\phi _{\bar{q}} |_U$ is injective and thus gives an isomorphism between the differential function field of X and its image, completing the proof.    $\square$

Remark 3.12

The assumption that $tp(\alpha _{d+1} , \ldots , \alpha _n /F \langle \alpha _1, \ldots , \alpha _d \rangle )$ is orthogonal to the constants is rather difficult to verify in practice. On the other hand, it is folklore of the model theory of differential fields that most differential equation of some order ${\ge }1$ and degree ${\ge }2$ should be strongly minimal and trivial (which implies orthogonality to the constants). For specific instances of results of this nature, see [2, 13].

There is a considerable literature devoted to verifying this condition for various specific differential equations [2, 4, 13]; proving that a given strongly minimal differential equation has trivial forking geometry (and is thus orthogonal to the constants) is also the key to proving that differential closure is not minimal [16]. As far as we can tell, only the results of [4] provide examples which are defined over a differential transcendental, which is the only case pertinent to the differential Lüroth theorem. To give the reader an idea of the hypothesis, we will give a specific example, in the case of two variables in order to keep the technicalities minimal.

So, let

$$P_1/Q_1:=S(u'+u) + R(u'+u) \cdot ((u'+u)')^2 \text {,}$$

where

$$R(y) = \frac{y^2 - 1968 y + 2~654~208}{2 y^2 (y - 1728)^2} \text {,}$$

and

$$S(x)=\left( \frac{x''}{x'}\right) '-\frac{1}{2} \left( \frac{x''}{x'}\right) ^2 \text { }$$

is the Schwarzian derivative. Let

$$P_2/Q_2 = u'+u.$$

Then the type $tp (P_2/Q_2 / F \langle P_1/Q_1 \rangle )$ is the generic solution to the differential equation

$$S(x) + R(x) \cdot ((x)')^2 = P_1/Q_1.$$

By the results of [4], this type is strongly minimal. It follows that the type is nonorthogonal to the constants, since the equivalence relation of nonorthogonality refines transcedence degree on strongly minimal sets (the authors of [4] also prove that this set has trivial forking geometry).

Propositions 3.5 and 3.10 are effective in the sense that the degree of the elements which generate the differential field extension are bounded. We further conjecture the following mild strengthening of Pogudin’s primitive element theorem:

Conjecture 3.13

Assume F is a constant differential field. Suppose that $K= F \langle \alpha _1 , \ldots , \alpha _n \rangle$ is a finitely generated differential field extension of F such that the differential transcendence degree of K over F is $d >0$. Assume without loss of generality that $\alpha _1, \ldots , \alpha _d$ are a differential transcendence basis for K over F. Assume that at least one of $\alpha _{d+1} , \ldots , \alpha _n$ is a nonconstant. Then there is a polynomial $P \in \mathbb Q [ x_{d+1}, \ldots , x_n ]$ such that

$$K = F \langle \alpha _1 , \ldots , \alpha _d, P(\alpha _{d+1} , \ldots , \alpha _n ) \rangle .$$

The above conjecture is a direct consequence of the following stronger conjecture.

Conjecture 3.14

Assume F is a differential field which contains at least one nonconstant. Suppose that $K= F \langle \alpha _1 , \ldots , \alpha _n \rangle$ is a finitely generated differential field extension of F such that each $\alpha _i$ is differentially algebraic over F and at least one $\alpha _i$ is a nonconstant. Then there is a polynomial $P \in \mathbb Q [ x_{1}, \ldots , x_n ]$ such that

$$K =F \langle P(\alpha _{1} , \ldots , \alpha _n ) \rangle .$$

The above conjecture is not true if all $\alpha _i$ are constants. Similar to Example 3.9, if x, y are constants which are independent algebraic indeterminates, $\mathbb Q(t)\langle x,y\rangle \ne \mathbb Q(t)\langle P(x,y)\rangle$ for any $P\in \mathbb Q[x,y]$.

In the current work in progress we hope to establish the conjecture in an effective form (bounding the degree of P); bounding the degree of P might then be used to improve the bounds for the degree of a Lüroth generator while working over a constant differential field.

4 Improving the Bounds in the Differential Lüroth Theorem

In this section, we explain how the results of the previous section can be applied to improve the degree bound for the differential Lüroth theorem.

4.1 The Nonconstant Case

We will work first in the case where F contains some nonconstant element, proving Theorem 2.4. The analysis is simpler in this case and the primitive element style analysis of the previous section yields improved bounds.

Suppose that

$$K= F \langle P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u)\rangle $$

where $P_j, Q_j\in F \{ u \}$ are relatively prime differential polynomials with order satisfying $e=\max \{{\mathrm{ord}}(P_i),{\mathrm{ord}}(Q_j)\}\ge 1$ and total degree bounded by d such that $P_j/Q_j\notin F$ for every $1\le j \le n$.

When $x \in \mathbb N$, let $\lfloor {x}{\rfloor }, \lceil {x}{\rceil }$ denote the standard floor and ceiling functions, respectively. Let

$$K_1 = F \langle P_1(u)/Q_1(u), \ldots , P_{\lfloor {n/2}{\rfloor }}(u)/Q_{\lfloor {n/2}{\rfloor }}(u) \rangle $$

and consider the differential field extension

$$K=K_1 \langle P_{\lfloor {n/2}{\rfloor }+1}(u)/Q_{\lfloor {n/2}{\rfloor }+1}(u), \ldots , P_{n}(u)/Q_{n}(u) \rangle .$$

Since each $P_i(u)/Q_i(u)\notin F$ and $\mathrm{{d.tr.deg}}\,K/F=1$, $\mathrm{{d.tr.deg}}\,K/K_1=0$. Apply Proposition 3.4 to the extension K over $K_1$ with $K_1$ playing the role of F in Proposition 3.4, then we obtain a generator $\beta$ for K over $K_1$ which is an F-linear combination of $P_{\lfloor {n/2}{\rfloor }+1}(u)/Q_{\lfloor {n/2}{\rfloor }+1}(u), \ldots , P_n(u)/Q_n(u)$. Note that this $\beta$ has order at most e and degree at most $\lceil {n/2}{\rceil } \cdot d$. Specifically, the total degree of $\beta$ is bounded by the sum of the degrees of $P_{\lfloor {n/2}{\rfloor }+1}(u)/Q_{\lfloor {n/2}{\rfloor }+1}(u), \ldots , P_n(u)/Q_n(u)$. Here, It may happen that $K=K_1$ and in this case, the obtained $\beta$ may be contained in F. If this happens, we reset $\beta =P_{n}(u)/Q_{n}(u)$.

Now we have $K=K_1\langle \beta \rangle =F\langle P_1(u)/Q_1(u), \ldots , P_{\lfloor {n/2}{\rfloor }}(u)/Q_{\lfloor {n/2}{\rfloor }}(u),\beta \rangle$. Clearly, $\mathrm{{d.tr.deg}}\,K/F \langle \beta \rangle =0$. Applying Proposition 3.4 to the differential field extension

$$K=F \langle P_1(u)/Q_1(u), \ldots , P_{\lfloor {n/2}{\rfloor }}(u)/Q_{\lfloor {n/2}{\rfloor }}(u), \beta \rangle $$

over $F \langle \beta \rangle$, then there is an F-linear combination of

$$P_1(u)/Q_1(u), \ldots , P_{\lfloor {n/2}{\rfloor }}(u)/Q_{\lfloor {n/2}{\rfloor }}(u)$$

which generates K over $F \langle \beta \rangle$. Call this element $\alpha$ and note that $\alpha$ has order at most e and degree at most $\lfloor {n/2}{\rfloor } \cdot d$. Specifically, the total degree of $\alpha$ is bounded by the sum of the total degrees of $P_1(u)/Q_1(u), \ldots , P_{\lfloor {n/2}{\rfloor }}(u)/Q_{\lfloor {n/2}{\rfloor }}(u)$.

Now, we have obtained $K=F\langle \alpha ,\beta \rangle$ with $\max \{\mathrm{ord}(\alpha ),\mathrm{ord}(\beta )\}=e_1\le e$ and $\deg (\alpha ),\deg (\beta )\le \lceil {n/2}{\rceil }\cdot d$. Note it may happen that $e_1=0$. In the following, we show that applying Theorem 2.3 to the differential field extension $F \langle \alpha , \beta \rangle$ over F, the degree of a Lüroth generator is bounded by $(\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$.

Lemma 4.1

The degree of a Lüroth generator of $F \langle \alpha , \beta \rangle$ over F is bounded by $(\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$.

Proof

If $e_1\ge 1$, applying Theorem 2.3 directly to the differential field extension $F \langle \alpha , \beta \rangle$ over F, the bound can be obtained.

Now suppose $e_1=0$ and $\alpha =R_1(u)/S_1(u), \beta =R_2(u)/S_2(u)\in F(u)$. Let $u=z'$, the first derivative of a new element z. Since u is differentially transcendental over F, z is differentially transcendental over F too. Thus, $K=F\langle R_1(u)/S_1(u),R_2(u)/S_2(u)\rangle =F\langle R_1(z')/S_1(z'), R_2(z')/S_2(z')\rangle \subset F\langle z\rangle$. With respect to the new differential indeterminate z, $\max \{{\mathrm{ord}}(R_i,z),{\mathrm{ord}}(S_i,z)\}=1$ which satisfying the conditions in Theorem 2.3. Thus, there exists coprime paris $(P(z), Q(z))\in F\{z\}^2$ with $\mathrm{ord}(P,z), \mathrm{ord}(Q,z) \le 1$ and $\deg (P),\deg (Q) \le (\lceil {n/2}{\rceil } \cdot d+1)^{(1+1)2}\le (\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$ such that $K=F\langle P(z)/Q(z)\rangle$. Since $K=F\langle R_1(u)/S_1(u),R_2(u)/S_2(u)\rangle$ has a Lüroth generator $T_1(u)/T_2(u)$ and by [8, p. 359], the two Lüroth generators P(z) / Q(z) and $T_1(u)/T_2(u)$ are related by the formula $P(z)/Q(z)=(aT_1(u)/T_2(u)+b)/(cT_1(u)/T_2(u)+d)$ for some $a,b,c,d\in F$, Thus, $P(z), Q(z)\in F[z']$. Replacing $z'$ by u in P and Q, we get a Lüroth generator $P_0(u), Q_0(u)$ satisfying ${\mathrm{ord}}({P}_0,u), {\mathrm{ord}}(Q_0,u)=0$ and $\deg (P_0),\deg (Q_0) \le (\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$.    $\square$

Combining the degree bound given in Theorem 2.3 with Lemma 4.1, we obtain the degree of a Lüroth generator of $K= F \langle P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u)\rangle$ over F is bounded by

$$\min \{(\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}, (d+1)^{(e+1)n} , (nd(e+1)+1)^{2e+1} \}.$$

This establishes Theorem 2.4.

Remark 4.2

The first quantity of the minimum taken above is $(\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$. This is almost always smaller than the second quantity $(d+1)^{(e+1)n}$, the only pertinent exceptional case being $n=3, e=1,$ and $d=1$. When any of the inputs is larger, $(\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$ is less than $(d+1)^{(e+1)n}$. It is also true that $(\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$ is very often smaller than $(nd(e+1)+1)^{2e+1}$, though there are infinitely many exceptional cases (essentially by picking n or d to be sufficiently large compared to e). From a practical standpoint, examples with low order, degree, and number of variables are of particular interest; when $n \le 10$ and $d \le 10$, $(\lceil {n/2}{\rceil } \cdot d+1)^{(e+1)2}$ is the smallest of the above bounds (excluding the exceptional case $n=3, e=1,$ and $d=1$).

4.2 The Constant Case

In this subsection, we assume F is a field of constants. Let

$$K= F \langle P_1(u)/Q_1(u), \dots , P_n(u)/Q_n(u)\rangle $$

where each $P_i(u)/Q_i(u)\notin F$.

If $tp(P_2(u)/Q_2(u), \dots , P_n(u)/Q_n(u) / F \langle P_1 (u) /Q_1 (u) \rangle )$ is orthogonal to the constants, then the analysis from the previous subsection works completely analogously with Proposition 3.11 in place of Proposition 3.5. The criterion also applies with $P_i (u ) /Q_i (u)$ exchanging roles with $P_1 (u) / Q_1 (u)$, for any $i=2, \ldots ,n$. In the following, we do not assume such conditions on the generators $P_i/Q_i$.

To give the main theorem in this section, we first need several lemmas.

Lemma 4.3

[10, Theorem 18]. Let $\mathcal {I}$ be a prime differential ideal of differential dimension d in ${F}\{y_1,\ldots ,y_n\}$, and $\mathcal {A}=\{A_{1},\ldots ,A_{n-d}\}$ a characteristic set of $\mathcal {I}$ under an arbitrary ranking. Then $\mathrm {ord}(\mathcal {I})$ is bounded by the Jacobi number of $\mathcal {A}$. That is,

$${\mathrm{ord}}(\mathcal {I})\le \max _\sigma \sum _{i=1}^{n-d} {\mathrm{ord}} (A_i,y_{\sigma (i)}),$$

where $\sigma$ runs among all injective maps from $\{1,\ldots ,n-d\}$ to $\{1,\ldots ,n\}$.

Lemma 4.4

[5, Theorem 2.11]. Let $\mathcal {I}$ be a prime differential ideal in ${F}\{y_1,\ldots ,y_n\}$. Then $\mathrm{ord}(\mathcal {I})$ is the maximum of all the relative orders of $\mathcal {I}$, that is,

$$\mathrm{ord}(\mathcal {I})=\max _\mathrm{U }\, \mathrm{ord}_\mathrm{U}( \mathcal {I}),$$

where U is any parametric set of $\mathcal {I}$, that is, U is a maximal subset of variables $\{y_1,\ldots ,y_n\}$ such that $\mathcal {I}\cap F\{U\}=\{0\}$.

With the above preparations, we now prove Theorem 2.5. That is, to show when F is a field of constants, then the total degree of a Lüroth generator is bounded by

$$\min \{(d(n+e-1)+1)^{2(e+1)}, (nd(e+1)(n+e-1)+1)^{2e+1} \}.$$

Proof of Theorem 2.5 . Let $\alpha _i=P_i(u)/Q_i(u)\,(i=1,\ldots ,n)$ and $K=F\langle \alpha _1,\ldots ,\alpha _n\rangle$. Clearly, the differential transcendence degree of K over F is 1. Also, by the hypothesis, each $\alpha _i\in F\langle u\rangle \backslash F$. Suppose h is the order of the prime differential ideal $\mathbb I ((\alpha _2,\ldots ,\alpha _n))$ over $F\langle \alpha _1\rangle$. By Proposition 3.10, there exist $c_{jk}\in \mathbb Q$ such that for $\eta =\sum _{j=2}^n(\sum _{k=0}^{h}c_{jk}\alpha _1^k)\alpha _j$, $K=F\langle \alpha _1,\eta \rangle$. The problem is reduced to the case $n=2$. The order of $\eta$ is still bounded by e. The degree of $\eta$ is bounded by $d(n+h-1)$.

It suffices to give a bound for h. Consider the prime differential ideal

$$\mathcal {I}=\mathbb I ((u,\alpha _1,\alpha _2,\ldots ,\alpha _n))\subset F\{y_1,\ldots ,y_n,z\}.$$

It is easy to show that

$$\mathcal {A}:=Q_1(z)y_1-P_1(z),\ldots ,Q_n(z)y_n-P_n(z)$$

is a characteristic set of $\mathcal {I}$ w.r.t. the elimination ranking $z<y_1<\ldots <y_n$. Since the orders of $P_i, Q_i$ is bounded by e, the order matrix $(s_{ij})_{n\times (n+1)}$ of $\mathcal {A}$ satisfies $s_{ii}={\mathrm{ord}}(A_i,y_i)=0$, $s_{ij}=-\infty \,(i\ne j\le n)$ and $s_{i,n+1}={\mathrm{ord}}(A_i,z)\le e$. So the Jacobi number of $\mathcal {A}$, $\max _\sigma \{\sum _{i=1}^ns_{i\sigma (i)}\}$ for $\sigma$ running through injective maps from $\{1,\ldots ,n\}$ to $\{1,\ldots ,n+1\}$, is bounded by e. Thus, by Lemma 4.3, $\text {ord}(\mathcal {I})\le$ $e$. Let $\mathcal {I}_1=\mathbb I ((\alpha _1,\alpha _2,\ldots ,\alpha _n))=\mathcal {I}\cap F\{y_1,\ldots ,y_n\}$. The Kolchin polynomials of $\mathcal {I}$ and $\mathcal {I}_1$ have the following relations: for sufficiently large t,

$$\begin{aligned} \omega _{\mathcal {I}}(t)= & {} {\mathrm{{tr.deg}}}\,F\big (u^{(k)}, \alpha _i^{(k)}: k\le t, i=1,\ldots ,n\big )/F \\= & {} (t+1)+{\mathrm{ord}}(\mathcal {I}) \\= & {} (t+1)+{\mathrm{ord}}(\mathcal {I}_1) \\&+\; {\mathrm{{tr.deg}}}\, F\big (u^{(k)}, \alpha _i^{(k)}: k\le t, i=1,\ldots ,n\big )/F\big (\alpha _i^{(k)}: k\le t, i=1,\ldots ,n\big ) \end{aligned}$$

Thus, $\mathrm{ord}(\mathcal {I}_1)\le \mathrm{ord}(\mathcal {I})$ $\le e$. Note that $h={\mathrm{{tr.deg}}}\,F\langle \alpha _1,\alpha _2,\ldots ,\alpha _n\rangle /F\langle \alpha _1\rangle$ is also equal to the relative order of $\mathcal {I}$ w.r.t. the parametric set $\{y_1\}$, by Lemma 4.4, $h\le \mathrm{ord}(\mathcal {I})\le$ $e$.

So the degree of $\eta$ is bounded by $d(n+e-1)$. Hence, by Theorem 2.3, the degree of a Lüroth generator is bounded by

$$\min \{(d(n+e-1)+1)^{2(e+1)}, (2d(e+1)(n+e-1)+1)^{2e+1} \}.$$

   $\square$

Remark 4.5

In most of the cases, especially when either n or e is large, we have

$$(d(n+e-1)+1)^{2(e+1)}<(2d(e+1)(n+e-1)+1)^{2e+1}$$

and

$$(d(n+e-1)+1)^{2(e+1)}<(nd(e+1)+1)^{2e+1}.$$

Hence, the degree bound given in Theorem 2.5 is smaller than that in Theorem 2.3.

As an experiment to compare the two bounds, we have computed more than 10,000 randomly generated tuples (n, d, e) simulating the pertinent cases of the bounds when each of the variables is less than 30, and our bound gives the better result approximately 94.3 % of the time.

In future work, we hope to prove Conjecture 3.13 in an effective manner and use the result to improve the bounds for the degree in the effective differential Lüroth theorem for the case that the base field is constant.