# A classification of first order differential equations

Partha Kumbhakar, Ursashi Roy, Varadharaj R. Srinivasan  Indian Institute of
Science Education and Research Mohali, India.
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###### Abstract.

Let $k$ be a differential field of characteristic zero with an algebraically
closed field of constants. In this article, we provide a classification of
first order differential equations over $k$ and study the algebraic dependence
of solutions of a given first order differential equation. Our results
generalize parts of the work of Noordman et al. [20] and complements the work
of Freitag et al. [4].

##  1\. Introduction

Let $k$ be a differential field of characteristic zero having an algebraically
closed field of constants $C$. Let $f\in k[Y,Z]$ be an irreducible polynomial
involving the variable $Z.$ By a _solution_ of the differential equation
$f(y,y^{\prime})=0,$ we mean an element $u$ in a differential field extension
of $k$ having $C$ as its field of constants such that $f(u,u^{\prime})=0.$ If
$f$ is defined over the field of constants $C$ then $f(y,y^{\prime})=0$ is
called an _autonomous first order differential equation_. The field of
fractions $k(f)$ of the ring $k[Y,Z]/\langle f\rangle$ has a natural
derivation induced by $f$ that maps $y$ to $z$ where $y,z$ are the images of
$Y,Z$ in $k[Y,Z]/\langle f\rangle$ under the canonical projection. It then
follows that $f(y,y^{\prime})=0$ has a nonalgebraic solution if and only if
the differential field $k(f)$ has $C$ as its field of constants.

In this article, we study transcendence degree $1$ differential subfields of
iterated strongly normal extensions. As a consequence of our study, we obtain
a classification of first order differential equations $f(y,y^{\prime})=0$
over $k$ into the following types:

  1. (I)

(_algebraic type_) All solutions of $f(y,y^{\prime})=0$ are algebraic over $k$
or equivalently, $k(f)$ with the induced derivation has a nonalgebraic
constant.

  2. (II)

(_Riccati type_) There are a nonalgebraic solution of $f(y,y^{\prime})=0,$ a
finite algebraic extension $\tilde{k}$ of $k$ and an element
$t\in\tilde{k}(f)$ such that $\tilde{k}(f)$ is a finite algebraic extension of
$\tilde{k}(t)$ and that $t$ is a solution of a Riccati differential equation:

(1.0.1) $t^{\prime}=a_{2}t^{2}+a_{1}t+a_{0},\ \text{with}\ \
a_{0},a_{1},a_{2}\in\tilde{k},\ \text{not all zero}.$

  3. (III)

(_Weierstrass type_) There are a nonalgebraic solution of $f(y,y^{\prime})=0,$
a finite algebraic extension $\tilde{k}$ of $k$ and an element
$t\in\tilde{k}(f)$ such that $\tilde{k}(f)$ is a finite algebraic extension of
$\tilde{k}(t,t^{\prime})$ and that $t$ is a solution of a Weierstrass
differential equation:

(1.0.2) $(t^{\prime})^{2}=\alpha^{2}(4t^{3}-g_{2}t-g_{3}),\ g_{2},g_{3}\in
C,\alpha\in\tilde{k}\ \text{and}\ 27g^{2}_{3}-g^{3}_{2}\neq 0.$

  4. (IV)

(_general type_) The differential equation $f(y,y^{\prime})=0$ is not of the
above types.

When $k=C,$ we obtain the classification of autonomous differential equation
due to Noordman et al. [20, page 1655]. If a differential equation is of
Riccati or Weierstrass type then it is easy to see that $k(f)$ can be embedded
in an iterated strongly normal extension. The following theorem contains the
main results (Theorem 5.2, Theorem 6.3 and Corollary 6.4) of this article.

######  Theorem 1.1.

Let $k$ be a differential field with an algebraically closed field of
constants $C$ and $f(y,y^{\prime})=0$ be a differential equation over $k.$
Suppose that $f(y,y^{\prime})=0$ has a nonalgebraic solution in an iterated
strongly normal extension $E$ of $k$ or equivalently, the differential field
$k(f)$ can be embedded in an iterated strongly normal extension $E$ of $k.$
Then the following statements hold.

  1. (i)

The differential equation $f(y,y^{\prime})=0$ is of either Riccati or
Weierstrass type.

  2. (ii)

If $k=C$ and $f(y,y^{\prime})=0$ is of Riccati type then there is an element
$t\in C(f)$ such that

(1.0.3) $\text{either}\ t^{\prime}=1\ \ \text{or}\ \ t^{\prime}=ct\ \ \
\text{for some nonzero}\ c\in C.$

  3. (iii)

If $E$ is a strongly normal extension of $k$ and $f(y,y^{\prime})=0$ is of
Riccati type (respectively, Weierstrass type) then the finite algebraic
extension $\tilde{k}$ and the element $t\in\tilde{k}(f),$ as in the definition
of a Riccati type (respectively, Weierstrass type), can be chosen so that
$\tilde{k}(f)=\tilde{k}(t)$ (respectively,
$\tilde{k}(f)=\tilde{k}(t,t^{\prime})$).

We use the above theorem to produce a family of geometrically irreducible
plane algebraic curves $f$ over $k$ such that $f(y,y^{\prime})=0$ is of
general type (see Theorem 7.1). This family of examples of general type
equations includes Abel differential equations (see Example 7.4) of the form

$y^{\prime}=a_{n}y^{n}+\cdots+a_{2}y^{2},$

where $n\geq 3,$ $a_{i}\in k$ and both $a_{2}$ and $a_{3}$ have no
antiderivatives in $k.$ Note that the differential equation
$y^{\prime}=y^{3}-y^{2},$ that appears in [15] and [20], belongs to the above
family of Abel differential equations and therefore is of general type. We
also prove the following results concerning algebraic dependence of
nonalgebraic solutions (see Theorem 7.6):

  1. (i)

If a first order differential equation over $k$ has a nonalgebraic solution in
an iterated strongly normal extension of $k$ then any four nonalgebraic
solutions from a differential field extension of $k$ having $C$ as its field
of constants are $k-$algebraically dependent.

  2. (ii)

If an autonomous differential equation has a nonalgebraic solution in an
iterated strongly normal extension of $C$ then any two nonalgebraic solutions
from a differential field having $C$ as its field of constants are
$C-$algebraically dependent.

In a recent article [4], Freitag et al. study the algebraic dependence of
solutions of differential equations of any order using model theoretic
methods. They show that if a first order differential equation (respectively,
an autonomous differential equation) has four (respectively, two)
algebraically independent solutions then any $m$ distinct nonalgebraic
solutions are algebraically independent. Thus, it is natural to make the
following

######  Conjecture 1.2.

A first order differential equation over $k$ (respectively, An autonomous
differential equation over $C$) is not of general type if and only if it has
at most three (respectively, at most one) algebraically independent solutions
in any given differential field extension of $k$ (respectively, $C$) having
$C$ as its field of constants.

We provide some evidence for this conjecture in Section 7.3. We will not prove
any results concerning algebraic type differential equations. The articles
[12], [14] and [20] are directly related to our work. We shall now explain in
detail their connections to our main theorem.

###  1.1. Classification of autonomous differential equations [20].

Let $f\in C[Y,Z]$ be irreducible and $C(f)$ be the function field of the curve
defined by $f.$ Let $\Gamma$ be the projective nonsingular model whose
function field $C(\Gamma)$ is isomorphic to $C(f).$ Since $C(f)$ is a
differential field with the derivation induced by $f,$ we could also regard
$C(\Gamma)$ as a differential field with the induced derivation
${}^{\prime}:C(\Gamma)\to C(\Gamma)$ .

Let $Der_{C}(C(\Gamma))$ be the $C(\Gamma)-$module of $C-$derivations on
$C(\Gamma).$ Then, there is a natural $C(\Gamma)-$module isomorphism between
$Der_{C}(C(\Gamma))$ and
$Hom_{C(\Gamma)}\left(\Omega^{1}_{C(\Gamma)|C},C(\Gamma)\right)$ given by
$(a\mapsto a^{\prime})\mapsto\omega,$ where $\omega$ is the differential such
that $\Phi(\omega)=1.$

${C(\Gamma)}$${\Omega^{1}_{C(\Gamma)|C}}$${C(\Gamma)}$$\scriptstyle{\mathrm{d}}$$\scriptstyle{a\mapsto
a^{\prime}}$$\scriptstyle{\Phi}$

Thus, to an autonomous differential equation, one can associate the tuple
$(\Gamma,\omega).$ Using this identification, in [20], autonomous equations
are classified into four types:

  1. (1)

$\omega=\mathrm{d}g$ for some $g\in C(\Gamma)$ or equivalently, there is an
element $t\in C(\Gamma)$ such that $t^{\prime}=1;$ in this case the equations
are of _exact type_.

  2. (2)

$\omega=\mathrm{d}g/cg$ for some $g\in C(\Gamma)$ and nonzero constant $c$ or
equivalently, there is a $t\in C(\Gamma)$ such that $t^{\prime}=ct$ for some
nonzero constant $c;$ in this case the equations are of _exponential type_.

  3. (3)

$\omega=\mathrm{d}g/h,$ for $h,g\in C(\Gamma)$ and $h^{2}=g^{3}+ag+b$ with
$4a^{3}+27b^{2}\neq 0$ or equivalently, there is an element $t\in C(\Gamma)$
such that $t^{\prime 2}=t^{3}+at+b$ for $a,b\in C$ with $4a^{3}+27b^{2}\neq
0$; in this case the equations are of _Weierstrass type_.

  4. (4)

If the equation is not of the above three types then it is of _general type_.

Since $C$ is an algebraically closed field of constants, every nonconstant
solution must be nonalgebraic and every autonomous differential equation has a
nonalgebraic solution (see Section 7.3). Thus, from Theorem 1.1 (ii), we
obtain that our classification of differential equations coincides with the
one above.

In [20], an autonomous equation $(\Gamma,\omega)$ is called _new_ if
$(\Gamma,\omega)$ is not a proper pull back. Equations of type general and new
are shown to have the following interesting property [20, Theorem 2.1]: Any
number of distinct nonalgebraic solutions are $C-$algebraically independent.
Using this property, the authors prove that nonalgebraic solutions of
equations of general type cannot be found in any iterated Picard-Vessiot
extension [20, Proposition 7.1]. In fact, their arguments can be extended to
show that equations of general type do not have nonalgebraic solutions in any
iterated strongly normal extension as well. Thus, when $k=C,$ our theorem can
also be recovered from their work. We would also like to point out that the
_type_ of a rational autonomous equation; $y^{\prime}=f(y),$ where $f$ is a
nonzero rational function in one variable over $C,$ can be completely
classified using [18, Proposition 3.1]; for details see Remark 7.5.

###  1.2. Painlevé property and differential equations of Ricatti or
Weierstrass type [12] & [14]:

Let $k$ be a finite algebraic extension of the ordinary differential field
$\mathbb{C}(x)$ of rational functions over complex numbers with
$x^{\prime}=1.$ A differential equation $f(y,y^{\prime})=0$ over $k$ is said
to have the _Painlevé property_ if the set of all branch points and the set of
all essential singularities of the solutions form a discrete set. A
differential equation over $k$ of algebraic type has the Painlevé property if
and only if all of its solutions lie in a fixed finite algebraic extension of
$k$ (see [14, Theorem 2.7 and Example 2.9]).

Let $f(y,y^{\prime})=0$ have a nonalgebraic solution $u.$ Then the function
field $k(f)$ with the induced derivation is isomorphic to $k(u,u^{\prime})$ as
differential fields. Since $k(u,u^{\prime})$ has $\mathbb{C}$ as its field of
constants, from [12, Theorem 4.5], we deduce that $f$ has the Painlevé
property if and only if there exists a finite algebraic extension $\tilde{k}$
of $k$ and an element $t\in\tilde{k}(u,u^{\prime})$ such that
$\tilde{k}(u,u^{\prime})=\tilde{k}(t,t^{\prime})$ and that $t$ is a solution
of either a Riccati or a Weierstrass differential equation over $\tilde{k}.$
In the former case, $\tilde{k}(t,t^{\prime})$ can be embedded in a Picard-
Vessiot extension of $\tilde{k}$ whose Galois group is isomorphic to a closed
subgroup of $\mathrm{GL}_{2}(\mathbb{C})$ and in the latter case,
$\tilde{k}(t,t^{\prime})$ is an abelian extension of $\tilde{k}$ whose Galois
group is isomorphic to the $\mathbb{C}-$points of a nonsingular elliptic curve
defined over $\mathbb{C}$ (see 7.2 for details).

Now, from our Theorem 1.1(iii), we can make the following observation:

  * Let $k$ be a finite algebraic extension of the ordinary differential field $\mathbb{C}(x)$ of rational functions in one variable $x$ with the derivation $d/dx.$ Let $f(y,y^{\prime})=0$ be a first order differential equation with a nonalgebraic solution. Then, $f(y,y^{\prime})=0$ has a nonalgebraic solution in a strongly normal extension of a finite algebraic extension of $k$ if and only if $f(y,y^{\prime})=0$ has the Painlevé property.

There are differential equations of Ricatti type that do not have the Painlevé
property. For example, the differential equation

$y^{\prime}=-\frac{1}{2x}y^{3}$

has $u_{c}=1/\sqrt{\ln(x)+c}$ as a one-parameter family of solutions, where
the parameter $c\in\mathbb{C}$. Note that
$u_{c}\in\mathbb{C}(x)(\ln(x))\left(\sqrt{\ln(x)+c}\right)$ and that

$\mathbb{C}(x)\subset\mathbb{C}(x)(\ln(x))\subset\mathbb{C}(x)(\ln(x))\left(\sqrt{\ln(x)+c}\right).$

Clearly, $\mathbb{C}(x)(\ln(x))(\sqrt{\ln(x)+c})$ is an iterated (Picard-
Vessiot) strongly normal extension of $\mathbb{C}(x).$ Since
$\ln(x)\in\mathbb{C}(x)(u_{c})$ and $(\ln(x))^{\prime}=1/x\in\mathbb{C}(x),$
we see that $y^{\prime}=-\frac{1}{2x}y^{3}$ is of Ricatti type. However, the
set of all branched points of these solutions, namely the set $\mathbb{C},$ is
not discrete.

### Organisation of the paper.

In Section 2, we gather facts from the theory of strongly normal extensions
for easy reference. In Section 3, we provide an outline of the proof of our
main theorem and in Section 4, using a work of André Yves on solution
algebras, we prove a structure theorem for differential subfields of
liouvillian extensions. The proof of Theorem 1.1 can be found in Sections 5
and 6. In Section 7, we prove results concerning the algebraic dependence of
solutions of first order differential equations. We also show that Conjecture
1.2 holds for rational autonomous differential equations.

##  2\. Differential Galois Theory

Here, we shall set up notations and put together the necessary definitions and
results from the theory of strongly normal extensions.

###  2.1. Strongly normal extensions [7], [8], [9], [10].

A differential field extension $E$ of $k$ is called a _strongly normal
extension_ if $E$ is finitely differentially generated over $k$ and if every
differential $k-$isomorphism $\sigma$ of $E$ satisfies the following
conditions: $E\sigma E=EC(\sigma)=\sigma EC(\sigma)$, where $C(\sigma)$ is the
field of constants of the compositum $E\sigma E$ and that $\sigma(c)=c$ for
all constants $c$ of $E$. It can be shown that every strongly normal extension
of $k$ is a finitely generated field extension of $k$ and has $C$ as its field
of constants ([9, Propositions 12.2, 12.4]). A differential field extension
$E$ of $k$ is called an _iterated strongly normal extension_ if there is a
tower of differential fields

$k=E_{0}\subseteq E_{1}\subseteq E_{2}\subseteq\cdots\subseteq E_{n}=E$

such that $E_{i}$ is a strongly normal extension of $E_{i-1}$ for each
$i=1,\dots,n.$ When the field of constants is algebraically closed, strongly
normal extensions exist and are unique up to differential isomorphisms.

The group of all $k-$differential automorphisms of a differential field
extension $E$ of $k$ is called the _differential Galois group_ of $E$ over $k$
and is denoted by $\mathscr{G}(E|k)$ or simply by $\mathscr{G},$ when there is
no ambiguity. If the field of constants of $E$ is $C$ and for any $x\in
E\setminus k$ there is $\sigma\in\mathscr{G}(E|k)$ such that $\sigma(x)\neq x$
then $E$ is called a _weakly normal extension_ of $k.$

For a strongly normal extension $E$ over $k,$ the elements of
$\mathscr{G}(E|k)$ can be canonically identified with the $C-$points of a
reduced group scheme of finite type defined over $C$ [9, Sections 33-36 ].
Thus, $\mathscr{G}(E|k)$ can be thought of as an algebraic group111Since $C$
is of characteristic zero and algebraically closed, group schemes of finite
type over $C$ are smooth. (though not necessarily affine) defined over the
field of constants $C$. The fundamental theorem of strongly normal extension
([9, Theorem 36.3]) provides a bijective correspondence between the
intermediate differential subfields and the Zariski closed subgroups (closed
subschemes) of $\mathscr{G}(E|k)$. If $\mathscr{H}$ is a closed subgroup of
$\mathscr{G}(E|k)$ and $K$ is an intermediate differential field then the
bijective correspondence is given by the maps

$\displaystyle K\to\mathscr{G}(E|K):=\left\\{\sigma\in\mathscr{G}(E|k)\ |\
\sigma(u)=u\ \ \text{for all}\ \ u\in K\right\\}$ $\displaystyle\mathscr{H}\to
E^{\mathscr{H}}:=\left\\{u\ \in E\ |\ \sigma(u)=u\ \ \text{for all}\
\sigma\in\mathscr{H}\right\\}.$

In this correspondence, the field fixed by the group $\mathscr{G}(E|k)$ is the
base field $k$, that is $E^{\mathscr{G}(E|k)}=k$. Let $K$ be a differential
field intermediate to $E$ and $k$. Then $K$ is a strongly normal extension of
$k$ if and only if $\mathscr{G}(E|K)$ is a closed normal subgroup of
$\mathscr{G}(E|k)$; in which case, the differential Galois group
$\mathscr{G}(K|k)$ is isomorphic to the quotient group
$\mathscr{G}(E|k)/\mathscr{G}(E|K)$. The algebraic closure of $k$ in $E$ is a
finite Galois extension, which we denote by $E^{0}$.

A strongly normal extension $E$ over $k$ has the following structure ([10,
Proposition 12.2]), which will play a crucial role in our proofs:

(2.1.1) $k\subseteq E^{0}\subseteq F\subseteq E,$

where $E^{0}$ is the relative algebraic closure of $k$ in $E$ and that $E^{0}$
is also a finite Galois extension of $k,$ $F$ is an _abelian extension_ of
$E^{0},$ that is, $F$ is strongly normal over $E^{0}$ such that
$\mathscr{G}(F|E^{0})$ is isomorphic to an abelian variety defined over $C,$
and $E$ is a strongly normal extension of $F$ such that $\mathscr{G}(E|F)$ is
a connected linear algebraic group defined over $C.$ Such an extension $E$ is
a Picard-Vessiot extension of $F$ (see [8, Theorem 2] or [10, Sections 8-11]).

###  2.2. Picard-Vessiot theory [21]

A _Picard-Vessiot ring_ $R$ for a matrix differential equation
$Y^{\prime}=AY,$ where $A\in M_{n}(k),$ is a differential ring containing $k$
and having the following properties:

  1. (a)

$R$ is a simple differential ring, that is, $R$ has no nontrivial differential
ideals.

  2. (b)

There is a _fundamental matrix_ $F$ for $Y^{\prime}=AY$ with entries in $R,$
that is, a matrix $F\in\mathrm{GL}_{n}(R)$ such that $F^{\prime}=AF.$

  3. (c)

$R$ is minimal with respect to these properties, that is, $R$ is generated as
a ring by the elements of $k,$ the entries of $F$ and inverse of the
determinant of $F.$

By a _differential $k-$module_ $(M,\partial),$ we mean a finite dimensional
$k-$module $M$ and an additive map $\partial:M\to M$ such that
$\partial(\alpha m)=\alpha^{\prime}m+\alpha\partial(m)$ for all $\alpha\in k$
and $m\in M.$ Let $M$ be a differential $k-$module. By fixing a $k-$basis
$e_{1},\dots,e_{n}$ of $M,$ we obtain a matrix $A=(a_{ij})\in M_{n}(k)$ such
that $\partial(e_{i})=-\sum_{j}a_{ji}e_{j}$ and a corresponding matrix
differential equation $Y^{\prime}=AY.$ Choosing any other basis will amount to
obtaining a differential equation of the form $Y^{\prime}=\tilde{A}Y,$ where
$\tilde{A}=B^{\prime}B^{-1}+BAB^{-1}$ for some $B\in\mathrm{GL}_{n}(k).$
Furthermore, if $R$ is a Picard-Vessiot ring for $Y^{\prime}=AY$ with
fundamental matrix $F$ then $(BF)^{\prime}=\tilde{A}BF$ and thus $R$ is also
the Picard-Vessiot ring for $Y^{\prime}=\tilde{A}Y.$ This observation allows
one to define a _Picard-Vessiot ring_ for a differential module $M$ to be a
Picard-Vessiot ring for a corresponding matrix differential equation
$Y^{\prime}=AY$ of $M.$ Picard-Vessiot rings are integral domains and their
field of fractions are called _Picard-Vessiot extensions_. Let $E$ be a
Picard-Vessiot extension of $k$ with Picard-Vessiot ring $R$ then from [21,
Corollary 1.38] it follows that $R$ is the $k-$algebra generated by all the
solutions in $E$ of all linear differential equations over $k.$

Let $M$ be a $k-$differential module with matrix differential equation
$Y^{\prime}=AY$ and $M^{\vee}$ be the dual of a differential module of $M.$
Then $Y^{\prime}=-A^{t}Y,$ where $A^{t}$ is the transpose of $A,$ is a matrix
differential equation corresponding to $M^{\vee}.$ Thus, if $R$ is a Picard-
Vessiot ring for $M$ with fundamental matrix $F\in\mathrm{GL}_{n}(R)$ then
$\left((F^{t})^{-1}\right)^{\prime}=-A^{t}(F^{t})^{-1}$ and thus $M$ and
$M^{\vee}$ have the same Picard-Vessiot ring $R$. Let $k[\partial]$ be the
ring of differential operators over $k$ and
$\mathscr{L}:=\partial^{n}+a_{n-1}\partial^{n-1}+\cdots+a_{0}\in k[\partial].$
Then for $M=k[\partial]/k[\partial]\mathscr{L},$ a matrix equation
corresponding to the dual $M^{\vee}$ is $Y^{\prime}=A_{\mathscr{L}}Y,$ where
$A_{\mathscr{L}}$ is the companion matrix of $\mathscr{L}.$ Therefore, if $R$
is a Picard-Vessiot ring for $M^{\vee}$ then the fundamental matrix
$F\in\mathrm{GL}_{n}(R)$ for $Y^{\prime}=A_{\mathscr{L}}Y$ is a Wronskian
matrix.

$A_{\mathscr{L}}=\begin{pmatrix}0&1&0&0&\cdots&0\\\ 0&0&1&0&\cdots&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&\cdots&1\\\
-a_{0}&-a_{1}&\cdots&\cdots&\cdots&-a_{n-1}\end{pmatrix},\qquad
F=\begin{pmatrix}y_{1}&y_{2}&\cdots&\cdots&y_{n}\\\
y^{\prime}_{1}&y^{\prime}_{2}&\cdots&\cdots&y^{\prime}_{n}\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
y^{(n-1)}_{1}&y^{(n-1)}_{2}&\cdots&\cdots&y^{(n-1)}_{n}\end{pmatrix}.$

Note that $y_{1},\dots,y_{n}$ are $C-$linearly independent and the $C-$vector
space $V$ spanned by $y_{1},\dots,y_{n}$ is the set of all solutions of
$\mathscr{L}(y)=0.$

##  3\. Outline of the proof of the main theorem

Here we shall discuss the strategies involved in the proofs of Theorem 1.1 (i)
and (iii). Let $f(y,y^{\prime})=0$ be a differential equation having a
nonalgebraic solution $u$ in a strongly normal extension $E$ of $k$ and let
$K:=k(u,u^{\prime}).$ Observe that $\mathrm{tr.deg}(K|k)=1.$

The proof of Theorem 1.1(iii) appears as Theorem 5.2. It involves decomposing
$E$ as in (2.1.1) and considering the following cases:

Case(i). Suppose that $\mathrm{tr.deg}(KF|F)=1.$ Since the differential Galois
group $\mathscr{G}(E|F)$ is a connected linear algebraic group,
$\mathscr{G}(E|F)$ is a union of its Borel subgroups. Now $F$ is algebraically
closed in $E$ implies that one can choose a Borel subgroup $\mathscr{B}$ such
that $\mathrm{tr.deg}(KE^{\mathscr{B}}|E^{\mathscr{B}})=1.$ Since
$\mathscr{B}$ is (connected) solvable, $E$ is a liouvillian Picard-Vessiot
extension of $E^{\mathscr{B}}.$ We classify differential subfields of
liouvillian Picard-Vessiot extensions (Theorem 4.4) and show that either
$KE^{\mathscr{U}}=E^{\mathscr{U}}(t),$ where $\mathscr{U}$ is the unipotent
radical of $\mathscr{B}$ and $t^{\prime}\in E^{\mathscr{U}}$ or
$KE^{\mathscr{B}}=E^{\mathscr{B}}(t),$ where $t^{\prime}/t\in
E^{\mathscr{B}}.$ We then use the fact that the geometric genus is an
invariant under base change by separable fields and obtain that $K$ is a genus
zero extension of the algebraic closure of $k$ in $K$ (see [2, Theorem 5]).
Thus, one can find a finite algebraic extension $\tilde{k}$ of $k$ so that
$\tilde{k}K=\tilde{k}(y).$ We then use the fact that $y=(at+b)/(ct+d)$ for
$a,b,c,d\in\tilde{k}E^{\mathscr{U}}$ and show that $y$ satisfies a Riccati
equation over $\tilde{k}.$

${E}$${KE^{\mathscr{U}}}$${E^{\mathscr{U}}}$${KE^{\mathscr{B}}}$${E^{\mathscr{B}}}$${KF}$${F}$${K}$${k}$

Case(ii). Suppose that $\mathrm{tr.deg}(KF|F)=0.$ Then since $F$ is
algebraically closed in $E,$ we have $K\subseteq F.$ Now $KE^{0}$ is an
intermediate differential field of a strongly normal extension, whose Galois
group is abelian. Therefore $KE^{0}$ itself is a strongly normal extension of
$E^{0}$ such that $\mathrm{tr.deg}(KE^{0}|E^{0})=1.$ Then from [8, Theorem 3],
we obtain a finite algebraic extension $\tilde{k}$ of $E^{0}$ such that
$\tilde{k}K=\tilde{k}(t,t^{\prime}),$ where $t$ is a solution of a Weierstrass
differential equation over $\tilde{k}.$

######  Remark 3.1.

In the event that $E$ is a Picard-Vessiot extension of $k,$ one can show that
$KE^{0}$ is a rational field generated by a solution of a Riccati differential
equation. To see this, we consider the connected group $\mathscr{G}(E|E^{0})$
and its codimension one closed subgroup $\mathscr{G}(E|KE^{0}).$ Then, the
function field of the homogeneous space
$\mathscr{G}(E|E^{0})/\mathscr{G}(E|KE^{0})$ is known to be rational; for
example, see [3, Theorem 4.4]. Since $KE^{0}$ is isomorphic (as fields) to
$E^{0}(\mathscr{G}(E|E^{0})/\mathscr{G}(E|KE^{0}))=E^{0}(x),$ we obtain that
$KE^{0}=E^{0}(y)$ [11, p.87]. Now, we have
$KE^{\mathscr{U}}=E^{\mathscr{U}}(t)=E^{\mathscr{U}}(y),$ where $t^{\prime}\in
E^{\mathscr{U}}$ or $KE^{\mathscr{B}}=E^{\mathscr{B}}(t)=E^{\mathscr{B}}(y),$
where $t^{\prime}/t\in E^{\mathscr{B}}.$ One then goes on to show that such a
$y$ must satisfy a Riccati differential equation over $E^{0}.$

Now we shall detail the strategies involved in proving Theorem1.1(i) (Lemma
6.2 and Theorem 6.3).

Step 1. Let $E$ be a strongly normal extension of $k$ and $\tilde{E}$ be a
finite algebraic extension of $E.$ Let $K$ be a differential field such that
$k\subset K\subseteq\tilde{E}$ and that $\mathrm{tr.deg}(K|k)=1.$ Then using
the facts that $KE$ is strongly normal over $K$ and that the differential
Galois group $\mathscr{G}(KE|K)$ injects as a closed subgroup of the
differential Galois group $\mathscr{G}(E|k)$ and that $KE$ is an algebraic
extension of $E,$ we obtain that $\mathrm{tr.deg}((K\cap E)|k)=1.$ Now from
the first part of the theorem, we conclude that there is a finite algebraic
extension $\tilde{k}$ of $k$ and an element $y\in\tilde{k}K$ transcendental
over $\tilde{k}$ such that $\tilde{k}(K\cap E)=\tilde{k}(y,y^{\prime})$ and
that $y$ is a solution of either a Riccati differential equation or a
Weierstrass differential equation over $\tilde{k}.$

Step 2. (A differential descent lemma) Let $E$ be an iterated strongly normal
extension of $k,$ $\overline{E}$ be an algebraic closure of $E$ and $K$ be a
finitely generated differential field intermediate to $k$ and $\overline{E}$
with $\mathrm{tr.deg}(K|k)=1.$

Then one can assume that

$k=E_{0}\subseteq E_{1}\subseteq\cdots\subseteq E_{n-1}\subseteq
E\subseteq\overline{E},$

where $E_{i}$ is a strongly normal extension of $E_{i-1},$ $E$ is
transcendental over $E_{n-1},$ and $\mathrm{tr.deg}(KE_{n-1}|E_{n-1})=1.$ Now
from Step 1, we have

$\overline{E}_{n-1}\subseteq\overline{E}_{n-1}(y_{n-1},y^{\prime}_{n-1})\subseteq\overline{E}_{n-1}K,$

where $y_{n-1}$ is a nonalgebraic solution of a Riccati differential equation
or a Weierstrass differential equation over $\overline{E}_{n-1}$. We first
show that $\overline{E}_{n-1}K$ is a weakly normal extension of
$\overline{E}_{n-2}K$ and that $y_{n-1}$ can be chosen so that
$\overline{E}_{n-1}(y_{n-1},y^{\prime}_{n-1})$ is stabilized by the group
$\mathscr{G}(\overline{E}_{n-1}K|\overline{E}_{n-2}K).$ Using these facts, we
then show that one can obtain a nonalgebraic solution
$y_{n-2}\in\overline{E}_{n-2}K$ of some Riccati differential equation or a
Weierstrass differential equation over $\overline{E}_{n-2}.$ A repeated
application of this “descent argument” proves the second part of the main
theorem.

${\overline{E}}$${\overline{E}_{n-1}K}$${\overline{E}_{n-2}K}$${\overline{E}_{n-1}\langle
y_{n-1}\rangle}$${\overline{E}_{n-2}\langle
y_{n-2}\rangle}$${\overline{E}_{n-1}}$${\overline{E}_{n-2}}$

##  4\. Differential Subfields of Liouvillian Picard-Vessiot Extensions

In [1], a Galois correspondence between certain differential subfields, called
the solution fields, and the observable subgroups of the differential Galois
group has been established. We shall now explain relevant definitions and
results from [1] that are needed to prove a structure theorem for the
intermediate differential fields of liouvillian Picard-Vessiot extensions.

### Solution algebras and solution fields

Let $k[\partial]$ be the usual ring of differential operators. Let $M$ be a
differential $k-$module of dimension $n$. A differential field extension $E$
of $k$ having the same field of constants as $k$ is called a _solution field_
for $M$ if there is a morphism $\psi:M\to E$ of $k[\partial]-$modules such
that $\psi(M)$ generates $E;$ in which case, $E$ is said to be generated by a
solution $\psi.$ Let $S$ be a differential $k-$algebra and an integral domain
such that its field of fractions has field of constants $C$. Then $S$ is
called a _solution algebra_ for $M$ if there is a morphism $\psi:M\to S$ of
$k[\partial]-$modules such that $\psi(M)$ generates $S.$ Let $E$ be a
differential field extension of $k$ having the same field of constants as $k.$
For $i=1,\dots,n$, let $y_{i}\in E$ and $\mathscr{L}_{i}\in k[\partial]$ such
that $\mathscr{L}_{i}(y_{i})=0.$ Let $R$ be the differential $k-$subalgebra of
$E$ generated by $y_{1},\dots,y_{n}.$ Then $R$ is a solution algebra. To see
this, consider the differential modules
$k[\partial]/k[\partial]\mathscr{L}_{i},$ where $\mathscr{L}_{i}(y_{i})=0.$
Then the images of the $k[\partial]-$morphisms
$\oplus^{n}_{i=1}\psi_{i}:\bigoplus^{n}_{i=1}k[\partial]/k[\partial]\mathscr{L}_{i}\to
E$ defined by $\psi_{i}(1)=y_{i}$ generates $R.$ Similarly, the differential
field generated by $y_{1},\dots,y_{n}$ is a solution field. Conversely, let
$S$ be a solution algebra generated by a solution $\psi.$ Then we have a map
$\psi:M\to S$ of $k[\partial]-$modules such that $\psi(M)$ generates $S$ as a
$k-$algebra. Note that $\psi(M)$ is a (finite dimensional) differential
$k-$module. Let $y_{1},\dots,y_{n}$ be a $k-$basis of $\psi(M).$ Then each
$y_{i}$ must satisfy a linear homogeneous differential equation over $k.$ We
shall summarize this observation in the following

######  Proposition 4.1.

Solution fields (respectively, Solution algebras) over $k$ are generated as a
field (respectively, as a $k-$algebra) by solutions of linear homogeneous
differential equations over $k.$

Let $M^{\vee}$ be the dual of $M.$ A differential field $E$ is called a
_Picard-Vessiot field_ if it has the same field of constants as $k$ and the
$C-$modules Sol$(M,E):=$ Hom${}_{k[\partial]}(M,E)$ and Sol$(M^{\vee},E):=$
Hom${}_{k[\partial]}(M^{\vee},E)$ are of dimension $n$ over $C$ and $E$ is
minimal with respect to these properties. As noted in Section 2, $M$ and
$M^{\vee}$ have the same Picard-Vessiot extension and thus it is readily seen
that the notions of a Picard-Vessiot field for $M$ and a Picard-Vessiot
extension for $M$ are one and the same. Observe that if $E$ is a Picard-
Vessiot field of a differential $k-$module $M$ of dimension $n$ then $E$ is a
solution field of $M^{n}.$

######  Theorem 4.2.

([1, Lemma 4.2.2, Theorem 1.2.2]) Let $\langle M\rangle^{\otimes}$ be the
tannakian category over $C$ generated by the differential $k-$module $M.$

  1. (i)

The quotient field of a solution algebra $S$ for $M$ is a solution field for
$M$.

  2. (ii)

Conversely, any solution field $K$ for $M$ is the quotient field of (non
unique) solution algebra $\mathcal{S}$ for $M$.

  3. (iii)

Any solution field for $N\in\langle M\rangle^{\otimes}$ embeds as differential
subfield of a Picard-Vessiot field for $M.$

  4. (iv)

Let $E$ be the Picard-Vessiot field for $M$ with Galois group $\mathscr{G}$,
then an intermediate differential field $k\subseteq K\subseteq E$ is a
solution field for $N\in\langle M\rangle^{\otimes}$ if and only if the
corresponding subgroup $\mathscr{H}\subseteq\mathscr{G}$ is observable (i.e.
$\mathscr{G}/\mathscr{H}$ is quasi-affine).

In view of the above theorem, it is natural to ask for a characterization
theorem of those Picard-Vessiot extensions having only solution fields as
intermediate differential fields. Equivalently, characterize those linear
algebraic groups whose closed subgroups are observable. We shall provide such
a characterization theorem in the next proposition. But first, a few
definitions are in order.

A differential field extension $E$ of $k$ is called a _liouvillian extension_
of $k$ if there is a tower of differential fields

$k=E_{0}\subseteq E_{1}\subseteq\cdots\subseteq E_{n}=E$

such that $E_{i}=E_{i-1}(t_{i}),$ where $t_{i}$ is either algebraic over
$E_{i-1}$ or $t^{\prime}_{i}\in E_{i-1}$ or $t^{\prime}_{i}/t_{i}\in E_{i-1}.$
If $E$ is both a Picard-Vessiot extension and a liouvillian extension of $k$
then $E$ is called a _liouvillian Picard-Vessiot extension_ of $k.$ It is a
standard fact that liouvillian Picard-Vessiot extensions are precisely those
Picard-Vessiot extensions whose differential Galois groups have solvable
identity components.

######  Proposition 4.3.

Let $E$ be a differential field extension of $k$ having $C$ as its field of
constants. Then $E$ is a liouvillian Picard-Vessiot extension of $k$ if and
only if every differential field $K$ intermediate to $E$ and $k$ is a solution
field.

###### Proof.

Let $E$ be a liouvillian extension of $k,$ $K$ be an intermediate differential
subfield and $\mathscr{H}$:= $\mathscr{G}(E|K).$ We know that
$\mathscr{G}^{0}$, the identity component of the differential Galois group
$\mathscr{G}$ of $E$ over $k$, is a solvable linear algebraic group. From [5,
Pages 6, 12], we have the following facts:

  1. (i)

A closed subgroup $\mathscr{H}$ of an algebraic group $\mathscr{G}$ is
observable if and only if $\mathscr{H}\cap\mathscr{G}^{0}$ is observable in
$\mathscr{G}^{0}.$

  2. (ii)

If $\mathscr{G}$ is a solvable algebraic group then any closed subgroup
$\mathscr{H}$ of $\mathscr{G}$ is observable.

Since $\mathscr{G}^{0}$ is solvable, every closed subgroup of
$\mathscr{G}^{0}$ is observable. Thus in particular
$\mathscr{H}\cap\mathscr{G}^{0}$ is observable in $\mathscr{G}^{0}.$ This now
implies that our closed subgroup $\mathscr{H}$ is observable. Now by Theorem
4.2, $K$ must be a solution field. To prove the converse, we suppose that $E$
is not a liouvillian Picard-Vessiot extension of $k$. Then $\mathscr{G}^{0}$
is not solvable and therefore it contains a non-trivial Borel subgroup
$\mathscr{B}.$ Since $\mathscr{G}/\mathscr{B}$ is a projective variety, we
obtain from Theorem 4.2 that the differential field $K$ corresponding to
$\mathscr{B}$ is not a solution field. ∎

The following structure theorem for intermediate differential subfields of
liouvillian Picard-Vessiot extensions will be used in the proof of our main
theorem.

######  Theorem 4.4.

([16, Corollary 3.2]) Let $E$ be a liouvillian Picard-Vessiot extension of $k$
for $M$ and the differential Galois group $\mathscr{G}:=\mathscr{G}(E|k)$ be
connected. Let $K$ be a differential field intermediate to $E$ and $k$. Then
$K$ is a solution field generated by a solution $\psi:N\to K$ having elements
$t_{1},\dots,t_{n}\in\psi(N)$ such that for each $i,$
$t^{\prime}_{i}=a_{i}t_{i}+b_{i}$ for $a_{i}\in k$ and $b_{i}\in
k(t_{1},\dots,t_{i-1})$. Furthermore

  1. (i)

if $\mathscr{G}$ is a unipotent algebraic group then we may assume $a_{i}=0$
for each $i$ and that $t_{1},\dots,t_{n}$ are algebraically independent.

  2. (ii)

If $\mathscr{G}$ is a torus then we may assume $b_{i}=0.$

###### Proof.

To avoid triviality, we shall assume $k\neq K.$ We know from Proposition 4.3
that $K$ is a solution field. Let $\psi:N\to K$ be a $k[\partial]-$morphism
and $y_{1},\dots,y_{m}$ be a $k-$basis for the $k-$differential module
$\psi(N).$ Then, each $y_{i}$ is a solution of some linear homogeneous
differential equation over $k$ (see Proposition 4.1). Let $L$ be any
differential field such that $k\subseteq L\subsetneq K$. We claim that there
is a $y\in\psi(N)\setminus L$ such that $y^{\prime}=ay+b$ for some $a\in k,$
$b\in L$ and that $a$ can be taken zero if $\mathscr{G}$ is unipotent and that
$b$ can be taken to be zero if $\mathscr{G}$ is a torus.

Since $L\supsetneq K$, there is an $i$ such that $y_{i}\in\psi(N)\setminus L.$
Let $\mathscr{L}\in k[\partial]$ be of smallest positive degree $m$ such that
$\mathscr{L}(y)=0$ for some $y\in\psi(N)\setminus L$. Since $\mathscr{G}$ is
connected and solvable, the differential operator is linearly reducible [6,
Theorem 2 of Section 22]. So, $\mathscr{L}=\mathscr{L}_{m-1}\mathscr{L}_{1}$,
where $\mathscr{L}_{m-1},\mathscr{L}_{1}\in k[\partial]$ are of degrees $m-1$
and $1$ respectively. Let $\mathscr{L}_{1}=\partial-a$, $a\in k$. Observe that
$\mathscr{L}_{1}(y)\in\psi(N)$ and $\mathscr{L}_{m-1}(\mathscr{L}_{1}(y))=0$.
Therefore, from our choice of $m$, $\mathscr{L}_{1}(y)\in L$. Thus we have
found an element $y\in\psi(N)\setminus L$ such that $y^{\prime}=ay+b,$ where
$a\in k$ and $b\in L$.

If $\mathscr{G}$ is unipotent then $E=k(\eta_{1},\dots,\eta_{n}),$ where
$\eta_{i}^{\prime}\in k(\eta_{1},\dots,\eta_{i-1})$ and in this case
$\mathscr{L}$ admits a nonzero solution $\alpha\in k$ [19, Proposition 2.2].
Thus we may choose
$\mathscr{L}_{1}=\partial-\left(\alpha^{\prime}/\alpha\right).$ Then
$y/\alpha\in\psi(N)\setminus L$ and
$\left(y/\alpha\right)^{\prime}=b/\alpha\in L$. Moreover, such $y/\alpha$ must
be transcendental over $L$.

Now suppose that $\mathscr{G}$ is a torus. Then
$E=k\left(\xi_{1},\dots,\xi_{s}\right)$, where $\xi_{i}^{\prime}/\xi_{i}\in k$
for each $i$. If $b\neq 0$ then again from [19, Proposition 2.2] applied to
the extension $L\left(\xi_{1},\dots,\xi_{s}\right)$ of $L$ with
$\mathscr{L}_{1}(y)=y^{\prime}-ay=b,$ we obtain $\alpha\in L$ such that
$\alpha^{\prime}-a\alpha=b$. Now $y-\alpha\in\psi(N)\setminus L$ and
$(y-\alpha)^{\prime}/(y-\alpha)=a\in k.$ This proves the claim.

Taking $L=k,$ we obtain an element $t_{1}\in\psi(N)$ and $t_{1}\notin k$ such
that $t^{\prime}_{1}=a_{1}t_{1}+b_{1}$ for $a_{1},b_{1}\in k$. Likewise, we
find elements $t_{1},\dots,t_{n}\in\psi(N)$ and $t_{i}\notin
k(t_{1},\dots,t_{i-1})$ such that $t^{\prime}_{i}=a_{i}t_{i}+b_{i},$ where
$a_{i}\in k$ and $b_{i}\in k(t_{1},\dots,t_{i-1})$. Since $\psi(N)$ is finite
dimensional over $k$, there must be an integer $n$ such that $\psi(N)\subset
k(t_{1},\dots,t_{n}).$ Clearly, for that $n,$ we must have
$K=k(t_{1},\dots,t_{n}).$ ∎

######  Remark 4.5.

In the above theorem, the hypothesis that $\mathscr{G}\left(E|k\right)$ is
connected allowed us to factor the differential operator $\mathscr{L}$ in
$k[\partial]$, which is a crucial step in the proof. In fact, the assumption
that $\mathscr{G}\left(E|k\right)$ is connected cannot be dropped. For
example, consider the liouvillian extension
$E=\mathbb{C}(x)\left(\sqrt{x},e^{\sqrt{x}}\right),$ where the derivation is
${}^{\prime}:=d/dx.$ Then $E$ is a liouvillian Picard-Vessiot extension of
$\mathbb{C}(x)$ for the differential equation

$\mathscr{L}(y)=y^{\prime\prime}+\frac{1}{2x}y^{\prime}-\frac{1}{4x}y=0.$

The set
$V:=\text{span}_{\mathbb{C}}\left\\{e^{\sqrt{x}},e^{-\sqrt{x}}\right\\}$ is
the set of all solutions of $\mathscr{L}(y)=0$ in $E$. Since $E$ contains the
algebraic extension $\mathbb{C}(x)\left(\sqrt{x}\right),\
\mathscr{G}\left(E|k\right)$ is not connected. One can show that the
intermediate differential field $K:=\mathbb{C}(x)\langle
e^{\sqrt{x}}+e^{-\sqrt{x}}\rangle$ contains no elements satisfying a first
order linear differential equation over $\mathbb{C}(x)$ other than the
elements of $\mathbb{C}(x)$ itself [19, p. 376].

##  5\. Transcendence degree one subfields of strongly normal extensions

######  Proposition 5.1.

Let $E$ be a differential field extension field of $k$ having $C$ as its field
of constants. Let $L$ be a differential field intermediate to $E$ and $k.$
Suppose that there are two elements $y,t\in E$ each transcendental over $L$
such that $K=L(t)=L(y).$ Then $t$ satisfies a Riccati equation over $L$ if and
only if $y$ satisfies a Riccati equation over $L.$

###### Proof.

From Lüroth’s theorem, we know that there are elements $a,b,c,d\in L$ such
that $ad-bc\neq 0$ and that

$y=\frac{at+b}{ct+d}.$

We compute the derivative of the above equation and obtain

$y^{\prime}(ct+d)^{2}=\
ca^{\prime}t^{2}+(a^{\prime}d+b^{\prime}c)t+db^{\prime}-\left(c^{\prime}at^{2}+(bc^{\prime}+ad^{\prime})t+bd^{\prime}\right)+(ad-
bc)t^{\prime}.$

If $y^{\prime}=f(y)$ is a polynomial of degree $\leq 2$ then we see that
$y^{\prime}(ct+d)^{2}$ is a polynomial of degree $\leq 2.$ Since $0\neq ad-
bc\in L,$ we shall solve for $t^{\prime}$ and obtain $t^{\prime}=g(t),$ where
$g$ is a polynomial in one variable over $L$ of degree at most $2$. ∎

We recall the following facts about a strongly normal extension $E$ of $k.$
Since $\mathscr{G}:=\mathscr{G}(E|k)$ is an algebraic group, there is a chain
of subgroups (see [10, Proposition 12.2])

$\mathscr{G}\supseteq\mathscr{G}^{0}\supseteq\mathscr{H}\supseteq\left\\{1\right\\},$

where $\mathscr{G}^{0}$ is the identity component of $\mathscr{G}$,
$\mathscr{H}$ is a connected linear algebraic group as well as a closed normal
subgroup of $\mathscr{G}^{0}$ such that $\mathscr{G}^{0}/\mathscr{H}$ is an
abelian variety. From the fundamental theorem of strongly normal extensions,
we have a tower of fields

(5.0.1) $k\subseteq E^{0}\subseteq F\subseteq E,$

where $E^{0}$ is a finite Galois extension of $k$ with Galois group
$\mathscr{G}/\mathscr{G}^{0}$, $F$ is an abelian extension of $E^{0},$ that is
$\mathscr{G}(F|E^{0})\cong\mathscr{G}^{0}/\mathscr{H}$ is an abelian variety
and $E$ is a Picard-Vessiot extension of $F$ with
$\mathscr{H}\cong\mathscr{G}(E|F).$

######  Theorem 5.2.

Let $E$ be a strongly normal extension of $k$ and $K$ be a differential field
intermediate to $E$ and $k.$ If tr.deg$(K|k)=1$ then there is a finite
algebraic extension $\tilde{k}$ of $k$ and an element $t\in\tilde{k}K$ such
that one of the following holds:

  1. (i)

$\tilde{k}K=\tilde{k}(t)$ and $t$ is a solution of a Riccati equation over
$\tilde{k}.$

  2. (ii)

$\tilde{k}K=\tilde{k}(t,t^{\prime})$ and $t$ is a solution of a Weierstrass
differential equation over $\tilde{k}.$

###### Proof.

Decompose $E$ into a tower of fields as in Equation (5.0.1). Fix an algebraic
closure $\overline{E}$ of $E$ and extend the derivation of $E$ to a derivation
of $\overline{E}.$ Now we split the proof into two cases.

Case (i): Suppose that $KF\supsetneq F$. Choose a finite algebraic extension
$\tilde{k}\subset\overline{E}$ of $k$ so that there is a nonsingular
projective curve $\Gamma$ defined over $\tilde{k},$ the function field
$\tilde{k}(\Gamma)\cong\tilde{k}K$ and that $\Gamma$ has a $\tilde{k}-$point.
We first claim that $\Gamma$ is a rational curve, that is,
$\tilde{k}K=\tilde{k}(t).$

Since the compositum $\tilde{k}E$ remains a strongly normal extension of
$\tilde{k}$ (see [7, Theorem 5]), for convenience of notation, we shall
replace $\tilde{k}$ by $k.$ Choose an element $z\in KF\setminus F.$ Since
$\mathscr{G}(E|F)$ is a connected linear algebraic group and since a connected
linear algebraic group is a union of its Borel subgroups, by the fundamental
theorem, we obtain that

$\bigcap_{\mathscr{B},\ \text{Borel subgroups}}E^{\mathscr{B}}=F.$

Thus, there is a Borel subgroup $\mathscr{B}$ of $\mathscr{G}(E|F)$ such that
$z\in E\setminus E^{\mathscr{B}}.$ Note that
$\mathscr{B}=\mathscr{U}\rtimes\mathscr{T},$ where $\mathscr{U}$ is the
unipotent radical and $\mathscr{T}$ is a torus. Thus $E$ can be further
decomposed as follows:

$E\supseteq E^{\mathscr{U}}\supseteq E^{\mathscr{B}}\supseteq F,$

  1. (i)

$E$ is a Picard-Vessiot extension of $E^{\mathscr{U}}$ whose Galois group
$\mathscr{U}$ is unipotent.

  2. (ii)

$E^{\mathscr{U}}$ is a Picard-Vessiot extension of $E^{\mathscr{B}}$ whose
Galois group is isomorphic to the torus
$\mathscr{T}\cong\mathscr{B}/\mathscr{U}.$

Suppose that the compositum $E^{\mathscr{U}}K$ properly contains
$E^{\mathscr{U}}.$ Then since $\mathscr{U}$ is connected, $E^{\mathscr{U}}$ is
algebraically closed in $E^{\mathscr{U}}K$ and we obtain that
$\mathrm{tr.deg}(E^{\mathscr{U}}K|E^{\mathscr{U}})=1.$ Now we apply Theorem
4.4 and obtain $E^{\mathscr{U}}K=E^{\mathscr{U}}(y),$ where $y^{\prime}\in
E^{\mathscr{U}}.$ This, in particular, shows that the (geometric) genus of
$\Gamma$ is $0.$ Since $\Gamma$ has $k-$point, we obtain that $K=k(t)$ (see
[2, Sections 2 and 6 of Chapter 2]).

Thus $E^{\mathscr{U}}K=E^{\mathscr{U}}(t)=E^{\mathscr{U}}(y).$ From
Proposition 5.1, we then have $t$ is a solution of a Riccati equation over
$E^{\mathscr{U}}.$ Say, $t^{\prime}=g(t),$ where $g$ is a polynomial over
$E^{\mathscr{U}}$ of degree $\leq 2.$ Since $K=k(t)$ is a differential field,
we have $t^{\prime}=h_{1}(t)/h_{2}(t),$ where $h_{1}$ and $h_{2}$ are
relatively prime polynomials in $k[t]$ with $h_{2}$ being a monic polynomial.
Then $h_{1}$ and $h_{2}$ must remain relatively prime over any field extension
of $k$. Now since $g(t)h_{2}(t)=h_{1}(t)$, we must have $h_{2}=1$ and we
obtain that the coefficients of $g$ are in $k$.

Suppose that $E^{\mathscr{U}}K=E^{\mathscr{U}},$ that is $K\subseteq
E^{\mathscr{U}}.$ Since $\mathscr{T}$ is a commutative group,
$KE^{\mathscr{B}}$ must be a Picard-Vessiot extension of $E^{\mathscr{B}}$ and
since $\mathscr{T}$ is connected, $E^{\mathscr{B}}$ is algebraically closed in
$E^{\mathscr{U}}$ and therefore in $KE^{\mathscr{B}}$ as well. Now
$\mathrm{tr.deg}(KE^{\mathscr{B}}|E^{\mathscr{B}})=1$ implies that the
differential Galois group of $KE^{\mathscr{B}}$ over $E^{\mathscr{B}}$ is
isomorphic to $\mathrm{G_{m}}$ and it follows that
$E^{\mathscr{B}}K=E^{\mathscr{B}}(y)$ such that $y^{\prime}/y\in
E^{\mathscr{B}}$ (see [11, Example 5.24]). Thus we have again shown that
$\Gamma$ is a genus zero curve. Therefore $K=k(t)$ and we obtain
$E^{\mathscr{B}}K=E^{\mathscr{B}}(t)=E^{\mathscr{B}}(y)$ with $y^{\prime}/y\in
E^{\mathscr{B}}.$ Now a similar argument, as in the previous paragraph, shows
that $t$ satisfies a Riccati equation over $k.$ This completes the proof of
the theorem in this case.

Case (ii). Suppose that $K\subseteq F.$ Then since $F$ is an abelian extension
of $E^{0},$ every intermediate differential subfield is strongly normal over
$E^{0}.$ In particular, $KE^{0}$ is strongly normal over $E^{0}$ and
$\mathrm{tr.deg}(KE^{0}|E^{0})=1.$ Now we apply [8, Theorem 3] and obtain that
either $KE^{0}=E^{0}(y),$ where $y^{\prime}\in E^{0}$ or $y^{\prime}/y\in
E^{0}\setminus\\{0\\}$ or that $\bar{k}K=\bar{k}(t,t^{\prime}),$ where $t$ is
a solution of a Weierstrass differential equation over $\bar{k}.$ If
$KE^{0}=E^{0}(y)$ with $y^{\prime}\in E^{0}$ or $y^{\prime}/y\in
E^{0}\setminus\\{0\\}$ then $\mathscr{G}(KE^{0}|E^{0})\cong\mathrm{G_{a}}(C)$
or $\mathrm{G_{m}}(C)$ and in any event, we must then have a surjective
morphism from the abelian variety $\mathscr{G}(F|E^{0})$ to the linear
algebraic group $\mathscr{G}(KE^{0}|E^{0}),$ which is impossible. Therefore,
$\bar{k}K=\bar{k}(t,t^{\prime}),$ where $t$ is a nonalgebraic solution of a
Weierstrass differential equation over $\bar{k}.$ Since $K$ is finitely
generated over $k,$ there is a finite algebraic extension $\tilde{k}$ of $k$
such that $\tilde{k}K=\tilde{k}(t,t^{\prime}).$ ∎

##  6\. Transcendence degree one subfields of iterated strongly normal
extensions

######  Proposition 6.1.

Let $E$ be a strongly normal extension of $k$ with differential Galois group
$\mathscr{G}$ and $\tilde{E}$ be an algebraic extension of $E$. Suppose that
$K$ is a differential field intermediate to $k$ and $\tilde{E}$ such that
$\mathrm{tr.deg}\left(K|k\right)=1$. Then $\mathrm{tr.deg}(K\cap E|k)=1$ and
there is an algebraic extension $\tilde{k}$ of $k$ such that one of the
following holds:

  1. (i)

$\tilde{k}(K\cap E)=\tilde{k}(t),$ where $t$ satisfies a Riccati differential
equation over $\tilde{k}$.

  2. (ii)

$\tilde{k}(K\cap E)=\tilde{k}(t,t^{\prime}),$ where $t$ satisfies a
Weierstrass differential equation over $\tilde{k}$.

###### Proof.

From [7, Theorem 5], we have that $KE$ is a strongly normal extension of $K$
and that the natural restriction map from the differential Galois group
$\mathscr{G}(KE|K)$ to $\mathscr{G}$ is an isomorphism onto the subgroup
$\mathscr{G}(E|K\cap E)$ of $\mathscr{G}$. In particular, if
$\mathrm{tr.deg}(E|k)=n$ then $\mathrm{tr.deg}(E|K\cap E)=$
dim$(\mathscr{G}(E|K\cap E))=$ $\mathrm{tr.deg}(KE|K)=n-1.$ Thus,
$\mathrm{tr.deg}((K\cap E)|k)=1.$ Now, the rest of the proof follows from
Theorem 5.2. ∎

######  Lemma 6.2.

Let $L$ be a differential field extension of $\bar{k}$ having $C$ as its field
of constants. Suppose that $K$ and $E$ are differential subfields intermediate
to $L$ and $\bar{k}$ having the following properties: $K$ is finitely
generated over $\bar{k},$ $\mathrm{tr.deg}(K|\bar{k})=1,$
$\mathrm{tr.deg}(EK|E)=1,$ $E$ is an algebraically closed field, $EK$ is
weakly normal over $K$ and that the group of differential automorphisms
$\mathscr{G}(EK|K)$ stabilises $E.$

  1. (i)

If there is a Riccati equation over $E$ having a solution $t\in EK\setminus E$
then there is Riccati equation over $\bar{k}$ having a solution $v\in
K\setminus\bar{k}.$

  2. (ii)

If there is a Weierstrass differential equation over $E$ having a solution
$t\in EK\setminus E$ then there is a Weierstrass differential equation over
$\bar{k}$ having a solution $v\in K\setminus\bar{k}.$

###### Proof.

From our assumptions that $\mathscr{G}(EK|K)$ stabilizes $E,$
$\mathrm{tr.deg}(EK|E)=1$ and that $t$ is transcendental over $k,$ we have for
each $\sigma\in\mathscr{G}(EK|K)$, $\sigma(t)$ is algebraic over $E(t).$ Since
$K$ is finitely generated over $\bar{k},$ $EK$ is finitely generated over $E$
and we conclude that there are finitely many automorphisms
$\sigma_{1},\dots,\sigma_{n}\in\mathscr{G}(EK|K),$ where $\sigma_{1}$ is the
identity, such that $\mathscr{G}(EK|K)$ stabilizes the differential field
$E^{*}:=E(\sigma_{1}(t),\sigma_{1}(t)^{\prime},\dots,$
$\sigma_{n}(t),\sigma_{n}(t)^{\prime}).$

We now claim that $(E^{*}\setminus E)\cap K\neq\emptyset.$ Since
$\mathrm{tr.deg}(K|\bar{k})=$ $\mathrm{tr.deg}(EK|E)=1,$ there is an element
$u\in K\setminus\bar{k}$ such that $u$ is transcendental over $E.$ Let
$X^{m}+\alpha_{m-1}X^{m-1}+\cdots+\alpha_{0}\in E^{*}[X]$ be the irreducible
polynomial of $u$ over $E^{*}$.

Then for any $\sigma\in\mathscr{G}(EK|K)$, we have

$u^{m}+\sigma(\alpha_{m-1})u^{m-1}+\cdots+\sigma(\alpha_{0})=0$

and thus

$\left(\sigma(\alpha_{m-1})-\alpha_{m-1}\right)u^{m-1}+\cdots+\sigma(\alpha_{0})-\alpha_{0}=0.$

Therefore $\sigma(\alpha_{i})=\alpha_{i}$ for all $i=0,\dots,m-1.$ Since $EK$
is assumed to be weakly normal over $K,$ we obtain that $\alpha_{i}\in K$.
That is $\alpha_{i}\in E^{*}\cap K$ for all $i.$ Now $u$ is transcendental
over $E$ implies that there is at least one $i$ such that $s:=\alpha_{i}\in
E^{*}\setminus E$. This proves the claim.

${EK}$${\mathcal{E}}$${E^{*}}$${K}$${E(s,s^{\prime})}$${E}$${\bar{k}(s,s^{\prime})}$${\bar{k}}$$\scriptstyle{\mathscr{G}(EK|K)}$

Now we are ready to prove the lemma. Let $t^{\prime}=P(t)$ be a Riccati
equation over $E$ with $t\in EK\setminus E.$ Then for any
$\sigma\in\mathscr{G}(EK|K),$ $\sigma(t)^{\prime}=P_{\sigma}(\sigma(t))$ is
also a Riccati equation, where $P_{\sigma}\in E[X]$ is the polynomial obtained
by applying $\sigma$ to the coefficients of $P.$ Since each of
$\sigma_{1}(t),$ $\dots,\sigma_{n}(t)$ is a solution of some Riccati equation
over $E,$ one can construct a Picard-Vessiot extension $\mathcal{E}$ of $E$
containing $\sigma_{1}(t),\dots,\sigma_{n}(t).$ Now $E(s,s^{\prime})$ is a
differential subfield of $\mathcal{E}$ with
$\mathrm{tr.deg}(E(s,s^{\prime})|E)=1.$ Applying Theorem 5.2, we obtain
$E(s,s^{\prime})=E(y),$ where $y$ satisfies a Riccati differential equation
over $E$. Then $\bar{k}(s,s^{\prime})$ has genus $0$ and thus
$\bar{k}(s,s^{\prime})=\bar{k}(v).$ Applying Proposition 5.1, we obtain that
$v$ satisfies a Riccati equation over $\bar{k}.$ This proves (i).

Suppose that $t$ is a nonalgebraic solution of a Weierstrass differential
equation over $E;$ $t^{\prime 2}=\alpha^{2}(4t^{3}-g_{2}t-g_{3})$ with
$g_{2},g_{3}\in C.$ In this case, we claim that the differential field
$\overline{k}(s,s^{\prime})$ contains a nonalgebraic solution of a Weierstrass
differential equation over $\bar{k}$, which would then complete the proof of
this lemma. Observe that $\sigma(t)^{\prime
2}=\sigma(\alpha)^{2}(4\sigma(t)^{3}-g_{2}\sigma(t)-g_{3})$ for
$\sigma\in\mathscr{G}(EK|K).$ Then $E^{*}$, being a compositum of finitely
many strongly normal extensions $E(\sigma_{i}(t),\sigma_{i}(t)^{\prime})$ of
$E,$ is a strongly normal extension of $E.$ Also, $E^{*}\subseteq EK$ and
therefore $\mathrm{tr.deg}(E^{*}|E)=1.$ From the facts that $E(t,t^{\prime})$
is a genus one field extension of $E$ and that $E(t,t^{\prime})\subseteq
E^{*},$ we infer that $E^{*}$ must be a genus one abelian extension of $E.$
Now we consider the differential subfield $E(s,s^{\prime})$ of $E^{*}.$ As
noted in the proof of case (ii) of Theorem 5.2, $E(s,s^{\prime})$ is also a
genus one abelian extension of $E$. This implies $\bar{k}(s,s^{\prime})$ is a
genus one field extension of $\bar{k}.$

Let $\mathscr{C}_{1}$ be a nonsingular projective model for
$\bar{k}(s,s^{\prime}).$ Then $\mathscr{C}_{1}$ is also a model for
$E(s,s^{\prime}).$ By [8, Theorem 3], the group
$\mathscr{G}(E(s,s^{\prime})|E)$ is the $C-$points of a Weierstrass elliptic
curve $\mathscr{C}_{2}$ defined over $C:$

$\mathscr{C}_{2}=X^{2}_{2}X_{0}-4X^{3}_{1}+h_{2}X^{2}_{0}X_{1}+h_{3}X^{3}_{0},$

where $h_{2},h_{3}\in C.$ Furthermore, $E(s,s^{\prime})=E(x,x^{\prime}),$
where $(1:x:x^{\prime}/\alpha)$ is a point of $\mathscr{C}_{2}$ for some
$\alpha\in E.$ Thus $\mathscr{C}_{2}$ is also a nonsingular projective model
for $E(s,s^{\prime}).$ Then, $\mathscr{C}_{1}$ and $\mathscr{C}_{2}$ are
isomorphic over $E.$ Since both the curves are defined over $\bar{k}$ and the
$j-$ invariants $j(\mathscr{C}_{1})=j(\mathscr{C}_{2})\in C\subseteq\bar{k},$
the curves are isomorphic over $\bar{k}$ as well [17, Proposition 1.4]. Thus,
there is a $\bar{k}(s,s^{\prime})-$point $(1:\omega:\rho)$ of
$\mathscr{C}_{2}$ such that $\bar{k}(s,s^{\prime})=\bar{k}(\omega,\rho).$
Then, as in the proof of [8, Theorem 3], one can show that
$\bar{k}(s,s^{\prime})=\bar{k}(\omega,\omega^{\prime})$ and that
$(1:\omega:\omega^{\prime}/\beta)$ is a point on $\mathscr{C}_{2}$ for some
$\beta\in\overline{k}.$ This implies $\omega^{\prime
2}=\beta^{2}(4\omega^{3}-h_{2}\omega-h_{3}).$ ∎

######  Theorem 6.3.

Let $E$ be an iterated strongly normal extension of $k$ and $K$ be an
intermediate differential field having field transcendence degree
$\mathrm{tr.deg}(K|k)=1.$ Then, there is a finite algebraic extension
$\tilde{k}$ of $k$ and an element $t\in\tilde{k}K$ such that one of the
following holds:

  1. (i)

$\tilde{k}K$ is a finite algebraic extension of $\tilde{k}(t)$ and $t$ is a
solution of a Riccati differential equation over $\tilde{k}.$

  2. (ii)

$\tilde{k}K$ is a finite algebraic extension of $\tilde{k}(t,t^{\prime})$ and
$t$ is a solution of a Weierstrass differential equation over $\tilde{k}.$

###### Proof.

Let $k=:E_{0}\subseteq E_{1}\subseteq E_{2}\subseteq\cdots\subseteq E_{n+1}=E$
be a tower of strongly normal extensions, where we may assume that $E_{n+1}$
is an algebraic extension of $E_{n}$ and that $E_{n}$ is a transcendental
extension of $E_{n-1}.$ Let $\overline{E}$ be an algebraic closure of $E.$
Consider the following tower of algebraically closed differential fields

$\overline{E}_{0}\subseteq\overline{E}_{1}\subseteq\overline{E}_{2}\subseteq\cdots\subseteq\overline{E}_{n}=\overline{E}.$

Since the differential field $\overline{E}_{n-1}K$ is contained in
$\overline{E},$ by Proposition 6.1, we obtain an element
$y\in\overline{E}_{n-1}K$ transcendental over $\overline{E}_{n-1}$ such that
$y$ satisfies either a Riccati or a Weierstrass differential equation over
$\overline{E}_{n-1}.$

For the moment, let us assume that

  * (C1)

for all $i=1,\dots,n-1,$ $\overline{E}_{i}K$ is a weakly normal extension of
$\overline{E}_{i-1}K$ and that the group
$\mathscr{G}(\overline{E}_{i}K|\overline{E}_{i-1}K)$ stabilises
$\overline{E}_{i}.$

The differential field $K$, being a subfield of an iterated strongly normal
extension, is finitely generated over $k.$ Therefore, $\overline{E}_{i}K$ is
finitely generated over $\overline{E}_{i}.$ Now, a repeated application of
Lemma 6.2 will produce a solution $t\in\bar{k}K\setminus\bar{k}$ of a Riccati
or a Weierstrass differential equation over $\bar{k}.$ Indeed, one only needs
a finite algebraic extension $\tilde{k}$ so that the differential field
$\tilde{k}K$ contains both $t$ and the coefficients of the differential
equation of $t$. Thus the theorem is proved under the assumption (C1).

Now we shall verify the validity of (C1). Note that $\overline{E}_{i-1}E_{i}$
is strongly normal over $\overline{E}_{i-1}.$ Therefore the
$L^{*}:=\overline{E}_{i-1}KE_{i}$ is a strongly normal extension of
$L:=K\overline{E}_{i-1}.$ Since $\overline{E}$ and $k$ have the same field of
constants, the group $\mathscr{G}(L^{*}|L)$ stabilizes the strongly normal
extensions $E_{i}$ of $E_{i-1}$ and $\overline{E}_{i-1}E_{i}$ of
$\overline{E}_{i-1}.$ Next, we consider the set $I$ of all irreducible
polynomials whose coefficients are in $E_{i}.$ The group
$\mathscr{G}(L^{*}|L)$ acts on $I$ via the map $P\mapsto P_{\sigma},$ where
$P_{\sigma}$ is obtained by applying automorphism $\sigma$ to the coefficients
of $P.$ Now we lift each $\sigma\in\mathscr{G}(L^{*}|L)$ to a differential
automorphism of the splitting field $L^{*}(I)$ of $I$ contained in
$\overline{E}.$ Observe that
$L^{*}(I)=L^{*}\overline{E}_{i}=K\overline{E_{i}}.$ Since the fixed field of
the group $\mathscr{G}(K\overline{E}_{i}|L^{*})$ is the field $L^{*}$ and that
the fixed field of $\mathscr{G}(L^{*}|L)$ is $L=\overline{E}_{i-1}K,$ we
obtain that the fixed field of
$\mathscr{G}(\overline{E}_{i}K|\overline{E}_{i-1}K)$ is $\overline{E}_{i-1}K.$
This shows that $\overline{E}_{i}K$ is weakly normal over
$\overline{E}_{i-1}K.$ Every element of
$\mathscr{G}(\overline{E}_{i}K|\overline{E}_{i-1}K)$ stabilizes the strongly
normal extension $E_{i}$ of $E_{i-1}$ as well as the set $I$ of polynomials
over $E_{i}.$ Therefore the group stabilizes $\overline{E}_{i}.$ This shows
that (C1) always holds. ∎

######  Corollary 6.4.

If $E$ is an iterated strongly normal extension of the field of constants $C$
then every intermediate differential field $K$ with $\mathrm{tr.deg}(K|C)=1$
has an element $t$ such that either $t^{\prime}=1$ or $t^{\prime}=ct$ for some
nonzero $c\in C$ or $t$ is a solution of a Weierstrass differential equation
over $C.$

###### Proof.

We only need to consider the case when there is an element $y\in K\setminus C$
such that $y^{\prime}=ay^{2}+by+c$ for $a,b,c\in C.$ In this case, the
differential field $C(y)$ can be embedded in a Picard-Vessiot extension of
$C.$ Now since any Picard-Vessiot extension of $C$ has a connected abelian
differential Galois group, we obtain that $C(y)$ itself must be a Picard-
Vessiot extension of $C$ having a differential Galois group isomorphic to
$\mathrm{G_{a}}(C)$ or $\mathrm{G_{m}}(C).$ It follows that $C(y)=C(t),$ where
either $t^{\prime}=1$ or $t^{\prime}=ct$ for some nonzero constant $c.$ ∎

###### Proof of Theorem 1.1.

Follows from Theorem 5.2, Theorem 6.3 and Corollary 6.4. ∎

##  7\. nonalgebraic solutions of first order differential equations

###  7.1. Differential equations of general type

In the next theorem, we device a mechanism to generate differential equations
of general type.

######  Theorem 7.1.

Let $f\in\bar{k}[Y,Z]$ be an irreducible polynomial having the following
properties:

  1. (a)

$p:=(0,0)$ is a simple point of $f$ and $Z$ is the tangent line at $p.$

  2. (b)

With respect to the uniformizing parameter $Y,$ both the coefficients
$\lambda_{2}$ and $\lambda_{3}$ of the $Y-$adic expansion
$Z=\lambda_{2}Y^{2}+\lambda_{3}Y^{3}+\cdots$ do not have any antiderivatives
in $\bar{k}.$

Then the differential equation $f(y,y^{\prime})=0$ is of general type.

###### Proof.

Consider $\bar{k}(f)=\bar{k}(y,z)$ with the induced derivation
${}^{\prime}:\bar{k}(y,z)\to\bar{k}(y,z);$ $y^{\prime}=z.$ In order to show
that $f(y,y^{\prime})=0$ has a nonalgebraic solution, we only need to show
that the field of constants of $\bar{k}(y,y^{\prime})$ is the same as that of
$\bar{k}.$ Since $\bar{k}(y,y^{\prime})$ embeds as a differential field in
$\bar{k}((y)),$ for an element $w\in\bar{k}(y,y^{\prime})\setminus\bar{k},$ we
can write the $y-$adic expansion $w=\sum_{i=r}^{\infty}a_{i}y^{i},$
$a_{i}\in\bar{k}$ and $a_{r}\neq 0$ and obtain

$\displaystyle w^{\prime}$
$\displaystyle=a^{\prime}_{r}y^{r}+a^{\prime}_{r+1}y^{r+1}+\cdots+ra_{r}y^{r-1}(\lambda_{2}y^{2}+\lambda_{3}y^{3}+\cdots)+(r+1)a_{r+1}y^{r}(\lambda_{2}y^{2}+\cdots)+\cdots$
(7.1.1)
$\displaystyle=a^{\prime}_{r}y^{r}+(a^{\prime}_{r+1}+ra_{r}\lambda_{2})y^{r+1}+(a^{\prime}_{r+2}+ra_{r}\lambda_{3}+(r+1)a_{r+1}\lambda_{2})y^{r+2}+\cdots.$

We shall first prove that

  * (C2)

there is no $w\in\bar{k}(y,y^{\prime})\setminus\bar{k}$ such that
$w^{\prime}=\alpha w+\beta$ for any $\alpha,\beta\in\bar{k}.$

Suppose not. Then, substituting $w=\sum^{\infty}_{i=r}a_{r}y^{r},$ we have the
following cases to consider:

Case (i). ord${}_{p}(w)=0$. Here we have $a^{\prime}_{0}=\alpha a_{0}+\beta$
and if $m\geq 1$ is the least positive integer such that $a_{m}\neq 0$ then we
have $a^{\prime}_{m}=\alpha a_{m}$ and that
$a^{\prime}_{m+1}+ma_{m}\lambda_{2}=\alpha a_{m+1}.$ Thus, we obtain

$\left(-\frac{a_{m+1}}{ma_{m}}\right)^{\prime}=\lambda_{2},$

a contradiction to our assumption on $\lambda_{2}$.

Case (ii). ord${}_{p}(w)\geq 1$ or ord${}_{p}(w)\leq-2.$ Then
$a^{\prime}_{r}=\alpha a_{r}$ and that
$a^{\prime}_{r+1}+ra_{r}\lambda_{2}=\alpha a_{r+1}.$ This implies

$\left(-\frac{a_{r+1}}{ra_{r}}\right)^{\prime}=\lambda_{2}$

and as before we obtain a contradiction.

Case (iii). ord${}_{p}(w)=-1.$ Then $a^{\prime}_{-1}=\alpha a_{-1}$ and
$a^{\prime}_{1}=\alpha a_{1}+\lambda_{3}a_{-1}.$ This implies

$\left(\frac{a_{1}}{a_{-1}}\right)^{\prime}=\lambda_{3},$

which contradicts our assumption on $\lambda_{3}.$

Thus we have shown that there is no element
$w\in\bar{k}(y,y^{\prime})\setminus\bar{k}$ such that $w^{\prime}=\alpha
w+\beta$ for any $\alpha,\beta\in\bar{k}.$ In particular, we could choose
$\alpha=\beta=0$ to show that the field of constants of $\bar{k}$ is the same
as the field of constants of $\bar{k}(y,y^{\prime}).$

We shall now show that the equation $f(y,y^{\prime})=0$ is of general type.
Suppose that there is an element $w\in\bar{k}(y,y^{\prime})\setminus\bar{k}$
such that $w^{\prime}=b_{2}w^{2}+b_{1}w+b_{0}$ for
$b_{2},b_{1},b_{0}\in\bar{k}.$ Then we must have $b_{2}\neq 0.$ If
ord${}_{p}(w)\leq-1$ then from Equation (7.1.1) we obtain
ord${}_{p}(w^{\prime})\geq$ ord${}_{p}(w).$ But $b_{2}\neq 0$ and
ord${}_{p}(w)\leq-1$ implies ord${}_{p}(w^{\prime})=$
ord${}_{p}(b_{2}w^{2}+b_{1}w+b_{0})=2$ ord${}_{p}(w).$ Thus we have obtained
ord${}_{p}(w)\leq$ ord${}_{p}(w^{\prime})=2$ ord${}_{p}(w),$ a contradiction.
Next, if ord${}_{p}(w)\geq 1$ then $b_{0}=0$ and we obtain that
$(1/w)^{\prime}=-b_{1}(1/w)-b_{2},$ which contradicts (C2). Finally, if
ord${}_{p}(w)=0$ then $a_{0}\in\bar{k}$ is a solution of the Riccati equation
$a^{\prime}_{0}=b_{2}a^{2}_{0}+b_{1}a_{0}+b_{0}.$ A quick calculation shows

$\left(\frac{1}{w-a_{0}}\right)^{\prime}=\frac{-2b_{2}a_{0}-b_{1}}{w-a_{0}}-b_{2}$

and this again contradicts (C2). Thus we have shown that there is no element
$w\in\bar{k}(y,y^{\prime})\setminus\bar{k}$ satisfying a Riccati equation over
$\bar{k}.$

Now we suppose that there is a transcendental element
$w\in\bar{k}(y,y^{\prime})$ satisfying a Weierstrass differential equation:
$w^{\prime 2}=\alpha^{2}(4w^{3}-g_{2}w-g_{3})$ for $g_{2},g_{3}\in C$ and
$\alpha\in\bar{k}.$ Then from the $y-$adic expansion of $w$ and from Equation
(7.1.1), we have $w^{\prime 2}=a^{\prime 2}_{r}y^{2r}+\cdots$ and thus

$2\ \text{ord}_{p}(w)\leq\text{ord}_{p}(w^{\prime
2})=\text{ord}_{p}(\alpha^{2}(4w^{3}-g_{2}w-g_{3})).$

If ord${}_{p}(w)<0$ then ord${}_{p}(\alpha^{2}(4w^{3}-g_{2}w-g_{3}))=3\
\text{ord}_{p}(w)$ and we obtain $2$ ord${}_{p}(w)\leq 3$ ord${}_{p}(w),$ a
contradiction. If ord${}_{p}(w)>0$ then
ord${}_{p}(\alpha^{2}(4w^{3}-g_{2}w-g_{3}))\leq$ ord${}_{p}(w),$ and we obtain
$2\ \text{ord}_{p}(w)\leq$ ord${}_{p}(w),$ again a contradiction. Finally, let
ord${}_{p}(w)=0$ and in the $y-$adic expansion of $w,$ let $m$ be the least
positive integer such that $a_{m}\neq 0.$ Then from Equation (7.1.1), we have
the following equations

$\displaystyle w^{\prime 2}$ $\displaystyle=\alpha^{2}(4w^{3}-g_{2}w-g_{3})$
$\displaystyle a^{\prime 2}_{0}$
$\displaystyle=\alpha^{2}(4a^{3}_{0}-g_{2}a_{0}-g_{3})$ (7.1.2) $\displaystyle
2a^{\prime}_{0}a_{m}^{\prime}$
$\displaystyle=\alpha^{2}\left(12a^{2}_{0}a_{m}-g_{2}a_{m}\right).$

We assume for the moment that $a_{0}$ is a constant. Then $a_{0}$ must be one
of the distinct roots of the polynomial $4Y^{3}-g_{2}Y-g_{3}$ and in
particular, $a_{0}$ is not a root of $12Y^{2}-g_{2}.$ Now Equation (7.1.2)
becomes

$\alpha^{2}(12a^{2}_{0}-g_{2})a_{m}=0,$

which is absurd. Thus there is no element in
$\bar{k}(t,t^{\prime})\setminus\bar{k}$ satisfying a Weierstrass differential
equation over $\bar{k}.$

Now we shall show that $a_{0}$ is indeed a constant. Assume otherwise and
consider the nonsingular projective curve

(7.1.3) $X^{2}_{2}X_{0}-4X^{3}_{1}+g_{2}X^{2}_{0}X_{1}+g_{3}X^{3}_{0}.$

We have two nonconstant points of this curve, namely,
$(1:w:w^{\prime}/\alpha)$ and $(1:a_{0}:a^{\prime}_{0}/\alpha).$ If
$(1:\eta:\xi):=(1:w:w^{\prime}/\alpha)(1:a_{0}:a^{\prime}_{0}/\alpha)$ then

$\eta=-w-a_{0}+\frac{1}{4\alpha^{2}}\left(\frac{w^{\prime}-a^{\prime}_{0}}{w-a_{0}}\right)^{2}$

and thus $\eta\in\bar{k}(y,y^{\prime})\setminus\bar{k}.$ Now we apply [7,
Lemma 2] for the Weierstrass equations $w^{\prime
2}=\alpha^{2}(4w^{3}-g_{2}w-g_{3})$ and $a^{\prime
2}_{0}=(-\alpha)^{2}(4a^{3}_{0}-g_{2}a_{0}-g_{3})$ and obtain that
$\eta^{\prime}=0.$ This contradicts the fact that field of constants of
$\bar{k}(y,y^{\prime})$ is the same as the field of constants of $\bar{k}.$ ∎

######  Remark 7.2.

If we have an irreducible affine curve of the form $Z-F_{2}-F_{3}-\cdots-
F_{n},$ where $F_{i}$ are forms of degree $i,$ then one can quickly find the
coefficients $\lambda_{2}$ and $\lambda_{3}$ of the $Y-$adic expansion of $Z$
as follows: Let $F_{2}=x_{20}Y^{2}+x_{11}ZY+x_{02}Z^{2}$ and
$F_{3}=x_{30}Y^{3}+R_{3}.$ Then, on the curve, the value of $Z$ equals

$x_{20}Y^{2}+x_{11}ZY+x_{02}Z^{2}+x_{30}Y^{3}+R_{3}+F_{4}+\cdots+F_{n}.\\\ $

In the above expression we shall substitute back for $Z$ and obtain

$x_{20}Y^{2}+x_{11}\left(x_{20}Y^{2}+x_{11}ZY+\cdots\right)Y+x_{02}\left(x_{20}Y^{2}+x_{11}ZY+\cdots\right)^{2}+x_{30}Y^{3}+\cdots.\\\
$

Continuing this process one actually obtains the $Y-$adic expansion of $Z;$

$Z=x_{20}Y^{2}+\left(x_{11}x_{20}+x_{30}\right)Y^{3}+\cdots.$

######  Example 7.3.

Let $\mathbb{C}(x)$ be the ordinary differential field of rational functions
with the derivation $x^{\prime}=1.$ Then differential equation

$y^{\prime}-\frac{1}{x}y^{2}-xyy^{\prime}-\frac{1}{x+1}y^{3}+y(y^{\prime})^{2}=0$

is of general type as $1/x$ and $x(1/x)+1/(x+1)=1+1/(x+1)$ have no
antiderivatives in $\mathbb{C}(x).$

######  Example 7.4.

The differential equation

(7.1.4) $y^{\prime}=a_{n}y^{n}+\cdots+a_{3}y^{3}+a_{2}y^{2},$

where both $a_{2}$ and $a_{3}$ having no antiderivatives in $k,$ is readily
seen to be of general type. Thus, the differential equation
$y^{\prime}=y^{3}-y^{2}$ is of general type over $\mathbb{C}.$ In fact, any
autonomous equation of the kind (7.1.4) with $a_{2}$ and $a_{3}$ nonzero is of
general type.

######  Remark 7.5.

Let $C$ be an algebraically closed field with the zero derivation. From
Theorem 1.1 and from [18, Proposition 3.1], we have the following equivalent
statements:

  1. (i)

An autonomous differential equation over $C$ of the form $y^{\prime}=f(y);\
f\neq 0$ is not of general type, that is, the equation has a nonalgebraic
solution in an iterated strongly normal extension of $C.$

  2. (ii)

The differential field $C(y)$ has an element $z$ such that either
$z^{\prime}=1$ or $z^{\prime}=cz$ for some nonzero constant $c.$

  3. (iii)

There is an element $z\in C(y)$ and a nonzero constant $c\in C$ such that

$\frac{1}{f(y)}=\frac{\partial z}{\partial
y}\quad\text{or}\quad\frac{1}{f(y)}=\frac{1}{cz}\frac{\partial z}{\partial
y}.$

Thus the autonomous equation $y^{\prime}=f(y)$ is not of general type if and
only if either $1/f(y)$ has no residues at any element of $C$, that is, the
partial fraction expansion of $1/f(y)$ is of the form

(7.1.5) $h(y)+\sum_{i=1}^{n}\sum_{j=2}^{n_{i}}\frac{d_{ij}}{(y-c_{i})^{j}},$

or the partial fraction expansion of $1/f(y)$ is of the form

(7.1.6) $c\sum_{i=1}^{n}\frac{m_{i}}{(y-c_{i})},$

where $m_{i}$ are nonzero integers and $c$ is a nonzero constant.

The equations $y^{\prime}=y^{3}-y^{2}$ and $y^{\prime}=\dfrac{y}{y+1}$ are now
easily seen to be of general type.

###  7.2. Differential equations of nongeneral type.

In this section we shall study the algebraic dependence of first order
differential equations of nongeneral type. Let $L$ be a differential field
extension of $k$ having $C$ as its field of constants. It is easily seen that
any two nonzero solutions in $L$ of a homogeneous first order linear
differential equation $y^{\prime}=by,$ where $0\neq b\in k,$ are $C-$linearly
dependent. Suppose that $L$ has nonalgebraic solutions of the nonhomogeneous
differential equation $y^{\prime}=by+c$ for $b,c\in k$ and $c\neq 0.$ Consider
a Picard-Vessiot extension $E$ of $L$ for
$M=k[\partial]/k[\partial]\mathscr{L},$ where
$\mathscr{L}=\partial^{2}-(b+(c^{\prime}/c))\partial+b(c^{\prime}/c)-b^{\prime}$
is obtained by homogenizing $y^{\prime}=by+c.$ Let $V\subset E$ be the set of
all solutions of $\mathscr{L}(y)=0.$ Then dim${}_{C}V=2.$ The differential
field $\mathcal{E}:=k\langle V\rangle,$ generated by elements of $V$ and all
its derivatives, is a Picard-Vessiot extension of $k$ (for $M$). For any $u\in
L$ such that $u^{\prime}=bu+c,$ we see that $\mathscr{L}(u)=0$ and therefore
$u\in V\subset\mathcal{E}.$ Moreover, for any automorphism
$\sigma\in\mathscr{G}(\mathcal{E}|k),$ we have
$\mathscr{L}(\sigma(u))=\sigma\left(\mathscr{L}(u)\right)=0$ and that
$(\sigma(u)-u)^{\prime}=b(\sigma(u)-u).$ Thus, if we choose an automorphism
$\sigma\in\mathscr{G}(\mathcal{E}|k)$ such that $\sigma(u)\neq u$ then since
$\mathscr{L}(\sigma(u)-u)=\mathscr{L}(\sigma(u))-\mathscr{L}(u)=0,$ we have
$\\{\sigma(u)-u,u\\}$ to be a $C-$basis of $V.$ We therefore have
$\mathcal{E}=k(\sigma(u),u),$ where the fields $k(\sigma(u)-u)$ and $k(u)$ are
differential fields. This implies $\mathrm{tr.deg}(\mathcal{E}|k)\leq 2$ and
thus any three solutions in $L$ of $y^{\prime}=by+c,$ where $b,c\in k$ and
$c\neq 0,$ must be $k-$algebraically dependent.

Suppose that

(7.2.1) $y^{\prime}=ay^{2}+by+c\ \text{for}\ a,b,c\in k\ \text{with}\ a\neq 0$

has a nonalgebraic solution in $L.$ Let

$\mathscr{L}=\partial^{2}-\left(\frac{a^{\prime}}{a}+b\right)\partial+ac$

and $E$ be a Picard-Vessiot extension of $L$ for
$M=k[\partial]/k[\partial]\mathscr{L}.$ Let $V\subset E$ be the set of all
solutions of $\mathscr{L}(y)=0.$ Then $v\in V\setminus\\{0\\}$ if and only if
$-v^{\prime}/av$ is a solution of the Riccati equation (7.2.1). Note that
$\mathcal{E}:=k\langle V\rangle$ is a Picard-Vessiot extension of $k$ for $M.$
We first claim that every solution in $L$ of (7.2.1) belongs to $\mathcal{E}.$

Let $u\in L$ be a solution of the Riccati equation (7.2.1). Let $L^{*}$ be a
Picard-Vessiot extension of $E$ for the differential equation
$y^{\prime}=-auy.$ Then $L^{*}=E(x)$ for some nonzero $x$ such that
$x^{\prime}=-aux.$ Note that $V\subset\mathcal{E}\subseteq L^{*}$ and that $V$
is a two dimensional vector space over $C$. Since $C$ is the field of
constants of $L^{*}$ and $\mathscr{L}(x)=0,$ we must have $x\in
V\subset\mathcal{E}.$ Consequently, $u=-x^{\prime}/(ax)\in\mathcal{E}.$

Let $\Omega\subset\mathcal{E}$ be set of all nonalgebraic solutions of
Equation (7.2.1). Now we shall show that any four distinct elements of
$\Omega$ must be $k-$algebraically dependent. That is,
$\mathrm{tr.deg}(k(\Omega)|k)\leq 3.$

The differential Galois group $\mathscr{G}(\mathcal{E}|k)$ is a closed
subgroup of the algebraic group $\mathrm{GL}(V).$ Since dim
$\mathrm{GL}(V)=4,$ if $\mathscr{G}(\mathcal{E}|k)$ is a proper closed
subgroup then dim $\mathscr{G}(\mathcal{E}|k)\leq 3$ and we obtain
$\mathrm{tr.deg}(k(\Omega)|k)\leq$ $\mathrm{tr.deg}(\mathcal{E}|k)=$ dim
$\mathscr{G}(\mathcal{E}|k)\leq 3.$ Thus we only need to deal with the case
when $\mathscr{G}(\mathcal{E}|k)=\mathrm{GL}(V).$ In this case, we first
choose a $C-$basis $\\{y_{1},y_{2}\\}$ of $V$ and identify
$\mathscr{G}(\mathcal{E}|k)$ with $\mathrm{GL}(2,C).$ Next, we consider the
differential field $K=\mathcal{E}^{\mathcal{Z}},$ where $\mathcal{Z}$ is the
center of $\mathrm{GL}(2,C).$ Then for any nonzero solution
$v=c_{1}y_{1}+c_{2}y_{2}\in V$ and any automorphism $\tau\in\mathcal{Z},$ we
have $\tau(v)=c_{1}c_{\tau}y_{1}+c_{2}c_{\tau}y_{2}.$ Thus $\tau(v)=c_{\tau}v$
and we obtain $v^{\prime}/v\in K.$ This implies $\Omega\subset K.$ Since
$\mathcal{Z}$ is normal, $K$ is a Picard-Vessiot extension of $k$ with Galois
group

$\mathscr{G}(K|k)\cong\mathscr{G}(\mathcal{E}|k)/\mathcal{Z}\cong\
\mathrm{PGL}(2,C).$

Now $\mathrm{tr.deg}(K|k)=$ dim $\mathrm{PGL}(2,C)=3$ implies any four
distinct elements in $\Omega$ must be algebraically dependent over $k$. We
would like to remark that the Riccati equation $y^{\prime}=-y^{2}+x$ over the
differential field $\mathbb{C}(x)$ with the derivation $d/dx$ has exactly
three $\mathbb{C}(x)-$algebraically independent solutions in any Picard-
Vessiot extension of $\mathbb{C}(x)$ (see [11, Example 4.29] or [13]).

Suppose that $L$ has a nonalgebraic solution $u$ satisfying a Weierstrass
differential equation; $u^{\prime 2}=\alpha^{2}(4u^{3}-g_{2}u-g_{3}).$ Then
since $L$ has $C$ as its field of constants, from [7, Lemma 2], any other
nonalgebraic solution of the Weierstrass equation must belong to the field
$k(u,u^{\prime}).$ In fact, the points $(1:z:z^{\prime}/\alpha),$
$(1:y:y^{\prime}/\alpha)$ of the elliptic curve (7.1.3) has the following
relation; $(1:z:z^{\prime}/\alpha)=(1:y:y^{\prime}/\alpha)(1:c_{1}:c_{2}),$
where $(1:c_{1}:c_{2})$ is a $C-$point of the curve. With all these facts, we
shall now move on to prove the following

######  Theorem 7.6.

Let $L$ be a differential field extension of $k$ having $C$ as its field of
constants.

  1. (i)

If an autonomous differential equation over $C$ is not of general type then
there is at most one $C-$algebraically independent solution of the equation in
$L.$

  2. (ii)

If a first order differential equation over $k$ is not of general type then
there are at most three $k-$algebraically independent solutions of the
equation in $L.$

###### Proof.

Let $f(y,y^{\prime})=0$ be a differential equation of nongeneral type. We only
need to consider the case when the differential equation has a nonalgebraic
solution $u$ in an iterated strongly normal extension of $k.$ Let
$u_{1},u_{2},u_{3},u_{4}$ be distinct nonalgebraic solutions from $L.$ Replace
$k,$ if necessary, by a finite algebraic extension of $k$ so that
$k(u,u^{\prime})$ contains a nonalgebraic solution of a Riccati or a
Weierstrass differential equation over $k.$ Now for $i=1,2,3,4,$ consider the
natural differential embeddings

$\psi_{i}:k(u,u^{\prime})\to L;\ \
\psi_{i}(u)=u_{i},\psi_{i}(u^{\prime})=u^{\prime}_{i}.$

If there is a nonalgebraic solution $t\in k(u,u^{\prime})\setminus k$ of a
Riccati differential equation over $k,$ say
$t^{\prime}=a_{2}t^{2}+a_{1}t+a_{0},$ then
$\psi_{i}(t)^{\prime}=a_{2}\psi_{i}(t)^{2}+a_{1}\psi_{i}(t)+a_{0}$ for each
$i=1,2,3,4.$ Then as noted earlier,
$\mathrm{tr.deg}\left(k(\psi_{1}(t),\dots,\psi_{4}(t))|k\right)\leq 3.$ Since
$\psi_{i}(t)\in k(u_{i}),$ each $u_{i}$ is algebraic over
$k(\psi_{i}(t))\subseteq k(\psi_{1}(t),\dots,\psi_{4}(t)).$ This implies
$u_{1},\dots,u_{4}$ are $k-$algebraically dependent.

Similarly, if $t\in k(u,u^{\prime})\setminus k$ is a nonalgebraic solution of
a Weierstrass differential equation then we have $t^{\prime
2}=\alpha^{2}(4t^{3}-g_{2}t^{2}-g_{3})$ and for $i=1,2,$ we have
$(\psi_{i}(t))^{\prime
2}=\alpha^{2}(4(\psi_{i}(t))^{3}-g_{2}\psi_{i}(t)-g_{3}).$ As noted earlier,
we then have

$(1:\psi_{1}(t):\psi_{1}(t)^{\prime}/\alpha)=(1:\psi_{2}(t):\psi_{2}(t)^{\prime}/\alpha)(1:c_{1}:c_{2}),\
\ \text{where}\ c_{1},c_{2}\in C.$

That is, $\psi_{1}(k(t,t^{\prime}))=\psi_{2}(k(t,t^{\prime}))$ and we obtain
that $u_{1}$ and $u_{2}$ are algebraically dependent over $k.$ Thus, in this
case, any two nonalgebraic solutions of $f(y,y^{\prime})=0$ are
$k-$algebraically dependent.

Now we shall consider the special case when $f(y,y^{\prime})=0$ is an
autonomous differential equation having a nonalgebraic solution $u$ in an
iterated strongly normal extension of $C.$ In this case, we know from
Corollary 6.4 that there is an element $t\in C(u,u^{\prime})$ such that
$t^{\prime}=1$ or $t^{\prime}=ct$ for some nonzero constant $c.$ Then,
$\psi_{i}(t)^{\prime}=1$ or $\psi_{i}(t)^{\prime}=c\psi_{i}(t).$ Thus
$\psi_{1}(t)=\psi_{2}(t)+e$ for some $e\in C$ or $\psi_{1}(t)=e\psi_{2}(t)$
for some nonzero $e\in C.$ This shows that $\psi_{1}(C(t))=\psi_{2}(C(t))$ and
we obtain that $u_{1}$ and $u_{2}$ are algebraically dependent over $C.$ ∎

###  7.3. Rational autonomous differential equations

We shall now verify Conjecture 1.2 for the class of rational autonomous
differential equations $y^{\prime}=h(y),$ where $h$ is a nonzero rational
function over $C.$ Let $f(y,y^{\prime})=y^{\prime}-h(y).$ We first make the
following observations.

  1. (a)

The differential field $C(f)=C(y)$ with $y^{\prime}=h(y),$ has $C$ as its
field of constants. This is easily seen by noting that if $u^{\prime}=0$ for
$u\in C(y)\setminus C$ then $y$ is algebraic over the differential field
$C(u),$ which contains only constants. An easy calculation using the monic
irreducible polynomial of $y$ over $C(u)$ shows that $y^{\prime}=0.$ This
implies $h=0$, a contradiction. Thus every rational autonomous differential
equation admits a nonalgebraic solution.

  2. (b)

If $h$ has no zero in $C$ then $1/h$ is a polynomial over $C$ and has the form
(7.1.5) described in Remark 7.5. Thus if $C(y)$ is a transcendental extension
of $C$ with $y^{\prime}=h(y)$ then $C(y)$ has a nonzero element $z$ such that
$z^{\prime}=1.$ This shows that rational autonomous equations;
$y^{\prime}=h(y)$ with $h$ having no zeros in $C$, are not of general type.

Suppose that $\alpha\in C$ is a zero of $h.$ Consider the rational function
$g\in C(y)$ defined by $h(y):=g(y-\alpha).$ Then, we have

$(y-\alpha)^{\prime}=y^{\prime}=h(y)=g(y-\alpha).$

Thus, given a nonzero rational autonomous differential equation
$y^{\prime}=h(y)$ with $h$ having a zero in $C,$ it does no harm to assume
that $h$ has a zero at $y=0.$ In view of Theorem 7.6 the conjecture is
verified once we prove the following: If an autonomous differential equation

$y^{\prime}=h(y),\quad h\neq 0\quad\text{and}\quad h(0)=0$

has at most one nonalgebraic solution in any given differential field
extension $L$ of $C$ having $C$ as its field of constants then the equation is
not of general type.

Consider the purely transcendental differential field extension $C(t,y)$ of
$C,$ where $t^{\prime}=h(t)$ and $y^{\prime}=h(y).$ From our hypothesis, there
is an element $u\in C(t,y)\setminus C(t)$ such that $u^{\prime}=0$. Then we
have following equations

$y^{\prime}=h(y)=\sum^{\infty}_{i=m}c_{i}y^{i},\qquad
u=\sum^{\infty}_{i=p}b_{i}y^{i},$

where $m\geq 1,$ $c_{i}\in C$ for all $i\geq m$ and $c_{m}\neq 0,$ $p$ is an
integer and $b_{i}\in\overline{C(t)}$ for all $i\geq p$ and $b_{p}\neq 0.$
Taking derivatives, we obtain

$0=u^{\prime}=\sum^{\infty}_{i=p}b^{\prime}_{i}y^{i}+\left(\sum^{\infty}_{i=m}c_{i}y^{i}\right)\left(\sum^{\infty}_{i=p}ib_{i}y^{i-1}\right).$

We observe that

  1. (i)

If $p=0$ and $m\geq 2$ then letting $l$ be the least positive integer such
that $b_{l}\neq 0,$ we get $b_{l}^{\prime}=0$ and that
$b^{\prime}_{l+m-1}=-lc_{m}b_{l}\in C\setminus\\{0\\}$.

  2. (ii)

If $p\neq 0$ and $m\geq 2$ then $b^{\prime}_{p}=0$ and
$b^{\prime}_{p+m-1}=-pb_{p}c_{m}\in C\setminus\\{0\\}.$

  3. (iii)

If $p=0$ and $m=1$ then letting $l$ be the least positive integer such that
$b_{l}\neq 0,$ we get $b^{\prime}_{l}/b_{l}=-lc_{1}\in
C\setminus\left\\{0\right\\}$.

  4. (iv)

If $p\neq 0$ and $m=1$ then $b^{\prime}_{p}/b_{p}=-pc_{1}\in
C\setminus\\{0\\}.$

Thus, in the event that (i) or (ii) holds, we obtain an element $z\in C(t)$
such that $z^{\prime}=1$ and in the event that (iii) or (iv) holds, we obtain
an element $z\in C(t)\setminus\\{0\\}$ such that $z^{\prime}=cz$ for some
$c\in C\setminus\\{0\\}.$ Now from Remark 7.5, the equation $y^{\prime}=h(y)$
is not of general type. Thus, we have verified the conjecture for the class of
rational autonomous differential equations.

## References

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# An explicit finite $B_{k}$-sequence

Igor S. Sergeev111e-mail<EMAIL_ADDRESS>
###### Abstract

For any $n$ and $k$, we provide an explicit (that is, computable in polynomial
time) example of integer $\mathcal{B}_{k}$-sequence of size $n$ consisting of
elements bounded by $n^{k+o(k)}$.

dedicated to the memory of Vladimir Evgen’evich Alekseev (1943–2020)

Introduction. Recall that a set $B$ in some commutative group is a
$\mathcal{B}_{k}$-sequence if all $k$-element sums in $B$ are different, that
is, the equality

$a_{1}+\ldots+a_{k}=b_{1}+\ldots+b_{k},\qquad a_{i},b_{j}\in B,$

holds iff the multisets of summands coincide:
$\\{a_{1},\ldots,a_{k}\\}=\\{b_{1},\ldots,b_{k}\\}$.

$\mathcal{B}_{2}$-sequences are also known as Sidon sequences. Very often, the
notion of Sidon sequence stands as a synonym for $\mathcal{B}_{k}$-sequence in
general.

Easy to check, if $\mathbb{Z}_{N}$ contains a size-$n$
$\mathcal{B}_{k}$-sequence, then $N\geq\binom{n+k-1}{k}$. We want to consider
only satisfactorily dense size-$n$ $\mathcal{B}_{k}$-sequences, say, for
$N=(n+k)^{O(k)}$, avoiding trivial examples like $\\{k,k^{2},\ldots k^{n}\\}$
with exponentially large elements. Also, we interest in explicit
constructions, that is, those that can be computed in polynomial time with
respect to the binary size222That is, the length of the binary code
representing the elements of the set..

History. The most famous explicit examples of the optimal density integer
Sidon sequences are: a size-$(q+1)$ set in $\mathbb{Z}_{q^{2}+q+1}$ due to J.
Singer [10], a size-$q$ set in $\mathbb{Z}_{q^{2}-1}$ due to R. C. Bose [3],
and a size-$(p-1)$ set in $\mathbb{Z}_{p^{2}-p}$ due to V. E. Alekseev [2].
Here $p$ and $q$ stay for any prime number and prime power, respectively. The
latter set is attributed to I. Ruzsa [8] almost everywhere.

The classical example of a nearly dense-optimal $\mathcal{B}_{k}$-sequence was
proposed by Bose and S. Chowla in [4]. Let us recall this construction that
generalizes [3]. Let $GF(q)=\\{\alpha_{1},\ldots,\alpha_{q}\\}$, and $x$ be a
primitive element in $GF(q^{k})$. It can be easily verified that the set

$D[q,k]=\\{d_{i}\mid x^{d_{i}}=x+\alpha_{i},\;1\leq d_{i}<q^{k}\\}$

is a size-$q$ $\mathcal{B}_{k}$-sequence in $\mathbb{Z}_{q^{k}-1}$.

There are known also a number of similar constructions including another
$\mathcal{B}_{k}$-sequence from [4] generalizing [10]. H. Derksen [5] proposed
even more general constructions considering quotient polynomial rings
$GF(q)[x]/(P(x))$ instead of pure fields in the examples from [4]. C. A. Gómez
Ruiz and C. A. Trujillo Solarte [6] extended an example [2] to
$\mathcal{B}_{k}$-sequences in $\mathbb{Z}_{p^{k}-p}$.

Discussion. All these examples of $\mathcal{B}_{k}$-sequences may be
considered explicit only for constant or extremely slowly growing $k$’s with
respect to $n$, since they imply computation of discrete logarithms in groups
of generally non-smooth order. Indeed, probabilistic or greedy constructions
that we haven’t mentioned are even less explicit. It looks like we lack easily
computable and dense enough examples of $\mathcal{B}_{k}$-sequences that could
be useful in some specific situations, e.g. for proving explicit lower bounds
in computational complexity [9]. Thus, we intend to close this gap.

We follow the general idea of previous constructions: computing an additive
numeric $\mathcal{B}_{k}$-sequence as an image of some simple multiplicative
$\mathcal{B}_{k}$-sequence from an appropriate group. All we need to make
computations easy is to choose a basic multiplicative group of smooth order.
Note that in doing this, we will partially sacrifice the density.

Construction. Further, $p_{1},p_{2},\ldots$ denote odd prime numbers written
in growing order. Let $r=1+\lceil k\log p_{n}\rceil$. The set of odd numbers-
residues from 1 to $2^{r}-1$ constitutes the multiplicative group
$\mathbb{Z}^{*}_{2^{r}}$ of the ring $\mathbb{Z}_{2^{r}}$. For $r\geq 3$, this
group is a direct product of cyclic groups of orders 2 and $2^{r-2}$, namely,
$\mathbb{Z}^{*}_{2^{r}}\cong\langle-1\rangle_{2}\langle 5\rangle_{2^{r-2}}$
with $-1$ and $5$ being generating elements. Therefore, any odd number $x$ has
a unique representation $x\equiv(-1)^{j}\cdot 5^{h}\;(\bmod\;2^{r})$, where
$0\leq j\leq 1$ and $0\leq h<2^{r-2}$. For details, see e.g. [11].

Consider the number set

$H[n,k]=\\{h_{i}\mid p_{i}\equiv\pm 5^{h_{i}}\;(\bmod\;2^{r}),\,0\leq
h_{i}<2^{r-2},\,i=1,\ldots,n\\}.$

Let us check that the given set is a $\mathcal{B}_{k}$-sequence in
$\mathbb{Z}_{2^{r-2}}$. By the choice of $r$, for different tuples of indices
$1\leq i_{1}\leq\ldots\leq i_{k}\leq n$, all numbers $\pm
p_{i_{1}}\cdot\ldots\cdot p_{i_{k}}$ are different and do not exceed
$2^{r-1}-1$ by absolute value. Hence, all residues
$5^{h_{i_{1}}+\,\ldots+\,h_{i_{k}}\;}(\bmod\;2^{r})$ are different, and all
sums $h_{i_{1}}+\,\ldots+\,h_{i_{k}\;}(\bmod\;2^{r-2})$ are different as well.

The set $H[n,k]$ is not as dense as $D[q,k]$ or similar constructions. Still,
its density is satisfactorily in asymptotic sense:
$2^{r-2}<p_{n}^{k}<(2n\log(n+2))^{k}$ due to the well-known facts about
distribution of prime numbers, see e.g. [7].

We are left to confirm explicitness: that the set $H[n,k]$ requires
${(n+k)^{O(1)}}$ time to be constructed. First, we need to obtain the list of
prime numbers. Second, we have to compute discrete logarithms333Here, we don’t
resort to the commonly used notation ${\rm ind}_{g}\,x$. $\log_{5}(\pm p_{i})$
in $\mathbb{Z}_{2^{r}}$. For the first part, we may use Eratosthenes sieve or
any other known algorithm running in time $n^{O(1)}$. Discrete logarithm in
the cyclic group of order $2^{r-2}$ may be computed trivially by $O(r^{2})$
elementary arithmetic operations mostly consisting of squarings. Indeed, we
may determine binary digits of the number
$a=[a_{r-3},\,\ldots,a_{0}]_{2}=\log_{5}x\;(\bmod\;2^{r})$ sequentially as

$a_{0}=\log_{5^{2^{r-3}}}x^{2^{r-3}},\quad
a_{1}=\log_{5^{2^{r-3}}}(5^{-a_{0}}x)^{2^{r-4}},\quad\ldots,\\\
a_{r-3}=\log_{5^{2^{r-3}}}\big{(}5^{-2^{r-4}a_{r-4}-\ldots-2a_{1}-a_{0}}x\big{)}.$

Inner logarithms are performed in an order-2 subgroup with generating element
$5^{2^{r-3}}\equiv 2^{r-1}+1\;(\bmod\;2^{r})$ simply by comparing with 1 and
$2^{r-1}+1$. If both comparisons fail, then $x\notin\langle
5\rangle_{2^{r-2}}$.

Notes. In the above example, we intentionally used as smooth order for the
basic multiplicative group as possible. Instead, we can work in any ring
$\mathbb{Z}_{p^{r}}$ with an odd prime $p$. The multiplicative group
$\mathbb{Z}^{*}_{p^{r}}$ has order $(p-1)p^{r-1}$ and it is cyclic. The case
$p=3$ is especially attractive, since there we have 2 as a generating element
for the multiplicative group. With more care, we can consider residue rings of
some other smooth orders.

The choice of prime numbers for a “factor base” is also changeable. Say, we
can relax the condition of being prime to the condition of being pairwise
prime. Though, this relaxation alone doesn’t allow to substantially increase
the density of the set.

Essentially, the present text in an excerpt from [9].

## References

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  * [4] Bose R. C., Chowla S. Theorems in the additive theory of numbers. Commentarii Mathematici Helvetici. 1962. 37, 141-147.
  * [5] Derksen H. Error-correcting codes and $B_{h}$-sequences. IEEE Trans. Inform. Theory. 2004. 50(3), 476–485.
  * [6] Gómez Ruiz C. A., Trujillo Solarte C. A. Una nueva construcción de conjuntos $B_{h}$ modulares. Matemáticas: Enseñanza Universitaria. 2011. XIX(1), 53–62.
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  * [9] Sergeev I. S. On the additive complexity of some number sequences. Submitted to J. Applied and Industrial Math.
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