Proof Since the system is almost internal to the constants, there are generic solutions $(x_1,y_1),\dots , (x_m,y_m)$ over K of the system such that for any generic solution $(x,y)$ of the system over $M:=K(x_1,y_1,\dots , x_m,y_m),$ there are constants $c_1,\dots ,c_l$ such that x and y are algebraic over $M(c_1,\dots , c_l).$ We now show that $M(x,y)$ contains two M-algebraically independent constants. Let $c_1,\dots , c_r$ be an M-transcendence basis of constants for $M(c_1,\dots , c_l).$ Since $\mathrm {tr.deg}(M(c_1,\dots , c_l,x,y)|M)=r,$ we must have $\mathrm {tr.deg}(M(c_1,\dots , c_l,x,y)| M(x,y))=r-2.$ Therefore, there is a nonzero polynomial $P\in M(x,y)[X_1,\dots ,X_r]$ such that $P(c_1,\dots , c_r)=0.$ Since $c_1,\dots ,c_r$ are constants, it follows that such a polynomial P must belong to $C_{M(x,y)}[X_1,\dots ,X_r]$ [Reference Kolchin6, Section 14, Theorem 2]. As $\mathrm {tr.deg}(M(c_1,\dots ,c_l)|M)=r,$ it also follows that one of the coefficients of P must be a constant nonalgebraic over $M.$ If we call this constant $u_1$ and replace M by $M(u_1),$ then a similar argument shows the existence of another constant $u_2\in M(x,y),$ which is nonalgebraic over $M(u_1).$ Observe that either $u_1$ or $u_2$ must belong to $M(x,y)\setminus M(x).$

Let N be a differential field intermediate to $M(x,y)$ and M. Since $\mathrm {tr. deg}(N|M)=1,$ there is a nonzero polynomial $P\in C_N[X_1,X_2]$ such that $P(u_1,u_2)=0.$ Again, by the same argument as above, there is a constant $v\in N$ which is not algebraic over $M.$

Proof Let $x'=f(x)$ be internal to the constants. Then, there are generic solutions $x_1,\dots ,x_m$ over K such that for any generic solution x over $M:=K(x_1,\dots ,x_m),$ there are constants $c_1,\dots , c_l$ such that $x\in M(c_1,\dots , c_l)$ . From [Reference Kolchin7, Chapter 2, Corollary 2], we know that $M(x)$ is generated over M as a field by a set of constants. Therefore, we may assume that $M(x)=M(c_1,\dots ,c_l).$ We first claim that there exists an algebraic extension $\tilde {M}$ of M such that $\tilde {M}(x)=\tilde {M}(c)$ for some constant c.

Note that $C_{M(x)}$ and M are linearly disjoint over $C_M$ [Reference Kolchin7, Chapter 2, Corollary 1]. Therefore, $C_M(c_1,\dots , c_l)$ and M are linearly disjoint over $C_M.$ Since $M(c_1,\dots ,c_l)=M(x),$ we have $\mathrm {tr.deg}(M(c_1,\dots ,c_l)|M)=1$ and from linear disjointness, we obtain $\mathrm {tr.deg}(C_M(c_1,\dots , c_l)|C_M)=1.$ Now since $M(c_1,\dots ,c_l)=M(x)$ is a rational function field, there is a finite algebraic extension $\tilde {C}_M$ of $C_M$ such that $\tilde {C}_M(c_1,\dots , c_l)=\tilde {C}_M(c)$ [Reference Chevalley1, Theorem 5]. Any extension of a zero derivation on a field to its algebraic closure is again a zero derivation. Therefore, c must also be a constant. Let $\tilde {M}:=M\tilde {C}_M$ and observe that

$$ \begin{align*}\tilde{M}(x)=\tilde{M}(c_1,\dots,c_l)=\tilde{M}(c).\end{align*} $$

This proves the claim.

Since $\tilde {M}(x)=\tilde {M}(c),$ there exists $\alpha ,\beta ,\gamma , \delta \in \tilde {M}$ with $\alpha \delta -\beta \gamma \neq 0$ such that

$$ \begin{align*}c=\frac{\alpha x+\beta}{\gamma x+\delta}.\end{align*} $$

A simple calculation shows that $f(x)=x'=a_2x^2+a_1x+a_0$ for some $a_0,a_1,a_2\in \tilde {M}.$ Now since $f\in K(x),$ we obtain that $a_0, a_1, a_2\in K.$

To prove the converse, let $x, x_1,x_2,x_3$ be distinct generic solutions of $x'=a_2x^2+a_1x+a_0$ over $K.$ It is observed in [Reference Nagloo11, p. 102] that

$$ \begin{align*}c:=\frac{(x-x_2)(x_3-x_1)}{(x-x_1)(x_3-x_2)}\end{align*} $$

is a constant.Footnote ³ Since

$$ \begin{align*}x=\frac{x_2(x_3 - x_1) + c x_1(x_2 - x_3)}{(x_3 - x_1) + c (x_2 - x_3)},\end{align*} $$

we now obtain $x\in K(x_1, x_2, x_3,c).$ This proves that the equation $x'=a_2x^2+a_1x+a_0$ is internal to the constants.

Proof We know from Theorem 2 that $f(x)=a_2x^2+a_1x+a_0,$ where $a_0, a_1, a_2\in K.$ Suppose that the system (L) is almost internal to the constants. By Proposition 1, there exists a differential field extension M of K such that for any generic solution $(x,y)$ over $M,$ the differential field $M(x,y)$ contains two M-algebraically independent constants. Then, $M(x,y)$ contains a constant which is not in $M(x)$ and therefore the latter must contain an element z such that $z'=nxz$ for some positive integer n [Reference Magid10, Example 1.11]. Let $z=P/Q,$ where $P, Q\in M[x].$ We can assume that P and Q are relatively prime and Q is monic. Write $P=\alpha _px^p+\alpha _{p-1}x^{p-1}+\cdots , Q=x^q+b_{q-1}x^{q-1}+\cdots \in M[x]$ with $\alpha _p\neq 0.$ Then,

$$ \begin{align*}P'Q-Q'P=nxPQ\end{align*} $$

and we obtain $(p-q)a_2=n.$ Since $n\neq 0, a_2=n/(p-q)$ is a nonzero rational number.

For the converse, suppose that $f(x)=a_2x^2+a_1x+a_0,$ where $a_0, a_1\in K$ and $a_2=m_1/m_2, (m_1, m_2)\in \mathbb Z^2$ a nonzero rational number. Let $(x,y)$ be a generic solution of the system (L). We shall now find generic solutions $(x_1,y_1)$ and $(x_2, y_2)$ of the system and constants $c_1,c_2$ so that $x,y$ are algebraic over the field $K(y_1,x_1,y_2,x_2,c_1,c_2).$ This will then complete the proof of the theorem.

Let v be an element algebraic over the field $K(x,y)$ satisfying $v^{m_2}=y^{-m_1}.$ Then, $v'=-a_2x v, (a_2x)'=a_2x'=(a_2x)^2+a_1(a_2x)+a_2a_0$ and $v"=a_1v'-a_2a_0v.$ Consider the rational function field $K(v,v')(v_1,v^{\prime }_1,v_2,v^{\prime }_2)$ and extend the derivation on $K(v,v')$ to $K(v,v')(v_1,v_2,v^{\prime }_1,v^{\prime }_2)$ by declaring $v^{\prime \prime }_1=a_1v^{\prime }_1-a_2a_0v_1$ and $v^{\prime \prime }_2=a_1v^{\prime }_2-a_2a_0v_2$ . Let $N=K(v_1,v^{\prime }_1,v_2,v^{\prime }_2).$ Note that $v_1,v_2$ are $C_U$ -linearly independent and that the Wronskian of $v, v_1, v_2$ is zero. Therefore, $v=c_1v_1+c_2v_2,$ for some constants $c_1,c_2.$ This implies that $v\in N(c_1,c_2)$ and since $a_2x=-v'/v\in N(c_1,c_2),$ we obtain that $x\in N(c_1,c_2).$

Let $y_1$ and $y_2$ be elements algebraic over the field N such that $v^{m_2}_1=y^{-m_1}_1$ and $v^{m_2}_2=y^{-m_1}_2.$ Let $x_1:=-v^{\prime }_1/a_2v_1$ and $x_2:=-v^{\prime }_2/a_2v_2.$ Then, it is easy to see that $(x_1,y_1)$ and $(x_2,y_2)$ are generic solutions of the system (L). Now we consider the field $K(y_1,x_1,y_2,x_2,c_1,c_2).$ Clearly, $v_1$ and $v_2$ are algebraic over $K(y_1,x_1,y_2,x_2,c_1,c_2)$ and since $v=c_1v_1+c_2v_2,$ we obtain that v is also algebraic over $K(y_1,x_1,y_2,x_2,c_1,c_2).$ Since $x=-v'/a_2v,$ we obtain that x is algebraic over $K(y_1,x_1,y_2,x_2,c_1,c_2).$ Finally, $v^{m_2}=y^{-m_1}$ implies that y is algebraic over $K(y_1,x_1,y_2,x_2,c_1,c_2).$

Proof Let $(x,y)$ be a generic solution of the system (A) and $F=C(x,y).$ We first make the following observations. Let $z\in F\setminus C$ such that $z'=0.$ Then, $z\notin C(x);$ otherwise, $f(x)\frac {\partial z}{\partial x}=0,$ which is a contradiction. Therefore, $z\in F\setminus C(x).$ As noted in the proof of Theorem 4, there exists a positive integer m and an element $v\in C(x)$ such that $v'=mg(x)v.$ Therefore, $f(x)\frac {\partial v}{\partial x}=mg(x)v$ and we obtain

$$ \begin{align*}\frac{mg(x)}{f(x)}=\frac{1}{v}\frac{\partial v}{\partial x}.\end{align*} $$

Thus, if $z\in F\setminus C$ such that $z'=0$ then $z\in F\setminus C(x)$ , and in that case, the condition (2) holds with $b=0$ .

Along with these observations, we suppose that the system (A) is almost internal to the constants. Then, by Proposition 1, there exists a differential field extension M of C such that $M(x,y)$ contains two M-algebraically independent constants. Furthermore, since $M(x)$ is a differential subfield of $M(x,y)$ with $\mathrm {tr. deg}(M(x)|M)=1,$ it follows that $\mathrm {tr. deg}(C_{M(x)}|C_M)=1.$ By [Reference Rosenlicht13, Proposition], $C(x)$ contains an element u such that either $u'=1$ or $u'=cu$ for some nonzero $c\in C.$ Then, $\frac {1}{f(x)}$ is either $\frac {\partial u}{\partial x}$ or $\frac {1}{f(x)}=\frac {c}{u}\frac {\partial u}{\partial x}.$ This implies that condition (1) holds.

To prove that condition (2) also holds, we consider the following two cases. Suppose that $C_F\neq C$ . Then, by the observation made in the first paragraph, the condition (2) holds. Now suppose that $C_F=C.$ Since there exists a differential field M such that M and F are free over C and $MF$ contains two M-algebraically independent constants, by [Reference Kumbhakar and Srinivasan9, Proposition 4.2], there is a differential field L intermediate to C and F such that $\mathrm {tr. deg}(L|C)=2$ and L can be embedded in a strongly normal extension of $C.$ Since L is contained in the purely transcendental extension $C(x,y),$ by [Reference Kumbhakar and Srinivasan9, Theorem 1.2], L can be embedded in a Picard–Vessiot extension of $C.$ Footnote ⁵ The differential Galois group of any Picard–Vessiot extension of C is a connected commutative linear algebraic group over C. Therefore, by the fundamental theorem of differential Galois theory, L must also be a Picard–Vessiot extension of C with a differential Galois group isomorphic to $\mathbb {G}_a\times \mathbb {G}_m$ or $\mathbb {G}_m\times \mathbb {G}_m.$ This implies that $L=C(\alpha , \beta )$ where $\alpha , \beta $ are C-algebraically independent, $\beta '=c_2\beta $ for some nonzero $c_2\in C$ and either $\alpha '=1$ or $\alpha '=c_1\alpha $ for some nonzero $c_1\in C.$ Consequently, there exists an element $\zeta \in C(x,y)$ such that $\zeta '= a\zeta $ for some nonzero $a\in C.$ By the Kolchin–Ostrowski theorem, there exist integers $m,n$ such that $h:=y^m\zeta ^n\in C(x).$ Then,

$$ \begin{align*}h'=f(x)\frac{\partial h}{\partial x}=my^m\zeta^ng(x)+nay^m\zeta^n.\end{align*} $$

From the above equation, we obtain

$$ \begin{align*}\frac{mg(x)+b}{f(x)}=\frac{1}{h(x)}\frac{\partial h}{\partial x}, \quad na=:b\in C.\end{align*} $$

Thus, we have shown that the condition (2) holds in both cases.

To prove the converse, let $(x,y)$ be a generic solution of the system (A). From conditions (1) and (2), we obtain that $C(x)$ contains an element s such that $s'=1$ or $s'=cs$ for some nonzero $c\in C$ and that there is an element $t\in C(x,y)\setminus C(x)$ such that $t'=bt$ for some $d\in C.$ Note that $s=u$ and $t=y^m/v.$ Now let $(x_1,y_1)$ be another generic solution of the system (A). Consider the differential C-isomorphism $\phi :C(x,y)\rightarrow C(x_1,y_1),$ where $\phi (x)=x_1$ and $\phi (y)=y_1.$ Let $c_2:=t/\phi (t).$ If $s'=1,$ then let $c_1:=s-\phi (s)$ and if $s'=cs,$ then let $c_1:=s/\phi (s).$ In any event, note that $c_1, c_2$ are constants and that $s,t\in C(x_1,y_1,c_1,c_2).$ Since $x,y$ are algebraic over $C(s,t),$ they are also algebraic over $C(x_1,y_1,c_1,c_2)$ . This shows that the system (A) is almost internal to the constants.