Fact 2.2 [Reference JacobsonJac64, Theorem 18, Section IV.7]

Suppose $L/K$ is a separable field extension. If $\delta :K\to L$ is a derivation, then $\delta $ can be extended to a derivation $L\to L$ .

Proof Since $\delta (K(E))\subseteq K(E\cup \delta (E))$ we may apply Fact 2.2 to the derivation $\delta |_{K(E)}:K(E)\longrightarrow K(E\cup \delta (E))$ and get a derivation $\partial :K(E\cup \delta (E))\longrightarrow K(E\cup \delta (E))$ that restricts to $\delta $ on $K(E)$ . Assume then that E is finite. There is some nonzero $f\in K[E\cup \delta (E)]$ such that $f{\cdot } \partial (\delta (a))\in K[E\cup \delta (E)]$ for each $a\in E$ . Obviously f has the required property.

Proof Let $b\in S^n$ be such that S is the differential K-algebra generated by b and $A\subseteq K[b]$ . Let ${\mathfrak {p}}=\{f\in K\{x\}{\ \vert \ } f(b)=0\}$ be the differential vanishing ideal of b over K. Then ${\mathfrak {p}}$ is a separable prime differential ideal; separability is due to fact that $K\{x\}/{\mathfrak {p}}$ is K-isomorphic to S. By the differential basis theorem of Kolchin [Reference KolchinKol73, Corollary 4, Section III.5], there is a finite set $\Sigma \subseteq {\mathfrak {p}}$ that generates ${\mathfrak {p}}$ as a radical differential ideal. Take $d\geq 1$ such that each derivative of any $x_1,\ldots ,x_n$ occurring in some polynomial from $\Sigma $ has order $\leq d$ . Finally take

$$ \begin{align*}E=\{\delta^kb_i{\ \vert \ } i\in \{1,\ldots,n\},\ k\leq d\}\subseteq S.\end{align*} $$

By possibly taking a larger d, a result of Kolchin appearing in [Reference KolchinKol73, Lemma 1, Section III.2] tells us that $S/K(E)$ is separable. By Corollary 2.3, there are $f\in K[E\cup \delta (E)]$ and a derivation $\partial $ of $B:=K[E\cup \delta (E)]_f$ that restricts to $\delta $ on $K[E]$ . Then $\partial ^kb_i=\delta ^kb_i$ for all $i\in \{1,\ldots ,n\},\ k\leq d$ and therefore b is a solution to $\Sigma =0$ in $(B,\partial )$ . Consequently, the identity map of $K\cup \{b_1,\ldots ,b_n\}$ extends to a differential K-algebra homomorphism $\phi :S\longrightarrow (B,\partial )$ . By choice of b, the map $\phi $ restricts to the identity map of A.

Proof By assumption, there is a solution of $\Sigma =0$ in a differentially finitely generated K-subalgebra S of L. Now apply Proposition 2.4 to S and $A=K$ .

Observation 2.7 Let K be a differential field and let $f\in K\{x\}$ for x a single differential variable. Let $n=\mathop {\operatorname {ord}}\nolimits (f)\geq 0$ and let $a\in K^{n+1}$ with $f(a)=0$ and $s_f(a)\neq 0$ . Then there is an irreducible factor h of f with $\mathop {\operatorname {ord}}\nolimits (h)=n$ , $h(a)=0$ and $s_h(a)\neq 0$ .

Proof Let $f_0,f_1\in K\{x\}$ , with $f_0$ irreducible, $f=f_0{\cdot } f_1$ and $\mathop {\operatorname {ord}}\nolimits (f_0)=n$ . Then

$$\begin{align*}(*)\qquad\qquad s_f=\frac{\partial f}{\partial x_n}=\frac{\partial f_0}{\partial x_n}\,{\cdot}\, f_1+f_0\,{\cdot}\, \frac{\partial f_1}{\partial x_n}.\qquad\qquad \end{align*}$$

If $f_0(a)=0$ , then $(*)$ implies $s_{f_0}(a)=\frac {\partial f_0}{\partial x_n}(a)\neq 0$ . If $f_0(a)\neq 0$ , then $f_1(a)=0$ and $(*)$ shows $s_{f_1}(a)=\frac {\partial f_1}{\partial x_n}(a)\neq 0$ ; hence also $\mathop {\operatorname {ord}}\nolimits (f_1)=n$ and in this case we may replace f by $f_1$ and proceed by induction.

Proof (i) $\Rightarrow $ (iii). Let $f,g\in K\{x\}$ with $\mathop {\operatorname {ord}}\nolimits (f)\geq 1$ and $\mathop {\operatorname {ord}}\nolimits (f)\geq \mathop {\operatorname {ord}}\nolimits (g)$ and assume

$$\begin{align*}(\dagger)\qquad\qquad f(x)=0 \ \&\ g(x)\cdot s_f(x)\neq 0\qquad\qquad \end{align*}$$

has an algebraic solution in K. Let $n=\mathop {\operatorname {ord}}\nolimits (f)$ . By Observation 2.7, we may assume that f is irreducible. Let ${\mathfrak {p}}=[f]:s_f^\infty $ . Since $s_f\neq 0$ , Theorem 3.1(2) of [Reference Ino and SánchezIL23] says that ${\mathfrak {p}}$ is a separable prime differential ideal of $K\{x\}$ . We write $a=x{\mathop {\operatorname {\,mod\,}}\nolimits }\, {\mathfrak {p}}$ . Now, an algebraic solution of $f(x)=0 \ \&\ s_f(x)\neq 0$ in K is a smooth K-rational point of

$$\begin{align*}K[x_0,\ldots,x_n]/(f)\cong_KK[a,\ldots,a^{(n)}]. \end{align*}$$

The largeness of K yields that K is e.c. in $K(a,\ldots ,a^{(n)})$ . Since the latter is equal to the differential field $K\langle a\rangle $ generated by a over K, differential largeness implies that $(K,\delta )$ is e.c. in $(K\langle a\rangle , \delta )$ .

Since $\mathop {\operatorname {ord}}\nolimits (f)\geq \mathop {\operatorname {ord}}\nolimits (g)$ and $(\dagger )$ has an algebraic solution in K, Lemma 3.6(1) of [Reference Ino and SánchezIL23] implies that $g\cdot s_f\notin {\mathfrak {p}}$ . Hence a is a differential solution of $(\dagger )$ in $K\langle a\rangle $ . As $(K,\delta )$ is e.c. in $(K\langle a\rangle , \delta )$ also K has a differential solution $\alpha $ of $(\dagger )$ . To argue that there are infinitely many solutions, note that $g\cdot (x-\alpha )$ has again order at most $\mathop {\operatorname {ord}}\nolimits (f)$ . By largeness of K and the assumption $\mathop {\operatorname {ord}}\nolimits (f)\geq 1$ , there is an algebraic solution of the new system where we replace g with $g\cdot (x-\alpha )$ . It follows, by repeating the above argument, that there are infinitely many differential solutions of $(\dagger )$ in K.

(iii) $\Rightarrow $ (ii) It suffices to show that K is large as a field. By [Reference JardenJar11, Lemma 5.3.1, p. 67], a field K is large if and only if for every absolutely irreducible polynomial $F(X,Y)\in K[X,Y]$ , if there is a point $(a,b)\in K^2$ with $F(a,b) = 0$ and $\frac {\partial F}{\partial Y}(a,b)\neq 0$ , then there are infinitely many such points.

So take an absolutely irreducible polynomial $F(X,Y)\in K[X,Y]$ and some $(a,b)\in K^2$ with $F(a,b) = 0$ and $\frac {\partial F}{\partial Y}(a,b)\neq 0$ . Consider the differential polynomial $f(x)=F(x,x')$ . Then $f(x)=0\ \&\ s_f(x)\neq 0$ has an algebraic solution in K, namely $(a,b)$ . By (iii) there are infinitely many differential solutions in K. But then there are infinitely many solutions to $F(X,Y) = 0$ and $\frac {\partial F}{\partial Y}(X,Y)\neq 0$ in K as well.

(ii) $\Rightarrow $ (i). To prove differential largeness, let F be a differential field extension of K such that K is e.c. in F as a field. Note that then $F/K$ is separable. We need to show that K is e.c. in F as a differential field. Let $\Sigma $ be a system of differential polynomials in n differential variables over K and assume that $\Sigma =0$ has a solution $a\in F^n$ . We may assume that $F=K\langle a\rangle $ . By 2.5 applied to F, we may assume that F is differentially algebraic over K (and $F/K$ remains separable).

Condition (ii) guarantees that $[K:C_K]$ is infinite; hence, by the differential primitive element theorem [Reference KolchinKol73, Proposition 9, Section II.8, p. 103], the differential field F is differentially generated over K by a single element $b\in F$ . Let ${\mathfrak {p}}$ be the prime differential ideal of $K\{x\}$ associated with b. Note that ${\mathfrak {p}}$ is separable (over K).

Then, by Theorem 3.1(1) of [Reference Ino and SánchezIL23], ${\mathfrak {p}}=[f]:s_f^\infty $ for $f\in {\mathfrak {p}}$ irreducible of minimal rank. Write $a=(a_1,\dots ,a_n)$ and let $f_i,g\in K\{x\}$ with $a_i=\frac {f_i(b)}{g(b)}$ . By the differential division algorithm [Reference KolchinKol73, Section I.9] there are $h\in K\{x\}$ reduced with respect to f and some $r\geq 0$ with

$$\begin{align*}(i_fs_f)^rg\equiv h\quad {\mathop{\operatorname{\,mod\,}}\nolimits}[f]. \end{align*}$$

Since $f(b)=0$ and $i_f(b){\cdot } s_f(b)\neq 0$ we get $i_f^r(b)s_f^r(b)g(b)=h(b)\neq 0$ . Hence, we may replace g by h and $f_i$ by $(i_fs_f)^r{\cdot } f_i$ if necessary and assume that g is reduced with respect to f. Notice that $a_i\in K\{b\}_{g(b)}$ .

Now, since K e.c. in F as a field, the system $f(x)=0 \ \&\ s_f(x)\neq 0$ has an algebraic solutions in K. By condition (ii), the set

$$ \begin{align*}\{f=0\}\cup \{q\neq 0{\ \vert \ } q\in K\{x\} \text{ is nonzero and } \mathop{\operatorname{ord}}\nolimits (q)<\mathop{\operatorname{ord}}\nolimits (f)\}\end{align*} $$

is finitely satisfiable in the differential field K. Hence there is an elementary extension L of the differential field K having a differential solution c to $f(x)=0$ such that $q(c)\neq 0$ for all $q\in K\{x\}$ with $\mathop {\operatorname {ord}}\nolimits (q)<\mathop {\operatorname {ord}}\nolimits (f)$ . Since f is irreducible, it follows that $q(c)\neq 0$ for all $q\in K\{x\}$ that are reduced with respect to f.

In particular $f(c)=0\ \&\ g(c)\neq 0$ . Since $K\prec L$ there is some $d\in K$ with $f(d)=0\ \&\ g(d)\neq 0$ . This means there is a differential K-homomorphism $(K\{x\}/{\mathfrak {p}})_{g{\mathop {\operatorname {\,mod\,}}\nolimits } {\mathfrak {p}}}\longrightarrow K$ . By choice of ${\mathfrak {p}}$ we have $(K\{x\}/{\mathfrak {p}})_{g{\mathop {\operatorname {\,mod\,}}\nolimits } {\mathfrak {p}}}\cong K\{b\}_{g(b)}$ as differential K-algebras. Since $K\{a_1,\ldots ,a_n\}\subseteq K\{b\}_{g(b)}$ , we obtain a differential K-algebra homomorphism $K\{a_1,\ldots ,a_n\}\longrightarrow K$ and this corresponds to a differential solution of $\Sigma =0$ in $K^n$ .

Proof We verify 2.8(iii). Take $f,g\in K\{x\}$ , x a single differential variable, with $\mathop {\operatorname {ord}}\nolimits (f)\geq 1$ and $\mathop {\operatorname {ord}}\nolimits (f)\geq \mathop {\operatorname {ord}}\nolimits (g)$ , and assume that $f(x)=0 \ \&\ g(x)\cdot s_f(x)\neq 0$ has an algebraic solution in K. Since L is differentially large, it has infinitely many differential solutions to $f(x)=0 \ \&\ g(x)\cdot s_f(x)\neq 0$ . But then each of these solutions is differentially algebraic over K. Hence all these solutions are in K.

Proof (i) $\Rightarrow $ (ii) Let $(a,b)$ be a K-generic point of W. Since $\pi _W:W\to V$ is a separable morphism, we obtain that a is K-generic in V and $K(a,b)/K(a)$ is a separable extension. Since $W\subseteq \tau V$ , there is a derivation $\delta :K(a)\to K(a,b)$ extending the one on K such that $\delta (a)=b$ . As $K(a,b)/K(a)$ is separable, by 2.2, we can extend the derivation to $K(a,b)\to K(a,b)$ . Then, for any nonempty Zariski-open $O_W\subseteq W$ over K, in the differential field extension $(K(a,b),\delta )$ we can find a solution to $x\in V$ and $(x,\delta x)\in O_W$ (namely, the tuple a). Since W has a smooth K-point, we get that K is e.c. in $K(W)=K(a,b)$ as a field. By differential largeness, $(K,\delta )$ is e.c. in $(K(a,b),\delta )$ , and so we can find the desired solution in K.

(ii) $\Rightarrow $ (iii) If we let $W=s(V)\subseteq \tau V$ , then the pair V and W satisfy the conditions of (ii) (note that if b is a smooth point of V then $(b,s(b))$ is a smooth point of W). If follows that the set of points in W of the form $(a,\delta a)$ with $a\in V(K)$ is Zariski dense in W. But then, as $W=s(V)$ , the set of points $a\in V$ such that $s(a)=(a,\delta a)$ must be Zariski dense in V.

(iii) $\Rightarrow $ (i) We verify 2.8(ii). Let $f,g\in K\{x\}$ with $\mathop {\operatorname {ord}}\nolimits (g)<\mathop {\operatorname {ord}}\nolimits (f)$ and g nonzero. Assume the system

$$\begin{align*}f(x)=0 \ \&\ s_f(x)\neq 0 \end{align*}$$

has an algebraic solution in K. In particular, $s_f\neq 0$ . By Observation 2.7, we may assume that f is irreducible. By Theorem 3.1(1) of [Reference Ino and SánchezIL23], ${\mathfrak {p}}=[f]:s_f^\infty $ is a separable prime differential ideal of $K\{x\}$ . Let $a=x+{\mathfrak {p}}$ in the fraction field of $K\{x\}/{\mathfrak {p}}$ . Letting $n=\mathop {\operatorname {ord}}\nolimits (f)$ , we see that $(a,\delta a,\dots ,\delta ^{n-1}a)$ is algebraically independent over K and $\delta ^n a$ is separably algebraic over $K(a,\dots ,\delta ^{n-1}a)$ . It follows that

$$ \begin{align*}\delta^{n+1}a=\frac{h(a,\delta a, \dots, \delta^n a)}{s_f(a)}\end{align*} $$

for some $h\in K[t_0,\dots ,t_n]$ . Let V be the localisation at $g\cdot s_f$ of the Zariski-locus of $(a,\delta a, \dots ,\delta ^n a)$ over K. From the assumptions (on existence of an algebraic solution in K), we see that V has a smooth K-rational point and that the morphism on V induced by

$$ \begin{align*}(t_0,t_1, \dots,t_n)\mapsto ((t_0,t_1, \dots,t_n),\left(t_1,t_2,\dots,t_n, \frac{h(t_0,t_1, \dots, t_n)}{s_f(t_0,t_1, \dots, t_n)}\right)\end{align*} $$

yields a regular algebraic map $s:V\to \tau V$ . This equips V with a D-variety structure. Then, the assumption of (iii) yields $\alpha \in V(K)$ such that $s(\alpha )=(\alpha ,\delta \alpha )$ . But then $\alpha $ is the desired differential solution of $f(x)=0\ \&\ g(x)\neq 0$ in K.

Proof By [Reference Sánchez and TresslLST24, 3.5] there is a differential K-algebra homomorphism $\psi :S\to K[[t]]$ . Applying 2.4 to $\psi (S)$ , we may then find a finitely generated K-subalgebra B of $K((t))$ , a derivation $\partial $ of B extending $\delta $ on K together with a differential K-algebra homomorphism $\phi :\psi (S)\longrightarrow (B,\partial )$ . By [Reference Sánchez and TresslLST24, 3.5] applied to $(B,\partial )$ and the inclusion map $B\hookrightarrow K((t))$ there is a differential K-algebra homomorphism $\gamma :B\to K[[t]]$ . Since B is a finitely generated K-algebra, the image of $\gamma $ is in $K[[t]]_{\mathrm {diffalg}}$ . Hence the map $\gamma \circ \phi \circ \psi :S\longrightarrow K[[t]]_{\mathrm {diffalg}}$ has the required property.

Proof Apply 3.3 to the differential coordinate ring of $\Sigma $ .

Proof (i) $\Rightarrow $ (ii) is a consequence of [Reference Sánchez and TresslLST24, 4.3(ii)], which says that K is e.c. in $K((t))$ as a differential field.

(ii) $\Rightarrow $ (i). By 3.4 one verifies that K is e.c. in $K((t))$ as a differential field. Hence by [Reference Sánchez and TresslLST24, 4.3], $(K,\delta )$ is differentially large

(iii) $\Rightarrow $ (i) We verify 2.12(iii). Let $(V,s)$ be a K-irreducible D-variety with a smooth K-point. Let $h\in K[V]$ nonzero. Then, there is an induced D-variety structure in the localization $K[V]_h$ . Denote this D-variety by $(W,t)$ . As K is large and V has a smooth K-point, we get that K is Zariski dense in V. Thus, W has a K-point. The assumption now yields a K-point b in W such that $s(b)=(b,\delta b)$ . As h was arbitrary, it follows that the set of points $\{a\in V(K): s(a)=(a,\delta a)\}$ is Zariski dense.

(i) $\Rightarrow $ (iii) Let $(V,s)$ be a K-irreducible D-variety with a K-point. Applying Proposition 3.3 with $S=K[V]$ and $L=K$ , we find a $K((t))$ -rational point b of V such that $s(b)=(b,\delta b)$ . As K is differentially large, it is e.c. in $K((t))$ as a differential field. Hence, we can find such a point in K.

Proof The proof is identical to the proof of [Reference Sánchez and TresslLST24, 5.1], except we use Proposition 3.3 in that proof instead of [Reference Sánchez and TresslLST24, 3.5].

Observation 3.7 Let K be a differential field. Then $K[[t]]_{\mathrm {diffalg}}$ is a Henselian valuation ring.

Proof We write $S=K[[t]]_{\mathrm {diffalg}}$ . Since $S=K[[t]]\cap K((t))_{\mathrm {diffalg}}$ , it is a valuation ring. Clearly the maximal ideal of S is $t{\cdot } S$ . To verify that S is Henselian it suffices to show that for all $\mu _2,\ldots ,\mu _n\in {\mathfrak {m}}$ there is some $f\in S$ with

$$\begin{align*}1+f+\mu_2 f^2+\ldots+\mu_n f^n=0. \end{align*}$$

As $K[[t]]$ is Henselian, there is such an f in $K[[t]]$ . Obviously, $f\in S$ .

Proof By Observation 3.7, $K_n[[t_n]]_{\mathrm {diffalg}}$ is a Henselian valuation ring. By [Reference PopPop10], $K_n((t_n))_{\mathrm {diffalg}}$ is a large field. We see that all assumptions of Proposition 3.6 are satisfied for the $K_n$ and the inclusion maps $K_n\hookrightarrow K_{n+k}$ . Now the argument for [Reference Sánchez and TresslLST24, 5.2(i)] can be copied, where we use 3.6 instead of [Reference Sánchez and TresslLST24, 5.1].

Counter example 3.10 If K is algebraically closed in Proposition 3.3 then a stronger conclusion holds, namely there is a differential K-algebra homomorphism $S\to K[[t]]$ whose image is constrained. The reason is that there is a differential homomorphism $\varepsilon :S\longrightarrow K^{\mathrm {diff}}$ and then one can apply Proposition 3.3 to obtain a differential embedding of the image of $\varepsilon $ into $K[[t]]$ .

However, if K is not algebraically closed then in general there is no differential K-algebra homomorphism $S\to K[[t]]$ whose image is constrained. To see an example, consider the ordered field ${\mathbb {R}}(z)$ where $z>{\mathbb {R}}$ and let K be its real closure. We furnish K with the unique derivation extending the standard derivation $\frac {\mathrm {d}}{\mathrm {d} z}$ on ${\mathbb {R}}(z)$ . Let x be a new transcendental element and let R be the real closure of the ordered field $K(x)$ with the ordering $x>K$ . Extend the derivation of K to R by setting $\delta (x)=0$ . Let y be a square root of $x-z$ in R and let S be the differential K-subalgebra of R generated by y, hence $S=K[y,y^{-1}]$ . Now if $\phi :S\longrightarrow K[[t]]$ is a differential K-algebra homomorphism, then $\phi (x)'=\phi (x')=0$ and $\phi (x)=\phi (y^2+z)=\phi (y)^2+z$ , hence $\phi (x)$ is a constant and $\phi (x)-z$ is a square. As $z>{\mathbb {R}}$ and ${\mathbb {R}}$ is the constant field of K, we see that $\phi (x)$ cannot be in K. Hence $\phi (x)$ is a new constant of $K[[t]]$ and therefore it is not constrained over K.

3.11 (On the canonicity of differentially algebraic solutions)

Let K be a differential field. If S is a differentially finitely generated K-algebra and $\phi :S\longrightarrow K$ is a K-algebra homomorphism, then by [Reference Sánchez and TresslLST24, 3.5] one can explicitly construct a differential K-algebra homomorphism $\psi :S\longrightarrow K[[t]]$ , namely one can take $\psi $ to be the twisted Taylor morphism $T^*_\phi $ associated with $\phi $ . Now, by Proposition 3.3 there is even a differential K-algebra homomorphism $\rho :S\longrightarrow K[[t]]_{\mathrm {diffalg}}$ and one might ask whether $\rho $ can also be obtained in some canonical form out of $\phi $ . However Gabriel Ng has shown that this is not possible. We refer to [Reference NgNg23, Proposition 7.11] for details.

Notation 4.1 Let K be a field (of characteristic zero). A differentially large problem of K is a pair $(f,g)$ of polynomials from $K\{x\}=K[x_0,x_1,\dots ]$ such that f is of order $n\geq 0$ , the order of g is strictly less than n and for which there is an element $(c_0,\ldots ,c_n)\in K^{n+1}$ such that

$$ \begin{align*} f(c_0,\ldots,c_n)=0\ \&\ s_f(c_0,\ldots,c_n)\neq 0. \end{align*} $$

WeFootnote ² call ${\bar c}$ an algebraic solution of the differentially large problem. Obviously a differentially large problem over K remains a differentially large problem over every field extension of K. If d is a derivation of K, then a solution of a differentially large problem of K in a differential field $(L,\delta )$ extending $(K,d)$ is an element $a\in L$ with $f(a)=0\ \&\ g(a)\neq 0$ , where polynomials are now evaluated as differential polynomials.

Proof Let ${\bar x}=(x_0,\ldots ,x_n)$ and let Z be the solution set in L of the system

$$\begin{align*}f({\bar x})=0\ \&\ s_f({\bar x})\neq 0. \end{align*}$$

Proof Let W be the variety defined by the two polynomials

$$\begin{align*}f({\bar x}),\ y{\cdot} s_f({\bar x})-1\in K[{\bar x},y]. \end{align*}$$

Write $h({\bar x},y)=y{\cdot } s_f ({\bar x})-1$ . Then any common zero $({\bar a},c)$ of f and h in the algebraic closure of L is a regular point of W, because $c{\cdot } s_f ({\bar a})-1=0$ implies $\frac {\partial f}{\partial x_n}({\bar a})\neq 0$ and obviously $\frac {\partial h}{\partial y}=s_f$ does not vanish at ${\bar a}$ . Hence the determinant of the matrix $\begin {pmatrix} \frac {\partial f}{\partial x_n} & \frac {\partial f}{\partial y} \\ \frac {\partial h}{\partial x_n} & \frac {\partial h}{\partial y} \end {pmatrix}$ is not zero at $({\bar a},c)$ . This shows that W is smooth.

Since $(f,g)$ is a differentially large problem of K we know that W has a K-rational point. By [Reference FehmFeh11, Theorem 1], using $\mathop {\operatorname {tr.deg}}\nolimits (L/K)\geq n=\dim (W)$ , there is a K-embedding $K(W)\longrightarrow L$ . A generic point of W in $K(W)$ is then mapped to a point $(a_0,\ldots ,a_n)\in Z$ with $\mathop {\operatorname {tr.deg}}\nolimits (a_0,\ldots ,a_{n}/K)=n$ .

As $s_f (a_0,\ldots ,a_n)\neq 0$ , $a_n$ is algebraic over $K(a_0,\ldots ,a_{n-1})$ . But now we see that $g(a_0,\ldots ,a_n)\neq 0$ as the order of g is strictly less than n, and $K_1:=K(a_0,\ldots ,a_n)$ is isomorphic to the quotient field of $K\{x\}/{\mathfrak {p}}$ , where ${\mathfrak {p}}=[f]:s_f^\infty $ . This induces a derivation $\delta $ on $K_1$ and this derivation has the required properties: $a=a_0$ solves the given differentially large problem.

Proof Let $\kappa =\mathop {\operatorname {card}}\nolimits (K)$ . By extending K and d we may assume that $\mathop {\operatorname {tr.deg}}\nolimits (L/K)=\kappa $ . Let $\{t_i{\ \vert \ } i<\kappa \}$ be a transcendence basis of L over K and let $(f_i,g_i)_{i\in \kappa }$ be a list of all differentially large problems of L; so here $f_i,g_i\in L\{x\}$ in the terminology of Notation 4.1.

For $i<\kappa $ we define a subfield $K_i$ of L and a derivation $d_i$ of $K_i$ such that

1. (a) $K_i$ contains $t_i$ , $\mathop {\operatorname {tr.deg}}\nolimits (K_i/K)$ is finite for finite i and $\mathop {\operatorname {tr.deg}}\nolimits (K_i/K)\leq \mathop {\operatorname {card}}\nolimits (i)$ for $i\geq \omega $ ,

2. (b) $(K_i,d_i)$ extends $(K_j,d_{j})$ for $j<i$ , and

3. (c) $(K_i,d_i)$ solves the differentially large problem $(f_i,g_i)$ .

Suppose $i<\kappa $ and $(K_j,d_j)$ has already been defined with properties (a)–(c); this also covers the case $i=0$ . Let ${\bar b}\subseteq L$ be finite with $f_i,g_i\in K(\bar b)\{x\}$ such that there is an algebraic solution of the differentially large problem $(f_i,g_i)$ in $K({\bar b})$ . Let $K_*$ be the field generated by $K(t_i,{\bar b})\cup \bigcup _{j<i}K_j$ and extend the derivation $\bigcup _{j<i}d_j$ to a derivation $d_*$ of $K_*$ arbitrarily. Obviously then $\mathop {\operatorname {tr.deg}}\nolimits (K_*/K)$ is finite if i is finite and $\leq \mathop {\operatorname {card}}\nolimits (i)$ otherwise.

Consequently $\mathop {\operatorname {tr.deg}}\nolimits (L/K_*)$ is infinite and we may apply Proposition 4.2 to the extension $K_*\subseteq L$ , the derivation $d_*$ and the differentially large problem $(f_i,d_i)$ . We obtain an extension $(K_{i},d_{i})$ of $(K_*,d_*)$ such that $K_{i}$ is a subfield of L that is finitely generated over $K_*$ . Clearly $(K_{i},d_{i})$ satisfies (a)–(c).

Then $L=\bigcup _{i<\kappa }K_i$ and by 2.8 the differential field $(L,\partial )$ with $\partial =\bigcup _{i<\kappa }d_i$ is differentially large.

Proof If L has infinite transcendence degree, then by Theorem 4.3 applied with $K=\mathbb {Q}$ shows that there is a derivation d of L such that $(L,d)$ is differentially large. For the converse assume, there is a derivation d of L such that $(L,d)$ is differentially large. By [Reference Sánchez and TresslLST24, 5.12], the algebraic closure $\overline {L}$ of L is differentially closed. We may then replace L by the differential closure of $\mathbb {Q}$ . By the non-minimality of the differential closure of $\mathbb {Q}$ ([Reference RosenlichtRos74]), there is an embedding $L\longrightarrow L$ that is not surjective. Hence L cannot have finite transcendence degree.

Examples 5.2

1. (i) If L is an algebraically closed field of infinite transcendence degree over K and S is a differentially finitely generated K-algebra, then $\mathop {\operatorname {Sped}}\nolimits _L(S)=\mathop {\operatorname {Sped}}\nolimits (S)$ , and L-locally closed is the same as being constrained in the sense of Kolchin [Reference KolchinKol74].

2. (ii) If K is real closed and L is an $|S|^+$ -saturated real closed field, then $\mathop {\operatorname {Sped}}\nolimits _L(S)$ is the subspace of differential prime ideals ${\mathfrak {p}}$ that are real, i.e., $-1$ is not a sum of squares in $S/{\mathfrak {p}}$ . When we are in this example we will say real constrained instead of L-locally closed.

3. (iii) Clearly, being constrained and real implies real-constrained. However, the converse does not always hold. For instance, consider the real closure $K=\mathbb {Q}(t)^{\mathrm {rcl}}$ , where $\mathbb {Q}(t)$ is equipped with the unique ordering such that $t>\mathbb {Q}$ and with the unique derivation extending $\frac {d}{dt}$ on $\mathbb {Q}(t)$ . Let $\alpha _2$ be a transcendental over K. In the formally real field $K(\alpha _2)$ define a derivation $\delta $ that extends the one on K such that $\delta (\alpha _2)=\frac {-1}{2\alpha _2}$ . Let $\alpha _1=t+\alpha _2^2$ . Then, $\delta (\alpha _1)=0$ and $\alpha _1$ is transcendental over K (as $\alpha _1>\mathbb {Q}^{\mathrm {rcl}}$ ). Now consider $\alpha =(\alpha _1,\alpha _2)$ . Clearly $\alpha $ is not constrained over K (as $\alpha _1$ is a constant which is not algebraic over $C_K=\mathbb {Q}^{\mathrm {rcl}}$ ). But $\alpha $ is real-constrained over K. Indeed, for any differential specialization $\beta =(\beta _1,\beta _2)$ of $\alpha $ over K with $K\langle \beta \rangle $ formally real, we see that $\beta _1$ is transcendental over K (for the same reason that $\alpha _1$ was); thus, the map $K(\alpha )\to K(\beta )$ fixing K and mapping $(\alpha _1,\alpha _2)\mapsto (\beta _1,\beta _2)$ is a differential isomorphism (not necessarily preserving the orderings).

Observation 5.3 Let L be a differentially large field and let $K\subseteq L$ be a differential subfield. Then for every L-simple K-algebra S there is a differential K-embedding $S\longrightarrow L$ .

Proof As S is L-simple, there is a K-algebra embedding $\phi :S\longrightarrow L$ . Now $S\otimes _KL$ is a finitely generated differential L-algebra and $\phi $ extends to an L-algebra homomorphism $S\otimes _KL\longrightarrow L$ . By differential largeness, [Reference Sánchez and TresslLST24, 4.3(iv)] says that there is a differential L-algebra homomorphism $\psi :S\otimes _KL\longrightarrow L$ . Composing $\psi $ with the natural map $S\longrightarrow S\otimes _KL$ gives a differential K-algebra homomorphism $S\longrightarrow L$ . Since S is L-simple, this map is an embedding.

Proof By Noetherianity of $\mathop {\operatorname {Sped}}\nolimits (S)$ , there are points of $\mathop {\operatorname {Sped}}\nolimits _L(S)$ that are maximal for inclusion in $\mathop {\operatorname {Sped}}\nolimits _L(S)$ . These points are even closed in $\mathop {\operatorname {Sped}}\nolimits _L(S)$ .

Now take an L-locally closed points ${\mathfrak {p}}$ of S. Hence there is some $q\in S$ such that the prime ideal ${\mathfrak {p}}$ is maximal for inclusion in $D(q)\cap \mathop {\operatorname {Sped}}\nolimits _L(S)$ . In other words the zero ideal is the unique element of $\mathop {\operatorname {Sped}}\nolimits _L(A)$ , where $A=(S/{\mathfrak {p}})_q$ ; in particular there is a K-algebra embedding $A\longrightarrow L$ .

By 2.4 there are $a\in A$ and a (not necessary differential) finitely generated K-subalgebra B of $A_a$ and a derivation $\partial $ of B together with a differential K-algebra homomorphism $f:A\longrightarrow (B,\partial )$ . In particular the kernel of f is in $\mathop {\operatorname {Sped}}\nolimits (A)$ . Since B is a K-subalgebra of $A_a$ , the kernel of f is even in $\mathop {\operatorname {Sped}}\nolimits _L(S)$ and so the kernel of f is $0$ . Thus f is an embedding and $\mathop {\operatorname {qf}}\nolimits (A)$ is a finitely generated field extension of K. Since A is a differentially finitely generated K-algebra we may then localize it at some element of A and see that A is finitely generated as a K-algebra.

5.5 (The étale open topology)

Let K be a field.

1. (i) We call a subset U of K standard étale open if it is the image of the projection $K^2\longrightarrow K$ onto the first coordinate of a set of the form $\{(a,b)\in K^2{\ \vert \ } P(a,b)=0\ \&\ Q(a,b)\neq 0\}$ , where $P,Q\in K[x,y]$ such that $\frac {\partial }{\partial y}P$ is invertible in the localization of $K[x,y]/(P)$ at Q. In the terminology of [Reference PoonenPoo17, Definition 3.5.38] these sets are precisely the images of K-rational points of standard étale morphisms defined over K with codomain ${\mathbb {A}}^1$ . The standard étale open sets form a basis of a topology on K which is the étale open topology, cf. [Reference PoonenPoo17, Definition 3.5.38], [Reference Johnson, Tran, Walsberg and YeJTWY24, p. 4037].

2. (ii) [Reference Johnson, Tran, Walsberg and YeJTWY24] The field K is large if and only if the étale open topology is not discrete. If K is algebraically closed, then the étale open topology is the Zariski topology. If K is real closed, then the étale open topology is the order topology. If K possesses a nontrivial henselian valuation v, then the étale open topology is the open ball topology of v.

Proof By L-simplicity, we may assume that S is a K-subalgebra of L. We write $F=\mathop {\operatorname {qf}}\nolimits (S)$ and $\delta $ for the derivation of F. Since S is L-simple, there is $g\in S$ such that the localisation $\mathop {\operatorname {Sped}}\nolimits _L(S_g)$ only consist of the zero ideal.

Now suppose $f=\frac {p}{q}\in F$ is a constant, thus $\delta (f)=0$ . We aim to show that f is in $C_K$ . We work in the localization $S_{g\cdot q}$ , and we view it as the coordinate ring of an affine variety V defined over K. Then f yields an algebraic map $f:V(L)\to L$ . The image $W=f(V(L))$ ^[ Footnote ^{3 ]} is K-definable in L in the language of rings.

Case 1. W is infinite.

Since W is an existentially L-definable set it must have nonempty interior for the étale open topology of L by [Reference Walsberg and YeWY23, Corollary A, p.613]. Since $C_K$ is dense in L, there is $\varepsilon \in V(L)$ —hence a K-algebra homomorphism $S_{g\cdot q}\longrightarrow L$ —such that $c:=\varepsilon (f)\in C_K$ and we claim that $f=c$ .

Since $\delta (f-c)=0$ , the ideal ${\mathfrak {a}}:=(f-c)$ of $S_{g\cdot q}$ is differential and contained in the kernel ${\mathfrak {q}}$ of $\varepsilon $ . Choose any extension of the derivation of K to L and let $T_\varepsilon ^*:S_{g\cdot q}\longrightarrow L[[t]]$ be the twisted Taylor morphism of $\varepsilon $ and that derivation. We write ${\mathfrak {q}}^\#$ for the kernel of $T_\varepsilon ^*$ and obtain a differential K-algebra embedding $S_{g\cdot q}/{\mathfrak {q}}^\#\hookrightarrow L[[t]]$ ^[ Footnote ^{4 ]}. It follows that the K-variety $V_1$ defined by $S_{g\cdot q}/{\mathfrak {q}}^\#$ has a smooth $L[[t]]$ -rational point. Since L is a large field, $V_1$ also has a smooth L-rational point. Since S is a K-subalgebra of L we see that $\mathop {\operatorname {tr.deg}}\nolimits (L/K)\geq \mathop {\operatorname {tr.deg}}\nolimits (S/K)\geq \mathop {\operatorname {tr.deg}}\nolimits ((S_{g\cdot q}/{\mathfrak {q}}^\#)/K)$ . By [Reference FehmFeh11, Theorem 1] we know that there is a K-algebra embedding $S_{g\cdot q}/{\mathfrak {q}}^\#\hookrightarrow L$ . We have shown that ${\mathfrak {q}}^\#$ is in $\mathop {\operatorname {Sped}}\nolimits _L(A)$ . But $(0)$ is the only differential L-prime ideal of $S_{g\cdot q}$ , thus ${\mathfrak {q}}^\#=(0)$ . On the other hand, ${\mathfrak {a}}$ is a differential ideal and contained in the kernel ${\mathfrak {q}}$ of $\varepsilon $ . This implies ${\mathfrak {a}}\subseteq {\mathfrak {q}}^\#=(0)$ , showing that $f-c=0$ as required.

Case 2. W is finite.

Since V is irreducible and defined over K the assumption that K is algebraically closed in L, implies that the variety $V\times _KL$ is also irreducible. Consequently $V(L)$ is an irreducible subset for the Zariski topology of the L-rational points of V. It follows that the image W of f seen as a map $V(L)\longrightarrow L$ is also irreducible. As W is finite, W is a singleton set. Hence f is a constant algebraic map and thus $f=c$ .

Proof (i) Since $(0)$ is a differential L-prime ideal, there is a differential L-prime ideal ${\mathfrak {p}}$ of $K\{a\}$ with $g\notin {\mathfrak {p}}$ and some $s\in K\{a\}\setminus {\mathfrak {p}}$ such that $S=(K\{a\}/{\mathfrak {p}})_{g(a){\cdot } s}$ is L-simple. In particular, S can be embedded as a K-algebra into L. By Proposition 5.6, the fraction field of S has constant field $C_K$ .

(ii) follows from (i) and Observation 5.3.

Proof Using Corollary 5.7 we adapt the strategy of the proof of Theorem 4.3. We may replace K by its algebraic closure of K in L, hence we need to find a derivation on L extending d with constant field C. Let $\kappa =\mathop {\operatorname {card}}\nolimits (K)$ . By extending K with sufficiently many differentially algebraic independent elements, we may assume that $\mathop {\operatorname {tr.deg}}\nolimits (L/K)=\kappa $ ; this will not extend the constants as one verifies without difficulty. Let $(f_i,g_i)_{i\in \kappa }$ be a list of all differentially large problems of L, where $f_0=x,g_0=1$ ; so here $f_i,g_i\in L\{x\}$ in the terminology of 4.1.

For $i<\kappa $ , starting with $K_0=K$ , we define a subfield $K_i$ of L that is algebraically closed in L and a derivation $d_i$ of $K_i$ such that

1. (a) $(K_i,d_i)$ extends $(K_j,d_{j})$ for $j<i$ .

2. (b) $\mathop {\operatorname {tr.deg}}\nolimits (K_i/K)\leq \max \{\aleph _0,\mathop {\operatorname {card}}\nolimits (i)\}$ .

3. (c) $\mathop {\operatorname {tr.deg}}\nolimits (L/K_i)$ is infinite (which is implied by (b) when L is uncountable).

4. (d) $f_i,g_i\in K_i\{x\}$ and $(K_i,d_i)$ solves the differentially large problem $(f_i,g_i)$ .

5. (e) $(K_i,d_i)$ has constant field C.

Suppose $0<i<\kappa $ and $(K_j,d_j)$ has already been defined for $j<i$ with properties (a)–(e). Let $K_*=\bigcup _{j<i}K_j$ with derivation $d_*=\bigcup _{j<i}d_j$ . Obviously then $\mathop {\operatorname {tr.deg}}\nolimits (K_*/K)\leq \max \{\aleph _0,\mathop {\operatorname {card}}\nolimits (i)\}$ . If i is infinite, then L is uncountable and of size $>\mathop {\operatorname {card}}\nolimits (i)$ , hence $\mathop {\operatorname {tr.deg}}\nolimits (L/K_*)$ is infinite. If i is finite, then $K_*=K_{i-1}$ and $\mathop {\operatorname {tr.deg}}\nolimits (L/K_*)$ is infinite as well.

Since $\mathop {\operatorname {tr.deg}}\nolimits (L/K_*)$ is infinite, there is a countable infinite set $T\subseteq L$ that is algebraically independent over $K_*$ such that $\mathop {\operatorname {tr.deg}}\nolimits (L/K_*(T))$ is infinite, such that $f_i,g_i\in K_*(T)^{\mathrm {alg}}\{x\}$ and such that there is an algebraic solution of the differentially large problem $(f_i,g_i)$ in $K_*(T)^{\mathrm {alg}}$ (here the superscript “ $\mathrm {alg}$ ” stands for the algebraic closure in L).

It follows that $\mathop {\operatorname {tr.deg}}\nolimits (L/K_*(T))\geq \mathop {\operatorname {card}}\nolimits (K_*(T))$ in either case. Now the field $K_*(T)$ is isomorphic to the fraction field $K_*\langle x\rangle $ of the differential polynomial ring $K_*\{x\}$ and therefore there is a derivation $\partial $ of $K_*(T)$ extending $d_*$ such that $(K_*(T),\partial )$ is $K_*$ -isomorphic to $K_*\langle x\rangle $ with its natural derivation. It follows that the constant field of $\partial $ is the constant field of $K_*$ , which is C by property (e) in the induction hypothesis. Hence we may extend $\partial $ to the algebraic closure $K_+$ of $K_*(T)$ in L without extending the constants.

Since $\mathop {\operatorname {tr.deg}}\nolimits (L/K_+)\geq \mathop {\operatorname {card}}\nolimits (K_+)$ we may now apply Theorem 4.3 and extend $\partial $ to a derivation on L such that $(L,\partial )$ is differentially large. Since C is dense in L it is also dense in $K_+$ . Since $(f_i,g_i)$ is a differentially large problem of $K_+$ by choice of T, we may apply Corollary 5.7(ii), which shows that $(f_i,g_i)$ has a differential solution a in $(L,\partial )$ such that $(K_+\langle a\rangle ,\partial )$ has constant field C. We may then define $K_i$ to be the algebraic closure of $(K_+\langle a\rangle ,\partial )$ in L and see that all conditions (a)–(e) are satisfied.

Finally, $L=\bigcup _{i<\kappa }K_i$ because for each $b\in L$ , the differentially large problem $(x-b,1)$ is solved in the union. Hence the theorem follows.

Proof If $(L,K)\models T_{\mathrm {pair}}$ , then a standard compactness argument shows that there is an elementary extension $(L',K')\succ (L,K)$ , such that $\mathop {\operatorname {tr.deg}}\nolimits (L'/K')$ is infinite. By the downwards Skolem–Löwenheim theorem there is a countable elementary restriction $(L",K")\prec (L',K')$ such that $\mathop {\operatorname {tr.deg}}\nolimits (L"/K")$ is infinite.

Now take $(L_1,K_1),(L_2,K_2)\models T_{\mathrm {pair}}$ . In order to show that the pairs $(L_1,K_1)$ and $(L_2,K_2)$ are elementary equivalent we may apply the argument above to each pair and assume that $L_i$ is countable and of infinite transcendence degree over $K_i$ . By Corollary 5.9, we may expand $L_i$ to a differentially large field $(L_i,\delta _i)$ with constant field $K_i$ . Hence the completeness of $T^\delta $ implies that $(L_1,\delta _1)$ and $(L_2,\delta _2)$ are elementary equivalent as differential fields and therefore the pairs $(L_1,K_1)$ and $(L_2,K_2)$ are elementary equivalent as well.

Proof By [Reference Sánchez and TresslLST24, Corollary 4.8(iii)], $T^\delta $ is complete and model complete (recall that $T^\delta $ is the theory T together with the theory of differentially large fields). In particular, Theorem 5.11 yields that $T_{\mathrm {pair}}$ is complete. Now, (i) follows from the fact that a large field of characteristic zero with a stable theory must be algebraically closed [Reference Johnson, Tran, Walsberg and YeJTWY24]. For (ii), by the argument in the proof of Theorem 5.11, it suffices to observe that $T^\delta $ is simple whenever T is simple. This appears in [Reference Mohamed and SánchezML24, Corollary 3.6(i)]. Similarly, from [Reference Mohamed and SánchezML24, Corollary 3.6(ii)] we get that $T^\delta $ is NSOP $_1$ whenever T is NSOP $_1$ , and so (iii) follows. By [Reference MohamedMoh24, Corollary 5.1], $T^\delta $ has NIP if T has NIP, hence we obtain (iv).

Proof We take $M\supseteq R$ by successively adjoining infinitely large elements $a_\alpha $ for $\alpha <\kappa $ . Hence $a_\alpha>R(a_\beta {\ \vert \ } \beta <\alpha )$ in M and M is algebraic over $R(a_\alpha {\ \vert \ } \alpha <\kappa )$ . Then R is Dedekind complete in M and M has transcendence degree $\kappa $ over R.

For N we may take any real closed subfield of $R((t^{\mathbb {R}}))$ of transcendence degree $\kappa $ over R. Such fields exist because of the following reason: Let $\Lambda $ be a basis of the $\mathbb {Q}$ -vector space R. Since R is real closed, the cardinality of $\Lambda $ is $\kappa $ . Then the set $\{\exp (\lambda {\cdot } t){\ \vert \ } \lambda \in \Lambda \}\subseteq R[[t]]$ is an algebraically independent subset of $R((t^{\mathbb {R}}))$ over R: this is a baby case of Ax’s positive solution to the functional Schanuel conjecture, but is not difficult to prove directly. Hence we may take N as the real closure of $R(\exp (\lambda {\cdot } t){\ \vert \ } \lambda \in \Lambda )$ in $R((t^{\mathbb {R}}))$ . Clearly N has transcendence degree $\kappa $ over R.

Since R is Dedekind complete in M and in N, any real closed field S containing R with $\mathop {\operatorname {tr.deg}}\nolimits (S/R)\leq 1$ in which R is Dedekind complete, can be embedded into M and into N. It remains to show that any real closed subfield S of M containing R that can be embedded into N over R is of transcendence degree at most 1 over R; note that R is Dedekind complete in S because R is Dedekind complete in M (and in N).

For a contradiction, suppose S has transcendence degree $2$ over R. We furnish M with the valuation whose valuation ring is the convex hull of R in M. Real closures are now taken in M throughout and this is indicated by the superscript $^{\mathrm {rcl}}$ . Take ${\bar a}=(a_{\alpha _1},\ldots ,a_{\alpha _n})$ , $\alpha _1<\ldots <\alpha _n$ such that $S\subseteq R({\bar a})^{\mathrm {rcl}}$ . Then by choice of the $a_\alpha $ the chain $R\subseteq R(a_{\alpha _1})^{\mathrm {rcl}}\subseteq \ldots \subseteq R(a_{\alpha _1},\ldots ,a_{\alpha _n})^{\mathrm {rcl}}$ witnesses that the value group of $R(a_{\alpha _1},\ldots ,a_{\alpha _n})^{\mathrm {rcl}}$ has height n, where height stands for the number of convex subgroups of the value group. Since $\mathop {\operatorname {tr.deg}}\nolimits (S/R)=2$ , there are $n-2$ elements $b_1,\ldots ,b_{n-2}$ from $\{a_{\alpha _1},\ldots ,a_{\alpha _n}\}$ that are algebraically independent over S. Since S can be embedded into $R((t^{\mathbb {R}}))$ we know that S has height $1$ : Crucially we use here that any such embedding preserves the valuations because the natural valuation on $R((t^{\mathbb {R}}))$ again has the convex hull of R in $R((t^{\mathbb {R}}))$ as its valuation ring. But now the chain $R\subseteq S\subseteq S(b_1)^{\mathrm {rcl}}\subseteq \ldots \subseteq S(b_1,\ldots ,b_{n-2})^{\mathrm {rcl}}=R(a_{\alpha _1},\ldots ,a_{\alpha _n})^{\mathrm {rcl}}$ witnesses that the value group of $R(a_{\alpha _1},\ldots ,a_{\alpha _n})^{\mathrm {rcl}}$ has height at most $n-1$ , which gives the desired contradiction.

Proof Suppose there is a prime model extension $\hat K$ of K for $\mathrm {CODF}$ but $\hat K$ is not algebraic over K. Let R be the algebraic closure of K in $\hat K$ . Then R is a differential subfield of $\hat K$ and $\hat K$ is also a prime model of R for $\mathrm { CODF}$ : If $R\subseteq M\models \mathrm {CODF}$ , then any K-embedding $\hat K\longrightarrow M$ must be the identity on R. Hence we may assume that K is real closed all along.

Choose real closed fields $M,N$ for K as in Proposition 6.1. By Theorem 4.3 there are extensions of the derivation of K to $M,N$ respectively such that $M,N$ furnished with these extensions are CODF s. Since $\hat K$ can be embedded into M and into N by assumption, Proposition 6.1 implies that $\hat K$ must be of transcendence degree $\leq 1$ over K and K is Dedekind complete in $\hat K$ . As $K\neq \hat K$ , we know that $\mathop {\operatorname {tr.deg}}\nolimits (\hat K/K)=1$ .

Since $\hat K$ is a $\mathrm {CODF}$ it follows that $\hat K$ has a positive infinitesimal element t with respect to K such that $t'=1$ (in particular $t\notin K$ ). Then $\hat K$ is a differential subfield of $K((t^{\mathbb {Q}}))$ (endowed with the derivation extending the one on K and satisfying $t'=1$ ). By [Reference Sánchez and TresslLST24, end of 5.3] we know that $t^{-1}$ has no integral in $K((t^{\mathbb {Q}}))$ . This contradicts the fact that $t^{-1}$ has an integral in the CODF $\hat K$ .