From algebra to analysis: new proofs of
theorems by Ritt and Seidenberg

Fix $1\leqslant i\leqslant n$. Since $a_{i}$ is $\Delta$-algebraic over $R$, there exists nonzero $f_{i}\in R[\Delta^{\infty}x]$ such that $f_{i}(a_{i})=0$. We will choose this $f_{i}$ so that its highest (w.r.t. (1)) derivative is minimal and, among such polynomials, the degree is minimal. We will call such $f_{i}$ a minimal polynomial for $a_{i}$.

We introduce variables $\lambda_{2},\ldots,\lambda_{m}$ algebraically independent over $A$ and extend the derivations from $A$ to $A[\lambda_{2},\ldots,\lambda_{m}]$ by $\delta_{i}\lambda_{j}=0$ for all $i=1,\ldots,m$ and $j=2,\ldots,m$. Consider a set of derivations $D:=\{\operatorname{d}_{1},\operatorname{d}_{2},\ldots,\operatorname{d}_{m}\}$ defined by

Consider any $1\leqslant i\leqslant n$. We rewrite $f_{i}$ in terms of $D$ replacing $\delta_{1}$ with $\operatorname{d}_{1}$ and $\delta_{j}$ with $\operatorname{d}_{j}-\lambda_{i}\operatorname{d}_{1}$ for $j=2,\ldots,m$. We denote the order of the highest derivative appearing in $f_{i}$ by $r_{i}$ and the partial derivative $\frac{\partial}{\partial(\operatorname{d}_{1}^{r_{i}}x)}f_{i}$ by $s_{i}$. We will show that $s_{i}(a_{i})\neq 0$. We write

Due to the minimality of $f_{i}$ as a vanishing polynomial of $a$ and the algebraic independence of $\lambda_{j}$, the latter expression does not vanish at $x=a_{i}$. So, $s_{i}(a_{i})\neq 0$.

Since, for every $1\leqslant i\leqslant n$, $s_{i}(a_{i})$ is a nonzero polynomial in $\lambda_{2},\ldots,\lambda_{m}$ over $A$, it is possible to choose the values $\lambda^{\ast}_{2},\ldots,\lambda_{m}^{\ast}\in\mathbb{Z}\subset R$ so that neither of $s_{i}(a_{i})$ vanishes at $(\lambda_{2}^{\ast},\ldots,\lambda_{m}^{\ast})$. Let $\Delta^{\ast}=\{\delta_{1}^{\ast},\ldots,\delta_{m}^{\ast}\}$ be the result of plugging these values to $D$. Then we have $\operatorname{sep}_{x}^{\Delta^{\ast}}f(a_{i})=\dfrac{\partial f_{i}}{\partial((\delta_{1}^{\ast})^{r_{i}}x)}(a_{i})\neq 0$ for every $i=1,\ldots,n$, so $a_{1},\ldots,a_{n}$ are $\Delta^{\ast}$-integral over $R$. ∎

We will prove the lemma by induction on the number $n$ of $\Delta$-generators of $A$. If $n=0$, then $A=R$ and $A$ is clearly finitely $\Delta_{0}$-generated.

Assume that the lemma is proved for all extensions $\Delta$-generated by less than $n$ elements. Applying the induction hypothesis to $\Delta$-algebra $A_{0}:=R[\Delta^{\infty}a_{1},\ldots,\Delta^{\infty}a_{n-1}]$, we obtain $b_{0}\in A_{0}$ such that $A_{0}[1/b_{0}]$ is a finitely $\Delta_{0}$-generated $R$-algebra.

Since $a_{n}$ is $\Delta$-integral over $R$, there exists $P(x)\in R[\Delta^{\infty}x]$ such that $P(a_{n})=0$, the highest derivative in $P$ is $\delta_{1}^{r}x$, and $b_{2}:=\operatorname{sep}^{\Delta}_{x}(P)(a_{n})\neq 0$. We claim that

where $\delta_{1}^{\leqslant r}a_{n}:=\{a_{n},\delta_{1}a_{n},\ldots,\delta_{n}^{r}a_{n}\}$. Since $A_{0}[1/b_{1}]$ is finitely $\Delta_{0}$-generated over $R$, this would imply that $A[1/(b_{1}b_{2})]$ is finitely $\Delta_{0}$-generated over $R$ as well.

In order to prove (3), it is sufficient to show that the images of $\{\delta_{1}^{\leqslant r}a_{n},1/b_{2}\}$ under $\delta_{1}$ belong to $B:=A_{0}[1/b_{1},\Delta_{0}^{\infty}(1/b_{2}),\Delta_{0}^{\infty}(\delta_{1}^{\leqslant r}a_{n})]$. This is clear for $\delta_{1}^{<r}a_{n}$, so it remains to show that $\delta_{1}^{r+1}a_{n},\delta_{1}(1/b_{2})\in B$:

• •

  For $\delta_{1}^{r+1}a_{n}$ we use Remark 2 to write

  $$\delta_{1}^{r+1}a_{n}=\dfrac{-q(a_{n})}{\operatorname{sep}^{\Delta}_{x}(P)(a_{n})}=\dfrac{-q(a_{n})}{b_{2}}\in B,\text{ where }q\in R[\Delta_{0}^{\infty}(\delta_{1}^{\leqslant r}x)].$$

• •

  For $\delta_{1}(1/b_{2})$, we observe that

  $$\delta_{1}\left(\dfrac{1}{b_{2}}\right)\in\dfrac{1}{b_{2}^{2}}A_{0}[\Delta_{0}^{\infty}(\delta_{1}^{\leqslant r}a_{n})]\subseteq B.\qed$$

Let $K$ be $\Delta$-generated by $\{b_{i}\}_{i=1}^{\infty}$. For every $i\geqslant 0$, denote by $S_{i}$ the set of singularities of $b_{i}$. By definition, $S_{i}$ is a nowhere dense subset of $\mathbb{C}^{m}$. Therefore, the union $S=\bigcup\limits_{i=1}^{\infty}S_{i}$ is a meagre set. As $W$ is a domain in $\mathbb{C}^{m}$, the difference $W\setminus S$ is non-empty. Choose any point $(w_{1},\ldots,w_{m})\in W\setminus S$. Then all the restrictions of $b_{i}$ to $\{z_{1}=w_{1}\}$ are holomorphic at $(w_{2},\ldots,w_{m})$ and meromorphic in some vicinity $V\subseteq W\cap\{t_{1}=w_{1}\}$ of $(w_{2},\ldots,w_{m})$. Since every element $f\in K$ is a rational function in $b_{i}$’s and their partial derivatives, its restriction to $\{z_{1}=w_{1}\}$ is also a well-defined meromorphic function on $V$. ∎

Suppose $\operatorname{difftrdeg}_{K}^{\Delta}\mathcal{M}er_{m}(U)=l<\infty$. Let $\tau_{1},\ldots,\tau_{l}$ be a $\Delta$-transcendence basis of $\mathcal{M}er_{m}(U)$ over $K$. Let $L$ be a field $\Delta$-generated by $K$ and $\tau_{1},\ldots,\tau_{l}$. Note that $L$ is still at most countably $\Delta$-generated and $\mathcal{M}er_{m}(U)$ is $\Delta$-algebraic over $L$. Choose an arbitrary point $c\in U$ and denote by $F$ a subfield of $\mathbb{C}$ generated by the values at $c$ of those elements of $L$ that are holomorphic at $c$. Clearly $F$ is at most countably generated and the transcendence degree of $\mathbb{C}$ over $F$ is infinite. Now any function in $\mathcal{M}er_{m}(U)$ that is holomorphic at $c$ and such that the set of values of its derivatives at $c$ is transcendental over $F$ is $\Delta$-transcendental over $L$, which contradicts to the assumption that $\mathcal{M}er_{m}(U)$ is $\Delta$-algebraic over $L$. ∎

We will first reduce the theorem to the case when $L$ is $\Delta$-algebraic over $K$. Assume that it is not, and let $u_{1},\ldots,u_{\ell}$ be a $\Delta$-transcendence basis of $L$ over $K$. Lemma 4 implies that there exist functions $f_{1},\ldots,f_{\ell}\in\mathcal{M}er_{m}(W)$ $\Delta$-transcendental over $K$. Let $\tilde{K}$ be a $\Delta$-field generated by $K$ and $f_{1},\ldots,f_{\ell}$. The embedding $K\to L$ can be extended to an embedding $\tilde{K}\to L$ by sending $f_{i}$ to $u_{i}$ for every $1\leqslant i\leqslant\ell$. Therefore, by replacing $K$ with $\tilde{K}$ we will further assume that $L$ is $\Delta$-algebraic over $K$.

We will prove the theorem by induction on the number $m$ of derivations. If $m=0$, then $L$ can be embedded into $\mathbb{C}$ by [12, Chapter V, Theorem 2.8].

Suppose $m>0$. Let $a_{1},\ldots,a_{n}$ be a set of $\Delta$-generators of $L$ over $K$. Let $A:=K[\Delta^{\infty}a_{1},\ldots,\Delta^{\infty}a_{n}]$. Since $A$ is $\Delta$-algebraic over $K$, by Lemma 1, there exist and invertible $m\times m$ matrix $M$ over $\mathbb{Q}$ such that, for a new set of derivations

$a_{1},\ldots,a_{n}$ are $\Delta^{\ast}$-integral over $K$. Due to the invertibility of $M$, every $\Delta^{\ast}$-embedding $L\to\mathcal{M}er_{m}(U)$ over $K$ yields a $\Delta$-embedding. Therefore, by changing the coordinates in the space $\mathbb{C}^{m}$ from $(z_{1},\ldots,z_{m})$ to $M^{-1}(z_{1},\ldots,z_{m})$, we can further assume that $\Delta=\Delta^{\ast}$, so $a_{1},\ldots,a_{n}$ are $\Delta$-integral over $K$.

Lemma 2 implies that there exists $a\in A$ such that $B:=A[a^{-1}]$ is finitely $\Delta_{0}$-generated. By $\Delta$-integrality of $a_{1},\ldots,a_{n}$ and Remark 2, for every $1\leqslant i\leqslant n$, there exists a positive integer $r_{i}$ and a rational function $g_{i}\in K(\delta^{\alpha}y\mid\alpha<_{\operatorname{grlex}}(r,0,\ldots,0))$ such that $a_{i}$ satisfies

Since $K$ is at most countably $\Delta$-generated, Lemma 3 implies that there exist $w_{1}\in\mathbb{C}$ and $V\subseteq W\cap\{z_{1}=w_{1}\}\subseteq\mathbb{C}^{m-1}$ such that the restriction to $\{z_{1}=w_{1}\}$ induces a $\Delta_{0}$-embedding $\rho\colon K\to\mathcal{M}er_{m-1}(V)$. We apply the induction hypothesis to $\Delta_{0}$-fields $\rho(K)$ and $\operatorname{Frac}(B)=L$. This yields $\Delta_{0}$-embedding $h:L\rightarrow\mathcal{M}er_{m-1}(\widetilde{V})$ for some $\widetilde{V}\subseteq V$.

Choose a point $v=(w_{2},\ldots,w_{m})\in\widetilde{V}$ such that all the $h(a_{i})$ are holomorphic at $v$ and all the $g_{i}(a_{i})$ are holomorphic at $w=(w_{1},w_{2},\ldots,w_{m})\in W$. Consider the Taylor homomorphism $T_{h,w}\colon A\rightarrow\mathcal{M}er_{m-1}(\widetilde{V})[[z_{1}]]$ defined as follows (see Definition 7):

Note that $T_{h,w}$ is a $\Delta$-homomorphism.

Fix $1\leqslant i\leqslant n$. Since $a_{1}$ is a solution of (4) and $T_{h,w}$ is a $\Delta$-homomorphism, $T_{h,w}$ is a formal power series solution of $\delta_{1}^{r_{i}}y=g_{i}(y)$ corresponding to holomorphic initial conditions

By the Cauchy-Kovalevskaya theorem, this solution is holomorphic in some vicinity $U_{i}$ of $w$. We set $U:=\bigcap_{i=1}^{n}U_{i}$. Thus, $T_{h,w}$ induces a non-trivial $\Delta$-homomorphism from $A$ to $\mathcal{M}er_{m}(U)$. Since $h$ is injective, $T_{h,w}$ is also injective, so it can be extended to a $\Delta$-embedding $L\to\mathcal{M}er_{m}(U)$ over $K$.

∎

Consider any non-zero (not necessarily differential) homomorphism $\psi:A\rightarrow F$ ($F\supseteq K$ is a field) and the corresponding Taylor homomorphism $T_{\psi,0}\colon A\rightarrow F[[z_{1},\ldots,z_{m}]]$, which is a $\Delta$-homomorphism. Since $A$ is $\Delta$-simple, the kernel of $T_{\psi,0}$ is zero. Therefore, $T_{\psi,0}$ is a $\Delta$-embedding of $A$ into $F[[z_{1},\ldots,z_{m}]]$. Since the latter does not contain zero divisors, the same is true for $A$.

Assume that $A$ has infinite $\Delta_{0}$-transcendence degree, that is, $\ell>0$. Since $a_{\ell+1},\ldots,a_{n}$ are $\Delta$-integral over $R:=K[\Delta^{\infty}a_{1},\ldots,\Delta^{\infty}a_{\ell}]$, Lemma 2 implies that there exists an element $b\in A$ such that $A_{0}:=A[1/b]$ is a finitely $\Delta_{0}$-generated algebra over $R$. Note that $A_{0}$ is also $\Delta$-simple. Let $A_{0}=R[\Delta_{0}^{\infty}b_{1},\ldots,\Delta_{0}^{\infty}b_{s}]$. For every $j\geqslant 0$, consider $\Delta_{0}$-algebra

For every $j\geqslant 0$, we have

This inequality implies that there exists $N$ such that, for every $j>N$, $\delta_{1}^{j}a_{1},\ldots,\delta_{1}^{j}a_{l}$ are $\Delta_{0}$-independent over $B_{j}$. Consider any non-zero $\Delta_{0}$-homomorphism

where $L\supseteq K$ is a $\Delta_{0}$-field. Due to the $\Delta_{0}$-independence of the rest of the elements $\delta^{j}_{1}a_{i}$ for $1\leqslant i\leqslant l$ and $j>N$, $\tilde{\varphi}$ can be extended to a homomorphism $\varphi\colon A_{0}\rightarrow L$ so that $\varphi(\delta_{1}^{j}a_{i})=0$ for every $1\leqslant i\leqslant l$ and $j>N$.

Consider a Taylor homomorphism $T_{\varphi,0}\colon A_{0}\to L[[z]]$ with respect to $\delta_{1}$. Since $\varphi$ was a $\Delta_{0}$-homomorphism, $T_{\varphi,0}$ is a $\Delta$-homomorphism. It remains to observe that the kernel of $T_{\varphi,0}$ contains $\delta_{1}^{N+1}a_{1},\ldots,\delta_{1}^{N+1}a_{l}$ contradicting to the fact that $A_{0}$ is $\Delta$-simple. ∎