An extension result for maps admitting an algebraic addition theorem

TL;DR

This paper extends Weierstrass's result from one dimension to multiple dimensions, showing that maps with an algebraic addition theorem can be extended to meromorphic maps with a rational addition theorem.

Contribution

It generalizes Weierstrass's extension theorem for algebraic addition theorems from one-dimensional to multi-dimensional complex maps.

Findings

Existence of a meromorphic extension with algebraic addition theorem

Extension to higher dimensions beyond Weierstrass's original result

The extended map admits a rational addition theorem

Abstract

We prove that if an analytic map $f:=(f_1,\ldots ,f_n):U\subset \mathbb{C}^n\rightarrow \mathbb{C}^n$ admits an algebraic addition theorem then there exists a meromorphic map $g:=(g_1,\ldots ,g_n):\mathbb{C}^n\rightarrow \mathbb{C}^n$ admitting an algebraic addition theorem such that $f_1,\ldots ,f_n$ are algebraic over $\mathbb{C}(g_1,\ldots ,g_n)$ on $U$ (this was proved by K. Weierstrass in dimension $1$). Furthermore, $(g_1,\ldots ,g_n)$ admits a rational addition theorem.

Equations

$$\mathbb{K}({\varphi }_1(u),\dots ,{\varphi }_n(u),{\varphi }_1(v),\dots ,{\varphi }_n(v)).$$

$$\begin{array}[]{l}\phi_{(u,v)}:=\big{(}\phi_{1}(u),\ldots,\phi_{m}(u),\phi_{1}(v),\ldots,\phi_{m}(v)\big{)}\in\mathcalboondox{M}_{\mathbb{K},2n}^{2m}.\\ \phi_{u+v}:=\big{(}\phi_{1}(u+v),\ldots,\phi_{m}(u+v)\big{)}\in\mathcalboondox{M}_{\mathbb{K},2n}^{m}.\\ \phi_{u+a}:=\big{(}\phi_{1}(u+a),\ldots,\phi_{m}(u+a)\big{)}\in\mathcalboondox{M}_{\mathbb{K},n}^{m}.\end{array}$$

$$W\subset \{a\in U_{\mathbb{K},n}(ϵ)|\varphi (a)\in {\mathbb{K}}^n\}$$

$$U:=\{a\in W|P(X_1,\dots ,X_n,\varphi (a);Y)\ne 0\}.$$

$$\{a\in U_{\mathbb{K},n}(ϵ)|P(\varphi (u),\varphi (a);Y)\in \text{mathcalboondox}{M}_{\mathbb{K},n+1}\text{ and }P(\varphi (u),\varphi (a);Y)=0\}$$

$$P_a(X_1,\dots ,X_n;Y)=\sum _{i,\mu \le N}{b}_{i,\mu ,a}{X}_1^{{\mu }_1}\dots X_n^{{\mu }_n}{Y}^i$$

$$E(P_a):=\sum _{i,\mu \le N}\Vert b_{i,\mu ,a}{\Vert }^2.$$

$$Q_a(X_1,\dots ,X_n;Y):=\sum _{i,\mu \le N}{c}_{i,\mu ,a}{X}_1^{{\mu }_1}\dots X_n^{{\mu }_n}{Y}^i,$$

$$c_{i,\mu ,a}:=\frac{b_{i,\mu ,a}}{\sqrt{E(P_a)}}.$$

$$\vec{v}(a):=(c_{i,\mu ,a}{)}_{i,\mu \le N}\in \{z\in {\mathbb{K}}^{(N+1{)}^{(n+1)}}|\Vert z\Vert =1\}.$$

$$\sum _{i,\mu \le N}{c}_{i,\mu ,a_k}{\varphi }_1(u{)}^{{\mu }_1}\dots {\varphi }_n(u{)}^{{\mu }_n}f(u,a_k{)}^i=0.$$

$$\sum _{i,\mu \le N}{c}_{i,\mu ,a_k}{\varphi }_1(u{)}^{{\mu }_1}\dots {\varphi }_n(u{)}^{{\mu }_n}\alpha (u,a_k{)}^i\beta (u,a_k{)}^{N-i}=0.$$

$$c_{i,\mu ,a}:=\underset{k\to \infty }{\lim }{c}_{i,\mu ,a_k}.$$

$$\sum _{i,\mu \le N}{c}_{i,\mu ,a}{\varphi }_1(u{)}^{{\mu }_1}\dots {\varphi }_n(u{)}^{{\mu }_n}\alpha (u,a{)}^i\beta (u,a{)}^{N-i}=0.$$

$$\sum _{i,\mu \le N}{c}_{i,\mu ,a}{\varphi }_1(u{)}^{{\mu }_1}\dots {\varphi }_n(u{)}^{{\mu }_n}f(u,a{)}^i=0$$

$$Q_a(X_1,\dots ,X_n;Y):=\sum _{i,\mu \le N}{c}_{i,\mu ,a}{X}_1^{{\mu }_1}\dots X_n^{{\mu }_n}{Y}^i$$

$$A_j(u,v)=A_j(u+a,v-a),\text{ for all }a\in U_{\mathbb{K},n}({ϵ}^{\prime }).$$

$$P_0(X)=X^{{\ell }_0+1}+\sum _{j=0}^{{\ell }_0}{A}_{0,j}(u,v)X^j$$

$$Q_0(X):=X^{{\ell }_0+1}+\sum _{j=0}^{{\ell }_0}{A}_{0,j}(u,v)X^j-X^{{\ell }_0+1}-\sum _{j=0}^{{\ell }_0}{A}_{0,j}(u+a_1,v-a_1)X^j$$

$$P_1(X)=X^{{\ell }_1+1}+\sum _{j=0}^{{\ell }_1}{A}_{1,j}(u,v)X^j$$

$${\mathcal{S}}_k:={\mathcal{S}}_{k-1}\cup \{a+a_k|a\in {\mathcal{S}}_{k-1}\}\cup \{a-a_k|a\in {\mathcal{S}}_{k-1}\},$$

$$U\subset \{a\in U_{\mathbb{K},n}({ϵ}^{\prime })|\varphi (a+c)\in {\mathbb{K}}^n\text{ for all }c\in \mathcal{C}\}$$

$$U\subset \{a\in U_{\mathbb{K},n}({ϵ}^{\prime })|A_0(u,a),\dots ,A_N(u,a)\in \text{mathcalboondox}{M}_{\mathbb{K},n}\}.$$

$$V:=\{a\in U_{\mathbb{K},n}({ϵ}^{\prime })|\Vert a-b\Vert <{ϵ}^{\prime \prime }\}\subset U.$$

$$A_j(u,a+b)=A_j(u+a,b)\text{ for all }a\in U_{\mathbb{K},n}({ϵ}^{\prime \prime }).$$

$${\mathbb{L}}_1:=\mathbb{K}((B_j{)}_{u+a}|a\in U_{\mathbb{K},n}({ϵ}^{\prime \prime }),0\le j\le N).$$

$${\mathbb{L}}_2:=\mathbb{K}((B_j{)}_{-u+a}|a\in U_{\mathbb{K},n}({ϵ}^{\prime \prime }),0\le j\le N).$$

$$\mathbb{L}:=\mathbb{K}((B_j{)}_{u+a},(B_j{)}_{-u+a}|a\in U_{\mathbb{K},n}({ϵ}^{\prime \prime }),0\le j\le N),$$

$$\mathbb{L}\subset \mathbb{K}({\varphi }_{u+c},{\varphi }_{-u+c}|c\in \mathcal{C})$$

$${\mathbb{L}}_1\subset \mathbb{K}({\varphi }_{u+c}|c\in \mathcal{C})$$

Full text

All the authors supported by Spanish GAAR MTM2011-22435 and MTM2014-55565. Second author also supported by a grant of the International Program of Excellence in Mathematics at Universidad Autónoma de Madrid.

1. Introduction

The aim of this paper is to study maps admitting an algebraic addition theorem, maps whose coordinate functions can be viewed as limitting (degenerate) cases of abelian functions. Let $\mathbb{K}$ be $\mathbb{C}$ or $\mathbb{R}$ and $\mathcalboondox{M}_{\mathbb{K},n}$ be the quotient field of $\mathcal{O}_{\mathbb{K},n}$, the ring of power series in $n$ variables with coefficients in $\mathbb{K}$ that are convergent in a neighborhood of the origin.

Definition. Let $u$ and $v$ be variables of $\mathbb{C}^{n}$. We say $(\phi_{1},\ldots,\phi_{n})\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admits an algebraic addition theorem (AAT) if $\phi_{1},\ldots,\phi_{n}$ are algebraically independent over $\mathbb{K}$ and if each $\phi_{i}(u+v)$, $i=1,\ldots,n$, is algebraic over

[TABLE]

The concept of AAT was introduced by K. Weierstrass during his lectures on abelian functions in Berlin in $1870$ (see [12]). He stated that the coordinate functions of a global meromorphic map admitting an AAT are either abelian functions or degenerate abelian functions. He proved it for dimension $1$ and F. Severi in [10] (see also Y. Abe [1, 2]) for dimension $n$. Weierstrass also proved the following extension result: the germ of an analytic function admitting an AAT can be transformed algebraically into the germ of a global function admitting an AAT; and he stated, without a proof, an $n$-dimensional version of it. As far as we know, no such proof existed in the literature so far. We prove it here as a consequence (Corollary 2) of the main result of the paper, that we now state.

Theorem 1** (Extension Theorem).**

Let $\phi:=(\phi_{1},\ldots,\phi_{n})\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admit an AAT. Then, there exist $\psi:=(\psi_{1},\ldots,\psi_{n})\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admitting an AAT and algebraic over $\mathbb{K}(\phi)$, and an additional meromorphic series $\psi_{0}\in\mathcalboondox{M}_{\mathbb{K},n}$ algebraic over $\mathbb{K}(\psi)$ such that,

1. $(1)$

For each f(u)\in\mathbb{K}\big{(}\psi_{0}(u),\ldots,\psi_{n}(u)\big{)},

1. $(a)$

f(u+v)\in\mathbb{K}\big{(}\psi_{0}(u),\ldots,\psi_{n}(u),\psi_{0}(v),\ldots,\psi_{n}(v)\big{)}* and* 2. $(b)$

f(-u)\in\mathbb{K}\big{(}\psi_{0}(u),\ldots,\psi_{n}(u)\big{)}. 2. $(2)$

Each $\psi_{0},\ldots,\psi_{n}$ is the quotient of two convergent power series whose complex domain of convergence is $\mathbb{C}^{n}$.

Corollary 2**.**

Any $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admitting an AAT is algebraic over $\mathbb{K}(\psi)$ for some $\psi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admitting an AAT and whose coordinate functions are the quotient of two convergent power series whose complex domain of convergence is $\mathbb{C}^{n}$.

We point out that this theorem gives not only an extension result, but also a uniform rational version of the AAT. In fact, given a $\phi$ admitting an AAT, we obtain the rational version in Theorem 1 (1a) through the coefficients of the polynomial asssociated to each $\phi_{i}(u+v)$. Then, we obtain the extension result of Theorem 1 (2) by considering the rational expression obtained in Theorem 1 (1a). In particular, this shows that any $\phi$ admitting an AAT can be analytically extended to a multivalued analytic map with a finite number of branches. Thus, we provide a new way of proving Weierstrass’ extension result in dimension $1$, whose classical proofs go the other way around (and do not provide a rational counterpart): first it is proved the finiteness of the number of branches of the extension of such analytic $\phi$ and then, making use of the coefficients of the relevant polynomials, it is given a global univaluated meromorphic function admitting an AAT.

The motivation of the results of this paper is to study abelian locally $\mathbb{K}$-Nash groups, for $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Charts at the identity of such groups admit an AAT. Locally Nash groups (i.e. for $\mathbb{K}=\mathbb{R}$) were studied by J.J. Madden and C.M. Stanton [8] and M. Shiota [11], mainly in dimension $1$. In particular, the Extension Theorem will allow us to reduce the study of simply connected abelian locally Nash groups to those whose charts are restrictions of (global) meromorphic functions admitting an AAT (see [3]).

The results of this paper are part of the second author’s Ph.D. dissertation.

2. The Extension Theorem.

For each $\epsilon>0$, let $U_{\mathbb{K},n}(\epsilon):=\{a\in\mathbb{K}^{n}\;|\;\|a\|<\epsilon\}$. We will only consider convergence over open subsets of $\mathbb{C}^{n}$, let $U_{n}(\epsilon):=U_{\mathbb{C},n}(\epsilon)$. We say that $(\phi_{1},\ldots,\phi_{m})\in\mathcalboondox{M}_{\mathbb{K},n}^{m}$ is convergent in $U_{n}(\epsilon)$ if each $\phi_{1},\ldots,\phi_{m}$ is the quotient of two power series convergent on $U_{n}(\epsilon)$.

As usual, by the identity principle for analytic functions, we identify $\mathcal{O}_{\mathbb{K},n}$ with the ring of germs of analytic functions at [math], and $\mathcalboondox{M}_{\mathbb{K},n}$ with its quotient field. We will use without mention properties of $\mathcal{O}_{\mathbb{K},n}$, see e.g. R.C. Gunning and H. Rossi [6] and J.M. Ruiz [9].

Let $\epsilon>0$. Let $\phi:=(\phi_{1},\ldots,\phi_{m})\in\mathcalboondox{M}_{\mathbb{K},n}^{m}$ be convergent on $U_{n}(\epsilon)$, let $a\in U_{\mathbb{K},n}(\epsilon)$ and let $(u,v):=(u_{1},\ldots,u_{n},v_{1},\ldots,v_{n})$ be a $2n$-tuple of variables. We will use the following notation:

[TABLE]

Given $\phi\in\mathcalboondox{M}_{\mathbb{K},p}^{n}$ and $\psi\in\mathcalboondox{M}_{\mathbb{K},p}^{m}$ we say that the tuple $\phi$ is algebraic over $\mathbb{K}(\psi):=\mathbb{K}(\psi_{1},\ldots,\psi_{m})$ if each component, $\phi_{1},\ldots,\phi_{n}$, is algebraic over $\mathbb{K}(\psi)$.

Thus, $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admits an algebraic addition theorem (AAT) if $\phi_{1},\ldots,\phi_{n}$ are algebraically independent over $\mathbb{K}$ and $\phi_{u+v}$ is algebraic over $\mathbb{K}(\phi_{(u,v)})$.

Note that if $\phi\in\mathcalboondox{M}_{\mathbb{R},n}$ admits an AAT then $\phi$ also admits an AAT when considered as an element of $\mathcalboondox{M}_{\mathbb{C},n}$.

We first prove two properties of maps admitting an AAT.

Lemma 3**.**

Let $\epsilon>0$ and let $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ be convergent on $U_{n}(\epsilon)$. If $\phi$ admits an AAT then $\phi_{u+a}$ is algebraic over $\mathbb{K}(\phi)$, for each $a\in U_{\mathbb{K},n}(\epsilon)$.

Proof.

Fix $j\in\{1,\ldots,n\}$ and let $f(u,v):=\phi_{j}(u+v)$. By hypothesis, there exists $P\in\mathbb{K}[X_{1},\ldots,X_{2n}][Y]$ such that $P(\phi(u),\phi(v);Y)\neq 0$ and $P(\phi(u),\phi(v);f(u,v))=0$. For any $a\in U_{\mathbb{K},n}(\epsilon)$ such that $P(\phi(u),\phi(a),Y)$ is not identically zero, we clearly obtain that $f(u,a)$ is algebraic over $\mathbb{K}(\phi)$. We have to consider those $a\in U_{\mathbb{K},n}(\epsilon)$ such that $P(\phi(u),\phi(a);Y)$ is identically zero.

We first check that there exists an open dense subset $U$ of $U_{\mathbb{K},n}(\epsilon)$ such that for each $a\in U$, $P(X_{1},\ldots,X_{n},\phi(a);Y)\in\mathbb{K}[X_{1},\ldots,X_{n}][Y]$ is a non-zero polynomial. Let $W$ be an open dense subset of $U_{\mathbb{K},n}(\epsilon)$ such that

[TABLE]

and $\phi:W\rightarrow\mathbb{K}^{n}$ is analytic. Let

[TABLE]

Since $W$ is an open dense subset of $U_{\mathbb{K},n}(\epsilon)$, it is enough to show that $W\setminus U$ is closed and nowhere dense in $W$. Clearly $W\setminus U$ is closed in $W$ because $\phi$ is continuous in $W$. To prove the density, we note that if $W\setminus U$ contains an open subset of $W$ then

[TABLE]

contains an open subset of $U_{\mathbb{K},n}(\epsilon)$ and therefore $P(\phi(u),\phi(v);Y)=0$, a contradiction.

To finish the proof we will show that for each $a\in U_{\mathbb{K},n}(\epsilon)$, there exists $Q_{a}\in\mathbb{K}[X_{1},\ldots,X_{n}][Y]$ such that $Q_{a}(\phi(u);Y)$ is not identically zero and $Q_{a}(\phi(u);f(u,a))=0$. We follow the proof of [5, Ch. IX. §5. Theorem 5]. For each $a\in U$, where $U$ is as above, let

[TABLE]

denote the polynomial $P(X_{1},\ldots,X_{n},\phi(a);Y)$. We have that $U$ is dense in $U_{\mathbb{K},n}(\epsilon)$ and $P_{a}\neq 0$ for all $a\in U$. For each $a\in U$, we define

[TABLE]

We note that $E(P_{a})>0$, for all $a\in U$. For each $a\in U$, let

[TABLE]

where

[TABLE]

Hence, for each $a\in U$, we have that $Q_{a}(\phi(u);Y)$ is not identically zero, $Q_{a}(\phi(u);f(u,a))=0$ and $E(Q_{a})=1$. We define

[TABLE]

Take $a\in U_{\mathbb{K},n}(\epsilon)\setminus U$. Since $U$ is an open dense subset of $U_{\mathbb{K},n}(\epsilon)$, there exists a sequence $\{a_{k}\}_{k\in\mathbb{N}}\subset U$ that converges to $a$. For each $a_{k}$, the identity $Q_{a_{k}}(\phi(u);f(u,a_{k}))=0$ holds, therefore

[TABLE]

By hypothesis there are $\alpha,\beta\in\mathcal{O}_{\mathbb{K},2n}$, $\beta\neq 0$, convergent on $U_{2n}(\epsilon)$, such that $f(u,v)=\frac{\alpha(u,v)}{\beta(u,v)}$ and $\beta(u,a)\neq 0$ for all $a\in U_{\mathbb{K},n}(\epsilon)$. In particular

[TABLE]

Since $\{z\in\mathbb{K}^{(N+1)^{(n+1)}}\;|\;\|z\|=1\}$ is compact, taking a suitable subsequence we can assume that the sequence $\{\vec{v}(a_{k})\}_{k\in\mathbb{N}}$ is convergent. For each $i,\mu\leq N$, we define

[TABLE]

Since $\alpha$ and $\beta$ are continuous, when $k$ tends to infinity equation (2.1) becomes

[TABLE]

So dividing by $\beta(u,a)^{N}$, we also have

[TABLE]

and hence the polynomial

[TABLE]

satisfies $Q_{a}(\phi(u),f(u,a))=0$. We note that $E(Q_{a})=\lim_{k\rightarrow\infty}E(Q_{a_{k}})=1$, so $Q_{a}\neq 0$. Since $\phi_{1},\ldots,\phi_{n}$ are algebraically independent over $\mathbb{K}$ and $Q_{a}(X_{1},\ldots,X_{n},Y)\neq 0$, we have $Q_{a}(\phi(u),Y)$ is not identically zero. ∎

Lemma 4**.**

Let $\phi,\psi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ and suppose that $\phi$ is algebraic over $\mathbb{K}(\psi)$. If $\phi$ admits an AAT then $\psi$ admits an AAT. The converse is also true, provided $\phi_{1},\ldots,\phi_{n}$ are algebraically independent over $\mathbb{K}$.

Proof.

Assume that $\phi$ admits an AAT, hence $\psi_{1},\ldots,\psi_{n}$ are algebraically independent over $\mathbb{K}$ because $\phi$ is algebraic over $\mathbb{K}(\psi)$. To check that $\psi_{u+v}$ is algebraic over $\mathbb{K}(\psi_{(u,v)})$ it is enough to show that $\psi_{u+v}$ is algebraic over $\mathbb{K}(\phi_{u+v})$, $\phi_{u+v}$ is algebraic over $\mathbb{K}(\phi_{(u,v)})$ and $\phi_{(u,v)}$ is algebraic over $\mathbb{K}(\psi_{(u,v)})$. The three conditions above are trivially satisfied because $\phi$ admits an AAT and both $\phi$ is algebraic over $\mathbb{K}(\psi)$ and $\psi$ is algebraic over $\mathbb{K}(\phi)$. The converse follows by symmetry because if $\phi_{1},\ldots,\phi_{n}$ are algebraically independent over $\mathbb{K}$ then $\psi$ is algebraic over $\mathbb{K}(\phi)$. ∎

Now, we adapt to our context a result on AAT due to H.A.Schwarz, see [7, Ch. XXI. Art. 389] for details.

Lemma 5**.**

Let $\epsilon>0$ and let $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ be convergent on $U_{n}(\epsilon)$ such that it admits an AAT. Then, there exist a finite subset $\mathcal{C}\subset U_{\mathbb{K},n}(\epsilon)$, with $0\in\mathcal{C}$ and $\mathcal{C}=-\mathcal{C}$, and $\epsilon^{\prime}\in(0,\epsilon]$ satisfying: each element of $\mathbb{K}(\phi_{u+a}\;|\;a\in\mathcal{C})$ is convergent on $U_{n}(2\epsilon^{\prime})$, and there exist $A_{0},\ldots,A_{N}\in\mathbb{K}(\phi_{(u+a,v+a)}\;|\;a\in\mathcal{C})$ convergent on $U_{2n}(2\epsilon^{\prime})$ such that $\phi_{u+v}$ is algebraic over $\mathbb{K}(A_{0},\ldots,A_{N})$ and, for each $j\in\{0,\ldots,N\}$,

[TABLE]

Proof.

Fix $i\in\{1,\ldots,n\}$. Let $\mathcal{S}_{0}:=\{0\}$ and $\mathbb{K}_{0}:=\mathbb{K}(\phi_{(u,v)})$. Let

[TABLE]

be the minimal polynomial of $\phi_{i}(u+v)$ over $\mathbb{K}_{0}$. If each $A_{0,j}$ satisfies equation (2.2) for $\epsilon^{\prime}=2^{-1}\epsilon$ then we are done for this $i$ letting $\epsilon^{\prime}:=2^{-1}\epsilon$, $\mathcal{C}:=\mathcal{S}_{0}$ and $A_{j}:=A_{0,j}$, for each $0\leq j\leq\ell_{0}$. Otherwise, there exists $a_{1}\in U_{\mathbb{K},n}(2^{-1}\epsilon)$ such that

[TABLE]

is not zero. Since $u+v=(u+a_{1})+(v-a_{1})$, we deduce that $\phi_{i}(u+v)$ is a root of $Q_{0}(X)$. Let $\mathcal{S}_{1}:=\mathcal{S}_{0}\cup\{a_{1},-a_{1}\}$ and $\mathbb{K}_{1}:=\mathbb{K}(\phi_{u+a,v+a}\;|\;a\in\mathcal{S}_{1})$. By definition $\mathbb{K}_{0}\subset\mathbb{K}_{1}$. Let

[TABLE]

be the minimal polynomial of $\phi_{i}(u+v)$ over $\mathbb{K}_{1}$. We note that the elements of $\mathbb{K}_{1}$ are convergent on $U_{2n}(2^{-1}\epsilon)$. If each $A_{1,j}$ satisfies equation (2.2) for $\epsilon^{\prime}=2^{-2}\epsilon$ then we are done for this $i$ letting $\epsilon^{\prime}:=2^{-2}\epsilon$, $\mathcal{C}:=\mathcal{S}_{1}$ and $A_{j}:=A_{1,j}$, for each $0\leq j\leq\ell_{1}$. Otherwise, we can repeat the process to obtain sets $\mathcal{S}_{2}$, $\mathcal{S}_{3}$ and so on, where the set $\mathcal{S}_{k}$ is obtained from the set $\mathcal{S}_{k-1}$ as

[TABLE]

for some $a_{k}\in U_{\mathbb{K},n}(2^{-k}\epsilon)$ such that $Q_{k-1}$ is not [math]. Similarly, we obtain $\mathbb{K}_{k}:=\mathbb{K}(\phi_{u+a,v+a}\;|\;a\in\mathcal{S}_{k})$ whose elements are convergent on $U_{2n}(2^{-k}\epsilon)$. Since in the $k$ repetition the degree of $P_{k}$ is smaller than that of $P_{k-1}$, this process eventually stops, say at step $s$. Letting $\epsilon^{\prime}:=2^{-s-1}\epsilon$, $\mathcal{C}:=\mathcal{S}_{s}$ and $A_{j}:=A_{s,j}$, for each $0\leq j\leq\ell_{s}$, we are done for this $i$. The elements $A_{0},\ldots,A_{\ell_{s}}$ are convergent on $U_{2n}(2\epsilon^{\prime})$ because they are elements of $\mathbb{K}_{s}$.

For each $i$, $1\leq i\leq n$, denote by $\epsilon^{\prime}_{i}$, $\mathcal{C}_{i}$ and $A_{0}^{i},\ldots,A_{N_{i}}^{i}$ the elements $\epsilon^{\prime}$, $\mathcal{C}$ and $A_{1},\ldots,A_{\ell_{s}}$ previously obtained for that choice of $i$. To complete the proof, take $\mathcal{C}:=\bigcup_{i}\mathcal{C}_{i}$, $\epsilon^{\prime}:=\min_{i}\{\epsilon^{\prime}_{i}\}$, and let $\{A_{0},\ldots,A_{N}\}$ be the union of the sets $\{A_{0}^{i},\ldots,A_{N_{i}}^{i}\}$. ∎

We need two additional lemmas before proving the Extension Theorem.

Lemma 6**.**

Let $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admit an AAT. Then, $\phi(-u)$ is algebraic over $\mathbb{K}(\phi(u))$.

Proof.

Take $\epsilon>0$ such that $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ is convergent on $U_{n}(\epsilon)$. Since $\phi$ admits an AAT, we know that $\phi(u+v)$ is algebraic over $\mathbb{K}(\phi(u),\phi(v))$. Taking into account transcendence degrees, it follows that $\phi(v)$ is algebraic over $\mathbb{K}(\phi(u+v),\phi(u))$. For some $a\in U_{\mathbb{K},n}(\epsilon)$, we may substitute $v$ by $-u+a$, so $\phi(-u+a)$ is algebraic over $\mathbb{K}(\phi(u))$. By Lemma 3, $\phi(-u)$ is algebraic over $\mathbb{K}(\phi(-u+a))$ and hence over $\mathbb{K}(\phi(u))$. ∎

Lemma 7**.**

Let $\epsilon>0$. Let $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ be convergent on $U_{n}(\epsilon)$ such that it admits an AAT. Then there exist $\epsilon_{1}\in(0,\epsilon]$ and $\Psi:=(\psi_{0},\ldots,\psi_{n})\in\mathcalboondox{M}_{\mathbb{K},n}^{n+1}$ convergent on $U_{n}(\epsilon_{1})$ and algebraic over $\mathbb{K}(\phi)$ satisfying $\psi:=(\psi_{1},\ldots,\psi_{n})$ admits an AAT, $\psi_{0}$ is algebraic over $\mathbb{K}(\psi)$ and, for each $f\in\mathbb{K}(\Psi)$, $f(-u)\in\mathbb{K}(\Psi(u))$, there exists $\delta\in(0,\epsilon_{1}]$ such that for each $a\in U_{\mathbb{K},n}(\delta)$, $f_{u+a}\in\mathbb{K}(\Psi)$ and $f_{u+a}$ is convergent on $U_{n}(\epsilon_{1})$.

Proof.

We will define a field $\mathbb{L}$ generated over $\mathbb{K}$ by certain elements of $\mathcalboondox{M}_{\mathbb{K},n}$, next we will prove that each $f\in\mathbb{L}$ satisfy the conclusion of the lemma and finally we find a primitive element $\Psi$ such that $\mathbb{L}=\mathbb{K}(\Psi)$.

Let $\epsilon^{\prime}\in(0,\epsilon]$, $\mathcal{C}\subset U_{\mathbb{K},n}(\epsilon)$ and $A_{0},\ldots,A_{N}\in\mathbb{K}(\phi_{(u+c,v+c)}\;|\;c\in\mathcal{C})$ be the ones provided by Lemma 5 for $\phi$. Let $U$ be an open dense subset of $U_{\mathbb{K},n}(\epsilon^{\prime})$ such that

[TABLE]

and

[TABLE]

In particular, $U\subset\{a\in U_{\mathbb{K},n}(\epsilon^{\prime})\;|\;\phi(a)\in\mathbb{K}^{n}\}$ because $0\in\mathcal{C}$. Since $U$ is open there exist $b\in U$ and $\epsilon^{\prime\prime}\in(0,\epsilon^{\prime}-\|b\|]$ such that

[TABLE]

Fix such $b$. Then, for each $a\in U_{\mathbb{K},n}(\epsilon^{\prime\prime})$, each $A_{j}(u,a+b)$, $j=1,\ldots,N$ is an element of $\mathcalboondox{M}_{\mathbb{K},n}$. We note that since each $A_{j}(u,v)$ is convergent on $U_{2n}(2\epsilon^{\prime})$ and by definition of $b$ and $\epsilon^{\prime\prime}$, each $A_{j}(u,a+b)$ is convergent on $U_{n}(\epsilon^{\prime})$, for each $a\in U_{\mathbb{K},n}(\epsilon^{\prime\prime})$. Also, since each $A_{j}$ satisfies the equation (2.2) of Lemma 5,

[TABLE]

For each $j\in\{0,\ldots,N\}$, we define $B_{j}(u):=A_{j}(u,b)$. Let

[TABLE]

Since, for each $a\in U_{\mathbb{K},n}(\epsilon^{\prime\prime})$, each $A_{j}(u,a+b)$ is convergent on $U_{n}(\epsilon^{\prime})$, by equation (2.3) all the elements of $\mathbb{L}_{1}$ are convergent on $U_{n}(\epsilon^{\prime})$ and in particular in $U_{n}(\epsilon^{\prime\prime})$. Let

[TABLE]

Note that all the elements of $\mathbb{L}_{2}$ are also convergent on $U_{n}(\epsilon^{\prime\prime})$. Hence, if we define

[TABLE]

all the elements of $\mathbb{L}$ are also convergent on $U_{n}(\epsilon^{\prime\prime})$.

Let us show that

[TABLE]

and that each element of $\mathbb{L}$ is algebraic over $\mathbb{K}(\phi)$.

We begin proving that

[TABLE]

and that each element of $\mathbb{L}_{1}$ is algebraic over $\mathbb{K}(\phi)$. Fix $j\in\{0,\ldots,N\}$ and $a\in U_{\mathbb{K},n}(\epsilon^{\prime\prime})$. We recall from Lemma 5 that $A_{j}(u,v)$ is convergent on $U_{2n}(2\epsilon^{\prime})$ and $A(u,v)\in\mathbb{K}(\phi_{(u+c,v+c)}\;|\;c\in\mathcal{C})$. Hence we can evaluate $A_{j}(u,v)$ at $v=a+b$ to deduce that $A_{j}(u,a+b)\in\mathbb{K}(\phi_{u+c}\;|\;c\in\mathcal{C})$. Thus, by equation (2.3), $A_{j}(u+a,b)\in\mathbb{K}(\phi_{u+c}\;|\;c\in\mathcal{C})$. Hence, $\mathbb{L}_{1}\subset\mathbb{K}(\phi_{u+c}\;|\;c\in\mathcal{C})$ and therefore, by Lemma 3, each element of $\mathbb{L}_{1}$ is algebraic over $\mathbb{K}(\phi)$. By symmetry of $\mathcal{C}$, $\mathbb{L}_{2}\subset\mathbb{K}(\phi_{-u+c}\;|\;c\in\mathcal{C})$ and each element of $\mathbb{L}_{2}$ is algebraic over $\mathbb{K}(\phi(-u))$. Therefore $\mathbb{L}\subset\mathbb{K}(\phi_{u+c,-u+c}\;|\;c\in\mathcal{C})$ and, since by Lemma 6 we have that $\phi(-u)$ is algebraic over $\mathbb{K}(\phi(u))$, we deduce that each element of $\mathbb{L}$ is algebraic over $\mathbb{K}(\phi(u))$, as required.

Next, we show that $\phi_{1}(u+b),\ldots,\phi_{n}(u+b)$ are algebraically independent over $\mathbb{K}$. Let $P\in\mathbb{K}[X_{1},\ldots,X_{n}]$ be such that $P(\phi_{u+b})=0$. By notation, for each $a\in U_{\mathbb{K},n}(\epsilon^{\prime\prime})$, we have that $P(\phi_{u+b}(a))=0$ if and only if $P(\phi(a+b))=0$. Hence,

[TABLE]

Since $V$ is open in $U_{\mathbb{K},n}(\epsilon)$, $P(\phi)=0$ by the identity principle. Since $\phi_{1},\ldots,\phi_{n}$ are algebraically independent over $\mathbb{K}$, $P=0$ and we are done.

Next, we show that $\mathbb{L}$ is finitely generated over $\mathbb{K}$ and its transcendence degree is $n$. Firstly, we note that $\phi$ is algebraic over $\mathbb{K}(\phi_{u+b})$ because the coordinate functions of $\phi_{u+b}$ are algebraically independent over $\mathbb{K}$ and $\phi_{u+b}$ is algebraic over $\mathbb{K}(\phi)$ by Lemma 3. Since $\phi_{u+v}$ is algebraic over $\mathbb{K}(A_{0},\ldots,A_{N})$, evaluating each $A_{j}(u,v)$ at $v=b$ we deduce that $\phi_{u+b}$ is algebraic over $\mathbb{K}(B_{0},\ldots,B_{N})$. Therefore, $\phi$ is algebraic over $\mathbb{K}(B_{0},\ldots,B_{N})$. On the other hand, $\mathbb{K}(B_{0},\ldots,B_{N})$ is a subset of $\mathbb{K}(\phi_{u+c}\;|\;c\in\mathcal{C})$ and the latter field is algebraic over $\mathbb{K}(\phi)$ by Lemma 3. Hence the three fields have transcendence degree $n$ over $\mathbb{K}$. Recall that $\mathcal{C}=-\mathcal{C}$, so $\mathbb{K}(\phi_{-u+c}\;|\;c\in\mathcal{C})=\mathbb{K}(\phi_{-u-c}\;|\;c\in\mathcal{C})$. We also note that $\phi(-u)$ is algebraic over $\mathbb{K}(\phi(u))$, so $\mathbb{K}(\phi_{u+c},\phi_{-u-c}\;|\;c\in\mathcal{C})$ has transcendence degree $n$ over $\mathbb{K}$. Now, $\mathcal{C}$ is finite and

[TABLE]

therefore, $\mathbb{L}$ is finitely generated over $\mathbb{K}$ and its transcendence degree is $n$.

Fix $f\in\mathbb{L}$ and let us check that $f(-u)\in\mathbb{L}$ and that there exists $\delta>0$ such that for every $a\in U_{\mathbb{K},n}(\delta)$, $f_{u+a}\in\mathbb{L}$ and $f_{u+a}$ is convergent on $U_{n}(\epsilon^{\prime\prime})$.

Since $f\in\mathbb{L}$, there exist $m,m^{\prime}\in\mathbb{N}$, $j(1),\ldots,j(m+m^{\prime})\in\{0,\ldots,N\}$ and $a_{1},\ldots,a_{m+m^{\prime}}\in U_{\mathbb{K},n}(\epsilon^{\prime\prime})$ such that $f$ is a rational function of

[TABLE]

In particular, $f(-u)$ is a rational function of

[TABLE]

so $f(-u)\in\mathbb{L}$. Take $\delta>0$ such that $\delta<\epsilon^{\prime\prime}-\max\{\|a_{1}\|,\ldots,\|a_{m+m^{\prime}}\|\}$. Then, for all $a\in U_{\mathbb{K},n}(\delta)$, $f_{u+a}\in\mathbb{L}$ and $f_{u+a}$ is convergent on $U_{n}(\epsilon^{\prime\prime})$.

Finally, take $\psi_{1},\ldots,\psi_{n}\in\mathbb{L}$ algebraically independent over $\mathbb{K}$ and $\psi_{0}$ algebraic over $\mathbb{K}(\psi_{1},\ldots,\psi_{n})$ such that $\mathbb{L}=\mathbb{K}(\psi_{0},\psi_{1},\ldots,\psi_{n})$. Now, since all the elements of $\mathbb{L}$ are algebraic over $\mathbb{K}(\phi)$, $\psi:=(\psi_{1},\ldots,\psi_{n})$ admits an AAT by Lemma 4. ∎

We now have all the ingredients to prove our main result.

Proof of the Extension Theorem.

Let $\phi:=(\phi_{1},\ldots,\phi_{n})\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admit an AAT. Take $\epsilon>0$ such that $\phi$ is convergent on $U_{n}(\epsilon)$. Applying Lemma 7 we obtain $\epsilon_{1}\in(0,\epsilon]$ and $\Psi:=(\psi_{0},\ldots,\psi_{n})\in\mathcalboondox{M}_{\mathbb{K},n}^{n+1}$ as in the lemma. We next check that this $\Psi$ satisfies the conditions of the theorem.

$(1)$ By Lemma 7, if $f\in\mathbb{K}(\Psi)$ then $f(-u)\in\mathbb{K}(\Psi)$, so we only have to check f(u+v)\in\mathbb{K}\big{(}\Psi_{(u,v)}\big{)}. Fix a non-constant $f\in\mathbb{K}(\Psi)$ and $\delta\in(0,\epsilon_{1}]$ such that $f_{u+a}\in\mathbb{K}(\Psi)$, for each $a\in U_{n}(\delta)$, as in Lemma 7. Let $0<\varepsilon<\delta$ be such that $f_{u+v}$ is convergent on $U_{2n}(\varepsilon)$. Let $U$ be an open connected subset of $U_{n}(\varepsilon)$ such that $\Psi(u)$ is analytic on $U$. In particular, $\Psi_{(u,v)}$ is analytic on $U\times U$. On the other hand, if for each $a\in U$ we have that $g(u,v):=f(u+v)$ is not analytic in $(a,a)$ then we would deduce that $f(u)$ is not analytic on an open subset of $U_{n}(\varepsilon)$, a contradiction. Therefore, shrinking $U$ we can assume that $g(u,v):=f(u+v)$ is also analytic on $U\times U$. By Lemma 7, we have that $g(u,a)\in\mathbb{K}(\Psi(u))$ and $g(a,v)\in\mathbb{K}(\Psi(v))$, for each $a\in U$. Hence, by Bochner [4, Theorem 3], $g(u,v)\in\mathbb{C}(\Psi_{(u,v)})$ on $U\times U$. Since $U\times U$ is an open subset of $U_{2n}(\varepsilon)$, it follows that $g(u,v)\in\mathbb{C}(\Psi_{(u,v)})$ on $U_{2n}(\varepsilon)$. Moreover, clearly $g(u,v)\in\mathbb{K}(\Psi_{(u,v)})$ on $U_{2n}(\varepsilon)$ since both $\Psi\in\mathcalboondox{M}_{\mathbb{K},n}^{n+1}$ and $f\in\mathbb{K}(\Psi)$. This concludes the proof of $(1)$.

$(2)$ We may assume that $\psi_{0}\neq 0$. Fix $i\in\{0,\ldots,n\}$. We have already shown that $\psi_{i}(u+v)\in\mathbb{K}(\Psi_{(u,v)})$. Let $A(u,v):=\psi_{i}(u+v)$. By Lemma 7 and taking a smaller $\epsilon>0$ if necessary, we may assume that $\Psi$ is convergent on $U_{n}(\epsilon)$ and $\mathbb{K}(\Psi_{u+a})\subset\mathbb{K}(\Psi)$, for each $a\in U_{\mathbb{K},n}(\epsilon)$. Let us show that there exists $c\in U_{\mathbb{K},n}(\epsilon)$ such that

[TABLE]

Take $\alpha,\beta\in\mathcal{O}_{\mathbb{K},2n}$, $\beta\neq 0$ such that $A(u,v)=\frac{\alpha(u,v)}{\beta(u,v)}$. Suppose by contradiction that $\beta(u+c,u-c)=0$ for all $c\in U_{\mathbb{K},n}(\epsilon)$. Then

[TABLE]

for all $a,b\in U_{\mathbb{K},n}(\epsilon/2)$. So $\beta(a,b)=0$, for all $(a,b)\in U_{\mathbb{K},n}(\epsilon/2)$, that is, $\beta=0$, which is a contradiction. Consequently,

[TABLE]

By induction we deduce that

[TABLE]

for each $N\in\mathbb{N}$. Hence since $\Psi(2^{-N}u)$ is convergent on $U_{n}(2^{N}\epsilon)$, $\Psi$ is also convergent on $U_{n}(2^{N}\epsilon)$. Thus each $\psi_{i}$ is the quotient of two power series convergent in all $\mathbb{C}^{n}$ (by Poincaré’s problem [6, Ch. VIII, §B, Corollary 10]). ∎

Proof of Corollary 2.

Let $\phi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admit an AAT. By Theorem 1, there exists $\psi\in\mathcalboondox{M}_{\mathbb{K},n}^{n}$ admitting an AAT whose coordinate functions are the quotient of two convergent whose complex domain of convergence is $\mathbb{C}^{n}$ and such that $\psi$ is algebraic over $\mathbb{K}(\phi)$. Since the coordinate functions of $\psi$ are algebraically independent, $\phi$ is algebraic over $\mathbb{K}(\psi)$. ∎

Acknowledgements

The second author thanks E. Pantelis for the support to attend “Summer School in Tame Geometry”, Konstanz, July 18-23, 2016, where the results of this paper were presented. The authors also would like to thank José F. Fernando for helpful suggestions on an earlier version of this paper and Mark Villarino for his comments.