2.1 Proof that Theorem 1.4 implies Theorem 1.3

In the statement of Theorem 1.4, choose $X=\overline {\mathbb {Q}}^m$ and $\mathbb {K}=\mathbb {Q}^*+i\mathbb {Q}$ . Write $S=\{u_1,u_2,\ldots \}$ and $\overline {\mathbb {Q}}^m/S=\{v_1,v_2,\ldots \}$ (one of them may be finite) and define

$$ \begin{align*} E_u:=\begin{cases} \overline{\mathbb{Q}}& \mbox{if } u\in S,\\ \mathbb{K}\cdot\pi^n & \mbox{if } u=v_n. \end{cases} \end{align*} $$

By Theorem 1.4, there exist uncountably many transcendental entire functions

$$ \begin{align*} f(z_1,\ldots,z_m)=\sum_{k_1\geq0,\ldots,k_m\geq0}c_{k_1,\ldots,k_m}z_1^{k_1}\cdots z_m^{k_m} \end{align*} $$

in $\mathbb {K}[[z_1,\ldots ,z_m]]$ such that $f(u)\in E_{u}$ for all $u\in \overline {\mathbb {Q}}^m$ . Define $\psi (z_1,\ldots ,z_m)$ as

$$ \begin{align*} \psi(z_1,\ldots,z_m):=\frac{f(z_1,\ldots,z_m)+\overline{f(\overline{z_1},\ldots,\overline{z_m})}}{2}. \end{align*} $$

By the properties of the conjugation of power series,

$$ \begin{align*} \psi(z_1,\ldots,z_m)=\sum_{(k_1,\ldots,k_m)\in \mathbb{Z}^m_{\geq 0}}\mathrm{Re}(c_{k_1,\ldots,k_m})z_1^{k_1}\cdots z_m^{k_m} \end{align*} $$

is a transcendental entire function in $\mathbb {Q}[[z_1,\ldots ,z_m]]$ since $\mathrm{Re} (c_{k_1,\ldots ,k_m})$ is rational and nonzero for all $(k_1,\ldots ,k_m)\in \mathbb {Z}^m_{\geq 0}$ by construction. (Here, as usual, $\mathrm{Re} (z)$ denotes the real part of the complex number z.)

Therefore, it suffices to prove that $S_{\psi }=S$ . In fact, since S is closed under complex conjugation, if $u\in S$ , then $\overline {u}\in S$ and thus $f(u)$ and $\overline {f(\overline {u})}$ are algebraic numbers and so is $\psi (u)$ . (Observe also that $f(0,\ldots ,0)=c_{0,\ldots ,0}\in \overline {\mathbb {Q}}$ .) In the case in which $u=v_n$ , for some n, we can distinguish two cases. When $v_n\in \mathbb {R}^{m}$ , then $\psi (u)=\mathrm{Re} (f(v_n))$ is transcendental, since $f(v_n)\in \mathbb {K}\cdot \pi ^n$ . For $v_n\notin \mathbb {R}^m$ , we have $\overline {v_n}=v_l$ for some $l\neq n$ . Thus, there exist nonzero algebraic numbers $\gamma _1, \gamma _2$ such that

$$ \begin{align*} \psi(v_n)=\frac{\gamma_1\pi^n+\gamma_2\pi^l}{2}, \end{align*} $$

which is transcendental, since $\overline {\mathbb {Q}}$ is algebraically closed and $\pi $ is transcendental. In conclusion, $\psi \in \mathbb {Q}[[z_1,\ldots ,z_m]]$ is a transcendental entire function whose exceptional set is S.

2.2 Proof of Theorem 1.4

Let us proceed by induction on m. The case $m=1$ is covered by Lemma 1.2. Suppose that the theorem holds for all positive integers $k\in [1,m-1]$ . That is, if $\mathbb {K}$ is a dense subset of $\mathbb {C}$ , X is a countable subset of $\mathbb {C}^k$ and $E_u$ is a dense subset in $\mathbb {C}$ for each $u\in X$ , then there exist uncountably many transcendental entire functions $f\in \mathbb {K}[[z_1,\ldots ,z_k]]$ such that $f(u)\in E_u$ for all $u\in X$ , for any integer $k\in [1, m-1]$ .

Now, let X be a countable subset of $\mathbb {C}^{m}$ and $E_u$ a fixed dense subset of $\mathbb {C}$ for all $u\in X$ . Without loss of generality, we can assume that $(0,\ldots ,0) \in X$ . In this case, by hypothesis, $\mathbb {K} \cap E_{(0,\ldots ,0)} \neq \emptyset $ . To apply the induction hypothesis, we consider the partition of X given by

$$ \begin{align*} X=\bigcup_{S\in \mathcal{P}_{m}}X_{S}, \end{align*} $$

where $\mathcal {P}_{m}$ denotes the powerset of $[1,m]=\{1,\ldots ,m\}$ and $X_S$ denotes the set of all $z=(z_1,\ldots ,z_{m})$ in $X\subseteq \mathbb {C}^{m}$ such that $z_i\neq 0$ if and only if $i\in S$ . In particular, $X_{\emptyset }=\{(0,\ldots ,0)\}$ and $X_{[1,m]} = X\cap (\mathbb {C} \setminus \{0\})^m$ .

Given $S=\{i_1,\ldots ,i_k\}$ in $\mathcal {Q}_{m}=\mathcal {P}_m \setminus \{\emptyset ,[1,m]\}$ and $z=(z_1,\ldots ,z_{m})$ in $\mathbb {C}^m$ , we denote by $z_{S}$ the element $(z_{i_1},\ldots ,z_{i_k})\in \mathbb {C}^k.$ To simplify the exposition, we will assume that $i_1<\cdots <i_k$ for all $S\in \mathcal {Q}_m$ . Our goal is to show that there exist uncountably many ways to construct a transcendental entire function $f \in \mathbb {K}[[z_1, \ldots , z_m]]$ given by

$$ \begin{align*} f(z_1,\ldots,z_{m})=a_0+\bigg(\sum_{S\in\mathcal{Q}_m}\bigg(\prod_{i\in S}z_i\bigg)f_S(z_S)\bigg)+f^*(z_1,\ldots,z_{m}), \end{align*} $$

where $a_0\in E_{(0,\ldots ,0)}\cap \mathbb {K}$ and, for each $S=\{i_1,\ldots ,i_k\}\in \mathcal {Q}_m$ , the function $f_S:\mathbb {C}^k\to \mathbb {C}$ is a transcendental entire function such that

$$ \begin{align*} f_S(u_S)\in \frac{1}{\alpha_{i_1}\cdots \alpha_{i_k}}\cdot(E_u-\Theta_{S,u}) \end{align*} $$

for all $u=(\alpha _1,\ldots ,\alpha _{m})\in X_S$ with

$$ \begin{align*} \Theta_{S,u}=a_0+\sum_{T\in\mathcal{Q}_m,T\neq S}\bigg(\prod_{i\in T}\alpha_i\bigg)f_T(u_T)\in\mathbb{C}. \end{align*} $$

By the induction hypothesis, $f_S$ exists for all $S\in \mathcal {Q}_m$ (noting that if $E_u$ is a dense subset of $\mathbb {C}$ , then $(\alpha _{i_1}\cdots \alpha _{i_k})^{-1}\cdot (E_u-\Theta _{S,u})$ is also a dense set). Moreover, we want the function $f^*(z_1,\ldots ,z_m)\in \mathbb {K}[[z_1,\ldots ,z_m]]$ to satisfy the condition

(2.1) $$ \begin{align} f^*(u)\in \bigg(E_u-a_0-\sum_{S\in\mathcal{Q}_m}\bigg(\prod_{i\in S}\alpha_i\bigg)f_S(u_S)\bigg) \end{align} $$

for all $u=(\alpha _1,\ldots ,\alpha _m)\in X_{[1,m]}$ , and $f^*(z_1,\ldots ,z_m)=0$ whenever $z_i=0$ for some i with $1\leq i\leq m$ . Under these conditions, it is easy to see that if $S\in \mathcal {Q}_m$ and $u\in X_S$ , then $f^*(u)=0$ and $f(u)\in E_u$ .

To construct the function $f^*:\mathbb {C}^m\to \mathbb {C},$ let us consider an enumeration $\{u_1,u_2,\ldots \}$ of $X_{[1,m]}$ , where we write $u_j=(\alpha _1^{(j)},\ldots ,\alpha _{m}^{(j)})$ . We construct a function $f^*\in \mathbb {K}[[z_1,\ldots ,z_m]]$ given by

$$ \begin{align*} f^*(z_1,\ldots,z_{m})=\sum_{n=m}^{\infty}P_n(z_1,\ldots,z_{m})=\sum_{i_1\geq1,\ldots,i_m\geq1}c_{i_1,\ldots,i_m}z_1^{i_1}\cdots z_m^{i_m}, \end{align*} $$

where $P_n$ is a homogeneous polynomial of degree n and the coefficients $c_{i_1,\ldots ,i_m}\in \mathbb {K}$ will be chosen so that $f^*$ will satisfy the desired conditions.

The first condition is

$$ \begin{align*} |c_{i_1,\ldots,i_m}|<s_{i_1+\cdots+i_m}:=\frac{1}{\binom{i_1+\cdots+i_m-1}{m-1}(i_1+\cdots+i_m)!}, \end{align*} $$

where $c_{i_1,\ldots ,i_n}\neq 0$ for infinitely many m-tuples of integers $i_1\geq 1, \ldots , i_m\geq 1$ . These conditions will be used to guarantee that $f^*$ is an entire function. Let $L(P)$ denote the length of the polynomial $P(z_1,\ldots ,z_m)\in \mathbb {C}[z_1,\ldots ,z_m]$ given by the sum of the absolute values of its coefficients. Since

$$ \begin{align*}|P_n(z_1,\ldots,z_m)|\leq L(P_n)\max\{1,|z_1|,\ldots,|z_m|\}^{n},\end{align*} $$

for all $n\geq m$ and $(z_1,\ldots ,z_m)$ belonging to the open ball $B(0,R)$ ,

$$ \begin{align*} |P_n(z_1,\ldots,z_m)|<\frac{\binom{n-1}{m-1}}{\binom{n-1}{m-1}n!}\max\{1,R\}^{n}=\frac{\max\{1,R\}^{n}}{n!}, \end{align*} $$

since $P_n(z_1,\ldots ,z_m)$ has at most $\binom {n-1}{m-1}$ monomials of degree n. Hence, the series $\sum _{n\geq m}P_n(z_1,\ldots ,z_m)$ converges uniformly in any of these balls. Thus, $f^*$ is a transcendental entire function such that $f^*(0,z_2,\ldots ,z_m)=f^*(z_1,0,z_3,\ldots ,z_m)=f^*(z_1,z_2,\ldots ,0)=0$ .

To obtain the coefficients $c_{i_1,\ldots ,i_m}\in \mathbb {K}$ such that $f^*$ satisfies the condition (2.1), we consider a hyperplane $\pi (n,j)$ for positive integers n and j with $1\leq j \leq n$ , given by

$$ \begin{align*} \pi(n,j): \mu_{n,1}^{(j)}z_1+\cdots+\mu_{n,m}^{(j)}z_m-\lambda_n^{(j)}=0, \end{align*} $$

and such that if $u_j$ , $u_{n+1}$ and the origin are noncollinear, then $\pi (n,j)$ is a hyperplane containing $u_j$ and parallel to the line passing through the origin and the point $u_{n+1},$ and, if $u_j$ , $u_{n+1}$ and the origin are collinear, then $\pi (n,j)$ is a hyperplane containing $u_j$ and perpendicular to the line passing through the origin and the point $u_{n+1}$ . Note that in both cases, $\lambda _n^{(j)}\neq 0$ and $u_{n+1}$ does not belong to any hyperplane $\pi (n,j)$ with $1\leq j\leq n$ .

Now, we define the polynomials $A_0(z_1,\ldots ,z_m):=z_1\cdots z_m$ and

$$ \begin{align*} A_n(z_1,\ldots,z_{m}):=\prod_{j=1}^{n}(\mu_{n,1}^{(j)}z_1+\cdots+\mu_{n,m}^{(j)}z_m-\lambda_n^{(j)}) \end{align*} $$

for all $n\geq 1$ . By the definition of $\pi (n,j)$ , we have $A_n(u_j)=0$ for $1\leq j\leq n$ . Since $u_{n+1}$ and the origin do not belong to $\pi (n,j)$ , we also have $A_n(0,\ldots ,0)\neq 0$ and $A_n(u_{n+1})\neq 0$ for all $n\geq 1$ . Thus, we can define the function

$$ \begin{align*} f_{1,0}^*(z_1,\ldots,z_m):=\delta_{1,0}A_0(z_1,\ldots,z_m)=\delta_{1,0}z_1\cdots z_m \end{align*} $$

such that $\Theta _1+f_{1,0}^*(u_1)\in E_{u_1}$ and $0<|\delta _{1,0}|<s_m/m$ , where

$$ \begin{align*} \Theta_j:=a_0+\sum_{S\in \mathcal{Q}_m}\bigg(\prod_{i\in S}\alpha_i^{(j)}\bigg)f_S(u_{j,S}), \end{align*} $$

and $u_{j,S}=(\alpha _{i_1}^{(j)},\ldots ,\alpha _{i_k}^{(j)})$ for $S=\{i_1,\ldots ,i_k\}$ , for all integers $j\geq 1$ .

Since $\mathbb {K}$ is a dense subset of $\mathbb {C}$ , we can choose $\delta _{1,1}$ such that the coefficient $c_{1,1,\ldots ,1}$ of $z_1\cdots z_m$ in the function

$$ \begin{align*} f_{1,1}^*(z_1,\ldots,z_m):=f_{1,0}^*(z_1,\ldots,z_m)+\delta_{1,1}z_1\cdots z_mA_1^{(1)}(z_1,\ldots,z_m) \end{align*} $$

belongs to $\mathbb {K}$ with $|c_{1,1,\ldots ,1}|<s_m$ . Therefore, we take

$$ \begin{align*} f_1^*(z_1,\ldots,z_m):=f_{1,1}^*(z_1,\ldots,z_m), \end{align*} $$

where $P_1(z_1,\ldots ,z_m)=c_{1,1,\ldots ,1}z_1\cdots z_m$ .

Recursively, we can construct a function $f^*_{n,0}(z_1,\ldots ,z_{m})$ given by

$$ \begin{align*} f^*_{n,0}(z_1,\ldots,z_m):=f^*_{n-1}(z_1,\ldots,z_m)+\delta_{n,0}z_1^nz_2\cdots z_mA_{n-1}(z_1,\ldots,z_m) \end{align*} $$

where we take $\delta _{n,0}\neq 0$ in the ball $B(0,s_{n+m-1}/(n+m-1))$ such that

$$ \begin{align*} \Theta_n+f^*_{n,0}(u_n)\in E_{u_n}. \end{align*} $$

This is possible since $E_{u_n}$ is a dense subset of $\mathbb {C}$ and all coordinates of $u_n$ are nonzero.

Since $\mathbb {K}$ is a dense subset of $\mathbb {C}$ , if we consider the ordering of the monomials of degree $n+m-1$ given by the lexicographical order of the exponents, then we can choose $\delta _{n,l}$ such that the coefficient $c_{j_1,\ldots ,j_m}$ of the lth monomial $z_1^{j_1}\cdots z_m^{j_m}$ in

$$ \begin{align*} f^*_{n,l}(z_1,\ldots,z_m):=f^*_{n,l-1}(z_1,\ldots,z_m)+\delta_{n,l}z_1^{j_1}\cdots z_m^{j_m}A_n(z_1,\ldots,z_m) \end{align*} $$

belongs to $\mathbb {K}$ with $|c_{j_1,\ldots ,j_m}|<s_{n+m-1}$ . Thus, we define

$$ \begin{align*} f^*_n(z_1,\ldots,z_m):=f^*_{n,L}(z_1,\ldots,z_m), \end{align*} $$

where $L=\binom {n+m-2}{m-1}$ is the number of distinct monomials of degree $n+m-1$ . Then $f^*_n(z_1,\ldots ,z_m)$ is a polynomial function such that $c_{j_1,\ldots ,j_m}\in \mathbb {K}$ for every m-tuple $(j_1,\ldots ,j_m)$ such that $j_1+\cdots +j_m\leq n+m-1$ .

Finally, this construction implies that the functions $f^*_n$ converge to a transcendental entire function $f^*\in \mathbb {K}[[z_1,\ldots ,z_m]]$ as $n\to \infty $ such that

$$ \begin{align*}f^*(u_j)=f^*_n(u_j)=f^*_j(u_j)\end{align*} $$

for all $n\geq j\geq 1$ . Let $f:\mathbb {C}^{m}\to \mathbb {C}$ be the entire function given by

$$ \begin{align*} f(z_1,\ldots,z_{m})=a_0+\bigg(\sum_{S\in\mathcal{Q}_m}\bigg(\prod_{i\in S}z_i\bigg)f_S(z_S)\bigg)+f^*(z_1,\ldots,z_{m}). \end{align*} $$

Then $f(u)\in E_u$ for all $u\in X\subset \mathbb {C}^{m}$ . Since f is an entire function that is not a polynomial, it follows that f is transcendental. Note that there are uncountably many ways to choose the constants $\delta _{n,j}$ . This completes the proof.