Algebraic independence of certain values implies algebraic independence of functions?

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It is quite general question.

Is it possible that some holomorphic functions $f_1,\cdots,f_m$ on a region $\Omega$ of $\mathbb C$ satisfy:

Whenever $(f_1(z), \cdots, f_m (z))$ is a zero of some polynomial $p \in \mathbb Q [x_1, \cdots, x_m]$ for some $z \in \Omega $, then $p (f_1,\cdots,f_m)=0$.

Constant functions satisfy this property obviously, so I wonder the existence of non-constant maps of certain property.

And what about 'continuous functions', not 'holomorphic'?

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$m$ functions $f_1,...,f_m\colon D\subseteq\mathbb{C}\to\mathbb{C}$ are algebraically dependent iff $f_1(z),...,f_m(z)$ are algebraically dependent for all $z\in D$, algebraically independent otherwise.

$m$ functions $f_1,...,f_m\colon D\subseteq\mathbb{C}\to\mathbb{C}$ are algebraically independent iff $f_1(z),...,f_m(z)$ are algebraically independent for a $z\in D$, algebraically dependent otherwise.

That has nothing to do with holomorphic or continuous.

In the general case, algebraic dependence of $f_1(z),...,f_m(z)$ for only some $z\in D$ is not sufficient for algebraic dependence of $f_1,...,f_m$.

But your requirement is met by constant functions and by functions where $\Omega$ is a singleton.