0
$\begingroup$

It is quite general and elementary question.

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

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-cobstant maps of certain property.

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

I asked the same one at MSE but I didn't get an answer.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

EDITED:

No, it is not possible if $f_1, \ldots, f_m$ are not all constant.

Suppose wlog $f_1$ is not constant. I'll ignore $f_2, \ldots, f_m$ and consider polynomials $p(f_1(z))$. For convenience I'll omit the subscript and call this $p(f(z))$. $f(\Omega)$ is an open subset of $\mathbb C$, so it contains $\alpha + \beta i$ for some $\alpha, \beta \in \mathbb Q$, i.e. there is some $z_0$ such that $f(z_0) = \alpha + \beta i$, and so $p(f(z_0)) = 0$ where $p(w) = (w-\alpha)^2 + \beta^2$.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Adjusting the constant term in $p$ may not preserve the rationality of its coefficients. Or do I misunderstand something? $\endgroup$
    – Yaakov Baruch
    Commented Oct 5, 2018 at 2:31
  • 1
    $\begingroup$ But here p is defined over $\mathbb Q $. How can we take $z_0$ that $g (z_0) $ is a rational number? $\endgroup$
    – LWW
    Commented Oct 5, 2018 at 2:36
  • $\begingroup$ Of course one can just pick $z_0$ so that $g(z_0)$ is rational... unless $z_0$ is part of the given data, in which case there are trivial exampkes $\endgroup$
    – Yaakov Baruch
    Commented Oct 5, 2018 at 2:43
  • $\begingroup$ Baruch's question is exactly I want to ask. Maybe he or she misunderstood the question. $\endgroup$
    – LWW
    Commented Oct 5, 2018 at 2:57
  • $\begingroup$ You're right. I'll edit. $\endgroup$
    – Robert Israel
    Commented Oct 5, 2018 at 6:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .