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The problem is: let $f: \mathbb{R}\to \mathbb{R}$ be an analytic transcendental function and let $\psi(x)=\frac{x}{2(1+x^2)}$. Is the function $f(\psi(x))$ transcendental?

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    $\begingroup$ Galois theory... $\endgroup$ Aug 26, 2014 at 0:38
  • $\begingroup$ Prof. Voloch, could you please explain me this better? How to use Galois theory? $\endgroup$
    – Diego
    Aug 26, 2014 at 2:34
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    $\begingroup$ By assumption, $\mathbb{C}(y,f(y))$ is transcendental over $\mathbb{C}(y)$. Setting $y=\psi(x)$ is merely making an algebraic extension of each, so $\mathbb{C}(\psi(x),f(\psi(x)))$ is transcendental over $\mathbb{C}(\psi(x))$. Hence the tower $\mathbb{C}(\psi(x),f(\psi(x)))\supset \mathbb{C}(\psi(x))\supset \mathbb{C}(x)$ is transcendental, so $f(\psi(x))$ is transcendental over $\mathbb{C}(x)$. [This does seem a bit elementary for MO, so don't be surprised if it's closed.] $\endgroup$ Aug 26, 2014 at 2:44
  • $\begingroup$ Thanks Prof. Silverman, so the composition of a transcendental function with a non constant algebraic function is transcendental. $\endgroup$
    – Diego
    Aug 26, 2014 at 4:04

1 Answer 1

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Yes. If there is a polynomial relation $P(x, f(\psi(x))=0$, then we also have $P(\psi^{-1}(y), f(y))=0$ (passing to an open set where $\psi^{-1}$ is defined and analytic.) But $\psi^{-1}(y)$ is algebraic over $\mathbb{C}(y)$, so this would show $f(y)$ algebraic over $\mathbb{C}(y)$, contradicting the assumption that it is transcendental.

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  • $\begingroup$ Unfortunately your proof is not correct. Because $P(\psi^{-1}(y),f(y))=0$, for all $y$ belonging to an interval $I\neq \mathbb{R}$ does not imply f to be algebraic. You should think that your "proof" is so strong that it could prove that the composition of a transcendental function with its inverse also would be transcendental. Sorry. $\endgroup$
    – Diego
    Aug 26, 2014 at 2:32
  • $\begingroup$ This argument doesn't apply to the composition of a transcendental function with its inverse at all; it is important that $\psi$ is algebraic. $\endgroup$ Aug 26, 2014 at 2:37
  • $\begingroup$ Yes, you are right. But unfortunately $P(\psi^{-1}(y),f(y))=0$, for all $y\in I\neq \mathbb{R}$ does not imply $f$ algebraic. $\endgroup$
    – Diego
    Aug 26, 2014 at 2:40
  • $\begingroup$ Yes it does. Let $K$ be the set of functions $I \to \mathbb{C} \cup \{ \infty \}$ which extend to meromorphic functions on some open set containing $I$. This is a field and, if two functions from $K$ agree on $I$, then they are equal. $\endgroup$ Aug 26, 2014 at 2:43
  • $\begingroup$ Within the field $K$, $\psi^{-1}(y)$ is algebraic over the field of rational functions $\mathbb{C}(y)$. And we are assuming that $f$ is algebraic over $\mathbb{C}(\psi^{-1}(y))$. So $f$ is algebraic over $\mathbb{C}(y)$. And, once we have a polynomial relation $Q(f,y)=0$, it extends to all of $\mathbb{R}$ because $f$ and $y$ are analytic functions. $\endgroup$ Aug 26, 2014 at 2:46

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