It is well-known (see Allouche monography for example) that if $f$ is an algebraic function over $K(X)$ then $f'$ is also algebraic. I wonder whether $f$ and $g$ are algebraically dependent, then $f'$ and $g'$ are also algebraically dependent. I think that it is false, but I do have no counterexample, neither a proof that this assertion is true.

Any hints or answers will be welcome. Thanks in advance

  • $\begingroup$ Every two algebraic functions of $x$ are evidently algebraically dependent. $\endgroup$ – Alexandre Eremenko Nov 25 '16 at 3:07
  • $\begingroup$ @Alexandre, the hypothesis seems to be that $f$ and $g$ are algebraically dependent, not that $f$ and $g$ are algebraic. $\endgroup$ – Gerry Myerson Nov 25 '16 at 4:32
  • $\begingroup$ Where do your functions live? $\endgroup$ – abx Nov 25 '16 at 7:30
  • $\begingroup$ @Gerry Meyerson: OK, with this interpretation, the answer is also trivial: see my answer. $\endgroup$ – Alexandre Eremenko Nov 25 '16 at 16:33

The answer is no.

Example. Let $f=\Gamma(z)$, Euler's Gamma function, and $g(z)=\Gamma^2(z)$. Evidently they are algebraicaly dependent. Then $g'=2ff'$. Suppose that $f'$ and $g'$ are algebraically dependent, that is there is an equation $F(f',g')=0$ where $F$ is a polynomial with constant coefficients. Then $F(f',2ff')=0$ but this is an algebraic differential equation, and we know from Holder that $\Gamma$ does not satisfy any.

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