Let $\mathbb{A}$ be the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$.

A collection of functions $F=\lbrace f_i:X \rightarrow\mathbb{C}\rbrace$ is said to be algebraically independent over $\mathbb{Q}$ at $x \in X$ if the $f_i(x)$ are distinct, at most one of the $f_i(x)$ is algebraic and $ \lbrace f_i(x) \rbrace - \mathbb{A} $ is algebraically independent over $\mathbb{Q}$. (The reason why I allow one $f_i(x)$ to be algebraic will become clear shortly.)

I would like to know whether there exists a 2-sided sequence of entire functions:

$\ldots, f^{(-2)},f^{(-1)},f^{(0)},f^{(1)},f^{(2)}, \ldots$

such that for every $n \in \mathbb{Z}$, $f^{(n)}(z)=\frac{d}{dz}f^{(n-1)}(z)$, and the collection $\{f^{(i)} \: | \: i \in \mathbb{Z}\}$ is algebraically independent over $\mathbb{Q}$ everywhere except possibly on a "small" set S. Ideally, I'd like S to be empty, or at least closed and discrete, but pretty much any smallness condition you like would help me at least get some intuition for how to deal with such a problem.

Note that unless S is dense, there is an open set $U$ in its complement. By the open mapping theorem, each $f^{(i)}(U)$ is open and hence contains algebraic numbers. So allowing one $f^{(i)}$ to be algebraic at each point is necessary.

My thoughts (feel free to ignore them if you have your own approach):

It's pretty easy with the Lindemann–Weierstrass theorem to construct $f^{(0)}$ such that the derivatives of $f^{(0)}$ at 0 are algebraically independent, and choosing integration constants appropriately for negative $n$ gives a sequence which works at 0. The problem is that evaluating the function anywhere other than 0 requires summing an infinite series, something which behaves poorly with respect to transcendence.

It is likely not plausible to prove that any specific function works, since transcendence theory doesn't have many general results. On the other hand, I suspect that if such a function exists, they will be generic, or at least common. There may be an easy way to approach the problem nonconstructively with functional analysis. I couldn't think of anything in this direction, though.

On the other hand, if such a function does not exist, there should be a good reason why not. I suspect it would be demonstrable with just a few derivatives based mostly on topology. By the open mapping theorem, on each open set $\mathbb{A}(f^{(i)}(z))$ takes every value possible for each $i$. This seems like a strong condition, but it also implies that one can't topologize the field extensions of $\mathbb{A}$ in any nice way, which poses a bit of an issue if one wants to continue on this route.

Logan M
  • 6,771
  • 3
  • 29
  • 45
  • 3
    Dumb question: why wouldn't $f(z)=\frac1{\Gamma(z)}$ work? – J. M. ain't a mathematician Apr 27 '12 at 02:35
  • 1
    It might (with a bit of modification since the 0th and first derivatives are both algebraic at z=0), but I don't think I can prove it. For one, this would require proving that $\gamma$ is transcendental, which is open. – Logan M Apr 27 '12 at 02:38
  • 3
    I think at $z=0$ this fails for the fifth order term. The second order term corresponds to adjoining $\gamma$ and taking the algebraic closure, the third to $\pi$, the fourth to $\zeta(3)$. Since $\zeta(4)=\frac{\pi^4}{60}$, the fifth order term is not algebraically independent of the second, third, and fourth. It still could work almost everywhere else, but I don't know how to approach that. – Logan M Apr 27 '12 at 02:52

1 Answers1


If any $f^{(n)}$ is constant, then $f^{(n+1)}=f^{(n+2)}=\cdots=0$, and algebraic independence over $\Bbb Q$ (in the sense defined above) fails everywhere. Assume then that $f^{(1)}$ is not constant, so for some $K\in\{1,2\}$, $g_K:=Kf^{(1)}+f^{(2)}$ is not constant. Then by the Open Mapping Theorem, for any $z_0\in\Bbb C$ and $\epsilon>0$, $g_K(B_\epsilon(z_0))$ is open, so it contains algebraic numbers. However, if $g_K(z)$ is algebraic, then either neither of $f^{(1)}(z)$ and $f^{(2)}(z)$ is algebraic, or both are, and if neither are then they are algebraically dependent. Therefore, for any sequence, algebraic independence over $\Bbb Q$ (in the sense defined above) fails at a set which is dense in $\Bbb C$.

More generally, for any polynomial $P$ with rational coefficients, unless $P(f^{(i_1)}(z), \dots, f^{(i_k)}(z))$ is constant, the Open Mapping Theorem can likewise be applied to it, meaning that there will be a dense set on which $\{f^{(i_1)}(z),\dots,f^{(i_k)}(z)\}$ are algebraically dependent over $\Bbb Q$ (in the usual meaning of the phrase algebraically dependent, not the sense defined above.) So, I don't think that anything like what the questioner hopes for will be possible. The best you can do is to have a collection of dense sets of various forms of algebraic dependence, permeating the complex plane.

David Moews
  • 13,830
  • 1
  • 29
  • 44