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This question asked yesterday got me thinking. While the derivatives of the tangent function span an infinite dimensional vector space over $\mathbb{C},$ the transcendence degree of the field generated by these derivatives is finite.

Here are two ways to see this. One is to observe that

$$X^2 - DX + 1 = 0$$

where $X$ denotes the tangent function. While a better way is to observe that via the sum/product rule, the field obtained by adjoining the set of derivatives of a function $f$ to $\mathbb{C}$ denoted $\mathbb{C}(\mathcal{D}f)$ is closed under differentiation and $\mathbb{C}(\mathrm{tan}) \subset \mathbb{C}(Exp) = \mathbb{C}(\mathcal{D}Exp)$

In fact, using the basic rules of differentiation, one can show that if the transcendence degrees over $\mathbb{C}$ of $\mathbb{C}(\mathcal{D}f)$ and $\mathbb{C}(\mathcal{D}g)$ are finite so too are the transcendence degrees of $\mathbb{C}(\mathcal{D}(f + g)),$ $\mathbb{C}(\mathcal{D}(fg))$ and $\mathbb{C}(\mathcal{D}(f \circ g)).$

Which brings me to my question:

Do there exist any functions $f$ which are meromorhic on $\mathbb{C}$ such that the transcendence degree of $\mathbb{C}(\mathcal{D}f)$ over $\mathbb{C}$ is infinite?

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I'm not familiar with the theory at all, but perhaps considering the analogous question over a general differential field would lead to a more straightforward approach (or indicate some special property of $\mathbb{C}$ that would need to be used). –  Zev Chonoles Jun 29 '11 at 19:38
    
Zev. Here's some motivation for why one would care about the above problem over $\mathbb{C}.$ Let $f$ be a meromorphic function such that the transcendence degree $\mathbb{C}(\mathcal{D}f)$ is finite. For each $n,$ consider the evaluation map from $\mathbb{C}[X_0,....,X_n]$ to $\mathbb{C}[f,....,D^nf].$ The Kernel of this map is some ideal $I_n$ Because the zeros of complex functions are discrete, this ideal must be prime. Hence, for each $n$ the first $n$ derivatives of $f$ define functions on varieties $V(I_n)...$ –  jspecter Jul 1 '11 at 2:29
    
We may then ask about what properties of $f$ are determined by the geometry of the $V(I_n).$ –  jspecter Jul 1 '11 at 2:31

1 Answer 1

up vote 3 down vote accepted

The answer is yes and there are many.

Recalling that the transcendence basis of $\mathbb{R}/\mathbb{Q}$ is infinite, we may choose a sequence of real numbers $\langle a_i \rangle_{i\in\omega}$ which are algebraically independent over $\mathbb{Q}$ and satisfy $0<a_i<\frac{1}{i!}.$ Consider the series $\displaystyle\sum_{i=0}^{\infty} a_iz^i.$ This series converges uniformly on $\mathbb{C}$ and therefore defines an entire function. We denote this function by $f.$

We claim that the transcendence degree of $\mathbb{C}(\mathcal{D}f)/\mathbb{C}$ is infinite. Let $p \in \mathbb{C}[X_1,...,X_n]$ such that $p(f,Df,D^2f,...,D^nf) = 0$ and $\alpha_1,...,\alpha_k$ be the coefficients of $p$ in $\mathbb{C}.$ Equating coefficients in the power series $p(f,Df,D^2f,...,D^nf)$ and the power series $0,$ we obtain for each $i\in\omega$ a polynomial $p_i\in\mathbb{Q}(\alpha_1,...,\alpha_k)[Y_0,...,Y_i]$ such that $p_i(a_0,...,a_i) = 0.$ As the transcendence degree of $\mathbb{Q}(\alpha_1,...,\alpha_k)(a_0,a_1,...)/\mathbb{Q}(\alpha_1,...,\alpha_k)$ is infinite, it must be the case that there exists an $N\in\mathbb{N}$ such that for all $i>N$ the polynomial $p_i(a_0,..,a_N,Y_{N+1},...,Y_i) = 0.$

Let

$$g(z) = a_0 + ... + a_Nz^N + \displaystyle\sum_{i=N+1}^{\infty} Y_{i}z^{i}\in \mathbb{C}[Y_i:N<i][[z]].$$

Then $p(g,Dg,...D^ng) = 0.$ It follows that if $p$ is non-zero, the transcendence degree of $\mathbb{C}(\mathcal{D}g)/\mathbb{C}$ is finite. But $\mathbb{C}(\mathcal{D}g)$ contains the ring $\mathbb{C}[\mathcal{D}g]$ which surjects onto $\mathbb{C}[Y_{i}:i>N]$ via evaluation at $0.$ As this latter ring has infinite krull dimension so too must $\mathbb{C}[\mathcal{D}g]$ and therefore the transcendence degree of $\mathbb{C}(\mathcal{D}g)/\mathbb{C}$ is infinite. We conclude $p = 0$ and hence transcendence degree of $\mathbb{C}(\mathcal{D}f)/\mathbb{C}$ is infinite.

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