Algebraic independence of exponentials

Question

First of all, a happy new year. Be it better than 2015, healthy, wealthy, fruitful and cross-fertilizing for you, familly and friends.

In order to cope with families of solutions of evolution equations, I had to prove the following lemma

Lemma: Let $Z=\{z_n\}_{n\in \mathbb{N}}$ be a set of indeterminates, then $[e^{z_0},e^{z_1}]$ is algebraically independent on $\mathbb{C}[Z]$ within $\mathbb{C}[[Z]]$. In other words, if a finitely supported sum $$ \sum_{n,m}P_{n,m}[(z_i)_{i\geq 0}]e^{nz_0}e^{mz_1} $$
is zero, then every polynomial $P_{n,m}\in \mathbb{C}[Z]$ is zero.

I cannot imagine it is unknown among specialists. My question is :

Does someone have a reference for this property ?

Thanks in advance.

On request, a proof below (I have withdrawn the - too long - previous one, using "orders of infinity".)

All relies on the following proposition which is characteristic free.

Proposition Let $(\mathcal{A},d)$ be a commutative differential ring without zero divisor, and $R=ker(d)$ be its subring of constants. Let $z\in \mathcal{A}$ such that $d(z)=1$ and $S=\{e_\alpha\}_{\alpha\in I}$ be a set of eigenfunctions of $d$ all different ($I\subset R$) i.e.

1. $e_\alpha\not=0$
2. $d(e_\alpha)=\alpha e_\alpha\ ;\ \alpha\in I$

Then the family $(e_\alpha)_{\alpha\in I}$ is linearly free over $R[z]$ (the subring generated by $R\cup \{z\}$, see remark).

From this, one can show the

Corollary Let $\mathcal{A}$ be a $\mathbb{Q}$-algebra (associative, commutative and unital) and $z$ an indeterminate, then $\{z,e^z\}\subset \mathcal{A}[[z]]$ are algebraically independent over $\mathcal{A}$.

Remark If $\mathcal{A}$ is a $\mathbb{Q}$-algebra or only of characteristic zero, i.e. $$ n1_\mathcal{A}=0\Rightarrow n=0 $$ then $d(z)=1$ implies that $z$ is transcendent over $R$. This is never the case in characteristic $p$ where $z^p$ is a constant. End of remark

One finishes the job proving, by successive extensions, that the sets
$$ \{e^{z_0},e^{z_1},\cdots e^{z_n},z_0,z_1,\cdots ,z_n\} $$
are algebraically independent over $\mathcal{A}$. Which is stronger than the desired result.

Answer 1

First of all, I am not technically answering your question because you ask specifically for a reference. However, here is a nice algebraic proof.

The result should follow immediately from the following lemma, assumed known for $k = \mathbb C$.

Lemma 1. Let $k$ be a field of characteristic $0$. Then the elements $z, e^z \in k[[z]]$ are algebraically independent over $k$.

Proof. If they are algebraically dependent, there exists a finitely generated (over $\mathbb Q$) subfield $k'\subseteq k$ over which they are algebraically dependent. But any finitely generated field of characteristic $0$ embeds into $\mathbb C$, contradicting the fact that the lemma is true over $\mathbb C$. $\square$

(This type of argument is known in algebraic geometry as the Lefschetz principle. I believe the first place where it appeared was Lefschetz's Algebraic Geometry textbook from the fifties, but I have not been able to verify that it didn't appear somewhere before. See also this answer on MO for a survey.)

Recall the following lemmata from transcendence theory.

Lemma 2. Let $\Omega$ be a big overfield, and let $k \subseteq \Omega$ be some small base field. Let $\alpha_1, \ldots, \alpha_n \in \Omega$. Then $\alpha_1, \ldots, \alpha_n$ are algebraically independent over $k$ if and only if $\operatorname{tr.deg} (k(\alpha_1,\ldots,\alpha_n)/k) = n$.

(Reference: this follows immediately from Theorem VIII.1.1 of Lang's Algebra.)

Lemma 3. Let $k \subseteq l \subseteq m$ be a tower. Then $\operatorname{tr.deg} (m/k) = \operatorname{tr.deg} (m/l) + \operatorname{tr.deg} (l/k)$.

(Reference: exercise in [loc. cit.].)

Proposition. The elements $z_0, \ldots, z_n, e^{z_0}, e^{z_1}$ are algebraically independent.

Proof. By Lemma 2, we have to prove that $\operatorname{tr.deg} (\mathbb C(z_0, \ldots, z_n, e^{z_0}, e^{z_1})/\mathbb C) = n+3$. We clearly have $$\operatorname{tr.deg}(\mathbb C(z_2,\ldots,z_n)/\mathbb C) = n-1,$$ since the $z_i$ are assumed to be algebraically independent. Letting $k = \mathbb C(z_2,\ldots,z_n)$, Lemma 1 tells us that $$\operatorname{tr.deg}(k(z_1, e^{z_1})/k) = 2.$$ Hence, by Lemma 3, we get $$\operatorname{tr.deg}(\mathbb C(z_1,\ldots,z_n,e^{z_1})/\mathbb C) = n+1.$$ Another application of Lemma 1 and Lemma 3 gives the result. $\square$

The result now follows, as you noted, because an algebraic dependence is defined over a finitely generated subextension.

Answer 2

Here is an analytic argument. Suppose that

$$ \sum_{j,k=0}^N P_{j,k}(z_0,z_1,z_2, \dotsc,z_n)e^{jz_0}e^{k z_1}=0. $$

Fix $z_2,\dotsc, z_n$ and $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bR}{\mathbb{R}}$ let $z_0=\ii t_0$, $z_1=\ii t_1$, $\ii=\sqrt{-1}$, $t_0,t_1\in\bR$. We obtain an equality of the form $(\vec{z}=(z_2,\dotsc, z_n)$)

$$\sum_{j,k}P_{j,k}(t_0,t_1,\vec{z}) e^{\ii (jt_0+kt_1)}=0, $$

for all $t_0,t_1\in\bR$. Fix $t_1$. The functions $t^me^{\ii jt}$, $m,n\in\mathbb{Z}_{\geq 0}$ are linearly independent.(You can use Wronskians to prove this.) We deduce that

$$\sum_k P_{j,k}(t_0,t_1,\vec{z})e^{\ii k t_1}=0, $$

for all $j$, $t_0,t_1\in\bR$, $\vec{z}\in\mathbb{C}^{n-2}$. The same argument implies

$$ P_{j,k}(t_0,t_1,\vec{z})=0, $$

for all $j,k$, $t_0,t_1\in\bR$, $\vec{z}\in\mathbb{C}^{n-2}$.