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 On request, a proof below (I have withdrawn the - too long - previous one, using "orders of infinity".)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. $e\_\alpha\not=0$
- $d(e\_\alpha)=\alpha e\_\alpha\ ;\ \alpha\in I$


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. 



Algebraic independence of exponentials
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](https://mathoverflow.net/a/90554/82179) 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.


