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On request, a proof below (I have withdrawn the - too long - previous one, using "orders of infinity".)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

- $e_\alpha\not=0$
- $d(e_\alpha)=\alpha e_\alpha\ ;\ \alpha\in I$

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.