Let $K$ be the field of fractions of 
$\mathbb{C}[[z]]\otimes\_{\mathbb{C}}\mathbb{C}[[w]]\subset \mathbb{C}[[z, w]].$ Given
a formal power series in $t, f\in \mathbb{C}[[t]],$ is there any simple criterion which
will conclude that $f(zw)$ does not belong to $K?$ I suspect that 
$f(zw)\in K$ if
and only if $f(t)$ is a rational function, i.e. belongs to $\mathbb{C}(t).$ I am
especially interested in proving that $e^{zw}\notin K.$ 



How to prove that $e^{zw}$ can not be written as a rational expression in functions in $z$ and in $w$?
**Theorem 1.** Let $f\in\mathbb{C}[[t]]$. Then $f(zw)\in \mathbb{C}[[z]]\otimes\_{\mathbb{C}}\mathbb{C}[[w]]$ if and only if $f\in\mathbb{C}[t].$


**Proof.** The "if" part is obvious. For the "only if" part assume that $f\notin\mathbb{C}[t]$ but
$$f(zw) = \sum\_{i=1}^n g\_i(z)h\_i(w),$$
where $g\_i\in\mathbb{C}[[z]]$ and $h\_i\in\mathbb{C}[[w]]$. We can assume that $n>0$ is minimal. If $g\_i(0)=h\_i(0)=0$ for all $i$, then $f(0)=0$ and we can divide both sides by $zw$ in the obvious way. Therefore $g\_i(0)\neq 0$ or $h\_i(0)\neq 0$ for some $i$. Without loss of generality, $g\_1(0)=1$, then 
$$h\_1(w)=f(0)-\sum\_{i=2}^n g\_i(0)h\_i(w),$$
so that
$$f(zw)=g\_1(z)f(0)+\sum\_{i=2}^n\bigl(g\_i(z)-g\_i(0)g\_1(z)\bigr)h\_i(w).$$
Taking derivative with respect to $w$ and then dividing by $z$ yields
$$f'(zw)=\sum\_{i=2}^n\frac{g\_i(z)-g\_i(0)g\_1(z)}{z}h\_i'(w).$$
Here $f'\notin\mathbb{C}[t]$, but the fractions are elements of $\mathbb{C}[[z]]$, contradicting the minimality of $n$.


**Theorem 2.** $e^{zw}\notin K$.


**Proof.** Assume there are $u\_i,f\_j\in\mathbb{C}[[z]]$ and $v\_i,g\_j\in\mathbb{C}[[w]]$ such that
$$e^{zw}\sum\_{i=1}^m u\_i(z)v\_i(w)=\sum\_{j=1}^n f\_j(z)g\_j(w),$$
where the sums are nonzero. We can assume that $n>0$ is minimal, and also that $m>0$ is minimal for this $n$. In particular, the $u\_i$'s are independent over $\mathbb{C}$.
We can assume that some $v\_i(0)$ or $g\_j(0)$ is nonzero, for otherwise we can divide both sides by $w$ in the obvious way. Then
$$ \sum\_{i=1}^m u\_i(z)v\_i(0)=\sum\_{j=1}^n f\_j(z)g\_j(0) $$
is a nonzero element of $\mathbb{C}[[z]]$, call it $h(z)$. In particular, $g\_j(0)\neq 0$ for some $j$. Without loss of generality $g\_1(0)=1$, then with the notation
$$\tilde u\_i(z):=u\_i(z)/h(z)\in\mathbb{C}((z))$$
$$\tilde f\_j(z):=f\_j(z)/h(z)\in\mathbb{C}((z))$$
$$e^{zw}\sum\_{i=1}^m \tilde u\_i(z)v\_i(w)=g\_1(w)+\sum\_{j=2}^n \tilde f\_j(z)\bigl(g\_j(w)-g\_j(0)g\_1(w)\bigr).$$
The sum on the right hand side is independent of $z$ for otherwise we could contradict the minimality of $n$ by taking the $z$-derivative on both sides and then multiplying by a large power of $z$ to turn elements of $\mathbb{C}((z))$ into elements of $\mathbb{C}[[z]]$. At any rate, the right hand side is a nonzero element $k\in\mathbb{C}[[w]]$, because $k(0)=g\_1(0)=1$. With the notation
$$ \tilde v\_i(w):=v\_i(w)/k(w)\in\mathbb{C}((w))$$
we conclude
$$e^{-zw}=\sum\_{i=1}^m \tilde u\_i(z)\tilde v\_i(w).$$
In particular, there is some integer $r>0$ such that
$$(zw)^r e^{-zw}\in \mathbb{C}[[z]]\otimes\_{\mathbb{C}}\mathbb{C}[[w]],$$
which contradicts Theorem 1. The proof of Theorem 2 is complete.


