$\mathbb{C}\left(z\right)\left[\left[w\right]\right]\cap\mathbb{C}\left(w\right)\left[\left[z\right]\right]\overset{?}{=}\mathbb{C}\left(z,w\right)$

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Consider the formal power series: $$f\left(z,w\right)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}c_{m,n}z^{m}w^{n}$$

for some complex coefficients $c_{m,n}$. Defining:

$$\alpha_{m}\left(w\right)\overset{\textrm{def}}{=}\sum_{n=0}^{\infty}c_{m,n}w^{n}$$ $$\beta_{n}\left(z\right)\overset{\textrm{def}}{=}\sum_{m=0}^{\infty}c_{m,n}z^{m}$$

we can write: $$f\left(z,w\right)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}c_{m,n}z^{m}w^{n}=\sum_{m=0}^{\infty}\alpha_{m}\left(w\right)z^{m}=\sum_{n=0}^{\infty}\beta_{n}\left(z\right)w^{n}$$

As such:

(1) If all the $\alpha_{m}$s and $\beta_{n}$s are rational functions of $w$ and $z$, respectively, is $f$ itself then necessarily a rational function of $z$ and $w$?

(2) (Generalization to power series in $d$ indeterminates) Working with a formal power series over $\mathbb{C}$ in $d$ indeterminates $z_{1},\ldots,z_{d}$:

$$f\left(z_{1},\ldots,z_{d}\right)=\sum_{n_{1}=0}^{\infty}\cdots\sum_{n_{d}=0}^{\infty}c_{n_{1},\ldots,n_{d}}z_{1}^{n_{1}}\cdots z_{d}^{n_{d}}$$

suppose that for every $j\in\left\{ 1,\ldots,d\right\}$, the series:

$$\sum_{n_{1}=0}^{\infty}\cdots\sum_{n_{j-1}=0}^{\infty}\sum_{n_{j+1}=0}^{\infty}\cdots\sum_{n_{d}=0}^{\infty}c_{n_{1},\ldots,n_{j-1},n,n_{j+1}\ldots n_{d}}z_{1}^{n_{1}}\cdots z_{j-1}^{n_{j-1}}z_{j+1}^{n_{j+1}}\cdots z_{d}^{n_{d}}$$

is a rational function of $ z_{1},\ldots,z_{j-1},z_{j+1}\ldots,z_{d}$ for all $n\geq0$. Is that sufficient to force $f$ to be a rational function of $z_{1},\ldots,z_{d}$?

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Look at $e^{zw}=\sum_{n=0}^{\infty}\frac{(zw)^n}{n!}$. Each coefficient, as a $z$ series and as a $w$ series, is rational, but $e^{zw}$ isn't.