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Let $k=\mathbb F_q(T)$. Can one prove (or disprove) that the series $\sum_{n\ge0}(1-TX^{q^n})Y^{q^n}\in k[[X,Y]]$ belongs to $k(X,Y)$? At first, it looked like it was simple. But in fact, I have no clue to attack this question. I thought about Dwork-Polya-Bertrandias theorem, but I did not find a several variables version of this theorem.

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If you set $T=0$ or $X=0$ then you get the series $\sum_{n\geq 0} Y^{q^n}$. This cannot be rational because a rational power series in one variable that is not a polynomial cannot have arbitrarily long sequences of 0 coefficients (since the coefficients satisfy a linear recurrence relation with constant coefficients).

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  • $\begingroup$ Is your argument still true in positive characteristic? $\endgroup$
    – joaopa
    Commented Jul 10, 2020 at 22:38
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    $\begingroup$ @joaopa It certainly is. $\endgroup$
    – R.P.
    Commented Jul 10, 2020 at 23:01
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    $\begingroup$ @joaopa letting $k$ be a field of characteristic $p$, the formal power series $f(Y) = \sum_{n \geq 0} Y^{q^n}$ in $k[[Y]]$, where $q$ is a power of $p$, is not in $k(Y)$ since $f^q - f - Y = 0$ and this relation is impossible for $f$ being in $k(Y)$: that $Z^q - Z - Y$ is monic in $Z$ and vanishes at $Z = f$ would force $f$ to be in $k[Y]$ with positive degree, but then $\deg(f^q - f) > 1 = \deg(Y)$, so $f^q - f - Y$ can't be $0$. $\endgroup$
    – KConrad
    Commented Jul 11, 2020 at 1:39
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    $\begingroup$ The argument that I gave works in any characteristic. $\endgroup$ Commented Jul 11, 2020 at 2:15
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    $\begingroup$ Incidentally, a formal power series in one variable over a finite field is rational if and only if its coefficients are eventually periodic. $\endgroup$ Commented Jul 11, 2020 at 23:30

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