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.


1 Answer 1


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).

  • $\begingroup$ Is your argument still true in positive characteristic? $\endgroup$
    – joaopa
    Jul 10, 2020 at 22:38
  • 1
    $\begingroup$ @joaopa It certainly is. $\endgroup$
    – RP_
    Jul 10, 2020 at 23:01
  • 3
    $\begingroup$ @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
    Jul 11, 2020 at 1:39
  • 4
    $\begingroup$ The argument that I gave works in any characteristic. $\endgroup$ Jul 11, 2020 at 2:15
  • $\begingroup$ Incidentally, a formal power series in one variable over a finite field is rational if and only if its coefficients are eventually periodic. $\endgroup$ Jul 11, 2020 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.