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I know that an algebraic function has a Puiseux series expansion around every point. What does it mean when such a Puiseux series doesn't have any nonintegral powers? Can we say anything about such an algebraic function?

M. Wang
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  • I would have thought it means you have a Laurent series – Henry Apr 14 '21 at 12:05
  • @Henry Thanks for the reply. So if you have a Laurent series around every point, does it mean the function is actually just a rational function (fraction of two polynomials)? – M. Wang Apr 14 '21 at 12:29
  • As usual the case $k=\Bbb{C}$ is easier to understand:

    For $f\in \overline{\Bbb{C}(z)}$, for every $a$ there is some $n=n_a$ such that $f(a+(z-a)^n)$ is meromorphic at $a$, it has a Laurent expansion, giving the Puiseux series in $(z-a)^{1/n}$ of $f$. If every $n_a=1$ then $f$ is in $\Bbb{C}(z)$, this is because it extends to a meromorphic function on the Riemann sphere. The conjugates of $f$ are its analytic continuations along closed loops.

     – reuns Apr 14 '21 at 12:41
  • @reuns Does this mean that if I can show that no nonintegral terms occur in the Puiseux series at every point, then $n_a = 1$ for every $a$ and so $f$ is meromorphic on the Riemann sphere? – M. Wang Apr 14 '21 at 12:47
  • I overlooked that every Puiseux series of $f$ must be meromorphic at $a$, for example $((1+z)^{1/2}+1)^{1/2}$ is meromorphic at $0$ in the branch where $(1+z)^{1/2}-1$ vanishes at $0$ but not in the branch where $(1+z)^{1/2}+1$ vanishes. It is unclear if there is an example of a non-rational algebraic function with some meromorphic branch for every $a$. – reuns Apr 14 '21 at 13:14

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