Van der Poorten conjectures [in "Power series representing algebraic functions," Sem. Th. Nombres Paris 1990-91] that if a power series over the rationals is the [complete] diagonal [of a rational power series in several variables], then it is algebraic over $\mathbb{Q}(X)$ iff it is algebraic [modulo almost every prime] $p$, [of degree] bounded independiently of $p$.

I want to know if there are [any] results [available in this direction].

[Remark: Not that I know of, except of course that the diagonal of a rational power series in two variables is always algebraic. -- V.D.]

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    $\begingroup$ Could you give some more background on this? $\endgroup$ – Piotr Achinger Aug 30 '12 at 19:02
  • $\begingroup$ In case when the series is a solution to some linear differential equation, this seems to be related to en.wikipedia.org/wiki/… . $\endgroup$ – Piotr Achinger Aug 30 '12 at 19:04
  • $\begingroup$ Do you want you power series to have integer coefficients so you can reduce it mod primes? $\endgroup$ – Keenan Kidwell Aug 30 '12 at 23:31
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    $\begingroup$ Every power series $\sum a_n x^n$ is the diagonal of a series in two variables, e.g., $\sum a_n x^ny^n$. It is easy to give a nonalgebraic power series over the integers that is a polynomial (and hence algebraic of degree one) modulo any prime, e.g., $2 + 2\cdot 3x + 2\cdot 3\cdot 5 x^2+\cdots$. Hence I don't think that you have correctly stated Van Der Poorten's conjecture. $\endgroup$ – Richard Stanley Aug 31 '12 at 0:33
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    $\begingroup$ Based on your answer below, I assume you read French. If so, you might be better off asking questions in French (I think this is acceptable). Because either your English is woefully deficient, or you are seriously not understanding what you are talking about, judging from what I have seen so far from your MO contributions. I am sorry to have to say this so bluntly and harshly. $\endgroup$ – Todd Trimble Sep 2 '12 at 4:56

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