Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f(z)=\sum_{n\geq 0}a_n z^n$ be a Taylor series with rational coefficients with infinitely non-zero $a_n$ which converges in a small neighboorhood around $0$. Furthermore, assume that \begin{align*} f(z)=\frac{P(z)}{Q(z)}, \end{align*} where $P(z)$ and $Q(z)$ are coprime monic complex polynomials. By developing $\frac{P(z)}{Q(z)}$ as a power sereis around $0$ and comparing it with $f(z)$ we obtain infinitely many polynomial equations in the roots of $P(z)$ and $Q(z)$ which are equal to rational numbers so this seems to force the roots of $P(z)$ and $Q(z)$ to be algebraic numbers.

Q: How does one prove this rigourously?

share|improve this question
1  
A minor observations: It seems some condition is needed to avoid 'degenerate' cases; e.g., what if P = Q, or P = S Q with S a rational polynomial. –  quid Jan 22 '11 at 18:33
    
Don't the coefficients of the Taylor series satisfy a recurrence relation (or difference equation)? –  Mark Bennet Jan 22 '11 at 18:48
1  
I think what is being asked is to prove that if a sequence of rational numbers satisfies a recurrence with complex coefficients, then it must satisfy a recurrence with rational coefficients. –  Gjergji Zaimi Jan 22 '11 at 18:51
    
Yes Gjergji, I think that you may rephrase the problem in these terms –  Hugo Chapdelaine Jan 22 '11 at 18:53
add comment

2 Answers 2

up vote 6 down vote accepted

Let there be two fields $k\subset K$, and let $f\in k[[x]]$ be a formal power series with coefficients in $k$. If $f\in K(x)$ (rational functions with coefficients in $K$) then $f\in k(x)$. A proof of this is given in J.S. Milne's notes on Etale Cohomology (lemma 27.9).

share|improve this answer
    
Thanks a lot for the reference. (Actually the proof was taken from Bourbaki Algebre IV) –  Hugo Chapdelaine Jan 22 '11 at 20:13
    
The result is classical, but I couldn't find a proper reference, even though I remember there is an article of Polya that discusses this. –  Gjergji Zaimi Jan 22 '11 at 20:18
add comment

Well, I think there is a simpler argument. For a power series $g(x)\in\mathbb{C}[[x]]$ and $\sigma\in Aut(\mathbb{C})$ (note that except for the complex conjugation or the identity $\sigma$ is not continuous!) we may define define the power series with coefficients twisted by $\sigma$ which we denote by $g^{\sigma}(x)$. Now an element in $Aut(\mathbb{C})$ respect finite sum and products so it follows from that, that for all $\sigma\in Aut(\mathbb{C})$ one has $$ f^{\sigma}(z)=\frac{P^{\sigma}(z)}{Q^{\sigma}(z)}. $$ From this (and the unique factorization of $\mathbf{C}[x]$) it follows that $P(z)$ and $Q(z)$ have rational coefficients.

share|improve this answer
    
Yes, and this works in Gjergji's more general setting, too. I would emphasize that $K[[x]]$ is a UFD, it is implicitly used in your proof. –  GH from MO Jan 23 '11 at 12:35
    
Yes, I did not check it but the proof might carry over to power series ring in many variables. –  Hugo Chapdelaine Jan 23 '11 at 14:54
    
There is something which bothers me. You use the fact that the subfield of $\C$ fixed by $Aut(\C)$ is $\Q$. Why is it true ? For exemple it is false if you replace $\C$ by $\R$. –  Auguste Hoang Duc Jan 27 '11 at 22:06
    
Yes you are right, I use the fact that the fixed field of $Aut(C)$ is $Q$. The fact that $Aut(R)={Id}$ is not a problem since you may work in a suitable algebraic closure and as you know $C$ is an algebraic closure of $R$. I guess that in general if you have a field $k\subseteq K$ then you want to show the existence of a field $L$ which contains $K$ such that $Aut_k(L)=k$. Once you have that the proof works. –  Hugo Chapdelaine Jan 28 '11 at 0:46
    
To Hugo: You probably want that $L^{Aut_k(L)}=k$. But such an $L$ only exists if the field extension $K/k$ is separable. –  ACL Nov 15 '12 at 22:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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