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

Let $R_\infty$ be the ring of power series in a single variable with rational coefficients that converge in the whole complex plane. Let $R_\rho$ be the subring of $R_\infty$ that defines holomorphic functions of order no more than $\rho$.

What is $\operatorname{Spec} R_\rho$? Can we describe the vanishing sets of $\sin x$ and other simple power series?

This question was vaguely inspired by the idea of transalgebraic Galois theory.

About all I can say is that if $\rho<\infty$ then irreducible polynomials in $\mathbb Q[x]$ remain irreducible in $R_\rho$, but it is not even obvious to me whether they remain prime ideals.

share|improve this question
    
A theorem of Iss'sa (see Serge Lang, Corps de fonctions méromorphes sur une surface de Riemann, Séminaire Bourbaki 1963/64; numdam.org/numdam-bin/fitem?id=SB_1964-1966__9__231_0) asserts that discrete valuation rings of $R_\infty$ correspond to points of the complex plane. The spectrum of $R_\infty$ will be much more complicated. For example, any ultrafilter $\mathcal U$ on a discrete subset of $\mathbf C$ will give rise to a point of the spectrum, corresponding to functions vanishing at $\mathcal U$-almost every point. –  ACL Feb 15 '13 at 23:57
    
Sorry, what does order mean here? It's not the order of a pole, since your functions are defined everywhere. Anyway any notion of order I know would be additive under multiplication, so not define a subring. –  Allen Knutson Feb 16 '13 at 0:28
    
@ACL: That's what I was thinking. I think those might be all the points, other than the generics? @Allen Knutson: It is a very poorly named concept, though not as bad as the closely related concept of the genus of an entire function. en.wikipedia.org/wiki/Entire_function#Order_and_growth I only know about it because it appears on two of the sample questions on the graduate student's guide to generals. –  Will Sawin Feb 16 '13 at 1:32
    
So if $p\in \mathbb{Q}[x]$, it seems to me that the ideal generated by $p$ in $R_\rho$ is just the set of functions which vanish at the zeroes of $p$ to at least the same order as $p$. This means $R_\rho/(p)\subset \mathbb{Q}[x]/(p) \otimes_\mathbb{Q} \mathbb{R}$. My guess is this is always an isomorphism. –  Kevin Ventullo Feb 16 '13 at 5:24
    
For example, take $p = x^2 -2$. Let $a_0, a_1, a_2, \ldots$ be a sequence of rationals which sum to $\sqrt{2}$, and which tends to zero as fast as you want. Let $f(x) = x + (a_0 + \frac{a_1}{2}x^2 + \frac{a_2}{4}x^4 + \frac{a_3}{8}x^6 +\ldots) \in R_\rho$. Then $f(-\sqrt{2})=0$ and $f(\sqrt{2})=2\sqrt{2}$. One can modify this example to get any pair of real numbers. Thus $R_\rho/(p) \cong \mathbb{Q}(\sqrt{2})\otimes \mathbb{R} \cong \mathbb{R} \times \mathbb{R}$. –  Kevin Ventullo Feb 16 '13 at 5:24

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.