5
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – ACL
    Commented Feb 15, 2013 at 23:57
  • $\begingroup$ 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. $\endgroup$
    – Allen Knutson
    Commented Feb 16, 2013 at 0:28
  • $\begingroup$ @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. $\endgroup$
    – Will Sawin
    Commented Feb 16, 2013 at 1:32
  • $\begingroup$ 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. $\endgroup$
    – Kevin Ventullo
    Commented Feb 16, 2013 at 5:24
  • $\begingroup$ 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 $R_\rho/(p) \cong \mathbb{Q}(\sqrt{2})\otimes \mathbb{R} \cong \mathbb{R} \times \mathbb{R}$. $\endgroup$
    – Kevin Ventullo
    Commented Feb 16, 2013 at 5:24

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .