Let $p$ be a prime number and $\zeta_{p^n}$ a primitive $p^n$-th root of unity. Find $f \in \mathbf Q_p[[X]]$ fulfilling $f(\zeta_{p^n}-1)=1/p^n$ for all sufficiently large $n$.
$\begingroup$
$\endgroup$
1
-
5$\begingroup$ Can assign any values for any $n \ge 1$. Let $D$ be rigid-analytic open unit disc over $\mathbf{Q}_p$, $S$ the subset of non-rat'l pts $\zeta_ {p^n} - 1$ for $n \ge 1$. Since $S$ meets each closed disc centered at 0 in finite subset, it's "discrete" in $D$, so a 0-dimensional analytic set. Let $I$ be the radical coherent ideal of $S$ in $D$. Since $D$ is rigid-analytic Stein space (Stein exhaustion by closed discs centered at 0), ${\rm{H}}^1(D,\cdot)$ vanishes on coherent sheaves, such as $I$. So $O(D) \rightarrow (O/I)(D) = \prod_ {n \ge 1} \mathbf{Q}_ p(\zeta_ {p^n})$ is surjective. QED $\endgroup$– BCnrdCommented Aug 11, 2010 at 14:39
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
As Brian pointed out, the principal parts problem always has a solution on the open unit disk, and Lazard's 1962 article "Les zéros des fonctions analytiques d’une variable sur un corps valué complet" gives a nice proof which is also rather explicit.
Your problem has additional symmetries so one can be a bit more explicit, as follows.
Let $H$ be the set of power series holomorphic on the open unit disk, let $\varphi: H \to H$ be the map defined by $\varphi(f)(X)=f((1+X)^p-1)$ and let $d$ be defined by $d(f)(X)=(1+X)df/dX$. Note that $d\varphi=p\varphi d$. You can check that if you have a function $f$ such that $\varphi(f)-pf = (1+X/2)\log(1+X)/X$, then $f(0) \neq 0$ and $f(\zeta_{p^n}-1) = p^{-n} f(0)$ so this answers your problem.
Therefore you need to be able to solve an equation of the form $\varphi(f)-pf = g$. By taking $d$ of both sides this gives $\varphi(df)-df=dg/p$. Now you can solve an equation of the form $\varphi(a)-a=b$ if $b \in X \cdot H$ by writing $a = \sum_{n \geq 0} \varphi^n(b)$, which should converge in $H$ for its Fréchet topology. We have $d((1+X/2)\log(1+X)/X) \in X \cdot H$ (this is what the $(1+X/2)$ was put in for), and once you know $df$, you get $f$ by integrating and adjusting the constant.
answered Aug 11, 2010 at 16:10