ring of adeles: ring of integers or valuation ring

Question

I have read different definitions of the ring of adeles:

The ring of adeles is defined as the restricted topological product of the completions $K_v$ of a number field $K$ either with respect to the ring of integers of $K_v$ or with respect to the valuation ring of $K_v$.

Do these definitions coincide? If not, what is the "correct" definition?

Answer 1

You may wonder whether the two definitions of integrality in a local field coincide. But an element $x\in K_v$ has a priori infinite degree over $\mathbf Q$, hence cannot be integral over $\mathbf Z$ . Precise definitions are in order : $K$ is a number field, $K_v$ is the completion of $K$ w.r.t. a non archimedean place $v$ . Just as in the answer of @Adam Hughes, define the "valuation ring" $R_v$ = {$x \in K_v : v(x)\ge 0$}, which is a discrete valuation ring, with $K_v$ as its quotient field. If $p$ is the characteristic of the residue field of $v$, then $K_v$ is a finite extension of $\mathbf Q_p$, and your question amounts to showing that $R_v$ coincides with the "ring of integers" $O_v$ of $K_v$, i.e. the integral closure of $\mathbf Z_p$ in $K_v$.

1) $O_v$ is contained in $R_v$ : let $x \in O_v$ be a root of a monic polynomial $g \in \mathbf Z_p [X]$, whose coefficients are symmetric functions of the conjugates of $x$ in an algebraic closure of $\mathbf Q_p$. In particular the constant term of $g$ is $\pm N(x)$, where $N$ denotes the norm of $K_v / \mathbf Q_p$. Then the classical formula $v(x) = \frac 1f v_p (N(x))$, where $f$ is the inertia index (see e.g. Serre’s « Local Fields », chap.2, §2) implies $v(x) \ge 0$

2) The converse inclusion can be proved by using the known theorem that $K_v$ is obtained from $\mathbf Q_p$ by taking an unramified extension $N/\mathbf Q_p$ followed by a totally ramified extension $K_v/N$. In each of these extensions, a primitive element is known : $N=\mathbf Q_p(\zeta)$, where $\zeta$ is a root of unity of order prime to $p$, and $K_v=N(\pi)$, where $\pi$ is a uniformizer of $K_v$ (see op. cit., chap.1, §6). It follows that $R_v = \mathbf Z_p[\zeta , \pi]$ is contained in $O_v$ [perhaps there could be a simpler proof ?]

NB : the ring of integers of $K$ itself is the intersection of all the $O_v$ ‘s

Addendum : Actually there is a simpler proof of 2). Let $x \in R_v$ and consider its minimal monic polynomial $g \in \mathbf Q_p [X]$. All the conjugates of $x$ in an algebraic closure of $\mathbf Q_p$ have the same valuation (Serre, chap.2, §2, coroll.3; this is the same principle as behind the norm formula recalled in 1)). The coefficients of $g$ are symmetric functions of these conjugates, hence have valuation $\ge 0$ because $v(ab)=v(a)+v(b)$ and $v(a+b) \ge min(v(a), v(b))$. So in fact $g \in \mathbf Z_p [X]$ .

Answer 2

For each (non-Archimedean) completion simply write an arbitrary element $x = \pi^v\cdot u$ for some unit, $u$, and power of the uniformizer, $\pi^v$. Then this is an integer precisely when $v\ge 0$. (recall the restricted product does not talk about the Archimedean completions).