Intersection of completions

Question

Let $K$ be a number field and $S$ its set of places (finite and infinite). For some place $v \in S$ we denote by $K_v$ the completion of $K$ at v. Is it true that $$ K = \bigcap_{v \in S} K_v \quad ? $$ This equation should be understood in the following sense: If a sequence $(a_n)_{n \in \mathbb{N}}$ of numbers $a_n \in K$ is a Cauchy-Sequence with respect to every $| \cdot |_v$, it is a constant sequence.

Answer 1

Let $p_k$ denote the $k$th prime. Let $\varphi$ denote the Euler phi function. Let

$$e_n = \prod_{k=1}^{n-1} \varphi(p^n_k).$$

By Euler's Theorem, $p^{e_n}_{n}$ is congruent to $1$ modulo $p^n_k$ for all primes $k < n$. It follows that the sequence

$$a_n = \frac{1}{p^{e_n}_{n}}$$

converges to $1 \in \mathbf{Q}_p$ for all primes $p$ but converges to $0 \in \mathbf{R}$. In particular, the answer to your question is no.

On the other hand, although the sequence $a_n$ converges for all primes $p$, it does not do so uniformly; however large $n$ may be, there will still be primes (in particular $p = p_n$) for which $a_n$ is very far from the limit, or even from $a_{n+1}$. This is not a defect of the example, but rather is forced by the product formula $\prod_{v} |x|_v = 1$ for non-zero $x$. (Take $x = a_{n+1} - a_{n}$.)