Consider a number field $K$, and let $v\_1, \cdots v\_n$ ($n \in \mathbb N$) be some finite (i.e. non-archimedean) places of $K$. Is the following true?


For every $\alpha \in K^\times$ there exists $\beta \in \mathcal{O}\_K$ for which $\alpha\beta \in \mathcal{O}\_K$ and
$$|\beta|\_{v\_j} = \frac{1}{\max \{1, |\alpha|\_{v\_j} \}} \text{ for every }j \in \{1, \cdots ,n\} \hspace{10mm} \cdots (1)$$ 


I could see that this was immediate if $\mathcal{O}\_K$ is a UFD (or equivalently a PID) for in that case, I could simply write $\alpha$ uniquely as $\alpha\_1/\alpha\_2$ where $\alpha\_1$ and $\alpha\_2$ are algebraic integers sharing no common prime factor and let $\beta:=\alpha\_2$. However, I couldn't verify this in the general case. Two things I tried are the following:


1. I tried to naturally generalize the above approach for UFD's, by factoring the principal fractional ideal $\alpha \mathcal O\_K$ uniquely into prime ideals and consider the "denominator" ideal of $\alpha \mathcal O\_K$ (that is, if $\alpha \mathcal O\_K = \prod\_{i=1}^k \mathfrak{p}\_i^{a\_i} \prod\_{j=1}^l \mathfrak{q}\_j^{-b\_j}$ where $\mathfrak{p}\_i$ and $\mathfrak{q}\_j$ are all distinct prime ideals and $a\_i, b\_j \in \mathbb N$ for all $i \in [k], j \in [l]$, then the ideal I'm talking about is $\mathfrak{a} := \prod\_{j=1}^l \mathfrak{q}\_j^{b\_j}$). This need not be a principal ideal but I could raise it to the power of its order in the ideal class group. However that would disturb the exponents, violating my requirement (1). Not sure if there's a work-around.....
2. I tried using the Strong Approximation Theorem, in an attempt to obtain $\beta$ so as to make the $v$-adic absolute values of the difference $\beta-\alpha^{-1}$ sufficiently small for $v \in \{v\_1, \cdots v\_n\}$ (so the set of absolute values $w$ for which I'm trying to make $|\beta-\alpha^{-1}|\_w$ sufficiently small are supersets of $\{v\_1, \cdots v\_n\}$), but that hasn't worked out so far.....


I haven't had any luck in finding a counterexample either. I would really appreciate any help, and would also like to know if there is any similar result along these lines.


**Edit 1:** Another thing I tried along the lines of Approach 1 was to write $\alpha$ as $\beta / \gamma$ (where $\beta$ and $\gamma$ are algebraic integers) and compare the aforementioned prime factorization of $\alpha \mathcal{O}\_K$ with those of $\beta \mathcal{O}\_K$ and $\gamma \mathcal{O}\_K$. What I obtained (after some careful exponent comparison) was the following:
$$\beta\mathcal{O}\_K = \big(\prod\_{i=1}^k \mathfrak{p}\_i^{\alpha\_i} \big) \mathfrak{a\_1}\mathfrak{a\_2}\mathfrak{a\_3}$$
$$\gamma \mathcal{O}\_K = \big( \prod\_{j=1}^l \mathfrak{q}\_j^{b\_j} \big) \mathfrak{a\_1}\mathfrak{a\_2}\mathfrak{a\_3}$$
where $\mathfrak{a\_1}$ is made up of prime factors among the $\mathfrak{p}\_i$'s, $\mathfrak{a\_2}$ is made up of prime factors among the $\mathfrak{q}\_j$'s and $\mathfrak{a\_3}$ is made up of primes not in the set $\{\mathfrak{p}\_i : 1 \leq i \leq k\} \cup \{\mathfrak{q}\_j : 1 \leq j \leq l\}$. Of course, one or more of the $\mathfrak{a}\_i$ could be trivial (i.e. the unit ideal) but I don't think that's necessary.


And there seems to lie the source of my problem in Approach 1 - I couldn't seem to get rid of the common prime factors between $\beta\mathcal{O}\_K$ and $ \gamma \mathcal{O}\_K$, in order to be able to generalize the PID approach. Anyway I don't expect that the "numerator" and "denominator" ideals $\prod\_{i=1}^k \mathfrak{p}\_i^{a\_i}$ and $\prod\_{j=1}^l \mathfrak{q}\_j^{b\_j}$ of $\alpha \mathcal{O}\_K$ to be principal, if they did, then I should've been able to obtain algebraic integers $\beta$ and $\gamma$ for which $\alpha = \beta / \gamma$ and the principal ideals $\beta\mathcal{O}\_K$ and $\gamma \mathcal{O}\_K$ would factor as $\prod\_{i=1}^k \mathfrak{p}\_i^{a\_i}$ and $\prod\_{j=1}^l \mathfrak{q}\_j^{b\_j}$ respectively, and life would've been a lot easier. 


**Edit 2 (More details on Approach 2):** As asked in a comment by @Arno Fehm, here are some more details on my second approach. We know that the number of places $w$ of $K$ for which $|\alpha|\_w>1$ or $|\alpha|\_w<1$ are both finite. As such, I can fix some $\epsilon \in (0, \min\{1, |\alpha|\_w^{-1} : w \in N\_K\})$ (where I use $N\_K$ to denote the set of non-archimedean places of $K$), and then use SAT to obtain a $\beta \in K$ such that $|\beta - \alpha^{-1}|\_w < \epsilon$ for all $w \in S:= \{v\_1, \cdots , v\_n\} \cup \{w \in N\_K: |\alpha|\_w>1\}$ and $|\beta|\_w \leq 1$ for all the other (remaining) non-archimedean places $w$ of $K$. This ensures that $|\alpha\beta|\_w, |\beta|\_w \leq 1$ for all the places $w \in N\_K \setminus S$, whereas for places $w \in S$, I have 
$$|\beta - \alpha^{-1}|\_w < \epsilon < \min\{1, |\alpha|\_w^{-1}\} = \frac{1}{\max\{1, |\alpha|\_w\}} \hspace{2mm} \cdots (2)$$
Now for the places $w \in S$ for which $|\alpha|\_w \geq 1$, I could show by means of the ultrametric inequality that (2) forces 
$$|\beta|\_w = \frac{1}{|\alpha|\_w} = \frac{1}{\max\{1, |\alpha|\_w\}} \leq 1$$
Problems start occurring for those places $w \in S$ for which $|\alpha|\_w<1$ (so $w$ has to be one of $v\_1, \cdots , v\_n$). In this case, (2) yields $|\beta - \alpha^{-1}|\_w<1$ which in fact again forces $|\beta|\_w = |\alpha|\_w^{-1}>1$, for otherwise the ultrametric inequality leads to the following contradiction
$$1>|\beta - \alpha^{-1}|\_w = \max\{|\beta|\_w, |\alpha^{-1}|\_w\} \geq |\alpha^{-1}|\_w = |\alpha|\_w^{-1} > 1$$ 
This means that $\beta$ in fact cannot be an algebraic integer if $|\alpha|\_{v\_j}<1$ for one of the $j \in [n]$ (not to mention that (1) clearly fails for such $j$ as well). 


If this approach looks promising, I would really like to know how I should choose my parameters $\epsilon$, $S$ etc. to make it work. 



Existence of algebraic integer with absolute value equal to reciprocal of maximum of $1$ and absolute value of a given algebraic number
The answer is already in the comments, but here again for completeness: It indeed follows easily from strong approximation:


Let $S=\{v\_1,\dots,v\_n\}\cup\{v:|\alpha|\_v>1\}$. By the strong approximation theorem one can find $\beta\in\mathcal{O}\_K$ that is close to $1$ at those $v\in S$ with $|\alpha|\_v<1$ and close to $\alpha^{-1}$ at the other $v\in S$. In particular, $|\beta|\_v=1$ for those $v\in S$ with $|\alpha|\_v<1$ and $|\beta|\_v=|\alpha|\_v^{-1}$ at the other $v\in S$.


Thus $|\beta|\_v=\max\{1,|\alpha|\_v\}^{-1}$ for $v\in S\supseteq \{v\_1,\dots,v\_n\}$,
and $|\alpha\beta|\_v\leq 1$ for all finite $v$, hence $\alpha\beta\in\mathcal{O}\_K$.


