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Let $K$ be a number field with ring of integers $\mathcal O_K$, let $\mathfrak m$ be an ideal in $\mathcal O_K$ and let $a \in \mathcal O_K$ such that $(a, \mathfrak m) = 1$.

Does there necessarily exist an $x \in a + \mathfrak m$ such that $x$ is totally positive? That is, for every real embedding $\sigma: \mathcal O_K \to \mathbb R$, $\sigma(x) > 0$.

Note that squares are always totally positive and therefore this is true whenever $a \in (\mathcal O_k/\mathfrak m)^2.$ This lets us assume that $(2, \mathfrak m) = 1$ by the Chinese remainder theorem.

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  • $\begingroup$ What is your definition when $K$ has no real embeddings? $\endgroup$
    – Mathmo123
    Jul 2, 2016 at 22:09
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    $\begingroup$ The result you're looking for is called the strong approximation theorem. $\endgroup$
    – user23365
    Jul 2, 2016 at 22:09
  • $\begingroup$ Thanks, just to confirm the answer is positive right? $\endgroup$
    – Asvin
    Jul 3, 2016 at 0:28

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