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From Wikipedia: given $a\in K^\times$, (a,b)=1 for all b [in K*] if and only if a is in K*ⁿ

So suppose that $(\frac{a\ ,\ K^\times\!}{p})\neq 1$ [assume $n$ above generates the prime ideal $p\unlhd{\cal O}_K$]. Given $h\in\Bbb N$, are there any hypotheses that would allow us to conclude that $(\frac{a\ ,\ b}{p})\neq 1$ for some $b\in U_p(h)$? (i.e. so $b-1\in p^h$)

To reflect wrigley's comment: this is not true for large $h$. Can we prove it for small $h$, such as $h=0$?

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  • $\begingroup$ In general $n$ (a random positive integer) doesn't generate a prime ideal. Which ring is this an ideal of anyway? Hensel's Lemma might tell you that if you're sufficiently congruent to 1 mod something then you're an $n$th power anyway so if my understanding of your question is correct then for large enough $h$ no $b$ will exist. $\endgroup$
    – wrigley
    Commented Mar 7, 2016 at 23:39
  • $\begingroup$ Oops. My original post had this, and then I accidentally edited it out. $\endgroup$
    – Alex
    Commented Mar 7, 2016 at 23:41
  • $\begingroup$ Sure the edited question is true for $h=0$ because the group generated by the non-zero elements of the integers of $K$ is just $K^\times$, so just think about the kernel of $(a,-/n)$. PS you still seem to have a prime ideal generated by a random not-necessarily-prime element. $\endgroup$
    – wrigley
    Commented Mar 7, 2016 at 23:43
  • $\begingroup$ Sorry, I posted my last comment before you edited it to give a full answer. You can just post this comment as an answer. $\endgroup$
    – Alex
    Commented Mar 8, 2016 at 0:08

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For $h=0$ there must be a witness to $(a,b/n)\not=1$ because if all integers $b$ satisfied $(a,b/n)=1$ then by multiplicativity all $b\in K^\times$ would.

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