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 $(\frac{a\ ,\ K^\times\!}{p})\neq 1$ [$p$ here $:=$ prime ideal generated by $n$ above]. 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 $b-1\in p^h$)

This seems like should be trivially true, at least for $h=0$; however, I cannot think of a reason why.

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 $h=0$?