TITLE: property of trace modulo $n$
QUESTION [8 upvotes]: I recently noticed an interesting (at least to me) property of the trace but have been unable to prove it.
Let ${\mathbb K}$ is an algebraic number field with ${\mathcal O}$ as its ring of integers, $n$ is any positive integer relatively prime to the discriminant of ${\mathcal O}$ and
$$
T_{n} = \left\{ x \in {\mathcal O}: {\rm Tr}(x) \equiv 0 \bmod n \right\}.
$$
Then the elements $y \in {\mathcal O}$ such that $xy \in T_{n}$ for all $x \in T_{n}$ seem to always be of the form a rational integer plus $n$ times an element of ${\mathcal O}$.
Does anyone know how to prove such a result (or a counterexample too, of course)?
References to such results in the literature would also be welcome.
Thank you.

REPLY [3 votes]: For a finite degree field extension $E/K$ the trace is a $K$-linear map $E\to K$ and it determines the symmetric bilinear form $Tr(xy)$ on the $K$ vector space $E$. If the extension is separable (= etale) then the trace is not identically zero, so its kernel has codimension one, so the orthogonal complement of its kernel is one-dimensional and contains $K$ and so must be exactly $K$.
What you're looking at is exactly this, except that instead of a field extension $E$ over $K$ it's a ring extension $\mathcal O/n$ over $\mathbb Z/n$. (It's etale because the discriminant is prime to $n$.) I believe the same principles basically apply.