property of trace modulo $n$

Question

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.

Answer 1

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.

Answer 2

I'd lay odds on a counterexample. This is going to be a bit vague, but I assume the reason you're looking at a coset is down to Pontryagin duality at the local field level. That it is the coset for a rational integer can be the result of solving some congruences by the Chinese remainder theorem. Now, my guess would be that if the ramified primes are not totally ramified, there is no special reason why you could do that with a rational integer: the residue field won't be a prime field.