Let $K_p$ be a finite extension of $\mathbb Q_p$. Then we have the trace map: $$ T:=\operatorname{Tr}_{K_p|\mathbb Q_p}:K_p\to\mathbb Q_p $$ Is there any characterization of the open set $T^{-1}(\mathbb Z_p)$?

I only know that if $\pi$ is a uniformizing element of $K_p$ then there exists a maximum positive integer $d$ such that $\pi^{-d}O_{K_p}\subseteq T^{-1}(\mathbb Z_p)$.

In other words is there a way to characterize the element of a local field having integral trace?


1 Answer 1


Your set can be characterized as the elements with an integral trace. It's a tautological characterization, but there's really not anything better to expect. Even in the number field case, do you know any nice characterization of the elements with integral trace that's any better than "the elements with integral trace"?

What you might really be interested in is the elements having integral trace after multiplication by all integers of $K_p$. That is related to the local different ideal, just like the description of the global different ideal.

For a number field $K$, the set $$ S(K) = \{\alpha \in K : {\rm Tr}_{K/\mathbf Q}(\alpha\mathcal O_K) \subset \mathbf Z\} $$ is a fractional ideal in $K$ that contains $\mathcal O_K$. Since $1 \in \mathcal O_K$, all elements of that fractional ideal have integral trace, but to be in that fractional ideal is a stronger condition than merely having integral trace by itself and nothing else. When $K = \mathbf Q(i)$, $$ S(\mathbf Q(i)) = \{\alpha \in K : {\rm Tr}_{K/\mathbf Q}(\alpha\mathcal O_K) \subset \mathbf Z\} = \frac{1}{2}\mathbf Z[i] $$ while $$ \{\alpha \in K : {\rm Tr}_{K/\mathbf Q}(\alpha) \in \mathbf Z\} = \frac{1}{2}\mathbf Z + \mathbf Q{i}. $$ The first set is a fractional ideal while the second set is some ugly thing with no worthwhile properties (well, it's an abelian group, but certainly not a $\mathbf Z[i]$-module: it contains $i/3$ but not $i(i/3) = -1/3$).

Since $S(K)$ is a fractional ideal containing $\mathcal O_K$, the inverse $S(K)^{-1}$ is a nonzero ideal contained in $\mathcal O_K$: it is an integral ideal. This is called the different ideal $\mathfrak D_{K/\mathbf Q}$ and it is closed related to ramification: its prime ideal factors are precisely the prime ideals in $K$ that ramify over $\mathbf Q$ (the multiplicity of the prime ideal factors can be tricky).

The local story is analogous: the set $$ \{\alpha \in K_p : {\rm Tr}_{K_p/\mathbf Q_p}(\alpha\mathcal O_{K_p}) \subset \mathbf Z_p\} $$ is a nonzero finitely generated ${\mathcal O}_{K_p}$-module in $K_p$ that contains $\mathcal O_{K_p}$, so its inverse is a nonzero ideal in $\mathcal O_{K_p}$ that is called the local different ideal. It is $\mathcal O_{K_p}$ if and only if $K_p$ is unramified. When $K_p/\mathbf Q_p$ is ramified and $\pi$ is a uniformizer in $K_p$, the local different is $\pi^d\mathcal O_{K_p}$ where $d \geq 1$. On the other hand, the set $$ T^{-1}(\mathbf Z_p) = \{\alpha \in K_p : {\rm Tr}_{K_p/\mathbf Q_p}(\alpha) \in \mathbf Z_p\} $$ is some ugly thing with no worthwhile features. For example, when $K_2 = \mathbf Q_2(i)$, $$ T^{-1}(\mathbf Z_2) = \{\alpha \in K_2 : {\rm Tr}_{K_2/\mathbf Q_2}(\alpha) \in \mathbf Z_2\} = \frac{1}{2}\mathbf Z_2 + \mathbf Q_2i. $$ and when $K_3 = \mathbf Q_3(i)$, $$ T^{-1}(\mathbf Z_3) = \{\alpha \in K_3 : {\rm Tr}_{K_3/\mathbf Q_3}(\alpha) \in \mathbf Z_2\} = \frac{1}{2}\mathbf Z_3 + \mathbf Q_3i = \mathbf Z_3 + \mathbf Q_3i. $$

I suspect you really want to work with the local different ideal (or its inverse), not $T^{-1}(\mathbf Z_p)$.


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