Lower bound for the valuation of the trace of an element in a $p$-adic cyclotomic extension.

Question

I'm reading Pierre Colmez's Fonctions d'une variable p-adique and he summons the following inequality without proof:

Let $F_{n} := \mathbb{Q}_{p}(\zeta_{p^{n}})$ where $p$ is prime. Let $x \in F_{n}$, and let $v_{p}$ be the standard $p$-adic valuation on $\mathbb{Q}_{p}$, then $$ v_{p}(\text{tr}_{F_{n} / \mathbb{Q}_{p}}(x)) \geq n + \left\lfloor v_{p}(x) - \frac{1}{p-1}\right\rfloor $$

I've spent a few minutes toying with traces in cyclotomic extensions to no avail. I'd ideally spend longer on this but I haven't much time so I was hoping for a hint (read: not a full answer unless it is totally trivial) to push me in the right direction.

Answer 1

EDIT: The answer I originally posted here was only supposed to give a "hint", but as pointed out in comments, whatever I thought back when I wrote it cannot have been thought through. It basically stopped after the first computation of traces of $p$-power roots of unity.

Now, user994373's "different" answer is obviously the best. But I amended some computations and ideas below which give a little more precise results in some cases, and maybe a tiny bit of insight from a different angle into what's going on here.

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For $p \neq 2$, one can show a stronger statement with the following facts:

Let $\zeta_{p^k}$ be any primitive $p^k$-th root of unity. For $n \ge 1$ and $0 \le k \le n$ one has:

$$\text{tr}_{F_{n} \vert \mathbb{Q}_{p}} (\zeta_{p^k}) = \begin{cases} p^{n-1}(p-1) \qquad\text{ if } k=0 \text{ i.e. } \zeta_{p^0}=1\\-p^{n-1} \qquad \qquad \; \text{ if } k=1\\ 0 \qquad \qquad \qquad \; \; \text{ if } k \ge 2 \end{cases}$$

Let's fix such choices of primitive $p^k$-th roots of unity in our field.

In the case $n=1$, it's known that $\xi_1 := \zeta_p-1$ has valuation $v_p(\xi_1)= \frac{1}{p-1}$. This means that $$1, \xi_1, \xi_1^2, ..., \xi_1^{p-2}$$

are not only a $\mathbb Q_p$-basis of $F_1$, but have the stronger property that if we write any $x \in F_1$ as

$$x_0 \cdot 1 + x_1 \cdot \xi_1+ x_2 \cdot \xi_1^2 + ... +x_{p-2} \cdot \xi_1^{p-2}$$

then $$v_p(x) = \min_{0 \le i \le p-2} \left(v_p(x_i) + \frac{i}{p-1}\right).$$

Let's call $m(x) := \min v_p(x_i)$ and $i(x) := \min \{i: v_p(x_i)=m\}$. Note that $$v_p(x) = m(x) + \dfrac{i(x)}{p-1},$$ so that the RHS of the formula becomes

$$1 + \left\lfloor v_{p}(x) - \frac{1}{p-1}\right\rfloor = \begin{cases} m(x) (= v_p(x_0)) \quad \text{ if } i(x)=0 \\ 1 +m(x) \qquad \qquad \text { if } i(x) \ge 1 . \end{cases}$$

On the other hand, using the binomial formula, linearity of the trace and the fact that $\zeta_p^j$ is still a primitive $p$-th root of unity for all $1\le j \le p-1$, we have

$$\text{tr}_{F_{n} \vert \mathbb{Q}_{p}} (\xi_1^k) = \begin{cases} p-1 \quad \;\text{ for } k=0 \\-p \qquad \; \text{ for odd } 1 \le k \le p-2 \\ p \qquad \quad \text{ for even } 2 \le k \le p-2 \end{cases}$$

consequently

$$\text{tr}_{F_{1} \vert \mathbb{Q}_{p}} (x_0 \cdot 1 + x_1 \cdot \xi_1+ x_2 \cdot \xi_1^2 + ... +x_{p-2} \cdot \xi_1^{p-2}) \\ = (p-1) \cdot x_0 + p \cdot (-x_1 + x_2 -x_3 + ... \pm x_{p-2})$$

consequently the LHS of the formula is

$$v_p(\text{tr}_{F_{1} \vert \mathbb{Q}_{p}} (x)) \begin{cases} = m(x) (= v_p(x_0)) \quad \text{ if } i(x)=0 \\ \ge 1 +m(x) \qquad \qquad \text { if } i(x) \ge 1 .\end{cases}$$

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In the case $n=2$ we repeat the idea with $\xi_2 := \zeta_{p^2}-1$ but use as basis of $F_2$:

$$1, \xi_2, \xi_2^2, ..., \xi_2^{p-1}, \\ \xi_1, \xi_1 \xi_2, \xi_1 \xi_2^2, ..., \xi_1 \xi_2^{p-1}, \\ \xi_1^2, \xi_1^2 \xi_2, \xi_1^2 \xi_2^2, ..., \xi_1^2 \xi_2^{p-1},\\ \vdots \\ \xi_1^{p-2}, \xi_1^{p-2} \xi_2, \xi_1^{p-2} \xi_2^2, ..., \xi_1^{p-2} \xi_2^{p-1}$$

That's $p-1$ rows with $p$ elements in each row. If we numerate them through with indices $i=0,1, ..., p(p-1)-1$, then the $i$-th element has $p$-adic valuation $\dfrac{i}{p(p-1)}$.

Again we write a general element $x \in F_2$ in that basis, let its $i$-coefficient be $x_i$. Again call $m(x) := \min v_p(x_i)$ and $i(x) := \min \{i: v_p(x_i)=m\}$. Note that $v_p(x) = m(x) + \dfrac{i(x)}{p(p-1)}$.

Now the RHS of the formula becomes

$$2 + \left\lfloor v_{p}(x) - \frac{1}{p-1}\right\rfloor = \begin{cases} 1 +m(x) \text{ if } 0 \le i(x) \le p-1 \\ 2 +m(x) \text{ else.} \end{cases}$$

As for the traces: It looks like a lot of basis elements, but their traces are very simple! Indeed, all elements of the first row have trace $= \pm\text{tr}_{F_{2} \vert \mathbb{Q}_{p}} (1) = \pm p(p-1)$; and then, all remaining elements (starting in the second row) have trace $= \pm\text{tr}_{F_{n} \vert \mathbb{Q}_{p}} (\xi_1) = \pm p^2$. So the trace of a general $x$ written in our basis is

$$\text{tr}_{F_{2} \vert \mathbb{Q}_{p}} (x)) \\ = p(p-1) \cdot (\pm x_0 \pm x_1 \pm ... \pm x_{p-1})+ p^2 \cdot (\pm x_{p} \pm x_{p+1} \pm ... \pm x_{p(p-1)-1})$$

so the LHS of the formula is

$$v_p(\text{tr}_{F_{2} \vert \mathbb{Q}_{p}} (x)) \begin{cases} \ge 1 +m(x) \text{ if } 0 \le i(x) \le p-1 \\ \ge 2 +m(x) \text{ else.} \end{cases}$$

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I claim this generalizes neatly to higher $n$. It becomes cumbersome to write down the basis, but the idea should be visible. The case distinction in the formula happens precisely when we get to the primitive $p$-th root of unity, which happens at index $i=p^{n-1}$, i.e.

$$v_p(\text{tr}_{F_{n} \vert \mathbb{Q}_{p}} (x)) \begin{cases} \ge n-1 +m(x) \text{ if } 0 \le i(x) \le p^{n-1}-1 \\ \ge n +m(x) \text{ else.} \end{cases}$$

and

$$n + \left\lfloor v_{p}(x) - \frac{1}{p-1}\right\rfloor = \begin{cases} n-1 +m(x) \text{ if } 0 \le i(x) \le p^{n-1}-1 \\ n +m(x) \text{ else.} \end{cases}$$

The case $p=2$ is left to the reader.

Answer 2

I am sure this ancient question is no longer of interest (I guess I only noticed due to the edit) and the OP is probably long gone. But the statement to be proved is a special case of the following standard result: if $L/K$ is a totally ramified extension of local fields of ramification degree $e = [L:K]$, and $v$ is the valuation on $K$ (extended to $L$) normalized so that $v(\pi_K) = 1$, then

$$v(\mathrm{Tr}_{L/K}(x)) \ge \lfloor v_p(x) + v_p(\delta_{L/K}) \rfloor$$

where $\delta^{-1}_{L/K}$ is the inverse different (a fractional ideal of $L$). Since the norm of the different is the discriminant, one can replace $v_p(\delta_{L/K})$ by $v_p(\Delta_{L/K})/e$. For cyclotomic fields, the discriminant has valuation $n p^{n-1}(p-1) - p^{n-1}$, which after dividing by the degree is equal to $n - 1/(p-1)$. Of course one can pull $n$ out of the floor function, and one recovers the claim in the original question,

The proof of the claim is to note that, for some unit $u$, we have

$$x = \pi^{e \cdot v_p(x)}_L u = \pi^{e \cdot v_p(x) + e \cdot v_p(\delta_{L/K})}_L u \cdot \pi^{-e \cdot v_p(\delta_{L/K})}_L$$

Since $(\pi_K) = (\pi^e_L)$, if $m = \lfloor v_p(x) + v_p(\delta_{L/K}) \rfloor$ it follows that $\pi^{e \cdot v_p(x) + e \cdot v_p(\delta_{L/K})}_L$ is divisible by $\pi^{e \cdot m}_L$ (this makes sense even for general non-zero $x \in L$ rather than $\mathcal{O}_L$ where divisible means that the ratio is integral) and thus also divisible by $\pi^m_K$. It follows that

$$x \in \pi^m_K \cdot \delta^{-1}_{L/K}.$$

By definition, the trace of any element in the inverse different lies in $\mathcal{O}_K$. (Equivalently, it is the dual of $\mathcal{O}_K$ under the trace pairing; from this description one can prove the claim above relating the different to the discriminant.) Since the trace is $\mathcal{O}_K$-linear, it follows that the trace of $x$ is divisible by $\pi^m_K$ which is the required claim.