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I'm looking for solutions $t$ of an equation of the form $$ t \sigma(t) \cdots \sigma^{n-1}(t) = v $$ in a field equipped with an automorphism $\sigma$ of order $n$. In this case, I call $t$ a "$\sigma$-semilinear $n$-th root of $v$".

I have a particular situation in mind, coming from the study of $\varphi$-modules (see the motivation provided at the end of the question). This means that we are working over a $p$-adic field, and $\sigma$ is the Frobenius automorphism.
Allow me then to make the question more specific and precise.

Let $K / \mathbb{Q}_p$ be a finite extension, with ring of integers $\mathcal{O}_K$. Choose a uniformiser $\varpi$ of $\mathcal{O}_K$.
Let $K_0$ denote the maximal unramified intermediate extension $K / K_0 / \mathbb{Q}_p$. Write $f = [K_0 : \mathbb{Q}_p]$ for the inertial degree of $K/\mathbb{Q}_p$.

We know that $K_0 / \mathbb{Q}_p$ is a cyclic extension of degree $f$. Its Galois group is generated by the Frobenius automorphism, which we'll call $\sigma \in \text{Gal}(K_0 / \mathbb{Q}_p)$.

Question: I am looking for an explicit formula for $t \in K_0 \otimes_{\mathbb{Q}_p} K$, say $$ t = \sum_i x_i \otimes y_i, $$ such that $$ \sum_i x_i y_i = \varpi$$ and $$ \sum_i \sigma^j(x_i) y_i = 1,$$ for all $j = 1, \ldots, f-1$.

Such an element is then necessarily a $\sigma$-semilinear $f$-th root of $1 \otimes \varpi$. There are others (for instance, $\sigma(t)$ is another one; or one could throw in some units), but this particular one seems to be the element that comes up (see the motivation at the end).

Is there a general formula for $t$ that works for all $K$? Note that such $t$ is in fact integral, meaning that it lives in $\mathcal{O}_{K_0} \otimes_{\mathbb{Z}_p} \mathcal{O}_K$. So there should be no need to introduce denominators.
If you know of a formula for the special case when $K$ is unramified, so that $K = K_0$ and $\varpi = p$, I'd also love to hear it.

Motivation: The motivation for the question comes from $\varphi$-modules. The $\varphi$-module attached (contravariantly) to the Lubin–Tate formal group $\text{LT}_{K,\varpi}$ is a free $K_0 \otimes_{\mathbb{Q}_p} K$-module of rank $1$ on which $\varphi^f$ acts by multiplication by $1 \otimes \varpi$. The element $t$ then provides a natural ($\sigma$-semilinear) $f$-th root of $\varpi$, i.e. $$ t \sigma(t) \cdots \sigma^{f-1}(t) = \varpi.$$ It should then be the case that the ($\sigma$-semilinear) action of $\varphi$ on the $\varphi$-module attached to $\text{LT}_{K,\varpi}$ is given by $t$. The question is thus of finding an explicit description of this $\varphi$-module.

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    $\begingroup$ What happens in the extreme cases, namely when $K$ is either totally ramified or unramified over $\Bbb Q_p$? $\endgroup$ – Lubin Apr 5 '16 at 17:41
  • $\begingroup$ @Lubin If $K/\mathbb{Q}_p$ is totally ramified, then $f = 1$ and we can take $t = \varpi$. I don't know how it works if $K$ is unramified. I am hoping there is some expression involving $(p^f-1)$-th roots of unity that could work, but I haven't found a sensible way to approach it yet. $\endgroup$ – Will Apr 5 '16 at 18:28
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I have worked out an explicit formula in the special case that $K_0/\mathbb{Q}_p$ is quadratic, i.e. $f=2$.
Let $\zeta = \zeta_{p^2-1}$ be a primitive $(p^2-1)$-th root of unity in $K_0$, so that $K_0 = \mathbb{Q}_p(\zeta)$.

Define then $t \in K_0 \otimes_{\mathbb{Q}_p} K$ by: $$ t = 1 \otimes \left ( \varpi + (1-\varpi) \frac{2 \zeta^{p+1} - \zeta (\zeta+\zeta^p)}{4\zeta^{p+1}-(\zeta+\zeta^p)^2 }\right ) + \zeta \otimes \left ( (1-\varpi) \frac{ -(\zeta+\zeta^p)+2\zeta}{4\zeta^{p+1}-(\zeta+\zeta^p)^2 }\right ).$$ Note that $\zeta^{p+1}$ and $(\zeta+\zeta^p)$ are both elements of $\mathbb{Z}_p$, because they are the norm (respectively, the trace) of $\zeta$. Moreover, $(\zeta+\zeta^p)^2-4\zeta^{p+1}$ is the discriminant of the extension $\mathbb{Q}_p(\zeta)/\mathbb{Q}_p$; as this extension is unramified, it is a unit. This means that the above formula in fact defines an integral element $t \in \mathcal{O}_{K_0} \otimes_{\mathbb{Z}_p} \mathcal{O}_K$.

Using the identities $\zeta^2 = (\zeta+\zeta^p)\zeta - \zeta^{p+1}$ and $\zeta^p = (\zeta+\zeta^p)-\zeta$, it is straightforward to check that indeed $t$ satisfies the formulas of the question.

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