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Let $L$ be a finite extension of $\mathbb Q_p$ with ring of integers $\mathcal{O}=\mathcal{O}_L$ and let $B_1(L):=\{z \in L \colon \vert z-1 \vert <1 \}$. Let $\widehat{\mathcal{O}}(L)_{\mathbb Q_p}$ denote the set of all locally $\mathbb Q_p$-analytic characters on $\mathcal{O}$ with values in $L$. It can be shown, that \begin{align} B_1(L) \otimes_{\mathbb Z_p} \text{Hom}_{\mathbb Z_p}(\mathcal{O}, \mathbb Z_p) & \to \widehat{\mathcal{O}}(L)_{\mathbb Q_p} \\ z \otimes \beta & \mapsto [l \mapsto z^{\beta(l)}] \end{align} is a bijection. Denote the character associated to $z \otimes \beta$ under this map by $\chi_{z,\beta}$. We also define the map \begin{align} d\colon \widehat{\mathcal{O}}(L)_{\mathbb Q_p} & \to \text{Hom}_{\mathbb Q_p}(L,L) \\ \chi & \mapsto d\chi, \end{align} where $d\chi(a):=\frac{d}{dt}\chi(t \cdot a)\vert_{t=0}$ for $a \in L$.

I want to show that

\begin{align} d\chi_{z,\beta}=log(z)\cdot \beta. \end{align}

In my calculations I keep coming up with $d\chi_{z,\beta}(a)=log(z)\cdot \beta'(0)\cdot a$, which is not right. I would appreciate someone showing me the right way to calculate this.

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  • $\begingroup$ There is a suggested edit, where the editor thinks that you should use $\mathcal{O}$ instead of $o$, $\mathcal{O}_L$ instead of $o_L$ et cetera. I gues the reason is that $\mathcal{O}$ often stands for the ring of integers of an extension field. Would you be happy with such an edit, or is your text using those smaller letters? $\endgroup$ – Jyrki Lahtonen Dec 25 '18 at 21:08
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    $\begingroup$ I have never studied this kind of characters, so I'm prepared to be very wrong. But, may be you are differentiating $\beta$ a bit wrong. Remember that $\beta$ is only linear over $\Bbb{Z}_p$, not over $o$. I think that the derivative of $\beta$ should be something not unlike a Frechét derivative from multivariable calculus. After all, it is a multivariable function. But, being linear, it is equal to its own derivative. $\endgroup$ – Jyrki Lahtonen Dec 25 '18 at 21:19
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    $\begingroup$ Same as Jyrki, we have the directional derivative $D_{v}\chi_{z,\beta}(a)=\lim_{n \to \infty} \frac{\chi_{z,\beta}(a+v p^n))-\chi_{z,\beta}(a)}{p^n}$. Letting $\mathcal{O} = \sum_{j=1}^{[L:\mathbb{Q}_p]} o_j \mathbb{Z}_p$ then it suffices to know $D_{o_j}\chi_{z,\beta}(a)$ as $D_v f(a)$ is $\mathbb{Z}_p$-linear in $v$, and so $D_{o_j} \chi_{z,\beta}(0)=log(z) D_{o_j} \beta(0)$ where $\beta( \sum_{j=1}^{[L:\mathbb{Q}_p]}u_j o_j) = \sum_{j=1}^{[L:\mathbb{Q}_p]} u_j D_{o_j} \beta(0)$ for $u_j \in \mathbb{Z}_p$ as $\beta$ is $\mathbb{Z}_p$ linear. $\endgroup$ – reuns Dec 26 '18 at 0:26
  • $\begingroup$ @JyrkiLahtonen, thanks for the hint, working that out will probably lead to the solution. I will try to do it in the comments here. As for the edit with the $o$, it's all right. $\endgroup$ – Layer Cake Dec 26 '18 at 12:00
  • $\begingroup$ @reuns I find your comment helpful too, but I have two questions: $1)$ How do you get the equality $D_{o_j} \chi_{z,\beta}(0)=log(z) D_{o_j} \beta(0)$? $2)$ Having all these equations for the directional derivatives, how do I conclude the desired equation for the total derivative? $\endgroup$ – Layer Cake Dec 26 '18 at 12:12

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