I want to show $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2} \pi^2+\pi^3 \mathbb{Z}_p[\zeta_p]$, for $p \neq 2$.

Question

Consider the cyclotomic extension $\mathbb{Q}_p(\zeta_p)$, its ring of integers $\mathbb{Z}_p[\zeta_p]$ and uniformizer $\pi$.

Also assume that $\log_p(\alpha) \in p \mathbb{Z}_p[\zeta_p]$ for $\alpha \neq 0$. Here $\log_p$ is $p$-adic logarithm.

I want to show $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2} \pi^2+\pi^3 \mathbb{Z}_p[\zeta_p]$, for $p \neq 2$, $a_0 \in \mathbb{Z}_p[\zeta_p]$.

proceeding, \begin{align} \log_p(\alpha+a_0 \pi) &=\log_p(\alpha)+\log_p(1+a_1 \pi), \ \text{where} \ a_1=\frac{a_0}{\alpha} \in \mathbb{Z}_p[\zeta_p]. \end{align} If we assume $a_1 \in \mathbb{Z}_p[\zeta_p]-\pi \mathbb{Z}_p[\zeta_p]$, then $a_1=\beta+a_2 \pi \mathbb{Z}_p[\zeta_p]$ with $\beta \in \{0,1,2, \cdots, p-1\}$ and $a_2 \in \mathbb{Z}_p[\zeta_p]$. Then, \begin{align} \log_p(\alpha+a_0 \pi) &=\log_p(\alpha)+\log_p(1+(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)) \\ &=\left[(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)-\frac{(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)^2}{2}+\frac{(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)^3}{3}-\cdots \right]+\log_p(\alpha) \end{align}

From the second term of $RHS$, we get $-\frac{\beta^2}{2} \pi^2$,

but how to manipulate and abolish other terms in order to get the conclusion.

If $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2} \pi^2+\pi^3 \mathbb{Z}_p[\zeta_p]$, for $p \neq 2$ doesn't exactly hold , what would be the nearby relation ?

I mean, I need $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2}\pi^2+(\cdots \? \cdots)$, for $p \neq 2$.

Any help please.

Answer 1

I think this boils down to $$\log_p(1+\beta\pi)\equiv-\frac{\beta^2}2\pi^2\pmod{\pi^3}.$$ Well, $$\log_p(1+\beta\pi)=\beta\pi-\frac{\beta^2}{2}\pi^2+\cdots+\frac{\beta^p}{p}\pi^p -\cdots.\tag{*}$$ You want the $\beta\pi$ and $\beta^p\pi^p/p$ terms to "cancel" appropriately. As you say, $\beta=b+\gamma\pi$ where $b\in\Bbb Z$. Also you can choose the uniformiser in the extension field to satisfy $\pi^{p-1}=-p$. In this case $$\frac{\beta^p}p\pi^p=-(b+\gamma\pi)^p\pi\equiv-b^p\pi \equiv-b\pi\pmod{\pi^{p+1}}$$ so that $\beta\pi+\beta^p\pi^p/p$ is certainly zero modulo $\pi^3$. I reckon that the terms in (*) not listed there have $\pi$-valuation at least $3$.

Answer 2

Counterexample: $\alpha:=1, a_0:= \dfrac{\zeta_p-1}{\pi}$, so that $\alpha + a_0 \pi =\zeta_p$ and hence $$(*) \qquad \log_p(\alpha+a_0\pi)= \log_p(\zeta_p)=0.$$

But following your notation, we have $a_1=a_0$ and obviously $a_0 \in \mathbb Z_p[\zeta_p]^*$ meaning $\beta \not \equiv 0$ mod $(\pi)$, so if we had $$ \log_p(\alpha+a_0\pi) \stackrel{?}\equiv -\dfrac{\beta^2}{2}\pi^2 \text{ mod } \pi^3$$ this would contradict $(*)$.

The error in the other answer is explained in my comment to it.