[Math] Why do we define the $\mathfrak{p}$-adic logarithm on a $\mathfrak{p}$-adic number field such that $\log(p) = 0$

algebraic-number-theorydefinitionp-adic-number-theory

Suppose we have a finite extension $K / \mathbb{Q}_p$ with valuation ring $\mathcal{O}$ and maximal ideal $\mathfrak{p}$.

One can define the $\mathfrak{p}$-adic logarithm on the group of principal units $U^{(1)}$ of the local field $K$ using the power series expansion $$\log(1 + x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} + \ldots.$$

One can then extend the definition to a map $\log\colon K^\ast \rightarrow K$ satisfying the properties $\log(xy) = \log(x) + \log(y)$ and $\log(p) = 0$.

My question is, why do we want $\log(p) = 0$?

With the usual logarithm over $\mathbb{R}$, the kernel of $\log$ is $\{1\}$, and I don't see an analogy where $p$ could correspond to something in $\mathbb{R}$. So what makes this particular choice of $\log(p)$ desirable, over for example some other choice like $\log(p) = e$, where $(p) = \mathfrak{p}^e$?

Best Answer

The main goal is to construct a continuous function $log_p: \mathbf C_p ^* \to \mathbf C_p$ s.t. $log_p (xy) = log_p (x) + log_p (y)$. Since $ \mathbf C_p ^* = p^\mathbf Q \times W \times U_1$, where $U_1$ is the group of principal units and $W$ the group of roots of $1$ of order prime to $p$, it suffices to define $log_p$ on each of the direct factors. On $U_1$ one has already the usual power series $log_p (1+x)$ whose radius of convergence is $1$. On $W$, one must have the nullity of $log_p$, since for any root of unity $w$ of order $n$, necessarily $n.log_p (w)= log_p (1) = 0$. It remains only to adjust the value $log_p (p)$.

The choice is not quite arbitrary, because any $\sigma \in G_\mathbf {Q_p} $can be extended to a continuous automorphism of $\mathbf C_p$, and it follows that $log_p (p) \in \mathbf Q_p$. Your suggested choice $log_p (p)=e$ is not good either because it depends on the ambient field $K$. Actually, most of the ramification problems in CFT are concentrated in $U_1$, as well as most of the calculations about $L_p$-functions , so the definitely most natural (which is also the most simple) choice is $log_p (p)=0$. It follows that Ker $log_p = p^\mathbf Q \times \mu$, where $\mu$ is the group of all roots of unity (of arbitrary order).

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