Logarithm in $2$-adic numbers.

algebraic-number-theorylogarithmsp-adic-number-theory

  1. How can we find the first $10$ digits of $\dfrac{\log 5}{\log 13}$ in the $2$-adic numbers?
    I mean how can we write down $\{ a_i\}_{i=0}^{10}$'s explicitly,
    in $\dfrac{\log 5}{\log 13}=:\sum_{i=0}^{\infty} a_i 2^i$.

Motivation for asking this question:

Let's consider the set of prime numbers that are congruent to $5$ modulo $8$:

$$\{5, 13; 29, 37; 53, 61; 101, 109; 149, 157; 173, 181; 197, \cdots\}.$$

For any prime $p$ in this set we have $v_2(p-1)=2$.
From now on we assume that $2 \leq n$.
With a little work,
we can show that $\displaystyle{{\rm order}_{2^n}p = 2^{n-2}}$.
This implies that $p$ is a generator for the index $2$ subgroup
$$H:=\{a\in \dfrac{\mathbb{Z}}{2^n\mathbb{Z}} \mid a \stackrel{4}{\equiv} 1\},$$
of $\dfrac{\mathbb{Z}}{2^n\mathbb{Z}}$.
Considering these facts,
we can find that $\overline{\langle p\rangle}=1+4\mathbb{Z}_2$,
in the $2$-adic topology.

Now let $p_1=5$ and $p_2=13$. I am not sure, but I think we can conclude that $5=13^z$, for some $z \in \mathbb{Z}_2$.

  1. I can not see why should we have $\dfrac{\log p}{\log q} \in \mathbb{Z}_2$,
    for arbitrary prime numbers $p,q \stackrel{8}{\equiv} 5$.

Best Answer

  1. You can do it by hand, but it is much easier using a calculator (just as you do calculations of real numbers).

Paste the following into this site and press "Evaluate" to see the result.

F= Zp(2)
print(F)
print(log(F(5))/log(F(13)))

Output:

2-adic Ring with capped relative precision 20
1 + 2 + 2^2 + 2^3 + 2^8 + 2^9 + 2^10 + 2^13 + 2^15 + 2^17 + O(2^18)
  1. Because $\log$ is an isometry from $1 + 4\Bbb Z_2$ to $4\Bbb Z_2$ (which is an easy exercise). Thus for every $p \equiv 5\mod 8$, we have $$v_2(\log(p)) = v_2(\log(p) - \log(1)) = v_2(p - 1) = 2,$$ and hence the quotient $\log(p)/\log(q)$ has $2$-adic valuation $0$.
Related Question