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For which $\mathbb{Z}_2$ is the logarithm defined?

I have that the log series $\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$ gives a convergent series of p-adic integers whenever $x=pz:z\in\mathbb{Z}_2$

Therefore $\log(x)$ is well-defined provided $x$ is a principal unit (of the form $1+pz:z\in\mathbb{Z}_2$), and $\log{a}+\log{b}=\log{ab}$ provided $a,b$ are principal units.

I'm trying to get an idea what these principal units look like in $\mathbb{Z}_2$.

Since they're of the form $1+2z$ I can't escape the ideas that a) these are therefore the numbers ending in a $1$ and b) that would make them the numbers of the form $\lvert x\rvert_2=1$. This would make them the unit group. But surely the unit group must be distinct from the principal units?

So it seems I've got myself in a tangle somewhere.

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  • $\begingroup$ But the units do end in $1$... $\endgroup$ – Kenny Lau Jan 8 '18 at 20:37
  • $\begingroup$ @KennyLau are you saying the unit group is one and the same as the principal units in $\mathbb{Z}_2$? $\endgroup$ – samerivertwice Jan 8 '18 at 20:40
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    $\begingroup$ @KennyLau so my misconception is that the unit group must be distinct from the principal units. $\endgroup$ – samerivertwice Jan 8 '18 at 20:41
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    $\begingroup$ As the characteristic is $2$, every unit ends in $1$. $\endgroup$ – Kenny Lau Jan 8 '18 at 20:42
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    $\begingroup$ It’s nothing but the fact that $(\Bbb Z/2\Bbb Z)^*=\{1\}$. $\endgroup$ – Lubin Jan 9 '18 at 5:36
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The units and principal units are exactly the same in $\mathbb{Z}_2$, so you aren't missing anything. The $2$-adic logarithm series is defined on all units of $\mathbb{Z}_2$. (However, some of the nice properties you would expect of the $2$-adic logarithm, such as being the inverse of the $2$-adic exponential, only work when you restrict its domain to elements of the form $1+4z$ for $z\in\mathbb{Z}_2$.)

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  • $\begingroup$ Thanks. This may be way off (and probably is!) but I'm looking for a natural way to order the 2-adics, analogous to identifying which of them are positive and which are negative. Is it possible that a pair of square roots of some number are always one $\equiv1$ and the other $\equiv3\mod 4$ and this gives is some way to choose a more natural candidate for the "positive" square root? The reason I ask is you mention the shortcoming of the logarithm of the $\equiv3\mod 4$ numbers and it makes me think of the shortcoming of logs of negative numbers in $\mathbb{R}$ (i.e. they're in $\mathbb{C}$)! $\endgroup$ – samerivertwice Jan 9 '18 at 21:32
  • $\begingroup$ Well, that is always true of the two square roots of any $2$-adic unit (since the square roots are negatives of each other). But defining $2$-adics that are $1$ mod $4$ to be "positive" will not have most of the familiar properties of positive numbers, and definitely does not give rise to a natural total order. $\endgroup$ – Eric Wofsey Jan 9 '18 at 21:35
  • $\begingroup$ I don't think the rule could apply to $\mathbb{Q}\subset\mathbb{Q}_2$ as the idea is for an extension of the natural ordering that already exists on those. So there would have to be some process of induction by which it's extended to $\mathbb{Q}_2\setminus \mathbb{Q}$ $\endgroup$ – samerivertwice Jan 9 '18 at 21:36
  • $\begingroup$ Actually it does preserve the multiplicative group $(+,-)$; at least in $\mathbb{Z}\subset\mathbb{Z}_2$ doesn't it? $\endgroup$ – samerivertwice Jan 9 '18 at 21:57
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The case $p=2$ is a bit different than primes $p>2.$ In general, the complete logarithm of a $p$-adic number $x$ consists of three pieces of data: $[v,n,y]\;$ such that $\;x=p^v z_p^n \exp(y)\;$ where $v\in Z,\;$ $z_p$ is a primitive root where $\;z_p^{p-1}=1$ (or $z_p=-1$ if $p=2$), and $y\in pZ_p$ (or $y\in 4Z_p$ if $p=2$).

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