Does writing the infinite binary strings with the radix two and one, still give an exact cover of the two-adic numbers?

Unless I'm mistaken, $\displaystyle \Bbb Z_2=\left\{\sum_{i=0}^\infty2^ia_i:a_i\in\{0,1\}\right\}$ is an exact cover of the 2-adic integers.

And we can substitute any 2-adic unit for one in the radix, so for example:

$\displaystyle \Bbb Z_2=\left\{\sum_{i=0}^\infty2^ia_i:a_i\in\{0,5\}\right\}$

But what about substituting for zero? Can we substitute any number equivalent to $0\pmod 2?$ It's not so clear to me whether it is also the case that $\displaystyle \Bbb Z_2=\left\{\sum_{i=0}^\infty2^ia_i:a_i\in\{2,1\}\right\}$? I don't see how that would work.


1 Answer 1


For any prime element $\pi \in \mathbb Z_p$ and every fixed set of representatives $R \subset \mathbb Z_p$ of $\mathbb Z_p/p\mathbb Z_p \simeq \mathbb Z_p/\pi \mathbb Z_p$, each element $x \in \mathbb Z_p$ has a unique representation $x = \sum_{i=0}^\infty a_i \pi^i$ with all $a_i \in R$.

So instead of $2$ in your series expression you can take e.g. $6$, or $-18$, or $2\sqrt{17}$ (for either choice of $\sqrt{17}$ that exists in $\mathbb Z_2$). You cannot take "any number equivalent to $0$ (mod $2$)", namely not $4$ or $-32$ or $100$ or anything else that is not a prime a.k.a. uniformizer, i.e. does not have $2$-adic absolute value $1$.

Those series expansions look strange at first. In the first part of my answer to Why does this generalized ring of Witt vectors not depend on a choice of a prime element?, I wrote $5$-adic numbers as series expansions with the prime element $10 \in \mathbb Z_5$ (but keeping the popular set of representatives $\{0,1,2,3,4\}$ as $R$).

  • 1
    $\begingroup$ Thank-you. This question arose because I was struggling to write $0$ using the radix $\{2,1\}$. So trying again with renewed confidence I came up with $\overline12_2=0$, correct? $\endgroup$ Jun 6 at 15:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .