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Every non-zero $p$ adic number ($p$ is a positive rational prime number) can be uniquely written in canonical form, formally as a Laurent series, $$\xi = a_{-m}p^{-m}+a_{-m+1}p^{-m+1}+\cdots+a_0+a_1p+a_2p^2+\cdots$$ where $m \in \mathbb{N}$ $0 \leqq a_k \leqq p-1$ , for all $k$ s, and at least one of the integers $a_k$ is positive. In addition, we can write: $0 = 0+0p+0p^2+\cdots$ The field $\mathbb {Q}_p$ of the $p$ adic numbers is the completion of the field $\mathbb {Q}$ with respect to its $p$ adic valuation; thus $\mathbb {Q}$ may be thought the subfield
(prime subfield) of $\mathbb {Q}_p$ We can call the elements of $\mathbb{Q}_p\!\setminus\!\mathbb{Q}$ , the proper $p$ adic numbers.
If, e.g., $p = 2$ we have the 2-adic or, according to G. W. Leibniz, dyadic numbers, for which every $a_k$ is 0 or 1. In this case we can write the sum expression for $\xi$ in the reverse order and use the ordinary positional binary (i.e., dyadic) figure system. Then, for example, we have the rational numbers $$-1 = ...111111,$$ $$1 = ...0001,$$ $$6.5 = ...000110.1,$$ $$\frac{1}{5}= ...00110011001101.$$ (You may check the first by adding 1, and the last by multiplying by 5 = ...000101.) All 2-adic rational numbers have periodic binary expansion. Similarly as the decimal (according to Leibniz: decadic) expansions of irrational real numbers are aperiodic, the proper 2-adic numbers also have aperiodic binary expansion, for example the 2-adic fractional number $$\alpha = ...1000010001001011.10111.$$
The 2-adic fractional numbers have some bits ``1'' after the dyadic point ``.'' (in continental Europe: comma ``$,$ '), the 2-adic integers have none. The 2-adic integers form a subring of the 2-adic field $\mathbb {Q}_2$ such that $\mathbb{Q}_2$ is the quotient field of this ring.
Every such 2-adic integer $\varepsilon$ whose last bit is ``1'', as $-3/7 = ...11011011011$ is a unit of this ring, because the division $1\colon\!\varepsilon$ , clearly gives as quotient a similar integer (by the way, the divisions of the binary expansions in practice go from right to left and are very comfortable!).
Those integers ending in a ``0'' are non-units of the ring, and they apparently form the only maximal ideal in the ring (which thus is a local ring). This is a principal ideal $\mathfrak{p}$ the generator of which may be taken $...00010 = 10$ (i.e., two). Indeed, two is the only prime number of the ring, but it has infinitely many associates, a kind of copies, namely all expansions of the form $...10 = \varepsilon\cdot
10$ The only non-trivial ideals in the ring of 2-adic integers are $\mathfrak{p},\, \mathfrak{p}^2,\, \mathfrak{p}^3,\, \ldots$ , They have only 0 as common element.
All 2-adic non-zero integers are of the form $\varepsilon\cdot 2^n$ where $n = 0,\,1,\,2,\,\ldots$ The values $n = -1,\,-2,\,-3,\,\ldots$ , here would give non-integral, i.e. fractional 2-adic numbers.
If in the binary representation of an arbitrary 2-adic number, the last non-zero digit ``1'' corresponds to the power $2^n$ then the 2-adic valuation of the 2-adic number $\xi$ is given by $$|\xi|_2 = 2^{-n}.$$
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