Lagmental Vicfred

A p-Adic Integer Is an Infinite Base-p Expansion by Vicfred

tags: mathematics, p-adic numbers, valuations, Hensel lemma

Digits extending indefinitely to the left converge because higher powers of p become smaller. The example is deliberately concrete; it is a test of the statement, not a substitute for it.

Statement

The valuation \(v_p(n)\) counts factors of \(p\), and the metric \(|x|_p=p^{-v_p(x)}\) reverses the usual sense of size. Hensel lifting turns approximate roots modulo \(p\) into exact \(p\)-adic roots.

$$ x=a_0+a_1p+a_2p^2+\cdots,\qquad0\le a_i<p $$

The formulas should not be merged too early. The datum \(\mathsf D\), the conclusion \(\mathsf C\), and the bridge \(\Longrightarrow\) have three different logical jobs.

$$ \mathbf Z_p\cong\varprojlim_n\mathbf Z/p^n\mathbf Z $$

Worked algebra

The following line is the smallest calculation that still exercises the mechanism. It keeps nested delimiters and the order of operations explicit.

$$ -1=(p-1)+(p-1)p+(p-1)p^2+\cdots,\qquad(1-p)\sum_{k\ge0}p^k=1 $$

Conceptual compression

The two-row display is also a debugging tool: if the conclusion changes when only notation changes, some hidden choice has entered the argument.

$$ \begin{aligned} \mathsf{D}\;&:\quad x=a_0+a_1p+a_2p^2+\cdots,\qquad0\le a_i<p,\\[5pt] \mathsf{C}\;&:\quad \mathbf Z_p\cong\varprojlim_n\mathbf Z/p^n\mathbf Z. \end{aligned} $$

Caveat

A congruence root lifts uniquely only in the simple-root case. Multiple roots need stronger inequalities and may split, disappear, or lift nonuniquely.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \mathbf Z_p\cong\varprojlim_n\mathbf Z/p^n\mathbf Z \end{gathered}} $$

The result is compact enough to reuse without pretending that the caveat has disappeared. The worked line remains the quickest consistency check.

This article was posted on Wed 29 January 2020. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.