Lagmental Vicfred

I-Adic Completion Is an Inverse Limit by Vicfred

A completed element is a compatible sequence of residues modulo all powers of an ideal. A small computation will anchor the general statement before the abstraction takes over.

Set-up

The \(I\)-adic filtration \(A\supset I\supset I^2\supset\cdots\) records increasing orders of vanishing. Completion replaces \(A\) by compatible residues modulo every \(I^n\).

$$ \widehat A^{\,I}=\varprojlim_nA/I^n $$

I read the first line as input and the second as output. The symbols \(\forall\) and \(\exists\) are not interchangeable, and neither may be upgraded silently to \(\Longleftrightarrow\).

$$ \widehat A^{\,I}=\{(a_n):a_{n+1}\equiv a_n\pmod{I^n}\} $$

The calculation

The middle display is intentionally dense: it is where signs, bounds, multiplicities, or normalising factors are most likely to be lost.

$$ \mathbf Z_p=\varprojlim_n\mathbf Z/p^n\mathbf Z,\qquad\begin{cases}a_{n+1}\bmod p^n=a_n,\\0\le a_n<p^n.\end{cases} $$

What survives abstraction

A good test for understanding is to change the presentation while keeping the invariant fixed. The aligned form makes that comparison unusually easy.

$$ \begin{aligned} \mathsf{D}\;&:\quad \widehat A^{\,I}=\varprojlim_nA/I^n,\\[5pt] \mathsf{C}\;&:\quad \widehat A^{\,I}=\{(a_n):a_{n+1}\equiv a_n\pmod{I^n}\}. \end{aligned} $$

The boundary

Completion and localisation answer different questions and do not commute without hypotheses. Completeness is topological data, not merely another quotient.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \widehat A^{\,I}=\{(a_n):a_{n+1}\equiv a_n\pmod{I^n}\} \end{gathered}} $$

The important habit is to remember what was fixed before the calculation began and what was proved only afterward. The final display preserves that order.

This article was posted on Mon 30 July 2018. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.