Adic topology ($I^n$)

For a two-sided ideal $I$ of a ring $R$, the powers $I^n$ form a neighborhood base at zero for the adic topology. For an $R$-module $M$, use $I^{n}M$ on the left or $M{I}^n$ on the right. Convergence means that for every $n$ the differences eventually belong to the corresponding submodule. The formal power series ring $k[[t]]$ and formal power series module $M[[t]]$ are complete for $I=(t)$; convergence fixes each finite list of coefficients eventually. This is formal convergence, without an analytic condition on the size of coefficients.

For a complete Noetherian ring with ideal of definition $I$, completed localization is the completion of the localization of a ring $A_f$ in the induced adic topology. It gives the sections of the formal spectrum over $D(f)$. For $A=k[t][[q]]$, its completed localization at $t$ is $k[t,t^{-1}][[q]]$. Ordinary localization allows a bounded pole order in $t$ across all coefficients; the latter ring also contains ${\sum }_{n\ge 0}{q}^n{t}^{-n}$, whose pole orders are unbounded.

// Target: ringed-space.bigb

The adic completion is the inverse limit of the quotient modules $M/I^{n}M$. An element is a compatible family of residue classes at every order. For $R=k[t]$ and $M=V{\otimes }_{k}k[t]$, this gives the formal power series module $V[[t]]$. The canonical map from $M$ need not be injective in general: its kernel is ${\bigcap }_n{I}^{n}M$. It is injective for this polynomial example because a nonzero polynomial has finite degree. Successive changes of coordinates in formal rigidity from vanishing second Hochschild cohomology converge in this completion because the order-$r$ change is the identity modulo $t^r$.

If $I=(a_1,\dots ,a_s)$, evaluation $X_i\mapsto a_i$ gives a ring map to the $I$-adic completion, with each series evaluated modulo $I^r$ using only terms of total degree below $r$. It is surjective: successively represent the residual class in $I^r/I^{r+1}$ by a homogeneous degree-$r$ polynomial in the $a_i$. If $R$ is a commutative Noetherian ring, Noetherianity of formal power series rings therefore makes its completion Noetherian.

For any ideal $I$ and surjective module map $f:M\to N$, compatible representatives $n_r$ modulo $I^{r}N$ can be lifted recursively. Having chosen $f(m_r)=n_r$, lift $n_{r+1}-n_r\in I^{r}N$ to ${\delta }_r\in I^{r}M$ and put $m_{r+1}=m_r+{\delta }_r$. These define the required element of the adic completion of a module. No finite-generation hypothesis on the modules is required.