# Ideals in the I-adic completion of a ring

Let $R$ be a commutative ring; does every ideal in the $I$-adic completion of $R$

$$\varprojlim_i R/I^i$$

arise as the $I$-adic completion of some ideal inside of the original ring $R$?

• only closed ones – Misha Verbitsky Jun 25 '14 at 8:41
• The necessity of closedness is OK when $R$ is noetherian, but probably not in general. – user27920 Jun 25 '14 at 14:10

No: take the ring $k[[t_1, ..., t_n, ...]]$ of power series of countably many variables, and let $I$ be the ideal generated by $r:=\sum t_i^i$. Since there is no invertible $u$ such that $r=uP$, where $P$ is a polynomial, the ideal $I$ does not come from any ideal in the polynomial ring.
• Even with finitely many variables we get counterexamples, using irreducible $f \in R = k[x_1,\dots,x_n]$ with $f(0) = 0$ (so $f$ is irreducible in the local ring at 0) such that $f$ is reducible in $k[\![x_1,\dots,x_n]\!]$. Consider the invertible ideal $J$ generated by an irreducible formal factor of $f$. If it comes from an ideal of $R$, hence of the local ring $R_0$ at $0$, by faithful flatness of completion that ideal in $R_0$ would have to be invertible and hence provide a proper factor of $f$ in $R_0$, contradiction. – user27920 Jun 25 '14 at 18:04
Since irreducibility of noetherian schemes is not local for the etale topology, one gets zillions of noetherian counterexamples from irreducible varieties which are not "analytically irreducible". To be specific, let $R$ be any noetherian local domain whose completion has reducible spectrum (e.g., $R = k[x,y]/(y^2 = x^2(x+1))$ for any field $k$ not of characteristic 2) and let $I$ be the maximal ideal. Then every minimal prime of $\widehat{R}$ lies over $(0)$ in $R$ due to faithful flatness of such completion, so no such prime can arise from $R$ as such primes are nonzero (since there is more than one of them).