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 $R$?
3
-
5$\begingroup$ only closed ones $\endgroup$– Misha VerbitskyCommented Jun 25, 2014 at 8:41
-
$\begingroup$ The necessity of closedness is OK when $R$ is noetherian, but probably not in general. $\endgroup$– user27920Commented Jun 25, 2014 at 14:10
-
$\begingroup$ @MishaVerbitsky - Am I correct in assuming that if all ideals of the completion are closed, then all ideals of the completion are completions of ideals of the original ring? $\endgroup$– Chris LearyCommented Jul 10, 2023 at 18:37
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
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.
answered Jun 25, 2014 at 8:46
-
1$\begingroup$ 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. $\endgroup$ Commented Jun 25, 2014 at 18:04
$\begingroup$
$\endgroup$
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).
