adic complete wrt a smaller ideal

Stream: maths

[Andrew Yang]

Let $R$ be a commutative ring, $I \le J$ be ideals of $R$. If $R$ is $J$-adically complete, is it also $I$-adically complete? Equivalently, is $\lim\limits_\longleftarrow R / I ^ n \to \lim\limits_\longleftarrow R / J ^ n$ injective?

It is true when $I$ is finitely generated, but I don't have a proof nor a counterexample when $I$ is not fg. I am attaching my proof in the fg case below in case someone can strengthen (or simplify) it.

Proof

Let $x \in \lim\limits_\longleftarrow R / I ^ n$. We can write $x$ as $x = \sum_{i = 0}^\infty x_i$ with $x_i \in I ^ i$.
Suppose the image of $x$ in $\lim\limits_\longleftarrow R / J ^ n$ is zero.
We want to show that $x$ is zero in $\lim\limits_\longleftarrow R / I ^ n$, i.e. $\sum_{i = 0}^{n-1} x_i \in I ^ n$ for all $n$.
Since $I$ is fg, so is $I^n$. Suppose $I^n$ is generated by $\{a_k\}$.
We can write $x_{n + i} \in I^{n + i}$ as $x_{n + i} = \sum_k a_k y_{i, k}$ with $y_{i, k} \in I ^ i$.
Then $y_k := \sum_{i = 0}^\infty y_{i, k}$ is a well defined element in $R$ because it is $J$-adically complete, and $\sum_{i = 0}^{n - 1} x_i + \sum_k a_k y_k = 0$ as elements of $R$ because its image in $\lim\limits_\longleftarrow R / J ^ n$ is equal to the image of $x$, which is zero.

Since $\sum_k a_k y_k \in I ^ n$, this implies $\sum_{i = 0}^{n - 1} x_i \in I ^ n$ and we're done.

[Junyan Xu]

de Jong couldn't do better: stacks#090T :)

[Kevin Buzzard]

I don't know about your "equivalently" claim but if R is the integers of Qpbar, J is the maximal ideal and I=0 then the map isn't injective

[Junyan Xu]

But it's assumed that R is J-adically complete.

[Junyan Xu]

I think the "equivalently" means if you have two functions $f$ and $g$ such that $f\circ g$ is bijective, then $g$ is bijective iff $f$ is injective. Here $g:R\to\lim\limits_\longleftarrow R / I ^ n$ and $f:\lim\limits_\longleftarrow R / I ^ n\to\lim\limits_\longleftarrow R / J ^ n$.

[Kevin Buzzard]

If there's some natural $N$ such that $J^N\subseteq I$ then I bet you're fine, and $J$-adic completeness implies that the intersection of $J^n$ is 0 so $I$ can't be strictly smaller than this intersection, so perhaps the place to look for a counterexample is $R_0=\R[X_1,X_2,X_3,\ldots]$, $J_0=(X_1,X_2,X_3,\ldots)$, $I_0=(X_1,X_2^2,X_3^3,\ldots)$ and then $R$ is the $J_0$-adic completion of $R_0$ and $I,J$ are the images of $I_0,J_0$. If this isn't a counterexample then maybe the proof that it isn't a counterexample will turn into a proof that it's true in general?

[Kevin Buzzard]

Oh no it seems deeper than this. stacks#05JA shows that I can't just blithely complete and expect that the result is complete :-/

[Antoine Chambert-Loir]

Exactly. It is well known that « completion » of rings are not well-behaved outside of the finitely generated case, and I have the impression that mathematicians try to flee away of these questions, or approach them differently, eg, via condensed sets, to get a better definition of complete, with better theorems.