ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Powers of an ideal give the same one-step adic completion

Example

Let $R$ be a commutative ring, let $I \subseteq R$ be an ideal, let $r \geq 1$ , and let $M$ be an $R$ -module. Then the $I$ -adic and $I^{r}$ -adic completions are canonically isomorphic: ${\hat{M}}^{I} ≅ {\hat{M}}^{I^{r}} .$ Since $I + I^{r} = I$ , the same completed module is also the completion for the combined ideal $I + I^{r} = I,$ namely with ${\hat{M}}^{I}$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I \subseteq R$ , an integer $r \geq 1$ , and an $R$ -module $M$ .

[L1]

If two adic filtrations dominate one another up to bounded shifts, then their completions are canonically isomorphic (Equivalent adic filtrations have canonically isomorphic completions).

Verification

technique · direct

1.1L1algebra

Apply [L1] with $J = I^{r}$ . The inclusions $I^{r} \subseteq I and I^{r} \subseteq I^{r}$ give the hypotheses with $c = r$ and $d = 1$ , so ${\hat{M}}^{I} \tilde{\to} {\hat{M}}^{I^{r}} .$

2.1step 1.1algebra

Since $I + I^{r} = I$ , the completion for the combined ideal is exactly ${\hat{M}}^{I}$ . Combining this identity with step 1.1 shows that the $I$ -adic, $I^{r}$ -adic, and $(I + I^{r})$ -adic one-step completions are canonically the same module.

3.1step 2.1∎

Thus passing from $I$ to the power $I^{r}$ , or replacing the pair $(I, I^{r})$ by the combined ideal $I + I^{r}$ , does not change the one-step completion.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Lemma 10.96.8 and the surrounding completion discussion (standard reference, not scraped)