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Kernel and universal property of adic completion
Statement
Let be a commutative ring, let be an ideal, and let be an -module.
- The kernel of the completion map is In particular, is injective exactly when is -adically separated.
- If is an -adically complete -module, then every -linear map factors uniquely through by a continuous map when has its -adic topology and has the inverse-limit topology whose basic neighbourhoods of are Thus there is a unique continuous -linear map with .
Facts & Assumptions
Given: A commutative ring , an ideal , an -module , and for part 2 an -adically complete -module together with an -linear map .
The -adic completion is with completion map (The -adic completion of a module).
An -adically complete module is separated and its canonical map to the inverse limit of its quotients is an isomorphism (Separated and complete filtered modules).
The -adic topology has the submodules as a neighbourhood basis of (The -adic topology on a module).
Compatible maps into an inverse limit factor uniquely through it (Universal property of an inverse limit of modules).
Proof
The completion map is the unique map into the inverse limit of [L1] whose th component, for , is the quotient map
An element lies in exactly when every quotient map , , kills it, equivalently when it belongs to every for . Since , this gives
Let be -linear with complete. Since is -linear, one has for every , so induces maps These maps are compatible with the quotient transition maps.
By [L4], the compatible family yields a unique map such that the th projection of is . These are also the coordinates of , so uniqueness in [L4] gives
The image of is dense for the inverse-limit topology. Indeed, for and , choose representing ; then . Now let be continuous and vanish on . If , separatedness of from [L2] gives with . The inverse image is an open neighbourhood of by [L3], so density gives some in it. But , which would put in , a contradiction. Hence . Applying this to the difference of two continuous extensions proves uniqueness.
The last equality shows that is injective exactly when , which is exactly the definition of being -adically separated.
Write and . Step 1.4 gives so for every . Hence is continuous for the two inverse-limit topologies. Because is complete, [L2] identifies under this isomorphism the preimage of is , so [L3] makes a homeomorphism from the -adic topology to the inverse-limit topology. Thus is continuous and satisfies .
Parts 1 and 2 are proved by steps 1.2, 1.5, 2.1, and 2.2.
Depends on
Used by
Dependency tree · two levels
6 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
- A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §22.1 and §22.5 (standard reference, not scraped)
- The Stacks Project, Lemma 10.96.5 (standard reference, not scraped)