Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Kernel and universal property of adic completion

Statement

Let R be a commutative ring, let IR be an ideal, and let M be an R-module.

  1. The kernel of the completion map κM ⁣:MM^ is n0InM. In particular, κM is injective exactly when M is I-adically separated.
  2. If N is an I-adically complete R-module, then every R-linear map f ⁣:MN factors uniquely through κM by a continuous map when N has its I-adic topology and M^ has the inverse-limit topology whose basic neighbourhoods of 0 are Kn:=ker(M^M/InM)(n1). Thus there is a unique continuous R-linear map f^ ⁣:M^N with f^κM=f.

Facts & Assumptions

Given: A commutative ring R, an ideal IR, an R-module M, and for part 2 an I-adically complete R-module N together with an R-linear map f ⁣:MN.

[L1]

The I-adic completion is M^=limM/InM with completion map m(mmodInM)n (The I-adic completion of a module).

[L2]

An I-adically complete module is separated and its canonical map to the inverse limit of its quotients is an isomorphism (Separated and complete filtered modules).

[L3]

The I-adic topology has the submodules InN as a neighbourhood basis of 0 (The I-adic topology on a module).

[L4]

Compatible maps into an inverse limit factor uniquely through it (Universal property of an inverse limit of modules).

Proof

technique · direct
1.1

The completion map κM is the unique map into the inverse limit of [L1] whose nth component, for n1, is the quotient map qn ⁣:MM/InM.

L1L4
1.2

An element lies in ker(κM) exactly when every quotient map qn, n1, kills it, equivalently when it belongs to every InM for n1. Since I0M=M, this gives ker(κM)=n1InM=n0InM.

L1algebra
1.3

Let f ⁣:MN be R-linear with N complete. Since f is R-linear, one has f(InM)InN for every n1, so f induces maps fn ⁣:M/InMN/InN. These maps are compatible with the quotient transition maps.

givenalgebra
1.4

By [L4], the compatible family (fn) yields a unique map f~ ⁣:M^limN/InN such that the nth projection of f~κM is fnqn. These are also the coordinates of κNf, so uniqueness in [L4] gives f~κM=κNf.

L1L4
1.5

The image of κM is dense for the inverse-limit topology. Indeed, for xM^ and n1, choose mM representing πnM(x)M/InM; then xκM(m)Kn. Now let g:M^N be continuous and vanish on κM(M). If g(x)0, separatedness of N from [L2] gives n with g(x)InN. The inverse image g1(g(x)+InN) is an open neighbourhood of x by [L3], so density gives some κM(m) in it. But gκM(m)=0, which would put g(x) in InN, a contradiction. Hence g=0. Applying this to the difference of two continuous extensions proves uniqueness.

L2L3choosealgebra
2.1

The last equality shows that κM is injective exactly when n0InM=0, which is exactly the definition of being I-adically separated.

step 1.2L2
2.2

Write πnM:M^M/InM and πnN:limN/InNN/InN. Step 1.4 gives πnNf~=fnπnM, so f~(Kn)ker(πnN) for every n1. Hence f~ is continuous for the two inverse-limit topologies. Because N is complete, [L2] identifies κN:NlimN/InN; under this isomorphism the preimage of ker(πnN) is InN, so [L3] makes κN a homeomorphism from the I-adic topology to the inverse-limit topology. Thus f^:=κN1f~:M^N is continuous and satisfies f^κM=f.

L2L3step 1.4algebra
3.1

Parts 1 and 2 are proved by steps 1.2, 1.5, 2.1, and 2.2.

step 1.2step 1.5step 2.1step 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