Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedprecheck passaudited 2026-08-26
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Localising Z/12Z kills exactly the torsion seen by the denominator set

Example

Let M=Z/12Z. If S2={2n:n∈N} and S3={3n:n∈N}, then ker⁡(M→S2−1M)={[0],[3],[6],[9]},ker⁡(M→S3−1M)={[0],[4],[8]}. So localisation kills exactly the torsion detected by the chosen denominator set.

Facts & Assumptions

Given: The module M=Z/12Z and the multiplicative sets S2={2n:n∈N} and S3={3n:n∈N}.

[L1]

A fraction in a localised module is zero exactly when one denominator kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L2]

Elements of S−1M are fractions m/s with s∈S (Localisation of a module at a multiplicative subset).

Verification

technique · direct
1.1L1algebra

By [L1], [a]∈ker⁡(M→S2−1M) exactly when 2r[a]=0 in Z/12Z for some r≥0. This happens for [0],[3],[6],[9], and for no other class, because 12∣2ra is possible exactly when the odd part of a is divisible by 3.

1.2L1algebra

Likewise, [a]∈ker⁡(M→S3−1M) exactly when 3r[a]=0 for some r≥0. This happens for [0],[4],[8], and for no other class, because 12∣3ra is possible exactly when the 2-primary part of a is divisible by 4.

2.1step 1.1step 1.2L2∎

Steps 1.1 and 1.2 give the two kernels, and [L2] interprets them as the elements killed by the respective localisation maps.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources