Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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.

The kernel of the Jacobi map and the subgroup of unit squares

Statement

Let n=i<rpiei be an odd positive integer in canonical prime factorisation, let Un=(Z/n)×, and let Un2={u2:uUn}. Then

Un2kerχn,[Un:Un2]=2r.

The Jacobi homomorphism χn is trivial if and only if n is a square. Consequently,

[kerχn:Un2]={2r,n is a square,2r1,n is not a square.

For n=1, one has r=0 and every group and index above is trivial and equal to one.

Facts & Assumptions

Given: An odd positive integer n=i<rpiei in canonical prime factorisation and its unit group Un.

[L1]

The assignment χn([a]n)=(an) is a group homomorphism Un{±1} (For fixed odd modulus, the Jacobi symbol is a homomorphism on the unit group).

[L2]

Every unit square modulo an odd positive integer has Jacobi symbol one (A unit square modulo an odd integer has Jacobi symbol one).

[L3]

For odd n=i<rpiei, the soluble unit congruence x2a(modn) has exactly 2r roots (The number of square roots of a unit modulo n is the product of the local counts).

[L4]

For every odd prime p and k1, the group (Z/pk)× is cyclic of even order pk1(p1) (For every odd prime p and k1, (Z/pkZ)× is cyclic of order pk1(p1)).

[L5]

The Chinese remainder map gives Uni<r(Z/piei)×, including the empty factorisation (For pairwise coprime positive moduli, the Chinese remainder bijection restricts to an isomorphism of unit groups).

[L6]

For a group homomorphism f:GH, one has G/kerfimf (First isomorphism theorem for groups: G/kerfimf).

[L7]

If G is finite and HG, then G=[G:H]H (Lagrange's theorem: G=[G:H]H for every subgroup H of a finite group G).

[L8]

For a finite-index subgroup, [G:H]=G/H (The coset set G/H and the index [G:H] of a subgroup).

[L9]

If KHG and G is finite, then [G:K]=[G:H][H:K] (For KHG with G finite, [G:K]=[G:H][H:K]).

[L10]

For an odd prime p, a unit is a square modulo pk if and only if its Legendre symbol modulo p is one (Unit square criterion and root count modulo odd prime powers).

[L11]

For a homomorphism f:GH, kerf={gG:f(g)=eH} and imf={f(g):gG} (The kernel and image of a group homomorphism).

[L14]

The Jacobi symbol is the product of the prime Legendre symbols with their canonical multiplicities (The Jacobi symbol, with its zero value and empty-product convention).

Proof

technique · direct
1.1

Since Un is abelian, the squaring map s:UnUn, s(u)=u2, is a homomorphism with image Un2. By [L2] and the kernel definition [L11], every element of Un2 lies in kerχn, so Un2kerχn.

L1L2L11algebra
1.2

The kernel of s is the root set of x2=1, which has 2r elements by [L3]. Applying [L6] to s and then [L7] and [L8] gives [Un:Un2]=kers=2r. This also holds for r=0, when [L5] identifies U1 with the one-element group.

L3L5L6L7L8algebra
1.3

By [L12] and [L13], n is a square exactly when every exponent ei is even. In that case [L14] makes every value of χn equal to 1. If some ej is odd, [L4] supplies a generator, hence a nonsquare, in (Z/pjej)×; [L10] gives it Legendre symbol 1. Combine it with identity elements in the other factors by [L5]. Formula [L14] gives the resulting global unit Jacobi value 1, so [L1] is surjective. Therefore χn is trivial exactly when n is a square.

L1L4L5L10L12L13L14choose
2.1

If n is a square, step 1.3 gives kerχn=Un, so step 1.2 yields [kerχn:Un2]=2r. Otherwise [L1], [L6], [L7], [L8], and [L11] give [Un:kerχn]=2; applying [L9] to Un2kerχnUn and using step 1.2 gives [kerχn:Un2]=2r1.

step 1.1step 1.2step 1.3L1L6L7L8L9L11algebra

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 174 results over 31 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources