Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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:u∈Un}. Then

Un2≤ker⁡χn,[Un:Un2]=2r.

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

[ker⁡χn:Un2]={2r,n is a square,2r−1,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 x2≡a(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 k≥1, the group (Z/pk)× is cyclic of even order pk−1(p−1) (For every odd prime p and k≥1, (Z/pkZ)× is cyclic of order pk−1(p−1)).

[L5]

The Chinese remainder map gives Un≅∏i<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:G→H, one has G/ker⁡f≅im⁡f (First isomorphism theorem for groups: G/ker⁡f≅im⁡f).

[L7]

If G is finite and H≤G, 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 K≤H≤G and G is finite, then [G:K]=[G:H][H:K] (For K≤H≤G 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:G→H, ker⁡f={g∈G:f(g)=eH} and im⁡f={f(g):g∈G} (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
1.1L1L2L11algebra

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

1.2L3L5L6L7L8algebra

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]=∣ker⁡s∣=2r. This also holds for r=0, when [L5] identifies U1 with the one-element group.

1.3L1L4L5L10L12L13L14choose

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.

2.1step 1.1step 1.2step 1.3L1L6L7L8L9L11algebra∎

If n is a square, step 1.3 gives ker⁡χn=Un, so step 1.2 yields [ker⁡χn:Un2]=2r. Otherwise [L1], [L6], [L7], [L8], and [L11] give [Un:ker⁡χn]=2; applying [L9] to Un2≤ker⁡χn≤Un and using step 1.2 gives [ker⁡χn:Un2]=2r−1.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

78 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