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The kernel of the Jacobi map and the subgroup of unit squares
Statement
Let be an odd positive integer in canonical prime factorisation, let , and let . Then
The Jacobi homomorphism is trivial if and only if is a square. Consequently,
For , one has and every group and index above is trivial and equal to one.
Facts & Assumptions
Given: An odd positive integer in canonical prime factorisation and its unit group .
The assignment is a group homomorphism (For fixed odd modulus, the Jacobi symbol is a homomorphism on the unit group).
Every unit square modulo an odd positive integer has Jacobi symbol one (A unit square modulo an odd integer has Jacobi symbol one).
For odd , the soluble unit congruence has exactly roots (The number of square roots of a unit modulo is the product of the local counts).
For every odd prime and , the group is cyclic of even order (For every odd prime and , is cyclic of order ).
The Chinese remainder map gives , including the empty factorisation (For pairwise coprime positive moduli, the Chinese remainder bijection restricts to an isomorphism of unit groups).
For a group homomorphism , one has (First isomorphism theorem for groups: ).
If is finite and , then (Lagrange's theorem: for every subgroup of a finite group ).
For a finite-index subgroup, (The coset set and the index of a subgroup).
If and is finite, then (For with finite, ).
For an odd prime , a unit is a square modulo if and only if its Legendre symbol modulo is one (Unit square criterion and root count modulo odd prime powers).
For a homomorphism , and (The kernel and image of a group homomorphism).
Canonical prime-factor exponents are determined by the integer (For and any injective list of primes containing every prime divisor of , one has ; the exponents are determined by , and for every prime outside the list).
For a prime and nonzero integers , ( for nonzero integers , and whenever , and are all nonzero).
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
Since is abelian, the squaring map , , is a homomorphism with image . By [L2] and the kernel definition [L11], every element of lies in , so .
The kernel of is the root set of , which has elements by [L3]. Applying [L6] to and then [L7] and [L8] gives . This also holds for , when [L5] identifies with the one-element group.
By [L12] and [L13], is a square exactly when every exponent is even. In that case [L14] makes every value of equal to . If some is odd, [L4] supplies a generator, hence a nonsquare, in ; [L10] gives it Legendre symbol . Combine it with identity elements in the other factors by [L5]. Formula [L14] gives the resulting global unit Jacobi value , so [L1] is surjective. Therefore is trivial exactly when is a square.
If is a square, step 1.3 gives , so step 1.2 yields . Otherwise [L1], [L6], [L7], [L8], and [L11] give ; applying [L9] to and using step 1.2 gives .
Depends on
- For fixed odd modulus, the Jacobi symbol is a homomorphism on the unit group
- A unit square modulo an odd integer has Jacobi symbol one
- Unit square criterion and root count modulo odd prime powers
- The number of square roots of a unit modulo $n$ is the product of the local counts
- For every odd prime $p$ and $k\ge1$, $(\mathbb Z/p^k\mathbb Z)^\times$ is cyclic of order $p^{k-1}(p-1)$
- For pairwise coprime positive moduli, the Chinese remainder bijection restricts to an isomorphism of unit groups
- The kernel and image of a group homomorphism
- The Jacobi symbol, with its zero value and empty-product convention
- For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n) = 0$ for every prime $q$ outside the list
- $v_p(ab) = v_p(a) + v_p(b)$ for nonzero integers $a, b$, and $v_p(a+b) \ge \min\{v_p(a), v_p(b)\}$ whenever $a$, $b$ and $a+b$ are all nonzero
- First isomorphism theorem for groups: $G/\ker f\cong\operatorname{im}f$
- Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$
- The coset set $G/H$ and the index $[G:H]$ of a subgroup
- For $K\le H\le G$ with $G$ finite, $[G:K]=[G:H][H:K]$
Used by
Nothing in the library uses this result yet.
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Sources
- V. Shoup, A Computational Introduction to Number Theory and Algebra, 2nd ed., Exercise 12.3 (standard reference, not scraped)