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For every , the congruence-class ring is the quotient ring
Statement
For every , the congruence-class ring is the quotient ring .
This includes and .
Facts & Assumptions
Given: A natural number , viewed as a nonnegative integer.
The canonical quotient map is a surjective ring homomorphism (The canonical projection is a surjective ring homomorphism with kernel ).
The additive quotient is (For every , the congruence-class group is the quotient group ).
Congruence classes carry the published modular ring operations (For every natural , is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).
is a commutative ring with identity (The integers form a commutative ring).
Proof
Map to ; [L2] makes this an equality of additive quotient sets.
The quotient product maps to , exactly the modular product in [L4], and the identities agree.
Therefore the congruence-class ring is literally the quotient ring, including at .
Depends on
- The canonical projection $R\to R/I$ is a surjective ring homomorphism with kernel $I$
- For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$
- For every natural $n$, $(\mathbb{Z}/n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold
- Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$
- The integers form a commutative ring
Used by
- Localising at a zero divisor need not be injective: inverting 3 in ℤ/6 kills 2 Counterexample
- Outside a domain, the nonzero elements need not be multiplicative: 2·3=0 in ℤ/6 Counterexample
- Over Fₚ, xᵖ-x and the zero polynomial induce the same function but are distinct polynomials Counterexample
- Over ℤ/4, a nonconstant polynomial can be a unit and product degree can drop Counterexample
- Quadratics can have four roots over ℤ/6 and ℤ/8 Counterexample
- The total quotient ring of a nondomain need not be a field: Q(ℤ/6)≅ℤ/6 Counterexample
- For every integer n>1, nℤ is a maximal ideal of ℤ if and only if n is prime Example
- In characteristic 2, x²+1=(x+1)² has zero derivative and a repeated root Example
- Reduction modulo 2 proves x³+17x+391 irreducible over ℚ Example
- The four-element field (ℤ/2)[x]/(x²+x+1) Example
- ℤ₍ₚ₎ consists of rationals with denominator not divisible by p, has maximal ideal pℤ₍ₚ₎, and residue field Fₚ Example
- FALSE: F_pⁿ is the ring ℤ/pⁿℤ False statement
- The product of primitive integer polynomials is primitive, and contents multiply Lemma
- Eisenstein criterion over the integers Theorem
- Irreducibility after reduction modulo a prime implies irreducibility over ℚ when the leading coefficient survives Theorem
Dependency tree · two levels
31 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
- Conrad, Modular Arithmetic (standard reference, not scraped)