quotient ring modulo prime ideal

Theorem. Let R be a commutative ring with non-zero unity 1 and 𝔭 an ideal of R. The quotient ringMathworldPlanetmath R/𝔭 is an integral domainMathworldPlanetmath if and only if 𝔭 is a prime idealMathworldPlanetmathPlanetmathPlanetmath.

Proof. 1oΒ―. First, let 𝔭 be a prime ideal of R. Then R/𝔭 is of course a commutative ring and has the unity 1+𝔭. If the product  (r+𝔭)⁒(s+𝔭) of two residue classesPlanetmathPlanetmath vanishes, i.e. equals 𝔭, then we have  r⁒s+𝔭=𝔭,  and therefore r⁒s must belong to 𝔭. Since 𝔭 is , either r or s belongs to 𝔭, i.e.  r+𝔭=𝔭  or  s+𝔭=𝔭.  Accordingly, R/𝔭 has no zero divisorsMathworldPlanetmath and is an integral domain.

2oΒ―. Conversely, let R/𝔭 be an integral domain and let the product r⁒s of two elements of R belong to 𝔭. It follows that  (r+𝔭)⁒(s+𝔭)=r⁒s+𝔭=𝔭. Since R/𝔭 has no zero divisors,  r+𝔭=𝔭  or  s+𝔭=𝔭. Thus, r or s belongs to 𝔭, i.e. 𝔭 is a prime ideal.

Title quotient ring modulo prime ideal
Canonical name QuotientRingModuloPrimeIdeal
Date of creation 2013-03-22 17:37:09
Last modified on 2013-03-22 17:37:09
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 7
Author pahio (2872)
Entry type Theorem
Classification msc 13C99
Related topic CharacterisationOfPrimeIdeals
Related topic QuotientRing