polynomial ring over integral domain


If the coefficient ring R is an integral domainMathworldPlanetmath, then so is also its polynomial ringMathworldPlanetmath R[X].

Proof.  Let f(X) and g(X) be two non-zero polynomialsMathworldPlanetmathPlanetmath in R[X] and let af and bg be their leading coefficients, respectively.  Thus  af0,  bg0,  and because R has no zero divisorsMathworldPlanetmath,  afbg0.  But the product afbg is the leading coefficient of f(X)g(X) and so f(X)g(X) cannot be the zero polynomialMathworldPlanetmath.  Consequently, R[X] has no zero divisors, Q.E.D.

Remark.  The theorem may by induction be generalized for the polynomial ring  R[X1,X2,,Xn].

Title polynomial ring over integral domain
Canonical name PolynomialRingOverIntegralDomain
Date of creation 2013-03-22 15:10:06
Last modified on 2013-03-22 15:10:06
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 10
Author pahio (2872)
Entry type Theorem
Classification msc 13P05
Related topic RingAdjunction
Related topic FormalPowerSeries
Related topic ZeroPolynomial2
Related topic PolynomialRingOverFieldIsEuclideanDomain
Defines coefficient ring