polynomial ring which is PID

Theorem.  If a polynomial ring D[X] over an integral domainMathworldPlanetmath D is a principal ideal domainMathworldPlanetmath, then coefficient ring D is a field. (Cf. the corollary 4 in the entry polynomial ring over a field.)

Proof.  Let a be any non-zero element of D.  Then the ideal  (a,X)  of D[X] is a principal idealMathworldPlanetmathPlanetmath (f(X)) with f(X) a non-zero polynomial (http://planetmath.org/ZeroPolynomial2).  Therefore,


with g(X) and h(X) certain polynomials in D[X].  From these equations one infers that f(X) is a polynomial c and h(X) is a first degree polynomial b0+b1X (b10).  Thus we obtain the equation


which shows that cb1 is the unity 1 of D.  Thus  c=f(X)  is a unit of D, whence


So we can write


where  u(X),v(X)D[X].  This equation cannot be possible without that a times the constant term of u(X) is the unity.  Accordingly, a has a multiplicative inverseMathworldPlanetmath in D.  Because a was arbitrary non-zero elenent of the integral domain D, D is a field.


  • 1 David M. Burton: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).
Title polynomial ring which is PID
Canonical name PolynomialRingWhichIsPID
Date of creation 2013-03-22 17:53:04
Last modified on 2013-03-22 17:53:04
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 8
Author pahio (2872)
Entry type Theorem
Classification msc 13P05
Related topic PolynomialRingOverFieldIsEuclideanDomain