evaluation homomorphism

Let R be a commutative ring and let R[X] be the ring of polynomials with coefficientsMathworldPlanetmath in R.

Theorem 1.

Let S be a commutative ring, and let ψ:RS be a homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. Further, let sS. Then there is a unique homomorphism ϕ:R[X]S taking X to s and taking every rR to ψ(r).

This amounts to saying that polynomial ringsMathworldPlanetmath are free objects in the category of R-algebrasMathworldPlanetmathPlanetmath; the theorem then states that they are projective. This is true in much greater generality; in fact, the property of being projective is intended to extract the essential property of being free.


We first prove existence. Let fR[X]. Then by definition there is some finite list of ai such that f=iaiXi. Then define ϕ(f) to be iψ(ai)si. It is clear from the definition of additionPlanetmathPlanetmath and multiplication on polynomialsMathworldPlanetmath that ϕ is a homomorphism; the definition makes it clear that ϕ(X)=s and ϕ(r)=ψ(r).

Now, to show uniqueness, suppose γ is any homomorphism satisfying the conditions of the theorem, and let fR[X]. Write f=iaiXi as before. Then γ(ai)=ψ(ai) and γ(s) by assumptionPlanetmathPlanetmath. But then since γ is a homomorphism, γ(aiXi)=ψ(ai)si and γ(f)=iψ(ai)si=ϕ(f). ∎

Title evaluation homomorphism
Canonical name EvaluationHomomorphism
Date of creation 2013-03-22 14:13:51
Last modified on 2013-03-22 14:13:51
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 6
Author mathcam (2727)
Entry type Theorem
Classification msc 13P05
Classification msc 11C08
Classification msc 12E05
Synonym substitution homomorphism
Related topic LectureNotesOnPolynomialInterpolation
Defines evaluation homomorphism