evaluation homomorphism

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

Theorem 1.

Let $S$ be a commutative ring, and let $\psi \mathrm{:}R\mathrm{\to }S$ be a homomorphism. Further, let $s\mathrm{\in }S$. Then there is a unique homomorphism $\varphi \mathrm{:}R\mathit{}\mathrm{[}X\mathrm{]}\mathrm{\to }S$ taking $X$ to $s$ and taking every $r\mathrm{\in }R$ to $\psi \mathit{}\mathrm{(}r\mathrm{)}$.

This amounts to saying that polynomial rings are free objects in the category of $R$-algebras; 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.

Proof.

We first prove existence. Let $f\in R[X]$. Then by definition there is some finite list of $a_i$ such that $f={\sum }_i{a}_i{X}^i$. Then define $\varphi (f)$ to be ${\sum }_i\psi (a_i)s^i$. It is clear from the definition of addition and multiplication on polynomials that $\varphi$ is a homomorphism; the definition makes it clear that $\varphi (X)=s$ and $\varphi (r)=\psi (r)$.

Now, to show uniqueness, suppose $\gamma$ is any homomorphism satisfying the conditions of the theorem, and let $f\in R[X]$. Write $f={\sum }_i{a}_i{X}^i$ as before. Then $\gamma (a_i)=\psi (a_i)$ and $\gamma (s)$ by assumption. But then since $\gamma$ is a homomorphism, $\gamma (a_i{X}^i)=\psi (a_i)s^i$ and $\gamma (f)={\sum }_i\psi (a_i)s^i=\varphi (f)$. ∎

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