evaluation homomorphism
Let be a commutative ring and let be the ring of polynomials with coefficients in .
Theorem 1.
Let be a commutative ring, and let be a homomorphism. Further, let . Then there is a unique homomorphism taking to and taking every to .
This amounts to saying that polynomial rings are free objects in the category of -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 . Then by definition there is some finite list of such that . Then define to be . It is clear from the definition of addition and multiplication on polynomials that is a homomorphism; the definition makes it clear that and .
Now, to show uniqueness, suppose is any homomorphism satisfying the conditions of the theorem, and let . Write as before. Then and by assumption. But then since is a homomorphism, and . ∎
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 |