explicit generators of a quotient polynomial ring associated to a given polynomial

Let k be a field and consider ring of polynomials k[X]. If F(X)k[X] and W(X)k[X], then we will write W(X)¯ to denote element in k[X]/(F(X)) represented by W(X).

Lemma. Assume, that a1,,ank are different elements and F(X)=(X-a1)(X-an). Let Wi(X)k[X] be given by Wi(X)=(X-a1)(X-ai-1)(X-ai+i)(X-an). Then there exist λ1,,λnk such that F(X) divides polynomialMathworldPlanetmathPlanetmathPlanetmath


Proof. Note, that Wi(ai)0 for any i. Thus we may define λi=(Wi(ai))-1. Then for any i we have λiWi(ai)=1, therefore


is such that V(ai)=1 for any i. In particular U(ai)=V(ai)-1=0 and thus (X-ai) divides U(X) for any i. This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Corollary. Under the same assumptionsPlanetmathPlanetmath as in lemma, we have that ideal (W1(X)¯,,Wn(X)¯) in k[X]/(F(X)) is equal to k[X]/(F(X)).

Proof. Indeed, all we need to show is that we can generate 1¯. Lemma implies, that there is V(X)k[X] such that


Now, after aplying quotientPlanetmathPlanetmath homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to both sides we have


This completes the proof.

Remark. This gives us an explicit formula for generatorsPlanetmathPlanetmathPlanetmath of k[X]/(F(X)). In particular the dimension over k of this ring is at most degF(X). It can be shown that actualy it is equal, even if F(X) is arbitrary.

Title explicit generators of a quotient polynomial ring associated to a given polynomial
Canonical name ExplicitGeneratorsOfAQuotientPolynomialRingAssociatedToAGivenPolynomial
Date of creation 2013-03-22 19:10:01
Last modified on 2013-03-22 19:10:01
Owner joking (16130)
Last modified by joking (16130)
Numerical id 6
Author joking (16130)
Entry type Derivation
Classification msc 11C08
Classification msc 12E05
Classification msc 13P05