polynomial functions vs polynomials

Let k be a field. Recall that a function


is called polynomial function, iff there are a0,,ank such that


for any xk.

The ring of all polynomial functions (together with obvious addition and multiplication) we denote by k{x}. Also denote by k[x] the ring of polynomials (see this entry (http://planetmath.org/PolynomialRing) for details).

There is a canonical function T:k[x]k{x} such that for any polynomialPlanetmathPlanetmath


we have that T(W) is a polynomial function given by


(Although we use the same notation for polynomials and polynomial functions these conceptsMathworldPlanetmath are not the same). This function is called the evaluation map. As a simple exercise we leave the following to the reader:

PropositionPlanetmathPlanetmath 1. The evaluation map T is a ring homomorphismsMathworldPlanetmath which is ,,onto”.

The question is: when T is ,,1-1”?

Proposition 2. T is ,,1-1” if and only if k is an infiniteMathworldPlanetmathPlanetmath field.

Proof. ,,” Assume that k={a1,,an} is a finite fieldMathworldPlanetmath. Put


Then for any xk we have that x=ai for some i and


which shows that WKerT although W is nonzero. Thus T is not ,,1-1”.

,,” Assume, that


is a polynomial with positive degree, i.e. n1 and an0 such that T(W) is a zero function. It follows from the Bezout’s theorem that W has at most n roots (in fact this is true over any integral domainMathworldPlanetmath). Thus since k is an infinite field, then there exists ak which is not a root of W. In particular


ContradictionMathworldPlanetmathPlanetmath, since T(W) is a zero function. Thus T is ,,1-1”, which completesPlanetmathPlanetmathPlanetmathPlanetmath the proof.

This shows that the evaluation map T is an isomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath only when k is infinite. So the interesting question is what is a kernel of T, when k is a finite field?

Proposition 3. Assume that k={a1,,an} is a finite field and


Then T(W)=0 and if T(U)=0 for some polynomial U, then W divides U. In particular


Proof. In the proof of proposition 2 we’ve shown that T(W)=0. Now if T(U)=0, then every ai is a root of U. It follows from the Bezout’s theorem that (x-ai) must divide U for any i. In particular W divides U. This (together with the fact that T(W)=0) shows that the ideal KerT is generated by W. .

Corollary 4. If k is a finite field of order q>1, then k{x} has exactly qq elements.

Proof. Let k={a1,,aq} and


By propositions 1 and 3 (and due to First Isomorphism TheoremPlanetmathPlanetmath for rings) we have that


But the degree of W is equal to q. It follows that dimension of k[x]/(W) (as a vector space over k) is equal to


Thus k{x} is isomorphic to q copies of k as a vector space


This completes the proof, since k has q elements.

Remark. Also all of this hold, if we replace k with an integral domain (we can always pass to its field of fractionsMathworldPlanetmath). However this is not really interesting, since finite integral domains are exactly fields (Wedderburn’s little theorem).

Title polynomial functions vs polynomials
Canonical name PolynomialFunctionsVsPolynomials
Date of creation 2013-03-22 19:18:03
Last modified on 2013-03-22 19:18:03
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 13A99