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Mathematics LibreTexts

17.1: Polynomial Rings

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Throughout this chapter we shall assume that R is a commutative ring with identity. Any expression of the form

f(x)=ni=0aixi=a0+a1x+a2x2++anxn,

where aiR and an0, is called a polynomial over R with indeterminate x. The elements a0,a1,,an are called the coefficients of f. The coefficient an is called the leading coefficient. A polynomial is called monic if the leading coefficient is 1. If n is the largest nonnegative number for which an0, we say that the degree of f is n and write degf(x)=n. If no such n exists—that is, if f=0 is the zero polynomial—then the degree of f is defined to be . We will denote the set of all polynomials with coefficients in a ring R by R[x]. Two polynomials are equal exactly when their corresponding coefficients are equal; that is, if we let

p(x)=a0+a1x++anxnq(x)=b0+b1x++bmxm,

then p(x)=q(x) if and only if ai=bi for all i0.

To show that the set of all polynomials forms a ring, we must first define addition and multiplication. We define the sum of two polynomials as follows. Let

p(x)=a0+a1x++anxnq(x)=b0+b1x++bmxm.

Then the sum of p(x) and q(x) is

p(x)+q(x)=c0+c1x++ckxk,

where ci=ai+bi for each i. We define the product of p(x) and q(x) to be

p(x)q(x)=c0+c1x++cm+nxm+n,

where

ci=ik=0akbik=a0bi+a1bi1++ai1b1+aib0

for each i. Notice that in each case some of the coefficients may be zero.

Example 17.1

Suppose that

p(x)=3+0x+0x2+2x3+0x4

and

q(x)=2+0xx2+0x3+4x4

are polynomials in Z[x]. If the coefficient of some term in a polynomial is zero, then we usually just omit that term.

Solution

In this case we would write p(x)=3+2x3 and q(x)=2x2+4x4. The sum of these two polynomials is

p(x)+q(x)=5x2+2x3+4x4.

The product,

p(x)q(x)=(3+2x3)(2x2+4x4)=63x2+4x3+12x42x5+8x7,

can be calculated either by determining the cis in the definition or by simply multiplying polynomials in the same way as we have always done.

Example 17.2

Let

p(x)=3+3x3andq(x)=4+4x2+4x4

be polynomials in Z12[x].

Solution

The sum of p(x) and q(x) is 7+4x2+3x3+4x4. The product of the two polynomials is the zero polynomial. This example tells us that we can not expect R[x] to be an integral domain if R is not an integral domain.

Theorem 17.3

Let R be a commutative ring with identity. Then R[x] is a commutative ring with identity.

Proof

Our first task is to show that R[x] is an abelian group under polynomial addition. The zero polynomial, f(x)=0, is the additive identity. Given a polynomial p(x)=ni=0aixi, the inverse of p(x) is easily verified to be p(x)=ni=0(ai)xi=ni=0aixi. Commutativity and associativity follow immediately from the definition of polynomial addition and from the fact that addition in R is both commutative and associative.

To show that polynomial multiplication is associative, let

p(x)=mi=0aixi,q(x)=ni=0bixi,r(x)=pi=0cixi.

Then

[p(x)q(x)]r(x)=[(mi=0aixi)(ni=0bixi)](pi=0cixi)=[m+ni=0(ij=0ajbij)xi](pi=0cixi)=m+n+pi=0[ij=0(jk=0akbjk)cij]xi=m+n+pi=0(j+k+l=iajbkcl)xi=m+n+pi=0[ij=0aj(ijk=0bkcijk)]xi=(mi=0aixi)[n+pi=0(ij=0bjcij)xi]=(mi=0aixi)[(ni=0bixi)(pi=0cixi)]=p(x)[q(x)r(x)]

The commutativity and distribution properties of polynomial multiplication are proved in a similar manner. We shall leave the proofs of these properties as an exercise.

Proposition 17.4

Let p(x) and q(x) be polynomials in R[x], where R is an integral domain. Then degp(x)+degq(x)=deg(p(x)q(x)). Furthermore, R[x] is an integral domain.

Proof

Suppose that we have two nonzero polynomials

p(x)=amxm++a1x+a0

and

q(x)=bnxn++b1x+b0

with am0 and bn0. The degrees of p(x) and q(x) are m and n, respectively. The leading term of p(x)q(x) is ambnxm+n, which cannot be zero since R is an integral domain; hence, the degree of p(x)q(x) is m+n, and p(x)q(x)0. Since p(x)0 and q(x)0 imply that p(x)q(x)0, we know that R[x] must also be an integral domain.

We also want to consider polynomials in two or more variables, such as x23xy+2y3. Let R be a ring and suppose that we are given two indeterminates x and y. Certainly we can form the ring (R[x])[y]. It is straightforward but perhaps tedious to show that (R[x])[y]R([y])[x]. We shall identify these two rings by this isomorphism and simply write R[x,y]. The ring R[x,y] is called the ring of polynomials in two indeterminates x and y with coefficients in R. We can define the ring of polynomials in n indeterminates with coefficients in R similarly. We shall denote this ring by R[x1,x2,,xn].

Theorem 17.5

Let R be a commutative ring with identity and αR. Then we have a ring homomorphism ϕα:R[x]R defined by

ϕα(p(x))=p(α)=anαn++a1α+a0,

where p(x)=anxn++a1x+a0.

Proof

Let p(x)=ni=0aixi and q(x)=mi=0bixi. It is easy to show that ϕα(p(x)+q(x))=ϕα(p(x))+ϕα(q(x)). To show that multiplication is preserved under the map ϕα, observe that

ϕα(p(x))ϕα(q(x))=p(α)q(α)=(ni=0aiαi)(mi=0biαi)=m+ni=0(ik=0akbik)αi=ϕα(p(x)q(x)).

The map ϕα:R[x]R is called the evaluation homomorphism at α.


This page titled 17.1: Polynomial Rings is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Thomas W. Judson (Abstract Algebra: Theory and Applications) via source content that was edited to the style and standards of the LibreTexts platform.

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