17.1: Polynomial Rings
( \newcommand{\kernel}{\mathrm{null}\,}\)
Throughout this chapter we shall assume that R is a commutative ring with identity. Any expression of the form
f(x)=n∑i=0aixi=a0+a1x+a2x2+⋯+anxn,
where ai∈R and an≠0, 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 an≠0, 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 i≥0.
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=i∑k=0akbi−k=a0bi+a1bi−1+⋯+ai−1b1+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+0x−x2+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)=2−x2+4x4. The sum of these two polynomials is
p(x)+q(x)=5−x2+2x3+4x4.
The product,
p(x)q(x)=(3+2x3)(2−x2+4x4)=6−3x2+4x3+12x4−2x5+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)=m∑i=0aixi,q(x)=n∑i=0bixi,r(x)=p∑i=0cixi.
Then
[p(x)q(x)]r(x)=[(m∑i=0aixi)(n∑i=0bixi)](p∑i=0cixi)=[m+n∑i=0(i∑j=0ajbi−j)xi](p∑i=0cixi)=m+n+p∑i=0[i∑j=0(j∑k=0akbj−k)ci−j]xi=m+n+p∑i=0(∑j+k+l=iajbkcl)xi=m+n+p∑i=0[i∑j=0aj(i−j∑k=0bkci−j−k)]xi=(m∑i=0aixi)[n+p∑i=0(i∑j=0bjci−j)xi]=(m∑i=0aixi)[(n∑i=0bixi)(p∑i=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 am≠0 and bn≠0. 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 x2−3xy+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 α∈R. 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)=(n∑i=0aiαi)(m∑i=0biαi)=m+n∑i=0(i∑k=0akbi−k)αi=ϕα(p(x)q(x)).
The map ϕα:R[x]→R is called the evaluation homomorphism at α.