Abstract Algebra Theory and Applications - Computer Science ...

260 CHAPTER 15 POLYNOMIALSProof. Suppose that we have two nonzero polynomialsandp(x) = a m x m + · · · + a 1 x + a 0q(x) = b n x n + · · · + b 1 x + b 0with a m ≠ 0 and b n ≠ 0. The degrees of p and q are m and n, respectively.The leading term of p(x)q(x) is a m b n x m+n , which cannot be zero since R isan 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, suchas x 2 − 3xy + 2y 3 . Let R be a ring and suppose that we are given twoindeterminates x and y. Certainly we can form the ring (R[x])[y]. It isstraightforward but perhaps tedious to show that (R[x])[y] ∼ = R([y])[x]. Weshall 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 indeterminatesx and y with coefficients in R. We can define the ring of polynomialsin n indeterminates with coefficients in R similarly. We shall denotethis ring by R[x 1 , x 2 , . . . , x n ].Theorem 15.3 Let R be a commutative ring with identity and α ∈ R. Thenwe have a ring homomorphism φ α : R[x] → R defined byφ α (p(x)) = p(α) = a n α n + · · · + a 1 α + a 0 ,where p(x) = a n x n + · · · + a 1 x + a 0 .Proof. Let p(x) = ∑ ni=0 a ix i and q(x) = ∑ mi=0 b ix i . It is easy to showthat φ α (p(x) + q(x)) = φ α (p(x)) + φ α (q(x)). To show that multiplication ispreserved under the map φ α , observe thatφ α (p(x))φ α (q(x)) = p(α)q(α)( n∑) ( m)∑= a i α i b i α ii=0i=0( i∑)a k b i−kk=0=m+n∑i=0= φ α (p(x)q(x)).The map φ α : R[x] → R is called the evaluation homomorphismat α.α i□