Abstract Algebra Theory and Applications - Computer Science ...

266 CHAPTER 15 POLYNOMIALSProof. Suppose thatp(x) = b 0c 0+ b 1c 1x + · · · + b nc nx n ,where the b i ’s and the c i ’s are integers. We can rewrite p(x) asp(x) =1c 0 · · · c n(d 0 + d 1 x + · · · + d n x n ),where d 0 , . . . , d n are integers.d 0 , . . . , d n . ThenLet d be the greatest common divisor ofp(x) =dc 0 · · · c n(a 0 + a 1 x + · · · + a n x n ),where d i = da i and the a i ’s are relatively prime. Reducing d/(c 0 · · · c n ) toits lowest terms, we can writep(x) = r s (a 0 + a 1 x + · · · + a n x n ),where gcd(r, s) = 1.□Theorem 15.9 (Gauss’s Lemma) Let p(x) ∈ Z[x] be a monic polynomialsuch that p(x) factors into a product of two polynomials α(x) and β(x) inQ[x], where the degrees of both α(x) and β(x) are less than the degree ofp(x). Then p(x) = a(x)b(x), where a(x) and b(x) are monic polynomials inZ[x] with deg α(x) = deg a(x) and deg β(x) = deg b(x).Proof. By Lemma 15.8, we can assume thatα(x) = c 1d 1(a 0 + a 1 x + · · · + a m x m ) = c 1d 1α 1 (x)β(x) = c 2d 2(b 0 + b 1 x + · · · + b n x n ) = c 2d 2β 1 (x),where the a i ’s are relatively prime and the b i ’s are relatively prime. Consequently,p(x) = α(x)β(x) = c 1c 2d 1 d 2α 1 (x)β 1 (x) = c d α 1(x)β 1 (x),where c/d is the product of c 1 /d 1 and c 2 /d 2 expressed in lowest terms.Hence, dp(x) = cα 1 (x)β 1 (x).