Abstract Algebra Theory and Applications - Computer Science ...

268 CHAPTER 15 POLYNOMIALSwhere each factor is in Z[x] by Gauss’s Lemma. Hence,a + c = −2ac + b + d = 0ad + bc = 1bd = 1.Since bd = 1, either b = d = 1 or b = d = −1. In either case b = d and soad + bc = b(a + c) = 1.Since a + c = −2, we know that −2b = 1. This is impossible since b is aninteger. Therefore, p(x) must be irreducible over Q.Theorem 15.11 (Eisenstein’s Criterion) Let p be a prime and supposethatf(x) = a n x n + · · · + a 0 ∈ Z[x].If p | a i for i = 0, 1, . . . , a n−1 , but p̸ | a n and p 2̸ | a 0 , then f(x) is irreducibleover Q.Proof. By Gauss’s Lemma, we need only show that f(x) does not factorinto polynomials of lower degree in Z[x]. Letf(x) = (b r x r + · · · + b 0 )(c s x s + · · · + c 0 )be a factorization in Z[x], with b r and c s not equal to zero and r, s < n.Since p 2 does not divide a 0 = b 0 c 0 , either b 0 or c 0 is not divisible by p.Suppose that p̸ | b 0 and p | c 0 . Since p̸ | a n and a n = b r c s , neither b r nor c sis divisible by p. Let m be the smallest value of k such that p̸ | c k . Thena m = b 0 c m + b 1 c m−1 + · · · + b m c 0is not divisible by p, since each term on the right-hand side of the equationis divisible by p except for b 0 c m . Therefore, m = n since a i is divisible by pfor m < n. Hence, f(x) cannot be factored into polynomials of lower degreeand therefore must be irreducible.□Example 7. The polynomialp(x) = 16x 5 − 9x 4 + 3x 2 + 6x − 21is easily seen to be irreducible over Q by Eisenstein’s Criterion if we letp = 3.