Abstract Algebra II by Randall R. Holmes

By Randall R. Holmes

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Example text

Let f (x) be a nonconstant polynomial over Z[x]. (i) If f (x) factors over Q as f (x) = g(x)h(x), then it factors over Z as f (x) = g1 (x)h1 (x) with deg g1 (x) = deg g(x) and deg h1 (x) = deg h(x). (ii) If f (x) is irreducible over Z, then it is irreducible over Q. (iii) If f (x) is primitive and irreducible over Q, then it is irreducible over Z. Proof. (i) Let f (x) = g(x)h(x) be a factorization of f (x) with g(x), h(x) ∈ Q[x]. We may (and do) assume that the coefficients of g(x) all have the same denominator b ∈ Z.

Suppose that r has two factorizations, r = s1 s2 · · · sm and r = t1 t2 · · · tn with each si and each ti irreducible. We proceed by induction on m, assuming, without loss of generality, that m ≥ n. If m = 1, then s1 = r = t1 and the statement holds. Assume that m > 1. We have s1 s2 · · · sm = t1 t2 · · · tn , so sm | t1 t2 · · · tn . 3). Therefore, sm | tj for some j. By interchanging the factors tn and tj , if necessary, we may (and do) assume that sm | tn (for, if we prove the statement with this new ordering, then we can compose the permutation σ we get with the transposition (m, n) to get a permutation that works with the original ordering).

Since Z3 is a field, the theorem applies and it follows as in the preceding example that f (x) has a linear factor and hence a zero r ∈ Z3 = {0, 1, 2}. But f (0) = 2, f (1) = 1, and f (2) = 2, so f (x) has no zeros. We conclude that f (x) is irreducible as claimed. Other irreducibility criteria are given in Section 10. 6 F[x] is PID if F is field Let F be a field. 1 Theorem. The polynomial ring F [x] is a PID. Proof. We first check that F [x] is an integral domain. Any two monomials axi and bxj (a, b ∈ F ) commute since axi bxj = abxi+j = baxi+j = bxj ai .

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