By Patrice Tauvel
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4 Example Let R = Z8 , which is not an integral domain. The polynomial f (X) = X 3 has four roots in R, namely 0,2,4 and 6. 5 In the following sequence of problems, we review the Euclidean algorithm. Let a and b be positive integers, with a > b. Divide a by b to obtain a = bq1 + r1 with 0 ≤ r1 < b, then divide b by r1 to get b = r1 q2 + r2 with 0 ≤ r2 < r1 , and continue in this fashion until the process terminates: r 1 = r 2 q 3 + r 3 , 0 ≤ r 3 < r2 , .. rj−2 = rj−1 qj + rj , 0 ≤ rj < rj−1 , rj−1 = rj qj+1 .
Suppose that f is a polynomial in D[X] and that f can be factored as gh, where g and h belong to F [X]. Then there is a nonzero element λ ∈ F such that λg ∈ D[X] and λ−1 h ∈ D[X]. Thus if f is factorable over F , then it is factorable over D. Equivalently, if f is irreducible over D, then f is irreducible over F . Proof. The coefficients of g and h are quotients of elements of D. If a is the least common denominator for g (technically, the least common multiple of the denominators of the coefficients of g), then g ∗ = ag ∈ D[X].
For transitivity, suppose that (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). Then for some elements s and t in S we have s(ad − bc) = 0 and t(cf − de) = 0. Multiply the first equation by tf and the second by sb, and add the results to get std(af − be) = 0, which implies that (a, b) ∼ (e, f ), proving transitivity. If a ∈ R and b ∈ S, we define the fraction ab to be the equivalence class of the pair (a, b). The set of all equivalence classes is denoted by S −1 R, and is called (in view of what we are about to prove) the ring of fractions of R by S.