Question

to products of irreducibles in Z[c, y, z). You 5. Factor the following polynomials into products of irreducibles in Z[x, y, z). You need to show that the factors you write down are indeed irreducibles. (a) x3 + 8y3 – 18xy + 27. (b) x3 + 3x²y – 2y? (c) x2 – 3xy2 + 2y2 (d) x2 – 2zx – xy – 2yz. (e) 23 +2(y – z)x2 + (x – zby2 + (x + y)22 – 3xyz.

Answer 1

• Assume that g(x) = bmx m + · · · + b0, where m ≥ 0 and bm 6= 0. Consider the set S = {f(x) − g(x)s(x) : s(x) ∈ F[x]}. • If 0 ∈ S, then ∃ s(x) ∈ F[x] such that f(x) = g(x)s(x). Then we can take q(x) = s(x), r(x) = 0, and we are done. Proof of Theorem 23.1, continued • Assume that 0 6∈ S. Let r(x) be an element of minimal degree in S. We have f(x) = g(x)q(x) + r(x) for some q(x) ∈ F[x]. We need to show that deg r(x) < deg g(x). • Suppose that r(x) = ck x k + · · · + c0 with k ≥ m and ck 6= 0. Consider f(x) − g(x)(q(x) + ck /bmx k−m). • We have f(x)−g(x)q(x) − (ck /bm)x k−mg(x) = r(x) − (ck /bm)x k−mg(x) = (ck x k + · · · + c0) − (ck /bm)(bmx k + · · ·), whose degree is less than r(x). This contradicts to the assumption that r(x) is of minimal degree in S. Thus, deg r(x) must be less than deg g(x). Proof of Theorem 23.1, continued We now show that q(x) and r(x) are unique. • Suppose that f(x) = g(x)q1(x) + r1(x) f(x) = g(x)q2(x) + r2(x), where ri(x) either are the zero polynomial or satisfy deg ri(x) < deg g(x). • Then we have r1(x) − r2(x) = g(x)(q2(x) − q1(x)). • Now r1(x) − r2(x) is either zero or a polynomial of degree < deg g(x). However, if the right-hand side is not zero, then the degree is at least deg g(x). • Thus, the only possibility is that r1(x) = r2(x), and q1(x) = q2(x). This completes the proof.