Abstract Algebra II by Randall R. Holmes

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By Randall R. Holmes

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Therefore, r = ± 2, contradicting that 2 is irrational. We conclude that f (x) is irreducible over Q as claimed. 51 • We claim that the polynomial f (x) = x3 +x2 +2 ∈ Z3 [x] is irreducible. Suppose otherwise. 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.

We claim that the map f : {t + S | t ∈ T } → {ϕ(t) + ϕ(S) | t ∈ T } given by f (t + S) = ϕ(t) + ϕ(S) is a well-defined bijection. For t, t ∈ T , we have t + S = t + S ⇒ t − t ∈ S ⇒ ϕ(t) − ϕ(t ) = ϕ(t − t ) ∈ ϕ(S) ⇒ ϕ(t) + ϕ(S) = ϕ(t ) + ϕ(S), so f is well-defined. Let t, t ∈ T and suppose that f (t+S) = f (t +S). Then ϕ(t) + ϕ(S) = ϕ(t ) + ϕ(S), implying that ϕ(t − t ) = ϕ(t) − ϕ(t ) ∈ ϕ(S). Therefore, t − t ∈ ϕ−1 (ϕ(S)) = S, and so t + S = t + S. This shows that f is injective. That f is surjective is immediate, so the claim that f is bijective is established.

I (ii) Let m = deg f (x) and n = deg g(x) and write f (x) = m i=0 ai x and n g(x) = i=0 bi xi . The term with the highest power of x appearing in the product f (x)g(x) is am bn xm+n , and am bn = 0 since am , bn = 0 and R has no divisors of zero. Therefore, deg[f (x)g(x)] = m + n = deg f (x) + deg g(x). The assumption in force is that R is an integral domain. If this is changed so that R is allowed to have divisors of zero, then (ii) is no longer valid. For instance, if f (x) = 2x and g(x) = 3x, both polynomials over Z6 , then f (x)g(x) = 0, whence deg[f (x)g(x)] = −∞ < 2 = deg f (x) + deg g(x).

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