Abstract Algebra II by Randall R. Holmes

By Randall R. Holmes

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We claim that the polynomial f (x) = x5 + 5x2 − x − 2 is reducible over R. Since f (0) = −2 and f (1) = 3, and the polynomial function induced by f (x) (see Exercise 9–1) is continuous, the Intermediate Value Theorem implies that f (x) has a zero r (between 0 and 1). 4, x−r divides f (x), so that f (x) = (x−r)g(x) for some g(x) ∈ R[x]. 2, g(x) has degree 4, so this is a proper factorization of f (x). 50 Showing that a given polynomial is irreducible is often difficult. For irreducibility over a field one can sometimes use the following observation about a proper factorization, together with a proof by contradiction.

1 Theorem. If f (x+c) is irreducible over Z, then f (x) is irreducible over Q. Proof. Assume that f (x+c) is irreducible over Z. 1 to show that f (x) is irreducible over Z. The map ϕx+c : Z[x] → Z[x] given by ϕx+c (f (x)) = f (x + c) is an isomorphism (the proof that the evaluation map is a homomorphism easily generalizes to show that this map is a homomorphism as well, and this map is bijective since ϕx−c is an inverse). Since f (x+c) is irreducible over Z (which means after all that it is an irreducible element of the ring Z[x]), and since it corresponds to f (x) under the isomorphism, we conclude that f (x) is also irreducible over Z.

Prove that S + I is a subring of R. 7–2 Let R be a commutative ring with identity. Prove that R[x]/(x) ∼ = R. 7–3 Let R be a ring and let I = {(r, 0) | r ∈ R}. Prove that I is an ideal of R ⊕ R and (R ⊕ R)/I ∼ = R. 1 Theorem (Fundamental Theorem of Arithmetic). An integer greater than one can be factored as a product of prime numbers, and such a factorization is unique up to the order of the factors. ) This theorem has many uses in the study of the ring of integers. For instance, it makes possible the notions of the greatest common divisor and the least common multiple of a collection of integers greater than one.

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