By Randall R. Holmes

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**Extra resources for Abstract Algebra II**

**Example text**

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.

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. Because of its usefulness, we seek a generalization of this theorem to other rings. For convenience, we will restrict our search to integral domains. The theorem can be thought of as saying that prime numbers are the building blocks for the integers greater than one. We need a generalization of these building blocks to our arbitrary integral domain.

By part (i), x−r divides f (x) so that f (x) = q(x)(x − r) for some q(x) ∈ R[x]. If s ∈ R is a zero of f (x) and s = r, then s is a zero of q(x) since 0 = f (s) = q(s)(s−r) and s−r = 0. It follows that f (x) has at most one more zero than q(x) has. Now deg q(x) = deg f (x)−1 < deg f (x), so q(x) has at most deg q(x) zeros by the induction hypothesis. Therefore, f (x) has at most deg q(x) + 1 = deg f (x) zeros, and the proof is complete. 5 Irreducible polynomial Let R be an integral domain and let f (x) ∈ R[x].