By Randall R. Holmes

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**Additional resources for Abstract Algebra II**

**Example text**

B) Give an example to show that ϕ need not be surjective. (c) Prove that if R = Z2 , then R[x]/ ker ϕ ∼ = Z2 ⊕ Z2 . 9–2 Let R and R be commutative rings and let σ : R → R be a homomorphism. 8 is indeed a ring homomorphism as claimed. 1 Irreducible over Z implies irreducible over Q Let f (x) be a polynomial of degree n > 0 over Z. If f (x) does not factor as a product of two polynomials over Z each of degree strictly less than n, then it does not follow immediately that the same is true if the two polynomials are allowed to have coefficients in the larger ring Q.

For instance, the prime number 5 divides 30 and no matter how we express 30 as a product of integers, 5 always divides one of the factors: 30 = (2)(15) and 5|15, 30 = (−10)(−3) and 5| − 10, 30 = (6)(5) and 5|5, and so on. The nonprime 6 does not have this property, since 6|30, but in the factorization 30 = (3)(10), we have 6 3 and 6 10. So here is a second characterization: • a prime integer is an integer p, neither zero nor a unit, having the property that if p|mn (m, n ∈ Z), then either p|m or p|n.

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. Because of its usefulness, we seek a generalization of this theorem to other rings. For convenience, we will restrict our search to integral domains.