Abstract Algebra II by Randall R. Holmes

By Randall R. Holmes

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The factorization −120 = (−3)(2)(5)(−2)(−2), also exhibits −120 as a product of irreducible elements. This second factorization can be reordered as −120 = (−2)(−2)(2)(−3)(5) and comparing with the earlier factorization −120 = (−2)(2)(2)(3)(5) we see that corresponding factors are indeed associates. A field is a UFD (vacuously since a field has no nonzero nonunits). An example of an integral domain that is not a UFD is given in Exercises 8–1 and 8–3. 5 A PID is a UFD An integral domain R is a principal ideal domain, or PID for short, if every ideal of R is principal.

The following theorem says that the notions “quotient of R” and “homomorphic image of R” amount to the same thing. 1 Theorem. If R/I (I R) is a quotient of R, then R/I is a homomorphic image of R, namely, the image under the canonical epimorphism π : R → R/I. Conversely, the image im ϕ = ϕ(R) of R under a homomorphism ϕ : R → R is isomorphic to a quotient of R, namely R/ ker ϕ. Proof. 2. 1). 3 Second Isomorphism Theorem Let R be a ring, let S be a subring of R and let I be an ideal of R. By Exercise 7–1, S +I is a subring of R.

Since a and b are relatively prime, it follows that p divides the coefficient of every term of g (x)f (x). But this product is primitive by Gauss’s lemma, so this is a contradiction. Therefore, b = ±1 as claimed. Therefore, f (x) = g1 (x)h1 (x), where g1 (x) = (a/b)g (x) = ±ag (x) ∈ Z[x] and h1 (x) = h (x) ∈ Z[x]. Moreover, g1 (x) and h1 (x) have the same degrees as g(x) and h(x), respectively. (ii) We prove the contrapositive. Assume that f (x) is not irreducible over Q. Since f (x) is nonconstant, it is a nonzero nonunit, so it has a proper factorization f (x) = g(x)h(x) with g(x), h(x) ∈ Q[x].

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