By Randall R. Holmes

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**Extra info for Abstract Algebra II**

**Sample text**

Theorem. Let r ∈ R. (i) The element r is irreducible if and only if (r) is nonzero and maximal among the proper principal ideals of R. (ii) The element r is prime if and only if (r) is nonzero and prime. (iii) Every prime element of R is irreducible. (iv) If every ideal of R is principal, then every irreducible element of R is prime. Proof. 2. (i) Assume that r is irreducible. First, (r) is nonzero, since r is nonzero, and it is proper since r is not a unit. Let s be an element of R such that (r) ⊆ (s) and (s) is proper.

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. 37 These two characterizations of prime integer are equivalent and they describe the usual prime numbers together with their negatives: ±2, ±3, ±5, ±7, ±11, and so on. We alert the reader to a point of terminology.

What we have shown is that a nonzero nonunit of R that cannot be written as product of irreducible elements generates an ideal that is properly contained in the ideal generated by another nonzero nonunit of R that cannot be written as a 42 product of irreducible elements. We conclude that there exists a nonending sequence (r) (r1 ) (r2 ) · · · of principal ideals of R, each properly contained in the next. By Exercise 8–5, the union of these ideals is an ideal of R, and it is equal to (s) for some s ∈ R since R is a PID.