A Graduate Course in Algebra

Volume 2

Ioannis Farmakis, Martin Moskowitz

• Verlag: WSPC
• Jahr: 2017
• ISBN:
  9789813142695
• Thema: Sciences physiques
• Thema: Physique mathématique et informatique

This comprehensive two-volume book deals with algebra, broadly conceived. Volume 1 (Chapters 1–6) comprises material for a first year graduate course in algebra, offering the instructor a number of options in designing such a course. Volume 1, provides as well all essential material that students need to prepare for the qualifying exam in algebra at most American and European universities. Volume 2 (Chapters 7–13) forms the basis for a second year graduate course in topics in algebra. As the table of contents shows, that volume provides ample material accommodating a variety of topics that may be included in a second year course. To facilitate matters for the reader, there is a chart showing the interdependence of the chapters.

Chapter 9

Commutative Rings with Identity

This chapter is a continuation of the commutative branch of Chapter 5 of Volume I. Here, by ring we shall always mean a commutative ring with identity.

In the next sections we will present and discuss the interrelations of the following classes of rings: Integral domains ⊇ UFD’s ⊇ PID’s ⊇ Euclidean domains ⊇ Fields, as well as ACC and DCC (to be defined shortly). The last two inclusions having been proved in Chapter 5 of Volume I.

9.1Principal Ideal Domains

This section concerns the place of PID’s in the universe. For the reader’s convenience, we recall a definition from Chapter 5.

Definition 9.1.1. An ideal I in a ring R is a principal ideal if I = (a), i.e. it has a single ideal generator. If all the ideals I of the ring R are principal, we shall call R a principal ideal domain (or PID).

Proposition 9.1.2. A PID has no zero divisors.

Proof. Suppose ab = 0 and a ≠ 0. If a is a unit then b = 0 and we are done. Otherwise, Let I be a maximal ideal containing (a). Since R is a PID I = (p), where p is a prime. But a = pc for some c ∈ R. Hence 0 = ab = pbc. Therefore I = (0), a contradiction.

Proposition 9.1.3. Let R be a PID and p be a prime of R. If p | ab, then p | a or p | b (or both).

Proof. Suppose p does not divide a. Consider the ideal (p, a) = J of R. Evidently, J ⊇ (p). But J ≠ (p) because if it were p would divide a. On the other hand since p is a prime and R is a PID, Theorem 5.5.10 of Volume I tells us that (p) is a maximal ideal. Therefore J = R. Hence 1 = xp + ya for some x, y ∈ R. Hence b = bxp + yba. Since p | ab and also p | bxp, we see p | b.

We say R is a UFD if every non-zero element and every non-unit of R can be uniquely factored into primes (and a unit).

Theorem 9.1.4. If R is a PID, then R is a UFD.

Later we shall see another way to get this result.

Proof. Let a ∈ R, where a ≠ 0 and a is not a unit. If a is prime we already have a prime factorization. Otherwise a = bc, where neither b nor c is a unit. Now by the first isomorphism theorem for rings we get a commutative diagram

where π is surjective, but not injective (and similarly for c). This introduces a partial order a

b and a

c. By transfinite induction we assume b and c are each a product of a finite number of primes of R. Therefore so is a. This proves the existence of a prime factorization of each non-zero, non-unit of R.

Now suppose

. Then p₁ | q₁ · b, where b = a/q₁ up to a unit. By Proposition 9.1.3 p₁ | q₁ or p₁ | b. In the latter case by induction and repeated use of Proposition 9.1.3 p...