3.3 Noetherian Rings and Principal Ideal Domains

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Let R be a ring (commutative, remember?). We call R Noetherian if whenever

we have a chain

I1 _ I2 _ · · ·

of ideals, then there exists an integer N such that if n _ N, then In = IN.

If I _ R is an ideal, we say that I is finitely generated if there exist

a1, a2, . . . , ak 2 R such that I = (a1, a2, . . . , ak), i.e., if every element of I is

of the form

P

riai, ri 2 R.

The following is basic and quite useful.

Theorem 3.3.1 Let R be a commutative ring. The following are equivalent.

(i) R is Noetherian.

(ii) Every ideal in R is finitely generated.

(iii) Every collection of ideals has a maximal element with respect to inclusion.

The next result result is not only intrinsically interesting, it is also a fundamental

tool in the study of algebraic geometry and commutative algebra.

Theorem 3.3.2 (Hilbert Basis Theorem) If R is Noetherian, so is the

polynomial ring R[x].

An important, but rather “small” class of examples of Noetherian rings

is as follows. Let R be an integral domain. We say that R is a principal

ideal domain (p.i.d.) if every ideal of R is generated by a single element.

Thus if I _ R is an ideal, then there exists a 2 R such that I = (a). The

“canonical” examples are

(i) Z, and

(ii) The polynomial rings F[x], where F is a field.

However, that these really are examples is the result of their satisfying an

even stronger condition, as discussed in the next section.

Theorem 3.3.3 If R is a p.i.d., then R is Noetherian.

82 CHAPTER 3. ELEMENTARY FACTORIZATION THEORY

Theorem 3.3.4 If R is a p.i.d., then R is a u.f.d..

Exercises

1. Prove that if R is Noetherian, and if I is an ideal in R, then R/I is

Noetherian.

2. Let R _ S be integral domains, with R Noetherian. If s1, s2, . . . sr 2 S,

prove that R[s1, s2, . . . sr] _ S is Noetherian.

3. Let R be a ring and let x1, x2, . . . be infinitely many indeterminates

over R. Prove that the polynomial ring R[x1, x2, . . .] is not Noetherian.

4. Let R be a u.f.d. in which every prime ideal is maximal. Prove that

R is actually a p.i.d. (Start by showing that every prime ideal is

principal.)

5. Let R be a commutative ring and assume that the polynomial ring

R[x] is a p.i.d. Prove that R is, in fact, a field.

6. An integral domain R such that every non-unit a 2 R can be factored

into irreducibles is called an atomic domain. Prove that every

Noetherian domain is atomic. (This factorization might not be unique,

however. Note that the converse is not true: see Exercise 10 of Section

3.2 and Exercise 3, above. )

7. Show that in the ring R = Z[

p

−5], the ideal P = (3, 4 +

p

−5) _ R is

a non-principal prime ideal.

8. Let R be a Noetherian ring in which every pair of elements has a

greatest common divisor. Prove that R is a u.f.d. (Use Exercise 6

above and Exercise 6 of Section 3.2.)

9. Let R be a p.i.d.

(a) If a, b 2 R, with d = g.c.d.(a, b), show that there exist r, s 2 R,

such that ra + sb = d.

(b) If a, b 2 R with q = l.c.m.(a, b), show that (q) = (a) \ (b).

(c) Show that ((a) + (b))((a) \ (b)) = (a)(b).

(d) Which of the above are true if R is only assumed to be a u.f.d.?

3.3. NOETHERIAN RINGS AND PRINCIPAL IDEAL DOMAINS 83

10. Let R be a u.f.d. and assume that whenever a, b 2 R and are relatively

prime, then there exist elements s, t 2 R with sa + tb = 1. Prove that

every finitely generated ideal in R is principal. 1 In particular, if R

is Noetherian, then R is a p.i.d. (Hint: Let I _ R be an ideal and

let a, b 2 I. Let d be the greatest common divisor of a and b; thus

if a0 = a/d and b0 = b/d then a0 and b0 are relatively prime. Use the

condition to show that (a, b) = (d). Now use induction.)

11. Let R be a Noetherian ring and let Q _ R be a primary ideal (see

Exercise 11 of Section 3.1). If IJ _ Q and I 6_ Q, then there exists

n _ 1 such that Jn _ Q.

12. Let R be a Noetherian ring and let Q _ R be a P-primary ideal (see

Exercise 11a of Section 3.1). Show that there exists some n _ 1 such

that Pn _ Q.

1It is interesting to note that a slight variant of this result applies to the ring O(C),

introduced in Exercise 7 of Section 3.2, despite the fact that O(C) is not a u.f.d. The

relevant result is that if f, g are two holomorphic functions on C with no common zeros,

then there exist holomorphic functions s, t satisfying sf + tg = 1, identically on C. From

this fact, the reader should have no difficulty in showing that finitely generated ideals

in O(C) are principal. For details, consult R.B. Burckel’s An Introduction to Classical

Complex Analysis, Vol. 1, Birkh¨auser, Basel and Stuttgart, 1979, Corollary 11.42, p. 393.

84 CHAPTER 3. ELEMENTARY FACTORIZATION THEORY

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