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(a) Let a be an element of an integral domain R. Show that R=aR is an integral domain if, and only if, a is prime.
(b) Let a be an element of a principal ideal domain R. Show that R=aR is a a field if, and only if, a is irreducible.
(c) Show that a quotient R=I of a principal ideal domain R by a nonzero proper ideal I is an integral domain if, and only if, it is a field.
(d) Show that an ideal in a principal ideal domain is maximal if, and only if, it is prime.
6.5.23. Consider the ring of formal power series KŒŒx with coefficients in a field K.
(a) Show that the units of KŒŒx are the power series with nonzero constant term (i.e., elements of the form ˛0 C xf , where ˛0 ¤ 0, and f 2 KŒŒx).
(b) Show that KŒŒx is a principal ideal domain. Hint: Let J be an ideal of KŒŒx. Let n be the least integer such that J has an element of the form ˛nxn C Pj>n ˛j xj . Show that J D xnKŒŒx.
(c) Show that KŒŒx has a unique maximal ideal M , and KŒŒx=M Š K.
6.5.24. Fix a prime number p and consider the set Qp of rational numbers a=b, where b is not divisible by p. (The notation Qp is not standard.)
Show that Qp is a principal ideal domain with a unique maximal ideal M .
Show that Qp=M Š Zp.
6.6. Unique Factorization Domains
In the first part of this section, we discuss divisors in a unique factor ization domain. We show that all unique factorization domains share some of the familiar properties of principal ideal. In particular, greatest common divisors exist, and irreducible elements are prime.
Lemma 6.6.1. Let R be a unique factorization domain, and let a 2 R be a nonzero, nonunit element with irreducible factorization a D f1 fn.
If b is a nonunit factor of a, then there exist a nonempty subset S of f1; 2; : : : ; ng and a unit u such that b D u Qi2S fi.
Proof. Write a D bc . If c is a unit, then b D c 1a D c 1f1 fn, which has the required form. If c is not a unit, consider irreducible factorizations of b and c, b D g1 g` and c D g`C1 gm. Then a D g1 g`g`C1 gm is an irreducible factorization of a. By uniqueness i
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i of irreducible factorization, m D n, and the gi ’s agree with the fi ’s up to order and multiplication by units. That is, there is a permutation of f1; 2; : : : ; ng such that each gj is an associate of f.j /. Therefore b D g1 : : : g` is an associate of f.1/ f.`/.
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Lemma 6.6.2. In a unique factorization domain, any finite set of nonzero elements has a greatest common divisor, which is unique up to multiplication by units.
Proof. Let a1; : : : ; as be nonzero elements in a unique factorization domain R. Let f1; : : : ; fN be a collection of pairwise nonassociate irreducible elements such that each irreducible factor of each ai is an associate of some fj . Thus each ai has a unique expression of the form ai D
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u Q j .ai /
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j f , where u
j
i is a unit. For each j , let m.j / D mini fnj .ai /g.
Put d D Qj f m.j/. I claim that d is a greatest common divisor of
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fa1; : : : ; asg. Clearly, d is a common divisor of fa1; : : : ; asg. Let e be a common divisor of fa1; : : : ; asg. According to Lemma 6.6.1, e has the form e D u Qj f k.j/, where u is a unit and k.j /
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nj .ai / for all i and j . Hence for each j , k.j / m.j /. Consequently, e divides d .
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We say that a1; : : : ; as are relatively prime if 1 is a greatest common divisor of fa1; : : : ; asg, that is, if a1; : : : ; as have no common irreducible factors.
Remark 6.6.3. In a principal ideal domain R, a greatest common divisor of two elements a and b is always an element of the ideal aR C bR. But in an arbitrary unique factorization domain R, a greatest common divisor of two elements a and b is not necessarily contained in the ideal aRCbR. For example, we will show below that ZŒx is a UFD. In ZŒx, 1 is a greatest common divisor of 2 and x, but 1 62 2ZŒx C xZŒx.
Lemma 6.6.4. In a unique factorization domain, every irreducible is prime.
Proof. Suppose an irreducible p in the unique factorization R divides a product ab. If b is a unit, then p divides a. So we can assume that neither a nor b is a unit.
If g1 g` and h1 hm are irreducible factorizations of a and b, respectively, then g1 g`h1 hm is an irreducible factorization of ab.
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Since p is an irreducible factor of ab, by Lemma 6.6.1 p is an associate of one of the gi ’s or of one of the hj ’s. Thus p divides a or b.
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Corollary 6.6.5. Let R be a unique factorization domain domain. Consider the following properties of an nonzero, nonunit element p of R: pR is a maximal ideal.
pR is a prime ideal.
p is prime.
p is irreducible.
The following implications hold: pR maximal H) pR prime ” p prime ” p irreducible Proof. This follows from Lemma 6.5.18 and Lemma 6.6.4
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Example 6.6.6. In an UFD, if p is irreducible, pR need not be maximal.
We will show below that ZŒx is a UFD. The ideal xZŒx in ZŒx is prime but not maximal, since ZŒx=xZŒx Š Z is an integral domain, but not a field.
Polynomial rings over UFD’s
The main result of this section is the following theorem:
Theorem 6.6.7. If R is a unique factorization domain, then RŒx is a unique factorization domain.
It follows from this result and induction on the number of variables that polynomial rings KŒx1; ; xn over a field K have unique factorization; see Exercise 6.6.2. Likewise, ZŒx1; ; xn is a unique factorization domain, since Z is a UFD.
Let R be a unique factorization domain and let F denote the field of fractions of R. The key to showing that RŒx is a unique factorization domain is to compare factorizations in RŒx with factorizations in the Euclidean domain F Œx.
Call an element of RŒx primitive if its coefficients are relatively prime.
Any element g.x/ 2 RŒx can be written as g.x/ D dg1.x/;
(6.6.1)
where d 2 R and g1.x/ is primitive. Moreover, this decomposition is unique up to units of R. In fact, let d be a greatest common divisor of the (nonzero) coefficients of g, and let g1.x/ D .1=d /g.x/. Then g1.x/
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is primitive and g.x/ D dg1.x/. Conversely, if g.x/ D dg1.x/, where d 2 R and g1.x/ is primitive, then d is a greatest common divisor of the coefficients of g.x/, by Exercise 6.6.1. Since the greatest common divisor is unique up to units in R, it follows that the decomposition is also unique up to units in R.
We can extend this discussion to elements of F Œx as follows. Any element '.x/ 2 F Œx can be written as '.x/ D .1=b/g.x/, where b is a nonzero element of R and g.x/ 2 RŒx. For example, just take b to be the product of the denominators of the coefficients of '.x/. Factoring g.x/ as above gives '.x/ D .d=b/f .x/;
(6.6.2)
where f .x/ is primitive in RŒx. This decomposition is unique up to units in R. In fact, if .d1=b1/f1.x/ D .d2=b2/f2.x/; where f1 and f2 are primitive in RŒx, then d1b2f1.x/ D d2b1f2.x/. By the uniqueness of the decomposition 6.6.1 for RŒx, there exists a unit u in R such that d1b2 D ud2b1. Thus d1=b1 D ud2=b2.
Example 6.6.8. Take R D Z.
7=10 C 14=5x C 21=20x3 D .7=20/.2 C 8x C 3x3/; where 2 C 8x C 3x3 is primitive in ZŒx.
Lemma 6.6.9. (Gauss’s lemma). Let R be a unique factorization domain with field of fractions F .
(a) The product of two primitive elements of RŒx is primitive.
(b) Suppose f .x/ 2 RŒx. Then f .x/ has a factorization f .x/ D '.x/ .x/ in F Œx with deg.'/; deg. / 1 if, and only if, f .x/ has such a factorization in RŒx.
Proof. Suppose that f .x/ D P ai xi and g.x/ D P bj xj are primitive in RŒx. Suppose p is irreducible in R. There is a first index r such that p does not divide ar and a first index s such that p does not divide bs. The coefficient of xrCs in f .x/g.x/ is ar bs C Pi
P j
Suppose that f .x/ has the factorization f .x/ D '.x/ .x/ in F Œx with deg.'/; deg. / 1. Write f .x/ D ef1.x/, '.x/ D .a=b/'1.x/ and .x/ D .c=d / 1.x/, where f1.x/, '1.x/, and 1.x/ are primitive
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in RŒx. Then f .x/ D ef1.x/ D .ac=bd /'1.x/ 1.x/. By part (a), the product '1.x/ 1.x/ is primitive in RŒx. By the uniqueness of such decompositions, it follows that .ac=bd / D eu, where u is a unit in R, so f .x/ factors as f .x/ D ue'1.x/ 1.x/ in RŒx.
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Corollary 6.6.10. If a polynomial in ZŒx has a proper factorization in QŒx, then it has a proper factorization in ZŒx.
Corollary 6.6.11. The irreducible elements of RŒx are of two types: irreducible elements of R, and primitive elements of RŒx that are irreducible in F Œx. A primitive polynomial is irreducible in RŒx if, and only if, it is irreducible in F Œx.
Proof. Suppose that f .x/ 2 RŒx is primitive in RŒx and irreducible in F Œx. If f .x/ D a.x/b.x/ in RŒx, then one of a.x/ and b.x/ must be a unit in F Œx, so of degree 0. Suppose without loss of generality that a.x/ D a0 2 R. Then a0 divides all coefficients of f .x/, and, because f .x/ is primitive, a0 is a unit in R. This shows that f .x/ is irreducible in RŒx.
Conversely, suppose that f .x/ is irreducible in RŒx and of degree 1. Then f .x/ is necessarily primitive. Moreover, by Gauss’s lemma, f .x/ has no factorization f .x/ D a.x/b.x/ in F Œx with deg.a.x// 1 and deg.b.x// 1, so f .x/ is irreducible in F Œx.
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Proof of Theorem 6.6.7. Let g.x/ be a nonzero, nonunit element of RŒx.
First, g.x/ can be written as df .x/, where f .x/ is primitive and d 2 R; furthermore, this decomposition is unique up to units in R. The element d has a unique factorization in R, by assumption, so it remains to show that f .x/ has a unique factorization into irreducibles in RŒx. But using the factorization of f .x/ in F Œx and Gauss’s Lemma, we can write f .x/ D p1.x/p2.x/ ps.x/; where the pi .x/ are elements of RŒx that are irreducible in F Œx. Since f .x/ is primitive, it follows that pi .x/ are primitive as well, and hence irreducible in RŒx, by Corollary 6.6.11.
The uniqueness of this factorization follows from the uniqueness of irreducible factorization in F Œx together with the uniqueness of the factorization in Equation (6.6.2). In fact, suppose that f .x/ D p1.x/p2.x/ ps.x/ D q1.x/q2.x/ qr .x/;
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where the pi .x/ and qi .x/ are irreducible in RŒx. Since f .x/ is primitive, each pi .x/ and qi .x/ is primitive, and in particular of degree 1.
By Corollary 6.6.11, each pi .x/ and qi .x/ is irreducible in F Œx. By the uniqueness of the irreducible factorization in F Œx, after possibly renumbering the qi .x/, we have pi .x/ D ci qi .x/ for each i for some ci 2 F .
But then, by the uniqueness of the decompostion of Equation (6.6.2), each ci is actually a unit in R.
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A characteriziation of UFDs We are going to characterize unique factorization domains by two properties. One property, the so–called ascending chain condition for principal ideals, implies the existence of irreducible factorizations. The other property, that irreducible elements are prime, ensures the essential uniqueness of irreducible factorizations.
Definition 6.6.12. We say that a ring R satisfies the ascending chain condition for principal ideals if, whenever a1R a2R is an infinite increasing sequence of principal ideals, then there exists an n such that amR D anR for all m n.
Equivalently, any strictly increasing sequence of principal ideals is of finite length.
Lemma 6.6.13. A unique factorization domain satisfies the ascending chain condition for principal ideals.
Proof. For any nonzero, nonunit element a 2 R, let m.a/ denote the number of irreducible factors appearing in any irreducible factorization of a.
If b is a proper factor of a, then m.b/