3.2 Unique Factorization Domains

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In this section R is always an integral domain. If u 2 R is an element having

a multiplicative inverse, we call u a unit . The set U(R), with respect to

multiplication, is obviously an abelian group, called the group of units of R.

Let a, b 2 R, a 6= 0. If b = qa for some q 2 R, we say that a divides b,

and write a|b. Obviously, if u 2 U(R), and if b 2 R, then u|b. If a, b 2 R. Say

that a, b are associates if there exists u 2 U(R) such that a = ub. If a, b 2 R

and d is a common divisor of both, we say that d is a greatest common divisor

if any other divisor of both a and b also divides d. In the same vein, if l is

a multiple of both a and b, and if any multiple of both is also a multiple of

a and b, we say that l is a least common multiple of a and b. Note that if a

greatest common divisor exists, it is unique up to associates, and it makes

sense to write d = g.c.d.(a, b) for the greatest common divisor of a and b;

the same is true for a least common multiple, and we write l = l .c.m(a, b)

for the least common multiple of a and b.

Let p 2 R, p 62 U(R). We say that p is irreducible if p = ab implies that

a 2 U(R) or b 2 U(R). We say that p is prime if the ideal (p) _ R is a prime

ideal in R. The following is elementary.

Proposition 3.2.1 Let p 2 R. If p is prime, then p is irreducible.

The converse of the above fails; here’s an example. Let R = Z[

p

−5] =

{a + b

p

−5| a, b 2 Z}. Then 3 is easily checked to be irreducible, but (4 + p

−5)(4 −

p

−5) 2 (3), whereas (4 +

p

−5), (4 −

p

−5) 62 (3).

We say the integral domain R is a unique factorization domain (u.f.d.)

if

(i) Every irreducible element is prime.

(ii) If 0 6= a 2 R, and if a is not a unit, then there exist primes p1, p2, . . . , pn 2

R such that a = p1p2 · · · pn.

We remark that it is possible for either one of the above conditions to

hold in an integral domain without the other also being valid. For instance,

for a ring satisfying (ii) but not (i), see Exercise 4. For a ring satisfying (i)

but not (ii), consult Exercise 7, below.

Proposition 3.2.2 Let R be a unique factorization domain, and let a 2 R,

a 62 U(R). Then there exist unique (up to associates) primes p1, p2, . . . , pk 2

R, and unique exponents e1, e2, . . . , ek 2 N such that a = pe1

1 pe2

2 · · · pek

k .

3.2. UNIQUE FACTORIZATION DOMAINS 77

The above explains the terminology “unique” in unique factorization domain.

The reader should note that the ring of the above example is not a

unique factorization domain. One can show, however, that every non-unit

in R can be factored as a product of irreducibles. (See Exercise 4.) From

this example, I hope that the reader can get some idea of the subtlety of

unique factorization domains.

Let R be a u.f.d., and let a, b 2 R. In this setting the greatest common

divisor and least common multiple of a, b both exist in R, and can be

constructed as follows: If we factor a and b into primes:

a = pe1

1 pe2

2 · · · per

r , b = pf1

1 pf2

2 · · · pfr

r ,

(where possibly some of the ei’s or fj ’s are 0), we set

d = pt1

1 pt2

2 · · · ptr

r , ti = min{ei, fi}, i = 1, 2, . . . , r;

Then d is the greatest common divisor of a and b, denoted d = g.c.d.(a, b).

Likewise, if we set

q = ps1

1 ps2

2 · · · psr

r , si = max{ei, fi}, i = 1, 2, . . . , r,

then q is the least common multiple of a and b, and denoted q = l.c.m.(a, b)

It is assumed that the reader has already had a previous course in abstract

modern algebra, where one typically learns that two paradigm examples

of unique factorization domains are the ring Z of integers and the

polynomial ring F[x] over the field F. That these rings are both unique factorization

domains will be proved again in Section 3.4 below, independently

of the present section. For what follows, we shall use the fact that F[x] is a

u.f.d.

For the remainder of the section, we shall be concerned with the study

of the polynomial ring R[x], where R is an integral domain. Note that in

this case it is easy to see that R[x] is also an integral domain. It shall be

convenient to move back and forth between R[x] and F[x], where F = F(R) is

the field of fractions of R. As remarked above, F[x] is a unique factorization

domain. Note that U(R[x]) = U(R) the group of units of R.

Henceforth, we shall assume that R is a unique factorization domain.

Our goal is to supply the necessary ingredients to prove that R[x] is again

a unique factorization domain.

78 CHAPTER 3. ELEMENTARY FACTORIZATION THEORY

We say that the polynomial f(x) 2 R[x] is primitive if the greatest

common divisor of the coefficients of f(x) is 1. More generally, if If g(x) 2

R[x], and if c 2 R is the greatest common divisor of the coefficients of

g(x), then we may write g(x) = cf(x) where f(x) is a primitive polynomial

in R[x]. The element c 2 R is called the content of g(x); note that it is

well-defined, up to associates.

Lemma 3.2.3 (Gauss’ Lemma) If f(x), g(x) 2 R[x] are primitive polynomials,

then so is f(x)g(x).

Lemma 3.2.4 Let R be a u.f.d. with fraction field F. Assume that f(x), g(x) 2

R[x] are primitive polynomials and that f(x), g(x) are associates in F[x].

Then f(x), g(x) are associates in R[x].

Lemma 3.2.5 Let f(x) 2 R[x] be primitive, and assume that f(x) cg(x),

where c 2 R, and g(x) 2 R[x] is also primitive. Then f(x) g(x).

Lemma 3.2.6 If f(x) 2 R[x] is irreducible, then f(x) is still irreducible in

F[x].

With the above results in place, one can now prove the following:

Theorem 3.2.7 If R is a unique factorization domain, so is the polynomial

ring R[x].

In particular, it follows that if R is a unique factorization domain, and

if x1, x2, . . . , xr are indeterminates over R, then R[x1, x2, . . . , xr] is again a

unique factorization domain.

Before closing this section, I can’t resist throwing in the following batch

of examples. Let n be a positive integer and let _ = e2_i/n; set R = Z[_] _

C. The importance of these rings is that there exist early (but incorrect)

proofs of the famous Fermat conjecture (“Fermat’s Last Theorem;” recently

3.2. UNIQUE FACTORIZATION DOMAINS 79

proved by Andrew Wiles), based on the assumption that R is a unique

factorization domain for every value of n. Unfortunately, this assumption

is false; the first failure occurs for n = 23. (In the ring Z[e2_i/23], one has

that the number 2 is irreducible, but not prime.) This result came as a

shock to many mathematicians; on the other hand, it led to many new and

interesting avenues of research, leading to the development of “algebraic

number theory.” We shall touch on this area in the next chapter.

Exercises

1. Compute U(R) in each case below.

(a) R = Z.

(b) R = Z/(n).

(c) R = Z[i] = {a + bi| a, b 2 Z} (i2 = −1).

(d) R = Z[_] = {a + b_| a, b 2 Z} (_ = e2_i/3).

2. Let R = Z[

p

2] = {a + b

p

2| a, b 2 Z}. Prove that U(R) is infinite.

3. Let F be a field and let x be indeterminate over F. Prove that the ring

R = F[x2, x3] is not a u.f.d. (Consider the equation (x2)3 = (x3)2.)

4. Let R = Z[

p

−5] = {a + b

p

−5| a, b 2 Z}. Then every non-unit of R

can be factored as a product of irreducibles. (Hint: Define a “norm”

map on R by setting N(a + b

p

−5) = a2 + 5b2. Note that if r, s 2 R,

then N(rs) = N(r)N(s). So what?)

5. As above, let R = Z[

p

−5] = {a + b

p

−5| a, b 2 Z}. Show that it need

not happen that any pair of non-units in R has a greatest common

divisor. (In fact, one can show that 21 and 7(4+

p

−5) have no greatest

common divisor.)

6. Let R be an integral domain in which every pair of elements has

a greatest common divisor. Prove that every irreducible element is

prime. (Hint: Let p 2 R be irreducible and assume that p ab but that

p/ a. Then the elements ab and ap have a greatest common divisor d.

Thus p d and d ap, forcing d = pa0 for some divisor a0 a. Similarly

d = ab0 for some divisor b0 b. From pa0 = ab0 we get p = b0(a/a0);

since p is irreducible, we have b0 2 U0(R) or a/a0 2 U0(R) Now what?)

80 CHAPTER 3. ELEMENTARY FACTORIZATION THEORY

7. Consider the integral domain D = O(C) of holomorphic functions on

the complex plane. Prove that every irreducible element of D is prime,

but that D is not a u.f.d. More precisely, prove that

(a) The irreducibles in D are the functions of the form f(z) = z −z0

(and their associates), where z0 2 C.

(b) Irreducibles are primes.

(c) The function f(z) = sin z cannot be factored into irreducibles.

8. Prove Eisenstein’s Irreducibility Criterion. Namely, let R be a u.f.d.

and let f(x) 2 R[x]. Write

f(x) = anxn + an−1xn−1 + · · · + a1x + a0.

Assume that there exists a prime p 2 R such that

(a) p/ an;

(b) p ai, 0 _ i _ n − 1;

(c) p2 / a0.

Then f(x) is irreducible.

9. Let x1, x2, . . . , xn2 be indeterminates and consider the matrix

A =

2

6664

x1 x2 · · · xn

xn+1 xn+2 · · · x2n

...

...

...

· · · · · xn2

3

7775

.

Show that det A is an irreducible polynomial in C[x1, x2, . . . , xn2 ].

(Hint:

det

2

6664 y

0

0

·

·

·

0

x

1 y 0 · · · 0 0

...

...

...

0 0 0 · · · 1 y

3

7775

= yn ± x.

10. Let R be a u.f.d. and let x1, x2, . . . be indeterminates over R. Prove

that R[x1, x2, . . .] is a u.f.d.

3.3. NOETHERIAN RINGS AND PRINCIPAL IDEAL DOMAINS 81

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