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Let R be a ring x,y element of R. We say that x and y are associates if x = epsilony for some unit epsilon of R. A ring R having the property that every ideal is principal is called a principal ideal ring (PIR). If, in addition, R is an integral domain, then R is called a principal ideal domain (PID). We say that R is a unique factorization domain if the following two conditions are satisfied: (1) If x element of Rx is not a unit of R, then x can be written as a product of irreducible elements of R.and (2) If x element of Rx is not a unit of R, and if x = pi1 ... pis = lambda1 ... lambdat are two expressions of x as a product of irreducible elements, then s = t and it is possible to renumber pi1 ... pis so that pi1 and lambda1 are associates (1 < i < s). Let R be a ring with identity 1, and let a element of R. If a has an inverse with respect to multiplication, then we say that a is a unit of R. The set of integers. A commutative ring R is an integral domain if R contains no zero divisors. In other words, R is an integral domain if the product of any two nonzero elements of R is nonzero.

Arithmetic in a Principal Ideal Domain

We have seen that Z is a unique factorization domain and also a principal ideal domain. On the other hand, Z[square root of -5] is not a unique factorization domain and not a principal ideal domain. Our main result in this section will be the following theorem.

Theorem 1: Let R be a principal ideal domain. Then R is a unique factorization domain.

Let us begin by developing the theory of divisibility in an integral domain. Throughout this section, let R be an integral domain. If a,b element of R, then we say that a divides b (denoted a|b) if b = xa for some x element of R. If a does not divide b, we write a does not divide b. The elementary facts about divisibility are summarized in the following result:

Lemma 2: Let a,b,c element of R. Then:

(1) a|b, b|c implies a|c.

(2) a|b implies a|bc for all c element of R.

(3) a|b, a|c implies a|b+c.

(4) a|b, a|c implies a|bx + cy for all x,y element of R.

(5) Let b not equal 0. Then a|b, b|a implies a and b are associates.

Proof:

(1) b = xa, c = yb implies c =(xy)a implies a|c.

(2) b = xa implies bc = (xc)a implies a|bc.

(3) b = xa, c = ya implies b + c = (x + y)aimpliesa|b + c.

(4) By (ii) and (iii), a|bx, a|cy implies a|bx + cy.

(5) b = xa, a = yb implies b = (xy)b implies b(1-xy) = 0 implies 1-xy = 0 since bnot equal0 and since R is an integral domain implies xy = 1 implies x is a unit of R implies a a b are associates since b = xa.

It is possible to formulate the notion of divisibility in terms of ideals. For if a|b, then b = xa implies b element of (a) implies by element of (a) for all y element of R implies (b) subset of (a). Conversely, if (b) subset of (a), then b element of (a), so that b = xa and a|b. Thus we have

(1)
a|b if and only if (b) subset of (a).

Lemma 3: Let a,b element of R. Then a and b are associates if and only if (a) = (b).

Proof: implies If a and b are associates, a = epsilonb, where epsilon is a unit of R. Therefore, b = epsilon-1a, so that b|a and a|b. Thus by (1),

(a) subset of (b),     (b) subset of (a).
implies (a) = (b).

backwards implication Suppose that (a) = (b). Then by (1), we have a|b and b|a, so that a and b are associates by Lemma 2,part(5).

Definition 4: Let a,b element of R. A greatest common divisor (g.c.d) of a and b is an element c of R having the following properties:

(1) c|a, c|b.

(2) If d element of R is such that d|a and d|b, then d|c.

In the case R = Z, Definition 4 is essentially the same as the definition of g.c.d. that was given in the section on integers. There is one small difference, however. Earlier, we required that a g.c.d. be positive. In a general ring, the notion of positivity makes no sense, and therefore we cannot include the condition of positivity in Definition 4. Therefore, if a = 3, b = 6, then (a,b) = 3 using the earlier definition, whereas both 3 and -3 are g.c.d's in the sense of Definition 4. Note that 3 and -3 are associates. This is not an accident.

Lemma 5: Let a,b element of R, and let c and c' be two g.c.d.'s of a and b. Then c and c' are associates.

Proof: Since c|a and c|b, and since c' is a g.c.d. condition (2) of Definition 4 implies that c|c'. Reversing the roles of c and c', we see that c'|c. If c = 0, then c' = 0 and the lemma holds. If cnot equal0, then Lemma 2, part (5) implies that c and c' are associates.

Thus, we see that if c0 is a g.c.d. of a and b, then all g.c.d's of a and b are of the form epsilonc0, where epsilon is a unit of R.

Definition 6: Let a,b element of R. We say that a and b are relatively prime, denoted (a,b) = 1, if 1 is a g.c.d of a and b.

In an arbitrary integral domain R, it is not usually true that every pair of elements a,b element of R have a g.c.d. However, if R is a PID, then no pathologies occur and g.c.d's exist.

Proposition 7: Let R be a principal ideal domain and let a,b element of R, a and b not both 0. Then a and b have a g.c.d in R.

Proof: Let I = (a,b). Since R is a PID, I = (c) for some c element of R. Moreover, since a and b are not both 0, Inot equal(0), so that c not equal0. Let us show that c is a g.c.d. of a and b. Since a,b element of I, we see that a = cx, b = cyimpliesc|a, c|b implies (1). Suppose d element of R such that d|a and d|b. Since I = (a,b) = {ax + by | x,y element of R}, we have c = ax + by for some x,y element of R. But then by Lemma 2 part (4), we have d|ax+by implies d|c.

Corollary 8: Let R be a PID, and let a,b element of R, a, b not both 0. Then every g.c.d of a and b can be written in the form ax + by for some x,y element of R. In particular, if a and p are relatively prime, then there exist x,y element of R such that ax + by = 1.

In the case of integers, we proved that if p is prime and a and b are integers such that p|ab, then either p|a or p|b. Let us generalize this result to a PID.

Theorem 9: Let R be a principal ideal domain and let a,b element of R. Let pi be an irreducible element of R such that pi|ab. Then either pi|a or pi|b.

Proof: Let us assume that pi does not divide a. We must then show that pi|b. Since pi is irreducible, pinot equal0 and thus pi and a must have a g.c.d lambda by Proposition 7. But lambda|pi so that lambda is either an associate of pi or a unit of R. In the former case, pi|a since lambda|a. And this contradicts our assumption. Therefore, lambda is a unit of R and lambdalambda-1 is a g.c.d of pi and a, since lambda-1 is a unit of R. But then by Corollary 8, there exist x,y element of R such that

pix + ay = 1.

Thus we see that pi(bx) + (ab)y = b. However pi|pi and pi|ab. so that by Lemma 2, part(4), we see that pi|pibx + aby so that pi|b.

Corollary 10: Let R be a PID, a1,...,an element of R, pi and irreducible element of R. If pi|a1··· an, then pi|ai for some i.

We now come the the proof of Theorem 1. Our proof will be divided into two parts. First we will show that if x element of Rx, x not an element of UR, then x can be written as a product of irreducible elements. Second, we will show that the expression of x a product of irreducible elements is unique up to order and multiplication of the irreducible elements by units.

Theorem 11: Let x element of Rx, x not an element of UR. Then x can be written as a product of irreducible elements.

Proof: Let us reason by contradiction. Assume that x cannot be written a product of irreducible elements. In particular, x is not irreducible, so that we may write

x = a1b1,     a1,b2 not units.

Either a1 of b1 cannot be written as irreducible elements. For otherwise, x could be so written. Thus suppose that a1 is not a product of irreducible elements. Since x = a1b1, we see that a1|x and thus (a1) super set of (x) by (1). Moreover, since b1 is not a unit, (a1)not equal(x) by Lemma 3. Thus, (x) is not properly contained in (a1). Let us denote this fact by (a1) proper super set of (x). Note that a1 element of Rx, a1 not an element of UR, and a1 cannot be written as a product of irreducible elements. Thus we can apply the same reasoning to a1 as we did to x to find a2 element of Rx, a2 not an element of UR, a2 not equal to a product of irreducible elements, and (a2) proper super set of (a1). Let us proceed in this way and construct a3,a4, a5,.... We eventually arrive at the following chain of ideals

(2)
(a0) proper subset of (a1) proper subset of (a2) proper subset of ...,
where a0 = x. Set (3)
I = union of(ai).

Let us prove that I is an ideal of R. Let a,b element of I. Then a element of (ai), b element of (aj) for some i and j. Let j > i. Then (ai) subset of (aj) and a,b element of (aj). But since (aj) is an ideal a±b element of (aj), so that a±b element of I. Thus I is an additive subgroup of R. A similar argument shows that I is closed with respect to multiplication by elements of R. This shows that I is an ideal of R. Since R is a PID, I = (a) for some a element of R. Now a element of I, so that a element of (ak) for some k. Therefore, (a) subset of (ak). However, (ak)subset of I = (a), so that (a) = (ak). In particular, (ak+1) subset of (ak) by (3). But this is a contradiction to (2), which asserts that (ak)subset of (ak+1).

Theorem 12: Let x element of Rx, x not an element of UR and let

x = pi1...pis = lambda1...lambdat

be two expressions of x as a product of irreducible elements. Then s = t and by renumbering lambda1,...,lambdas, we can guarantee that pii and lambdai are associates (1 < i < s).

Proof: If s = 1, then x is irreducible and the assertion is clear. Assume that s > 1 and let us proceed by induction on s. Since pi1|x, we see that pi|lambda1...lambdat. Therefore, by Corollary 10, pi1|lambdai for some i (1 < i < t). by renumbering lambda1,...,lambdat, we may assume that i = 1, so that pi1|lambda1. But since lambda1 is irreducible and pi1 is not a unit we see that pi1 and lambda1 are associates, say epsilonpi1 = lambda1, epsilon element of UR. Therefore,

pi1...pis = epsilonpi1lambda2...lambdat,
pi1(pi2...pis-epsilonlambda2...lambdat) = 0,
impliespi2...pis = epsilonlambda2...lambdat since R is an integral domain.

Set lambda2' = epsilonlambda2, lambda3' = epsilonlambda3,,...,lambdat' = epsilonlambdat. Then lambdai' and lambdai are associates and pi2...pis = lambda2...lambdat' are two decompositions of pi2...pis into irreducible factors. Therefore by the induction hypothesis, s - 1 = t - 1 and after renumbering lambda2'...lambdat', we can guarantee pii and lambdai' are associates (2 < i < s). But then s = t and pii and lambdai are associates (1 < i < s). Thus the induction step is established.

We showed in the previous section that if F is a field and X is an indeterminate over F, then F[X] is a PID. Therefore by Theorem 1, we see that F[X] is a UFD. Let us see exactly what this says: The units of F[X] are just the nonzero constant polynomials. Therefore f element of F[X] is a nonzero, nonconstant polynomial, then f can be written in the form

(4)
f = f1...ft,

where fi element of F[X] is an irreducible polynomial. Moreover, the decomposition (4) is unique up to a rearrangement and multiplication of the fi by constants. Let f = anXn + an-1Xn-1 + ... + a0, annot equal 0> Then an is called the leading coefficient of f. If f has a leading coefficient 1, then we say that f is monic. It is clear that a nonzero polynomial g = bmXm + ... + b0, bmnot equal0 can be written as the product of a constant and a monic polynomial:

g = bm(Xm + bm-1bm-1Xm-1+...+bm-1b0).

Therefore we can write

(5)
f = af1*...ft*,

where a is a constant and fi* is monic and irreducible. By comparing leading coefficients on both sides of (5), we see that a = an.

Proposition 13: Let F be a field, X an indeterminate over F, f a nonzero, nonconstant polynomial in F[X], a the leading coefficient of f. Then f can be written in the form

f = af1*...ft*,

where fi* is an irreducible, monic polynomial. Moreover, this decomposition is unique up to the order of the fi*.

Proof: All that must be proved is the uniqueness. Let

f = af1*...ft* = ag1*...gs*,

be two decompositions of f, where fi* and gi* are irreducible and monic. Since F[X] is a UFD, we have s = t and may renumber f1*,...ft* so that fi* and gi* are associates (1 < i < s). But since fi* and gi* are both monic, fi* = gi*.