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The set of integers. The set of rational numbers. 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. 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).

Euclidean Domains

In the previous section we proved that a principal ideal domain possesses the property of unique factorization. Now we are faced with the problem of determining whether or not a given integral domain is a principal ideal domain. This is generally a very difficult question. However, a partial solution is available. Notice that we were able to prove that Z and F[X] are PIDs by using the division algorithm in Z and F[X] respectively. In this section let us introduce a class of integral domains in which an analogue of the division algorithm holds. We will show that every such ring is a PID.

Definition 1: Let R be an integral domain. Then R is called a Euclidean domain if there exists a function phi:RmapsZ such that

(1) phi(x) > 0 for all x element of R.

(2) phi(x) = 0 if and only if x = 0.

(3) phi(xy) = phi(x)phi(y) for all x,y element of R.

(4) Let x,y element of R, ynot equal0. Then there exist q,r element of R such that x = qy + r, 0 < phi(r) < phi(y). (This is a generalization of the division algorithm.)

Example 1: Let R = Z, phi(x) = |x|. Then conditions (1)-(3) are elementary properties of absolute values. Condition (4) is just the division algorithm in Z.

Example 2: Let R = F[X]. Set phi(x) = 2deg(x) (xnot equal0), and and = 0 (x = 0). Then (1)-(3) are easy to check. Moreover, by the division algorithm in F[X], given x,y element of R, y not equal 0, there exist q,r element of R such that x = qy + r, deg(r) < deg(y). But then 0 < phi(r) < phi(y) and condition (4) holds.

Theorem 2: Let R be a Euclidean domain. Then R is a principal ideal domain.

Proof: Let I be a nonzero ideal of R. It suffices to show that I is principal. Let a element of I be chosen so that anot equal0 and phi(a) is a small as possible. We assert that I = (a). It is clear that (a) subset of I. Let x element of I. Since R is a Euclidean domain, there exist q,t element of R so that

x = qa + r,    0 < phi(r) < phi(a).

Note that r = x - qa element of I If rnot equal0, then phi(r) < phi(a), r element of I, which contradicts the choice of a. Therefore, r = 0 and x = qa element of (a). Thus I subset of (a).

Let us now give an example of a Euclidean domain which is connected with Fermat's last theorem.

Theorem 3: Let zeta = (-1 + square root of 3i)/2. Then zeta is a primitive cube root of 1 and Z[zeta] is a Euclidean domain.

Before proving Theorem 3, it is necessary to describe Z[zeta] more explicitly. Note that zeta2 + zeta + 1 = 0, so that

zeta2 element of {a + bzeta | a,b element of Z}.

Also, zeta3 = -zeta2 - zeta, so that

zeta3 element of {a + bzeta | a,b element of Z}.

Proceeding by induction, we see that

zetan element of {a + bzeta | a,b element of Z}

for all n > 1. Therefore,

Z[zeta] = {a + bzeta | a,b element of Z}.

Moreover, every element of Z[zeta] can be written uniquely in the form a + bzeta, since

a + bzeta = a' + b'zeta implies a - a' = (b' - b)zeta
implies(b' - b)zeta element of Z
impliesb' - b = 0
impliesa - a' =0.

And we have proved the following

Lemma 4: Every element of Z[zeta] can be uniquely written in the form a + bzeta, a,b element of Z.

Proof of Theorem 3: If a,b element of Q, define phi(a + bzeta) = (a + bzeta)(a + bzeta conjugate) = a2 - ab + b2, where zeta conjugate denotes the complex conjugate of zeta. Since a + bzeta conjugate = a + bzeta conjugate, we see that phi(a + bzeta) = |a + bzeta|2 > 0. Moreover, phi(a + bzeta) = 0 equivalent a + bzeta = 0. Further, if a,b element of Z, phi(a + bzeta) = a2 - ab + b2 element of Z. We leave it as an exercise to verify that phi(alphabeta) = phi(alpha)phi(beta) for alpha,beta element of {a + bzeta | a,b element of Q}. Let us verify the division algorithm. This is where we use the special properties of zeta. Let x = a + bzeta, y = c + dzeta element of Z[zeta], ynot equal0. Then, in C, we have

division

where

division , division

belong to Q. There exist integers alpha and beta such that

|e - alpha| < 1/2,  |f - beta| < 1/2.

Set q = alpha + betazeta, r = [(e - alpha) + (f - beta)zeta]y. Then q element of Z[zeta] and

x = qy + r,

so that r = x - qy element of Z[zeta]. Finally

phi(r) = phi((e - alpha) + (f - beta)zeta)phi(y)
= {(e - alpha)2 - (e - alpha)(f - beta) + (f - beta)2}phi(y)
< {1/4 + 1/4 + 1/4}phi(y)
< phi(y).

Corollary 5: If zeta = (-1 + square root of 3i)/2 is a primitive cube root of 1, then Z[zeta] is a unique factorization domain.