Why Read This Book? - Index of

9.11 Euclidean Domains 327
Exercise 9.11.7 sheds a little more light on some things we already know. First,
since absolute value can serve as a valuation on the integers, we see that the
units in the integers are precisely the values of x for which |x| = |1|, namely, ±1.
Also, since every nonzero element of a field is a unit, any valuation d must satisfy
d(x) = d(e) for all nonzero x. Thus d must be constant. Conversely, if a valuation
on a Euclidean domain is constant, then every nonzero element is a unit, so that
the Euclidean domain is a field.
As we have progressed from rings to domains to UFDs to PIDs to Euclidean
domains, we have claimed that there exist examples of one structure that do not
qualify as an example of the next most restrictive structure. In every case up to
now, we have provided an example complete with proof, except for Z[t]. We have
shown that Z[t] is not a PID (Corollary 9.10.7), but we have not yet shown it is a
UFD. We will do this in the next section.
About now you should be asking, “Where is my example of a PID that’s not
Euclidean?” Well, this is the only place in our progression from more general to
more specialized rings where we are going to present an example of a PID that is
not Euclidean and will give only a loose explanation of how this would be shown.
The classic example of a non-Euclidean PID was first constructed by T. Motzkin in
1949—very recently indeed by mathematical standards. It is Z[(1 + √ −19)/2], the
set of all expressions of the form m + n(1 + √ −19)/2, where m and n are integers.
For convenience, let’s write α = (1 + √ −19)/2, and discuss how one goes about
showing Z[α] is a PID that is not Euclidean.
First, we address the fact that Z[α] is not Euclidean. We would do this by
showing that every Euclidean domain has a certain feature, then showing that
Z[α] does not have this feature. If D is any Euclidean domain that is not a field,
then there will exist nonzero, non-unit elements. If d is a valuation, then a nonzero,
non-unit element x will satisfy d(x) > d(e). Among all elements of D, let a be a
nonzero, non-unit element for which d(a) is minimal. Such an element is called
a universal side divisor. By the division algorithm on D, any nonzero x can be
written as x = aq + r, where either r = 0ord(r) < d(a). Since d(a) is minimal
among all nonzero, non-unit elements of D, we may say that r is either zero or a
unit. Every Euclidean domain that is not a field will have universal side divisors
because the valuation will not be constant.
Next we need to know what the units are in Z[α]. It turns out that the only
units in Z[α] are ±1. So let’s suppose Z[α] is a Euclidean domain and see how
this leads to a contradiction. Since the only units in Z[α] are ±1, Z[α] is not a
field. Thus there exists a universal side divisor a ∈ Z[α], and every x ∈ Z[α] can be
written as x = aq + r, where r ∈{0, ±1}. In particular, x = 2 must be writable in
this way. So aq = 2 − r, and the only possible values of 2 − r are {1, 2, 3}. Thus a
divides at least one of {1, 2, 3}, but since a is not a unit, a does not divide 1. With
some work, we could show that the only divisors of 2 are {±1, ±2} and the only
divisors of 3 are {±1, ±3}. Thus a ∈{±2, ±3}. But letting x = α, it turns out that
there is no q ∈ Z[α] for which α = aq + r, given that a ∈{±2, ±3} and r ∈{0, ±1}.
This is a contradiction, so Z[α] is not a Euclidean domain.
Now we address the fact that Z[α] is a PID. First, let’s take the defining characteristic
of a Euclidean domain and state it first in the original way and then in