Page 204 - DMTH403_ABSTRACT_ALGEBRA

Unit 20: Principal Ideal Domains

Then, for any a,b  Z \ {O], Notes
d(ab)= |ab| = |a| |b|  |a| (since | b| for b  0)
= d(a).

i.e., d(a)  d(ab).
Further, the division algorithm in Z says that if a, b  Z, b  0, then 3 q, r  Z such that
a = bq + r, where r = 0 or 0 < |r| < |b|.
i.e:, a = bq+r, where r = 0 or d(r) < d(b).
Hence, d is a Euclidean valuation and Z is a Euclidean domain.
Let us now discuss some properties of Euclidean domains. The first property involves the
concept of units. So let us define this concept. Note that this definition is valid for any integral
domain.
Definition: Let R be an integral domain. An element a  R is called a unit (or an invertible
element) in R, if we can find an element b  R, such that ab = 1, i.e., if a has a multiplicative
inverse.
For example, both 1 and -1 are units in Z since 1.1 = 1 and (-1).(-1) = 1.

!
Caution The difference between a unit in R and the unity in R. The unity is the identity
with respect to multiplication and is certainly a unit. But a ring can have other units ton, as
you have just seen in the case of Z.
Now, can we obtain all the units in a domain? You know that every non-zero element in a field
F is invertible. Thus, the set of units of F’ is F \ {0}. Let us look at some examples.

Example: Obtain all the units in F[x], where F is a field.

Solution: Let f (x)  P[ x] be a unit, Then  g(x)  F[x] such that f(x) g(x) = 1. Therefore,
deg f(x) g(x)) = deg(1) = 0, i.e.,
deg f(x) + deg g(x) = 0

Since deg f(x) and deg g(x) are non-negative integers, this equation can hold only if deg f(x) = 0
= deg g(x). Thus, f(x) must be a non-zero constant, i.e., an element of F\ {0}. Thus, the units of F[x]
are the non-zero elements of F. That is, the units of F and F[x] coincide.

Example: Find all the units in R =  a  b  5|a,b Z  .

5
Solution: Let a b  be a unit in R. Then there exists

5

c d   R such that
 a  b  5  c d   5 = 1

 (ac – 5bd) + (bc + ad)  = 1
5
 ac – 5bd = 1 and bc+ad = 0
 abc – 5b d = b and bc+ad = 0
2
 a(–ad) – 5b d = b, substituting bc = –ad.
2

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