Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

106
ADVANCED ABSTRACT ALGEBRA
Let p
≠
⊂ Q where Q = (a), a ∈ R .
Now since p is a prime, greatest common divisor of a and p is p or 1. If (a, p) = p then
a ∈ (p) = p
(a) ⊆ p
Q ⊆ p
But then Q = p which is not the case.
Therefore g.c.d. of a and p is one. Thus there exist x and y such that
1 = ax + py
Let us suppose that b ∈ R .
Now bpy ∈p ⊆ Q and bax ∈ Q
∴ b ∈ Q
R ⊂ Q
But Q being an ideal of R we have
Q ⊂ R .
Hence Q = R
This proves that p is maximal.
p | a and so
Cor. If D is a P.I.D and p is a prime, then (p) is a prime ideal, in fact, since for a commutative ring D
every maximal ideal is a prime ideal.
Euclidean Domain. An integral domain R is said to be a Euclidean domain (Euclidean ring) if there
exists a mapping φ of the set of non-zero members of R into the set of positive integers such that if a, b
be any two non-zero members of R then
(i)
there exists q, r ∈ R such that
a = bq + r
where either r = 0 or φ(r) < φ(b)
(ii)
φ(ab) ≥ φ(a) or φ(b).
Example 1. The domain I of integer is Euclidean, for the mapping φ defined by
φ (a) = |a|
satisfies the properties in question.
2. The domain K[x] of polynomials over a field K is Euclidean with the mapping defined by
deg ax
f(ax) = 2 where
a(x) ∈ K [x].
Theorem. Euclidean domain is a principal ideal domain.
Proof. Let D be any Euclidean domain and φ a mapping referred to in the definition. Let I be any ideal
of D. If I is zero ideal, then it is a principal ideal. Now suppose that I ≠ (0) so that it contains some nonzero
members.
Consider the set of φ images of the non-zero members of I which are all positive integers. Let a ≠ 0 be a
member of I so that φ(a) is minimal in all the φ images.
Let b be any arbitrary member of I. Then there exist two members q and r of D such that
b = qa + r
where either r = 0 or φ(r) < φ(a)