Euclidean Rings
Definition. An integral domain R is said to be a Euclidean ring or Euclidean domain if for every a(≠0)∈ R there is defined a non-negative integer d(a) such that
(1)for all a,b∈ R, a≠ 0,b≠ 0;d (a) < d (ab) and
(2)for any a,b∈R,b≠ 0 there exist q,r∈R such that a = bq +r where either r = 0 or d (r) <d(b) .
Note
- For any a(≠ 0)∈R,d (a)> 0.
- For the zero element 0 of R,d{0) is not defined. However, some authors defined d(0) = 0, integer.
- The property (2) in the above definition is called the division algorithm.
- From the above definition, we note that d: R- {0} -> Z is a mapping such that
(1) d (a) > 0 ∀ R— {0} .
(2) d (a) < d (a,b)∀1 a,b∈R- {0} and
(3) there exist q,r∈R so that a-bq+r where either r = 0 or d (r) < d (b) for any a b∈R and b≠0.
Subrings, Ideals, Quotient Rings, And Euclidean Rings Theorems
Theorem 1. Every field is a Euclidean ring
Proof. Let F be a field and F be the set of all non-zero elements of F. Since F is a field, F is an integral domain
Define the mapping d:FZ by d(a) = 0 (zero integer) for all a ϵ F
d (a) ≥0 ∀ a ϵF
Let a bϵF.
Then a,b, and ab are non-zero elements of F
d (a) = 0 and d (ab) = 0 => d (a) <,d (ab)
Let aϵF and bϵF.
Now a = a 1 where 1 is the unity element of F.
=a(b-1b) = (ab-1)b ( b-1b=1)
= (ab-1]) b+ 0 where ‘O’ is the zero element of the field F.
a = qb +r where q = ab-1,r = 0 1
Hence, for a ϵ F,b ϵ F there exists q,r ϵ F so that a = qb+r where r= 0. F is an Euclidean ring.
Note. We can prove the above theorem by defining d: F -> Z by d(a) = 1 (integer) ∀ a ϵ F
Theorems On Subrings And Their Applications
Theorem 2. Every Euclidean ring is a principal ideal ring (or) Every ideal of an Euclidean ring is a principal ideal
Proof. Let R be an Euclidean ring.
Let U = {0} where ‘O’ is the zero element of R. Then U = {0} is the ideal generated by QϵR.
U is a principal ideal of R.
Let U be an ideal of R.
Let u≠ {0}. there exists xϵU and x≠0 so that the set {d (x) | x≠ 0} is a non-empty set of nonnegative integers.
By well ordering principle there exists b≠0 ϵ U so that d(b)<d (x) where x≠0ϵU.
We now prove that U = (b). Let ‘a’ be any element of U.
By division algorithm, there exists q,rϵR so that a -bq +r where r = 0 or d (r) < d (b). bϵU,qϵR, and U is an ideal => bqϵU. aϵU,bq ϵU => a-bq =r ϵU
If r≠ 0 then d (r) < d (b) so that we have a contradiction as d (b) < d (x) V x≠ 0 ϵU
r = 0 and hence a=b q.
U- {bq| qϵR} = (b) is the principal ideal generated by b(≠ 0) ϵU. Hence every ideal U of R is a principal ideal.
R is a principal ideal ring.
Note 1. If U is an ideal of a Euclidean ring R then U is a principal ideal of R so that U = (b) = {bq |q ϵF}
For,’ the ring R = \(\left\{a+b\left(\frac{1+\sqrt{19 i}}{2}\right): a, b \in Z\right\}\) of complex numbers is a principal ideal ring but not Euclidean