# Principal Ideal Domain (P.I.D.) | Discrete Mathematics

Prerequisite :  Rings in Discrete Mathematics

Introduction :
Algebraic Structure : A non-empty set G equipped with 1 or more binary operations is called algebraic structure.
Example –

• (N,+) where N is a set of natural numbers and
• (R, *)  R is a set of real numbers.
Here ‘ * ‘ specifies a multiplication operation.

RING :
An algebraic structure that sets processing of two binary operations simultaneously is needed to form a Ring. A non-empty set R together with the operations multiplication & addition (Usually) is called a ring if :

```1. (R,+) is an Abelian Group (satisfies G1, G2, G3, G4 & G5)
2. (R, *) is a Semi Group. (satisfies G1 & G2)
3. Multiplication is distributive over addition :
(a) Left Distributive :  a*(b+c) = (a*b) + (a*c) ; ∀ a, b, c ∈ R
(b) Right Distributive : (b+c)*a = (b*a) + (c*a) ; ∀ a, b, c ∈ R```
1. GROUP –
An algebraic structure (G , o) where G is a non-empty set & ‘o’ is a binary operation defined on G is called a Group if the binary operation “o” satisfies the following properties :
G1. Closure : a ∈ G ,b ∈ G  => aob ∈ G ;  ∀ a,b ∈ G
G2. Associativity : (aob)oc = ao(boc) ; ∀ a,b,c ∈ G.
G3. Identity Element : There exists e in G such that aoe = eoa = a ; ∀ a ∈ G (Example – For addition, identity is 0)
G4. Existence of Inverse : For each element a ∈ G ; there exists an inverse(a-1)∈ G such that :  aoa-1 = a-1oa = e.
2. Abelian Group –
An algebraic structure (G , o) where G is a non-empty set & ‘o’ is a binary operation defined on G is called an abelian Group if it is a group (i.e. , it satisfies G1, G2, G3 & G4) and additionally satisfies :
G5 : Commutative: aob = boa ∀ a,b ∈ G
3. Semi Group –
An algebraic structure (G , o) where G is a non-empty set & ‘o’ is a binary operation defined on G is called a semi-group if it satisfies only 2 properties : G1 (closure ) & G2 (Associativity).
We usually write the ring structure as  : (R, +, *) or simply R.
Note : Additive identity 0 is unique & is called the zero element of the Ring R.
4. Commutative Ring –
(R,+, *)  is commutative : it means that the multiplication (*) is commutative.

Integral Domain :
A ring (R, +, *) is called an integral domain of :

```1. (R,+,*)  is commutative.
2. (R,+,*) is a ring with unit element.
3. It is a ring without zero divisors.```
1. (R,+, *)  is commutative –
means that the multiplication (*) is commutative.
2. (R,+,*) is a ring with a unit element –
It means that there exists a unit element, say 1∈ R, such that :
a*1 = 1*a = a  ∀ a ∈ R
3. R is a ring without zero divisors –
a*b = 0  =>a = 0 OR b = 0 where a, b ∈ R

Principal Ideal :
Let (R,+, *)  is a commutative ring with identity 1.
Let a ∈R, then the set = { ra : r ∈ R  is an ideal } called the Principal Ideal generated by a.

Principal Ideal Domain (P.I.D.) :
A ring (R,+, *)  is called a principal ideal domain if :

• R is an integral domain.
• Every ideal in R is principal.

If every one-sided ideal of a ring is ideal, it is termed a primary ideal ring. The principal ideal domain is a principal ring with no zero divisors.
Note : integrally closed domains  ⊂ integral domains ⊂ commutative rings ⊂ rings

Q. Showing that every field is a P.I.D.
Solution. Let F is a field. Therefore, F is an integral domain too. Also, F would have some unity element : a*1 = 1*a = a  ∀ a ∈ F. So, F is an integral domain with unity.
Every field has only 2 ideals. So F has 2 ideals : {0} & F where –
(i) {0} = 0*F
(ii) F = 1*F
So, F has only 2 ideals & they can be expressed in the form :  { f*a : f ∈ R  is an ideal and a ∈ F }
So, every F is a P.I.D.
Note : The converse may not be true.

Q. Show that Z,the ring of integers, is a P.I.D.
Answer. We know that Z, the set of integers, is an integral domain.
Let J be an ideal in Z. We show J is a principal ideal.
Case 1. – If J = {0}, then it is the principal ideal & hence the result.
Case 2. – If J ≠ {0}, Let 0 ≠ x ∈ J, then -x = (-1) x ∈ J for some positive x.
Hence, J contains at least one positive integer. Let a be the smallest positive integer in J.
We claim, J = { ra : r ∈Z }
For x  ∈ J , using the division algorithm,
x = qa + r ; 0 ≤ r ≤ a ; q ∈ Z
But J is an ideal and a ∈ J, q ∈ Z
Hence,  qa ∈ J and x – qa ∈ J ⇒ r ∈ J.
But, a is the smallest positive integer in J satisfying 0 ≤ r ≤ a. Hence, we must have r = 0.
So, x = qa , i.e.,
J = { qa : q ∈ Z}
Hence Z is a P.I.D.

Q. Prove :  A PID is a unique factorization domain.
Proof : The reverse inclusion relation in the set of nonzero ideals is well-founded in classical logic.
Let A denote the subset of ideals (a) that are products of a finite number(possibly zero) of maximal principal ideals . If every (t) correctly containing (x) can be factored into maximals for each valid ideal (x)(0), then so can (x). (Either (x) is maximal/irreducible, or it factors as (s)(t), where both s and t are non-units; by hypothesis, (s) and (t) factor into maximals, so does (x)). As a result, because A is an inductive set, it contains every ideal (x)(0), i.e., x may be factored into irreducible.

For the factorization’s uniqueness, we first note that if p is irreducible and p|ab, then p|a or p|b. (Because R/(p) is a field and hence a fortiori an integral domain, if ab≡0modp is true, then a≡0modp or b≡0modp is true as well.). p1 divides one of the irreducible if qi ,p1p2…pm=q1q2…qn are two factorizations into irreducible of the same element, then , in which case (p1)=(qi) and each is a unit times the other, meaning we can cancel p1 on both sides and argue by induction.

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