Algebraic Structure: A non-empty set G equipped with 1 or more binary operations is called an 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 the 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 q_i ,p₁p₂…p_m=q₁q₂…q_n are two factorizations into irreducible of the same element, then , in which case (p₁)=(q_i) and each is a unit times the other, meaning we can cancel p₁ on both sides and argue by induction. 

Solved Examples

Example 1: Show that Z (the ring of integers) is a PID.

Solution:

Take any ideal I in Z.

If I = {0}, then I is generated by 0, so I = <0>.

If I ≠ {0}, let a be the smallest positive integer in I.

We claim that I = <a>. To prove this:

Clearly, <a> ⊆ I since a ∈ I.

For any b ∈ I, we can use the division algorithm: b = qa + r, where 0 ≤ r < a.

r = b – qa ∈ I (since b ∈ I and a ∈ I)

But a is the smallest positive integer in I, so r must be 0.

Therefore, b = qa, meaning b ∈ <a>.

Thus, every ideal in Z is principal, so Z is a PID.

Example 2: Show that F[x] (the ring of polynomials over a field F) is a PID.

Solution:

Take any ideal I in F[x].

If I = {0}, then I = <0>, which is principal.

If I ≠ {0}, let f(x) be a non-zero polynomial of least degree in I.

We claim that I = <f(x)>. To prove this:

Clearly, <f(x)> ⊆ I since f(x) ∈ I.

For any g(x) ∈ I, we can use polynomial long division: g(x) = q(x)f(x) + r(x), where deg(r) < deg(f) or r = 0.

r(x) = g(x) – q(x)f(x) ∈ I (since g(x) ∈ I and f(x) ∈ I)

But f(x) has the least degree in I, so r(x) must be 0.

Therefore, g(x) = q(x)f(x), meaning g(x) ∈ <f(x)>.

Thus, every ideal in F[x] is principal, so F[x] is a PID.

Example 3: Show that Z[i] (the Gaussian integers) is a PID.

Solution:

Z[i] is an Euclidean domain with the norm function N(a + bi) = a² + b².

Every Euclidean domain is a PID. To see why:

Let I be a non-zero ideal in Z[i].

Choose α ∈ I with minimal norm N(α) > 0.

For any β ∈ I, divide β by α: β = qα + r, where N(r) < N(α).

r = β – qα ∈ I, but N(α) is minimal, so r must be 0.

Therefore, β = qα, meaning β ∈ <α>.

Thus, every ideal in Z[i] is principal, so Z[i] is a PID.

Example 4 : Prove that the ring R[x] (polynomials with real coefficients) is a PID.

Solution: This follows the same logic as F[x]. R is a field, so R[x] is a Euclidean domain with the degree function, and thus a PID.

Example 5 : Show that Z[√-5] is not a PID.

Solution: Consider the ideal I = <2, 1+√-5>. If Z[√-5] were a PID, I would be generated by a single element. However, 2 and 1+√-5 have no common divisor other than ±1, and neither generates the other. Thus, I is not principal, so Z[√-5] is not a PID.

Example 6 :Prove that any field F is a PID.

Solution: In a field, the only ideals are {0} and F itself. Both are principal: {0} = <0> and F = <1>. Thus, every field is a PID.

Example 7 : Show that Z[x] is not a PID.

Solution: Consider the ideal I = <2, x>. This ideal cannot be generated by a single element because any element of the form 2f(x) or xg(x) will always have even constant terms, while 2 and x are in I.

Example 8 : Prove that if R is a PID, then R[x] is a UFD (Unique Factorization Domain).

Solution: Every PID is a UFD. R[x] is a PID when R is a field. When R is a PID but not a field, R[x] is still a UFD (this is Gauss’s theorem).

Practice Problems

1).Prove that Z[√2] is a PID.

2).Show that Z[√-3] is not a PID.

3).Is the ring of continuous functions on R a PID? Justify your answer.

4).Prove or disprove: If R is a PID, then R[x, y] is a PID.

5).Show that in a PID, any two elements a and b have a greatest common divisor.

6).Prove that if R is a PID, then every ideal in R is finitely generated.

7).Is the ring Z[1/2] (integers with 2 inverted) a PID? Prove your answer.

8).Show that if R is a PID and I is a non-zero ideal of R, then R/I is a principal ideal ring.

9).Prove that if R is a PID, then every non-zero prime ideal of R is maximal.

10).Is the ring of all algebraic integers a PID? Justify your answer.

Summary:

A Principal Ideal Domain (PID) is an integral domain in which every ideal is principal, meaning it can be generated by a single element. PIDs are important in abstract algebra and number theory because they have several nice properties. Every PID is a Unique Factorization Domain (UFD), and in a PID, prime ideals are maximal. Common examples of PIDs include the integers (Z), polynomial rings over fields (F[x]), and Gaussian integers (Z[i]). PIDs are closely related to Euclidean domains, as every Euclidean domain is a PID.

FAQs on Principal Ideal Domain (P.I.D.)

What’s the difference between a PID and a UFD?

Every PID is a UFD, but not every UFD is a PID. In a UFD, every non-zero non-unit element can be factored uniquely into prime elements. In a PID, not only does this unique factorization hold, but every ideal is generated by a single element. For example, Z[x] is a UFD but not a PID.

Are all rings PIDs?

No, not all rings are PIDs. For example, Z[√-5] is not a PID. To be a PID, a ring must be an integral domain (no zero divisors) and satisfy the condition that every ideal is principal. Many common rings fail to meet these criteria.

How can I prove that a ring is a PID?

There are several approaches:

Show that the ring is a Euclidean domain (as all Euclidean domains are PIDs).

Directly prove that every ideal is principal by choosing an element of minimal norm or degree.

Use the structure of the ring to show that any ideal must be generated by a single element.

What are some applications of PIDs in mathematics?

PIDs have numerous applications:

In number theory, they’re used to study factorization properties of integers and algebraic integers.

In linear algebra, they’re used to study the structure of finitely generated modules over PIDs, leading to important results like the Fundamental Theorem of Finitely Generated Modules over a PID.

In algebraic geometry, they play a role in understanding the structure of certain types of rings and ideals.

In coding theory, PIDs are used in the construction and analysis of certain types of codes.