Relation between P.I.D and U.F.D
A Principal Ideal Domain is an integral domain in which every proper ideal can be generated by a single element.
Let R be a commutative ring with 1. If every ideal of R is a principal ideal, then R is called a principal ideal ring. An integral domain which is also a principal ideal ring is called a Principal Ideal Domain or P.I.D .
An integral domain D is called UFD or Unique Factorization Domain, if the following two conditions hold in D.
i) Every non-zero , non-unit element of D is expressed as a=p1 ,p2,… pn where p1 ,p2,… pn are irreducible element of D
ii) If a= p1 ,p2,… pn = q1 ,q2,… qn are two factorizations of a as a finite product of irreducible elements of D, then n=m and there is a permutation 𝛔 of {1,2….,n} . pi and q𝛔(i) are associates for all i=1,2,..,n
From various propositions deduced by these two concepts Principal Ideal Domain and Unique Factorization Domain we conclude the relation between the P.I.D and U.F.D.
“Every Principal Ideal Domain is a Unique Factorization Domain”
To prove this above theorem we need to show and prove the existence of a theorem (and lemma)
Lemma : Every Principal Ideal Domain satisfies the Ascending Chain Condition for Principal ideals (ACCP)
Let <a1> ⊆<a2> ⊆<a3> ….. be a chain of principal ideals in D. it can be easily verified that I=Ui⋲N<ai> is an ideal of D. Since D is a PID, there exists an element a⋲D such that I =<a>. Hence a⋲<an> for some positive integer n. Then I⊆<an> ⊆I . Therefore I=<an>. For t≥n, <at> ⊆I = <an> ⊆<at>. Thus<an> =<at> for all t≥n
Theorem1 : An integral domain D with the ACCP is a Factorization Domain.
Suppose D is not a Factorization Domain. Then there exists a non-zero, non-unit element a such that a does not have a factorization. Thus a is not irreducible and so a = a1b1 ,where a1 ,b1 ⋲D are non-trivial factors of a. Atleast one of a1 or b1 must not have a factorization, otherwise the factorization of a1 and b1 put together will produce a factorization of a. Suppose a1 does not have factorization, now a and a1 are not associates. Therefore <a> ⊂ <a1>. Since a1 does not have a factorization, we can express a1= a2b2 , where a2 ,b2⋲D are non-trivial factors of a1 . Atleast one of a2 or b2 does not have a factorization. Then <a> ⊂ <a1>⊂ <a2>. We now repeat the process with a2 . Thus we find there exists an infinite strictly ascending chain of principal ideals in D , a contradiction. Hence D is a Factorization Domain.
Theorem 2: Let R be a PID and p⋲R, if p is irreducible then p is prime.
Suppose p|ab , where a,b ⋲R. Then there exists r⋲R such that pr=ab. Now <p,b>=<d> for some d⋲R. Therefore there exists q⋲R and that p=dq. Since p is irreducible , either d or q must be a unit. If d is a unit, then <p,b>=<d>=R. Hence 1= sp+tb for s,t⋲R. Therefore a=asp+atb=asp+tpr= (as+tr)p. This implies that p|a. If on the other hand, q is a unit, then d=pq-1 ⋲<p>. Thus <d>⊆<p>⊆<p,b>=<d> so that <p>=<p,b>. Hence b⋲<p> and so p|b
Theorem 3: A Factorization Domain D is a UFD if and only if every irreducible element of D is a prime element.
Assume D is a unique factorization domain. Taking nonzero elements p,a,b∈D such that p is irreducible and p ∣ab. There exists a nonzero element h∈D such that hp=ab. Let H,A, B be factorizations for h,a, b, respectively. Hence ∏H⋅p∼∏A⋅∏B. By the unique factorization property there exists a bijection f:H⊔{p}→A⊔B such that q∼f(q) are associates for every q∈H⊔{p}. In particular, either f(p)∈ A or f(p)∈B. In the former case, p∼f(p)∣∏A∼a. In the latter case, p∼f(p)∣∏B∼b. Hence p is prime.
Conversely, assume every irreducible element of D is prime. Let S and T be factorizations of a common element. By induction on the cardinality of S, we have two cases:
· If S=∅, then ∏S=1, hence ∏T∈D∗, hence T=∅.
· If p∈S, then p∣∏T. Since p is prime, there exists q∈T such that p∣q. Since q is irreducible, p ∼q. Let S′=S∖{p} and T′=T∖{q}. Then p⋅∏S′∼q⋅∏T′. By the inductive hypothesis, S′ and T′ are equivalent, hence so are S and T.
Now combining the results of the above theorems(and lemma), it follows that every PID is a UFD.
But the converse is not true i.e every UFD is not a PID
For example: A ring R is a UFD if and only if the polynomial ring R[X] is one.
REFERENCE:
· FUNDAMENTALS OF ABSTRACT ALGEBRA by D.S Malik, John M. Mordeson, M.K Sen (by Mc Graw Hill Companies, Inc)
· Abstract Algebra by David S Dummit, Richard M. Foote
· Algebra by Artin (z-lib.org)
(For definition and better understanding of theories related to ring check https://phiwhyyy.medium.com/why-should-we-study-ring-theory-5ecc3ad069d)
