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[parent] every PID is a UFD (Theorem)
Theorem 1   Every Principal Ideal Domain (PID) is a Unique Factorization Domain (UFD).

The first step of the proof shows that any PID is a Noetherian ring in which every irreducible is prime. The second step is to show that any Noetherian ring in which every irreducible is prime is a UFD.

We will need the following

Lemma 2   Every PID $ R$ is a gcd domain. Any two gcd's of a pair of elements $ a,b$ are associates of each other.
Proof. Suppose $ a,b\in R$. Consider the ideal generated by $ a$ and $ b$, $ (a,b)$. Since $ R$ is a PID, there is an element $ d\in R$ such that $ (a,b)=(d)$. But $ a,b\in(a,b)$, so $ d\mid a, d\mid b$. So $ d$ is a common divisor of $ a$ and $ b$. Now suppose $ c\mid a, c\mid b$. Then $ (d)=(a,b)\subset (c)$ and hence $ c\mid d$.

The second part of the lemma follows since if $ c,d$ are two such gcd's, then $ (c)=(a,b)=(d)$, so $ c\mid d$ and $ d\mid c$ so that $ c,d$ are associates. $ \qedsymbol$

Theorem 3   If $ R$ is a PID, then $ R$ is Noetherian and every irreducible element of $ R$ is prime.
Proof. Let $ I_1\subset I_2\subset I_3\subset \ldots$ be a chain of (principal) ideals in $ R$. Then $ I_\infty = \cup_k I_k$ is also an ideal. Since $ R$ is a PID, there is $ a\in R$ such that $ I_\infty=(a)$, and thus $ a\in I_n$ for some $ n$. Then for each $ m>n$, $ I_m=I_n$. So $ R$ satisfies the ascending chain condition and thus is Noetherian.

To show that each irreducible in $ R$ is prime, choose some irreducible $ a\in R$, and suppose $ a=bc$. Let $ d=\gcd(a,b)$. Now, $ d\mid a$, but $ a$ is irreducible. Thus either $ d$ is a unit, or $ d$ is an associate of $ a$. If $ d$ is an associate of $ a$, then $ a\mid d\mid b$ so that $ a\mid b$ and $ c$ is a unit. If $ d$ is itself a unit, then we can assume by the lemma that $ d=1$. Then $ 1\in(a,b)$ so that there are $ x,y\in R$ such that $ xa+yb=1$. Multiplying through by $ c$, we see that $ xac+ybc=c$. But $ a\mid xac$ and $ a\mid ybc=ya$. Thus $ a\mid c$ so that $ b$ is a unit. In either case, $ a$ is prime. $ \qedsymbol$

Theorem 4   If $ R$ is Noetherian, and if every irreducible element of $ R$ is prime, then $ R$ is a UFD.
Proof. We show that any nonzero nonunit is $ R$ is expressible as a product of irreducibles (and hence as a product of primes), and then show that the factorization is unique.

Let $ \mathcal{U}\subset R$ be the set of ideals generated by each element of $ R$ that cannot be written as a product of irreducible elements of $ R$. If $ \mathcal{U}\neq\emptyset$, then $ \mathcal{U}$ has a maximal element $ (r)$ since $ R$ is Noetherian. $ r$ is not irreducible by construction and thus not prime, so $ (r)$ is not prime and thus not maximal. So there is a proper maximal ideal $ (s)$ with $ (r)\subsetneq (s)$, and $ s\mid r$.

Since $ (r)$ is maximal in $ \mathcal{U}$, it follows that $ (s)\notin \mathcal{U}$ and thus that $ s$ is a product of irreducibles. Choose some irreducible $ a\mid s$; then $ a\mid r$ and

$\displaystyle r=ab$
for some $ b\in R$. If $ (b)\notin \mathcal{U}$ (note that this includes the case where $ b$ is a unit), then $ b$ and hence $ r$ is a product of irreducibles, a contradiction. If $ (b)\in \mathcal{U}$ then $ (r)\subset (b)$ (since $ b\mid r$). $ (r)\neq (b)$ since $ a$ is not a unit, and thus $ (r)\subsetneq (b)$. This contradicts the presumed maximality of $ (r)$ in $ \mathcal{U}$. Thus $ \mathcal{U}=\emptyset$ and each element of $ R$ can be written as a product of irreducibles (primes).

The proof of uniqueness is identical to the standard proof for the integers. Suppose

$\displaystyle a = p_1\cdot \ldots \cdot p_n = q_1\cdot \ldots \cdot q_m$
where the $ p_i$ and $ q_j$ are primes. Then $ p_1\mid q_1\cdot\ldots\cdot q_m$; since $ p_1$ is prime, it must divide some $ q_j$. Reordering if necessary, assume $ j=1$. Then $ p_1=u\cdot q_1$ where $ u$ is a unit. Factoring out these terms since $ R$ is a domain, we get
$\displaystyle p_2\cdot\ldots\cdot p_n=u\cdot q_2\cdot\ldots\cdot q_m$
We may continue the process, matching prime factors from the two sides. $ \qedsymbol$



"every PID is a UFD" is owned by rm50.
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See Also: UFD, unique factorization and ideals in ring of integers


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Cross-references: sides, prime factors, matching, domain, terms, necessary, divide, integers, contradiction, maximal ideal, maximal element, product, expressible, unit, ascending chain condition, ideals, chain, irreducible element, Noetherian, divisor, ideal generated by, associates, gcd's, gcd domain, prime, irreducible, noetherian ring, unique factorization domain, principal ideal domain
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This is version 6 of every PID is a UFD, born on 2007-04-15, modified 2007-04-16.
Object id is 9196, canonical name is PIDsAreUFDs.
Accessed 1193 times total.

Classification:
AMS MSC: 13A15 (Commutative rings and algebras :: General commutative ring theory :: Ideals; multiplicative ideal theory)
 11N80 (Number theory :: Multiplicative number theory :: Generalized primes and integers)
 13G05 (Commutative rings and algebras :: Integral domains)
 16D25 (Associative rings and algebras :: Modules, bimodules and ideals :: Ideals)
 13F07 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Euclidean rings and generalizations)
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