every PID is a UFD

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. ∎

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. ∎

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

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

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

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. ∎

Title	every PID is a UFD
Canonical name	EveryPIDIsAUFD
Date of creation	2013-03-22 16:55:51
Last modified on	2013-03-22 16:55:51
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	9
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 13F07
Classification	msc 16D25
Classification	msc 11N80
Classification	msc 13G05
Classification	msc 13A15
Related topic	UFD
Related topic	UniqueFactorizationAndIdealsInRingOfIntegers