# every PID is a UFD

###### Theorem 1.

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