# 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 $a\mathrm{,}b$ 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 ${I}_{1}\subset {I}_{2}\subset {I}_{3}\subset \mathrm{\dots}$ be a chain of (principal) ideals
in $R$. Then ${I}_{\mathrm{\infty}}={\cup}_{k}{I}_{k}$ is also an ideal. Since $R$ is a PID, there is $a\in R$ such that ${I}_{\mathrm{\infty}}=(a)$, 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 $d=\mathrm{gcd}(a,b)$. 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 $\mathcal{U}\ne \mathrm{\varnothing}$, 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 $(r)\u228a(s)$, 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 $b\mid r$). $(r)\ne (b)$ since $a$ is not a unit, and thus $(r)\u228a(b)$. This contradicts the presumed maximality of $(r)$ in $\mathcal{U}$. Thus $\mathcal{U}=\mathrm{\varnothing}$ 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

$$a={p}_{1}\cdot \mathrm{\dots}\cdot {p}_{n}={q}_{1}\cdot \mathrm{\dots}\cdot {q}_{m}$$ |

where the ${p}_{i}$ and ${q}_{j}$ are primes. Then ${p}_{1}\mid {q}_{1}\cdot \mathrm{\dots}\cdot {q}_{m}$; 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

$${p}_{2}\cdot \mathrm{\dots}\cdot {p}_{n}=u\cdot {q}_{2}\cdot \mathrm{\dots}\cdot {q}_{m}$$ |

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 |