---
tag: GGEMJF449D
kind: statement
role: theorem
name: Nagata's criterion for factoriality
msc: [13F15, 13B30]
uses: [09S3PFBEPM, 0MVH4Y1CRZ, 0MNHXBSX3M, AN9SKEZ3VY, 4C1AE84N5N, 06YMXJMXYS]
about: [06YMXJMXYS, 4C1AE84N5N]
sources:
  - url: https://en.wikipedia.org/wiki/Unique_factorization_domain
---

## Statement

Let $A$ be a [[Noetherian|0MVH4Y1CRZ]] [[integral domain|09S3PFBEPM]].
Let $\mathcal{P}$ be a set of [[prime elements|0MNHXBSX3M]] of $A$.
Let $S$ be the [[multiplicative set|AN9SKEZ3VY]] generated by $\mathcal{P}$: the products of a unit of $A$ and finitely many elements of $\mathcal{P}$.
If the [[localization|4C1AE84N5N]] $S^{-1}A$ is a [[unique factorisation domain|06YMXJMXYS]], then $A$ is a unique factorisation domain.

## Remarks

Since $A$ is Noetherian, every nonzero nonunit of $A$ is a product of irreducible elements, so it suffices to show that every irreducible element $a$ of $A$ is prime.
If some prime $p \in \mathcal{P}$ divides $a$, then $a$ is associate to $p$ and so is prime.
Otherwise one checks that $a$ is prime in $S^{-1}A$ and that $aS^{-1}A \cap A = aA$, using that no prime of $\mathcal{P}$ divides $a$.
Then $aA$ is the contraction of a prime ideal, hence prime.
Example: let $A = \Z[x]$ and let $S$ be the set of nonzero integers.
$S$ is generated by the primes $p$, which are prime in $\Z[x]$ because $\Z[x]/(p) \cong \F_p[x]$ is a domain.
Then $S^{-1}A = \Q[x]$ is a principal ideal domain, hence a UFD, so $\Z[x]$ is a UFD.
The hypothesis on $S$ is necessary: for $A = \Z[\sqrt{-5}]$ and $S = A \setminus \set{0}$, $S^{-1}A$ is the field $\Q(\sqrt{-5})$, a UFD, while $A$ is not a UFD.

## Explanations

### L1 human

To prove that a Noetherian domain has unique factorization, it is enough to invert some prime elements and check unique factorization in the bigger ring.
This works because the inverted elements are prime, so they already factor well in $A$ and hide no failure of unique factorization.
So unique factorization can be tested in a simpler ring.

### L2 human

The criterion is a standard way to show that rings given by generators and relations are UFDs.
One finds a prime element $p$ such that $A[1/p]$ is visibly factorial, often a polynomial ring or a Laurent polynomial ring, and concludes for $A$.
It recovers the fact that the [[polynomial ring over a UFD is a UFD|0D6KH8Y59N]] for Noetherian coefficient rings, as in the example with $\Z[x]$.
For a [[Krull domain|CB9WJCN9BM]] $A$, the same idea shows that inverting prime elements does not change the [[divisor class group|WN8PK6RQZR]], since the height-one primes that meet $S$ are principal.