Nagata's criterion for factoriality

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.

Going deeper

For a student

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 for Noetherian coefficient rings, as in the example with $\Z[x]$. For a Krull domain $A$, the same idea shows that inverting prime elements does not change the divisor class group, since the height-one primes that meet $S$ are principal.