Lemma 10.120.7 (Nagata's criterion for factoriality). Let $A$ be a domain. Let $S \subset A$ be a multiplicative subset generated by prime elements. Let $x \in A$ be irreducible. Then

the image of $x$ in $S^{-1}A$ is irreducible or a unit, and

$x$ is prime if and only if the image of $x$ in $S^{-1}A$ is a unit or a prime element in $S^{-1}A$.

Moreover, then $A$ is a UFD if and only if every element of $A$ has a factorization into irreducibles and $S^{-1}A$ is a UFD.

**Proof.**
Say $x = \alpha \beta $ for $\alpha , \beta \in S^{-1}A$. Then $\alpha = a/s$ and $\beta = b/s'$ for $a, b \in A$, $s, s' \in S$. Thus we get $ss'x = ab$. By assumption we can write $ss' = p_1 \ldots p_ r$ for some prime elements $p_ i$. For each $i$ the element $p_ i$ divides either $a$ or $b$. Dividing we find a factorization $x = a' b'$ and $a = s'' a'$, $b = s''' b'$ for some $s'', s''' \in S$. As $x$ is irreducible, either $a'$ or $b'$ is a unit. Tracing back we find that either $\alpha $ or $\beta $ is a unit. This proves (1).

Suppose $x$ is prime. Then $A/(x)$ is a domain. Hence $S^{-1}A/xS^{-1}A = S^{-1}(A/(x))$ is a domain or zero. Thus $x$ maps to a prime element or a unit.

Suppose that the image of $x$ in $S^{-1}A$ is a unit. Then $y x = s$ for some $s \in S$ and $y \in A$. By assumption $s = p_1 \ldots p_ r$ with $p_ i$ a prime element. For each $i$ either $p_ i$ divides $y$ or $p_ i$ divides $x$. In the second case $p_ i$ and $x$ are associates (as $x$ is irreducible) and we are done. But if the first case happens for all $i = 1, \ldots , r$, then $x$ is a unit which is a contradiction.

Suppose that the image of $x$ in $S^{-1}A$ is a prime element. Assume $a, b \in A$ and $ab \in (x)$. Then $sa = xy$ or $sb = xy$ for some $s \in S$ and $y \in A$. Say the first case happens. By assumption $s = p_1 \ldots p_ r$ with $p_ i$ a prime element. For each $i$ either $p_ i$ divides $y$ or $p_ i$ divides $x$. In the second case $p_ i$ and $x$ are associates (as $x$ is irreducible) and we are done. If the first case happens for all $i = 1, \ldots , r$, then $a \in (x)$ as desired. This completes the proof of (2).

The final statement of the lemma follows from (1) and (2) and Lemma 10.120.5.
$\square$

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