localization preserving irreducible elements in an UFD

1. Proposition

Let \(A\) be a unique factorization domain and \(S \subseteq A\) a multiplicatively closed subset and \(\beta \in A \setminus S\) irreducible ¹

TFAE:

for any irreducible element \(\alpha \in A \setminus S\) it follows, that \(\alpha\) is irreducible in \(S^{-1}A\)
\(0 \not\in S\)

2. Proof

2.1. 1) \(\implies\) 2)

Suppose \(0 \in S\), then \(S^{-1}A = 0\) (cf. localization as zero ring) Thus \(\frac{\beta}{1} = 0\) and hence by definition, \(\beta\) is not irreducible

2.2. 2) \(\implies\) 1)

Let \(\alpha \mid \frac{r_1}{s_1} \cdot \frac{r_2}{s_2}\), then

\begin{align*} \alpha \cdot c =& \frac{1}{s_1 s_2} \cdot r_1 r_2 && \vert \cdot s_1 s_2 \\ \alpha \cdot \left(c \cdot s_1 \cdot s_2\right) =& r_1 \cdot r_2 \end{align*}

Thus \(\alpha \mid r_1 \cdot r_2\) in \(A\) and since irreducible elements are prime in an UFD, we conclude, that w.l.o.g. \(\alpha \mid r_1\) in \(R\) There, there exists a \(c \in R\) with \(\alpha \cdot c = r_1\) and thus

\begin{align*} \alpha \cdot \frac{c}{s_1} = \frac{\alpha \cdot c}{s_1} = \frac{r_1}{s_1} \end{align*}

hence in \(S^{-1}A\)

\begin{align*} a \mid \frac{r_1}{s_1} \end{align*}

Thus by prime implies irreducible, we conclude that

cheap trick - generalize

Footnotes:

Existence of a irreducible element is (possibly ?) required for 1) \(\implies\) 2)