Let $R$ be a commutative ring having regular elements and let $T$ be its total ring of fractions. If $\mathfrak{a}$ and $\mathfrak{b}$ are fractional ideals of $R$ then one can define two different quotients or residuals of $\mathfrak{a}$ by $\mathfrak{b}$

• $\mathfrak{a\!:\!b}\, := \{r\in R|\quad r\mathfrak{b} \subseteq \mathfrak{a}\}$
• $[\mathfrak{a\!:\!b}] := \{t\in T|\quad t\mathfrak{b} \subseteq \mathfrak{a}\}$

They both are fractional ideals of $R$ and the former in fact an integral ideal of $R$ It is clear that $$\mathfrak{a\!:\!b} = [\mathfrak{a\!:\!b}]\cap\!R.$$ In the special case that $R$ has non-zero unity and $\mathfrak{b}$ has the inverse ideal $\mathfrak{b}^{-1}$ we have $$[\mathfrak{a\!:\!b}] = \mathfrak{a}\mathfrak{b}^{-1},$$ in particular $$[R\!:\!\mathfrak{b}] = \mathfrak{b}^{-1}.$$

Some rules concerning the former type of quotient (the corresponding rules are valid also for the latter type):

1. $\mathfrak{a}\subseteq\mathfrak{b}\,\,\,\,\Rightarrow\,\,\, \mathfrak{a}:\mathfrak{c}\subseteq\mathfrak{b}:\mathfrak{c}\,\, \land\,\,\mathfrak{c}:\mathfrak{a}\supseteq\mathfrak{c}:\mathfrak{b}$
2. $\mathfrak{a}:(\mathfrak{b}\mathfrak{c}) = (\mathfrak{a}:\mathfrak{b}):\mathfrak{c}$
3. $\mathfrak{a}:(\mathfrak{b}+\mathfrak{c})= (\mathfrak{a}:\mathfrak{b})\cap(\mathfrak{a}:\mathfrak{c})$
4. $(\mathfrak{a}\cap\mathfrak{b}):\mathfrak{c} = (\mathfrak{a}:\mathfrak{c})\cap(\mathfrak{b}:\mathfrak{c})$

$\quad\quad \mathfrak{a}:(\mathfrak{b}\cap\mathfrak{c})= (\mathfrak{a}:\mathfrak{b})+(\mathfrak{a}:\mathfrak{c})$ $\quad\quad (\mathfrak{a}+\mathfrak{b}):\mathfrak{c} = (\mathfrak{a}:\mathfrak{c})+(\mathfrak{b}:\mathfrak{c})$ of the two last rules if the divisor ideals are finitely generated.