Let $R$ be a commutative ring containing regular elements and $\mathfrak{S}$ be the multiplicative semigroup of the non-zero fractional ideals of $R$ A fractional ideal $\mathfrak{a}$ of $R$ is called a cancellation ideal or simply cancellative, if it is a cancellative element of $\mathfrak{S}$ i.e. if $$\mathfrak{ab = ac}\, \Rightarrow\, \mathfrak{b = c} \quad\forall\,\,\mathfrak{b,\,c}\in\mathfrak{S}.$$

• Each invertible ideal is cancellative.
• A finite product $\mathfrak{a}_1\mathfrak{a}_2...\mathfrak{a}_m$ of fractional ideals is cancellative iff every $\mathfrak{a}_i$ is such.
• The fractional ideal $\mathfrak{a}/r := \{ar^{-1}\!:\,\,\,a\in\mathfrak{a}\}$ where $\mathfrak{a}$ is an integral ideal of $R$ and $r$ a regular element of $R$ is cancellative if and only if $\mathfrak{a}$ is cancellative in the multiplicative semigroup of the non-zero integral ideals of $R$
• If $r\in R$ then the principal ideal $(r)$ of $R$ is cancellative if and only if $r$ is a regular element of the total ring of fractions of $R$
• If $\mathfrak{a}_1\!+\!\mathfrak{a}_2\!+\!...\!+\!\mathfrak{a}_m$ , is a cancellation ideal and $n$ a positive integer, then $$(\mathfrak{a}_1\!+\!\mathfrak{a}_2\!+\!...\!+\!\mathfrak{a}_m)^n = \mathfrak{a}_1^n\!+\!\mathfrak{a}_2^n\!+\!...\!+\!\mathfrak{a}_m^n.$$ In particular, if the ideal $(a_1,\,a_2,\,...,\,a_m)$ , of $R$ is cancellative, then $$(a_1,\,a_2,\,...,\,a_m)^n = (a_1^n,\,a_2^n,\,...,\,a_m^n).$$

Bibliography

1

R. GILMER: Multiplicative ideal theory. Queens University Press. Kingston, Ontario (1968).

2

M. LARSEN & P. MCCARTHY: Multiplicative theory of ideals. Academic Press. New York (1971).