r-ideals in commutative rings

Question

I'm learning about r-ideals in commutative rings from a journal by Rostam Mohammadian.

"A proper ideal I in a ring R is called an r-ideals (resp., pr-ideal), if ab is an element in I with ann(a)=0 implies that b is an element in I for each a,b is an element in R"

I tried to find some examples for this, but until today.. ideal 0 was the only ideal that fulfill the definition. Do you think there is any example for this ideal?

Answer 1

If $A$ is a domain, then the ideal $A\times \{0\}$ in $R=A\times A$ satisfies the definition of r-ideal (I don't know about the definition of pr-ideal because it wasn't stated).

Answer 2

Let $A$ be a ring and $T(A)$ be the total ring of fractions of $A$, i.e. that is $T(A) := S^{-1}A$ with $S$ the multiplicative set of regular elements of $A$.

Lemma $I$ is an $r$-ideal iff $IT(A) \cap A = I$

Proof If $IT(A) \cap A =I$ then let $ab \in A$ with $\operatorname{Ann}(a) = 0$, i.e. $a$ is a regular element, then $b = (ab)/a \in IT(A) \cap A = I$ by assumption. Conversely, suppose that $I$ is an $r$-ideal and let $b \in IT(A) \cap A$. We aim to show $b \in I$. Since $b \in IT(A)$ we can write $b = \sum a_i/r_i$, with each $r_i$ a regular element. Letting $r = \prod r_i$ it is clear that $rb \in I$, $r, b \in A$. By the definition of an $r$-ideal, $b \in I$ as desired. $\square$

This is a helpful reformulation because it is very common that a ring (with zero divisors) has non-zero ideals $I$ such that $IB \cap A = I$ for every ring extension $A \subseteq B$.

Indeed, if $I$ is, for example, a pure ideal or a minimal prime ideal, then $I B \cap A = I$ for every extension of rings $A \subseteq B$, so a fortiori pure ideals and minimal primes are $r$-ideals.

Additionally this shows that if $A = T(A)$, i.e. the non-units of $A$ are zero divisors, then every ideal of $A$ is an $r$-ideal.