An example of ideal $I$ such that $I^{ec}\neq I$

Question

Let $A$ be a commutative ring, $S \subseteq A$ a multiplicative system and $i_S : A \rightarrow S^{-1}A$ the canonical morphism. Can you give me an example of ideal $I \unlhd A$ such that $I^{ec}\neq I$, where $I^e$ is the ideal generated by $i_S(I)$ and $J^c=i_S^{-1}(J)$?

Answer 1

Take $A=\mathbb Z$, $S=\mathbb Z\setminus \{0\}$ and $I=2\mathbb Z$. Then $I^{ec} =\mathbb Z$

Answer 2

An element of $I^{c}$ is of the form $a/s$, where $a\in A$ and $s\in S$. For $b\in A$, we have $b\in I^{ec}$ if and only if $b/1\in I^e$, that is, $$ b/1=a/s $$ for some $a\in I$ and $s\in S$, which means $$ t(bs-a)=0 $$ for some $t\in S$. In particular $bst=at\in I$.

Conversely, if $bs\in I$, for some $s\in S$, then $b/1=bs/s\in I^e$, so $b\in I^{ec}$.

Thus the condition $I=I^{ec}$ is the same as saying that “$bS\cap I\ne\emptyset$ implies $b\in I$”.

Usually one sets $(I:S)=\{a\in A: as\in I,\text{ for some }s\in S\}$, so the condition is $(I:S)=I$ (assuming $1\in S$, which is not restrictive).

It's also easy to see when $I^{ec}=A$: one just needs to see that $1\in(I:S)$, that is, $I\cap S\ne\emptyset$.

If $S=A\setminus P$, where $P$ is a prime ideal, for any ideal $I$ such that $I\not\subseteq P$, we have $I^{ec}=A$. In this case one can also see that, if $I\subseteq P$, then $I^{ec}=I$.

You can surely find several examples.