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Show by example that not every ideal of a subring $S$ of a ring $R$ need be of the form $I\cap S$ for some ideal $I$ of $R$

If $I$ is an ideal of $R$ and $S$ is a subring of $R$, then it can be easily shown that $I\cap S$ is an ideal of $S$. However, what is an example when this is not necessary.

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2 Answers 2

up vote 6 down vote accepted

Consider $R=\mathbb Q$ and $S=\mathbb Z$ and let $I$ be any non-trivial ideal on $\mathbb Z$.

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Let $K$ be a field and let $R\subseteq K$ be any subring which is not a field.

This works because there are exactly two ideals in $R$ which are intersections of ideals in $K$ with $R$, yet there are at least three ideals in $R$ because it is not a field.

(Notice that not every field contains rings which are not fields, so $K$ should be picked with a modicum of care here! For example, we may start with any domain $R$ which is not a field and take $K$ to be the fraction field of $R$.)

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