Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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 $\mathbb Z$.

share|improve this answer

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 $R$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.