If you find any mistakes, please make a comment! Thank you.

The intersection of an ideal and a subring is a subideal, but not necessarily vice versa

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.20

Solution: $I \cap S$ is a subring by Exercise 7.1.4, so it suffices to show absorption. If $s \in S$ and $x \in I \cap S$, then $rx, xr \in I$ since $I \subseteq R$ is an ideal, and $rx, xr \in S$ since $S$ is closed under multiplication. Thus $rx,xr \in I \cap S$, so that $I \cap S \subseteq S$ is an ideal.

Let $R = \mathbb{Q}$ and $S = \mathbb{Z}$. Clearly $S \subseteq R$ is a subring. Now consider the ideal $J = 2\mathbb{Z} \subseteq S$. Note that $\mathbb{Q}$ has only the ideals $0$ and $\mathbb{Q}$, and that $S \cap 0 = 0$ and $S \cap \mathbb{Q} = S$; neither of these is the ideal $J$.


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu