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

Let $R$ be a ring, let $I \subseteq R$ be an ideal, and let $S \subseteq R$ be a subring. Prove that $I \cap S \subseteq S$ is an ideal. Show by example that not every ideal of $S$ need be of this form.

$I \cap S$ is a subring by this previous exercise, 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 $J$.