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.

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