Let be a ring, let be an ideal, and let be a subring. Prove that is an ideal. Show by example that not every ideal of need be of this form.
is a subring by this previous exercise, so it suffices to show absorption. If and , then since is an ideal, and since is closed under multiplication. Thus , so that is an ideal.
Let and . Clearly is a subring. Now consider the ideal . Note that has only the ideals and , and that and ; neither of these is the ideal .