I've been thinking about the following Rotman's exercise, and just can't find an answer:
Give an example of a commutative ring $R$ containing proper ideals $I\subsetneq J\subsetneq R$ with $J$ finitely generated but with $I$ not finitely generated.
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3See http://math.stackexchange.com/questions/183199/commutative-non-noetherian-rings-in-which-all-maximal-ideals-are-finitely-genera – layman Mar 25 '15 at 16:48
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1Any example in a polynomial ring? – Crisp Mar 25 '15 at 18:59
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Recall that if $R$ is Noetherian then any ideal is finitely generated. Hence to write down a counterexample you need a ring that is not Noetherian. For my money, the simplest example of such a ring is a polynomial algebra $k[x_1, x_2, \dots]$ in countably many variables. $J = (x_1)$ is a finitely generated proper ideal, and it has a proper subideal
$$I = (x_1 x_2, x_1 x_3, x_1 x_4, \dots)$$
which I claim is not finitely generated. Do you see why?
Qiaochu Yuan
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Would another example be $J=(x_1, x_2)$ and $I=(x_1x_2, x_1x_3, \dots)+(x_2x_3, x_2x_4, \dots)$? – user5826 Jun 03 '20 at 15:39
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1@bananapeel: the minimum degree of an element of $I$ is $2$, and the degree-$2$ piece of $I$ is countable-dimensional. But if it were generated by $n$ elements, its degree-$2$ piece could be at most $n$-dimensional. – Qiaochu Yuan Feb 06 '23 at 18:46