Let $R$ be a commutative ring. If $R$ is not Noetherian, we can ask if some some ideals is finitely generated. For examples:
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Is intersection of finitely generated ideals finitely generated? No, see for instance here.
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Is radical $\sqrt{I}=\{x \in R \mid x^n \in I \text{ for some } n\ge 1\}$ of a finitely generated ideal $I$ finitely generated? No, see for instance here.
In the light of the previous two (sets of) counter-examples, I believe that claim
- If $\sqrt{I}$ is finitely generated, then $I$ is also finitely generated,
is also false, but I wasn't able to construct a counter-example. It would be interesting for me to see counter-examples of various nature.
Best Answer
The ideal $I=(x^2,xy_1,xy_2,xy_3,\ldots)$ of $\Bbb{Q}[x,y_1,y_2,y_3,\ldots]$ is not finitely generated, its radical is $(x)$.
(it is immediate that $I\subset (x), (x)\subset \sqrt{I}$ and $\sqrt{(x)}=(x)$ so $\sqrt{I}=(x)$)