Is an ideal finitely generated if its radical is finitely generated? Let $R$ be a commutative ring. If $R$ is not Noetherian, we can ask if some some ideals is finitely generated. For examples:

*

*Is intersection of finitely generated ideals finitely generated? No, see for instance here.


*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.
 A: Consider $A$ a reduced ring and let $M$ be a finitely generated $A$-module with a non-finitely generated submodule (namely, you may take $M=A$ and $A$ a reduced non-Noetherian ring). Call $R=A\boxed\times M$ the ring having support the set $A\times M$ and operations $(a,m)+(b,n)=(a+b,m+n)$ and $(a,m)\cdot(b,n)=(ab,an+bm)$. Let's identify $M=\{0\}\times M$ and $A=A\times \{0\}$. Notice that $(a,m)^k=(a^k,\text{stuff})$ and therefore $\sqrt{0}\subseteq M$. Also, $(0,n)(0,m)=(0,0)$, therefore $\sqrt 0\supseteq M$. $R$ acts by ring multiplication on subsets of $M$ essentially like $A$ acts by $A$-module action, since $(a,m)(0,n)=(0,an)$. Therefore, any non-finitely generated $A$-submodule $N$ of $M$ turns into a non-finitely generated ideal of $R$ such that $\sqrt N=M$ is finitely generated.
A: 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 $\sqrt{I}=(x)$)
