# 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.

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.
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 $$\sqrt{I}=(x)$$)