# 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.

• Hi Alex, appreciate that your question has received up votes. I think that when people visit this page, they may not know what finitely generated ideals are, or what the radical of an ideal is. The idea is that if you make your post self-contained and add some context in this regard, then your post will improve the site quality as well as attract attention. If you add the textbook/paper from which you procured this question, and/or which course you were taking at the time of this question, it would help others relate to the question and make it more self-contained.I look forward to your edits. – Teresa Lisbon Mar 8 at 18:10
• I've edit the question, by adding links to some "similar" questions with negative answers. There was no textbook/paper/course as a source of the question, just my own computation that has nothing to do with the question as it stands. Sometimes, isolated peaces of arguments/computations might be interesting on its own. – Alex Mar 9 at 7:14

## 2 Answers

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