Let $R$ be a commutative ring with identity and $I$ a finitely generated ideal of $R$; say $I = (a_1, \dots, a_n)$.
Question 1. Is $\sqrt I$ necessarily finitely generated?
Question 2. Is there any description of $\sqrt I$ in terms of $a_i$s?
Thank you.
The ring $$R=\mathbb{C}[x_1,x_2,\ldots]/(x_1,x_2,\ldots)^2\cong\mathbb{C}[\epsilon_1,\epsilon_2,\ldots]$$ is non-noetherian (where $\epsilon_i$ denotes the image of $x_i$ in the quotient), and the radical of the (obviously finite) zero ideal $I=(0)$ in $R$ is equal to the ideal $(\epsilon_1,\epsilon_2,\ldots)$ which is not finitely generated, much less finite.
It seems to me to be impossible to say something solely in terms of a generating set for $I$, the structure of $R$ is what determines $\sqrt{I\,\strut}$.
If $(D,M)$ is a one dimensional quasi-local domain, $M=\sqrt{(x)}$ for every $x\in M\backslash (0).$ This is because for every pair $x,y\in M\backslash \{0\}$ we have $x|y^{m}$ for some positive integer $m$ by Theorem 108 of [Kaplasky, Commutative Rings, 1974]. Now a one dimension quasi-local $(D,M)$ is Noetherian if and only if $M$ is finitely generated, by Cohen's theorem. So the radical of a finitely generated ideal may or may not be finitely generated.
Here are a couple of examples of one-dimensional quasi-local domains that are not Noetherian: (1) A non discrete rank one valuation domain, (2) A ring of formal power series the form $\mathcal{Q}$ $+X\mathcal{R}[[X]]$ where $% \mathcal{Q}$ is the field of rational numbers and $\mathcal{R}$ the field of real numbers.
Example 1. Take a rank one non-discrete valuation domain $V$. Note that the maximal ideal $M$ of $V$ is not finitely generated. Now take any nonzero non unit x in $V$ and consider $rad(xV)$. Because $V$ is one dimensional for each $y\in M$ we have $x|y^n$ for some $n$ [Kaplansky, Commutaive Rings Theorem 108]. That forces $rad(xV)=M$.
– mzafrullah Jul 30 '20 at 11:05