Is radical of finitely generated ideal finitely generated?

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.