Is radical of finitely generated ideal finitely generated?

1.2k Views Asked by At

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.

1

There are 1 best solutions below

0
On BEST ANSWER

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 $\sqrt{I\,\strut}$.