Proposition

Let $I$ be an ideal in $k[x_1, \cdots, x_n]$, where $k$ is an arbitrary field.

1. Show that $\sqrt{I}$ is a radical ideal.
2. Show that $I$ is radical if and only if $I = \sqrt{I}$.
3. Show that $\sqrt{\sqrt{I}} = \sqrt{I}$.

Solution

1

First, we will show that $\sqrt{I}$ is an ideal.

1. $0 \in \sqrt{I}$, so $\sqrt{I}$ is nonempty.
2. Let $f, g \in \sqrt{I}$. Then $f^m, g^n \in I$ for some $m, n \in \mathbb{N}$. Then $(f - g)^{m + n - 1} = \sum_{i=0}^{m + n - 1} (-1)^{m + n - i}f^ig^{m + n - i}$. Since for any $i$, $i \geq m$ or $m + n - i \geq n$, each term is in $I$. Thus $(f - g)^{m + n - 1} \in I$, so $f - g \in \sqrt{I}$.
3. Let $f \in \sqrt{I}, g \in k[x_1, \cdots, x_n]$. Then $f^m \in I$ for some $m \in \mathbb{N}$. $(gf)^m = g^mf^m \in I$, so $gf \in \sqrt{I}$. Therefore, $\sqrt{I}$ is indeed an ideal.

Let $f^m \in \sqrt{I}$. Then $(f^m)^n \in I$ for some $n \in \mathbb{N}$. $f^{mn} \in I$, so $f \in \sqrt{I}$.

Therefore, $\sqrt{I}$ is indeed a radical ideal.

2

Suppose $I$ is radical. Let $f \in I$. Then $f^1 \in I$, so $f \in \sqrt{I}$. Let $g \in \sqrt{I}$. Then $g^m \in I$ for some $m$. Since $I$ is radical, $g \in I$. Thus $\sqrt{I} = I$.

Suppose $\sqrt{I} = I$. In 1, we showed that $\sqrt{I}$ is radical. Thus $I$ is radical.

3

In 1, we showed that $\sqrt{I}$ is radical. By using the results of 2, $\sqrt{I} = \sqrt{\sqrt{I}}$.