Let $R$ be a ring. For a subset $X \subset R$ define the radical of $X$ with respect to $R$ by

$\sqrt{X} = \{x\in R|x^n \in X, n\ge 1\}$

Clearly $\sqrt{\cup_\alpha X_\alpha} = \cup_\alpha \sqrt{X_\alpha}$ for any family of subsets $X_\alpha$ of $R$.

Proposition: The set of zero-divisors of $R$ is equal to its radical $\cup_{x \ne 0} \sqrt{\mathrm{Ann} x}$

Proof: Let $D$ be the set of zero-divisors of $R$ so

$D = \cup_{x\ne 0} \mathrm{Ann}(x)$

We have $D \subset \sqrt{D}$. Suppose $y\in \sqrt{D}$. Then $y^k \in D$ for some $k\ge 1$, hence $y^k x = 0$ for some nonzero $x \in R$. If $k =1$ then $y\in D$, otherwise $y(y^{k-1}x) =0$. Then if $y^{k-1}x \ne 0$ then $y \in D$, otherwise if $y^{k-1}x = 0$ we have $y\in D$ by induction.

Thus

$D = \sqrt{D} = \sqrt{\cup_{x\ne 0} \mathrm{Ann}(x)} = \cup_{x\ne 0} \sqrt{\mathrm{Ann}(x)}$

Let $I\triangleleft R$. Then

$\array { \sqrt{I} &=& \{x\in R| x^n \in I\} \\ &=& \{ x\in R| I + x^n= I\} \\ &=& \{ x\in R| (I+x)^n=I \} \\ &=& \phi^{-1}(N_{R/I}) \triangleleft R }$

where $\phi : R \rightarrow R/I$ is the natural map and $N_{R/I}$ is the nilradical of $A/I$.

The following are easily verified for any ideals $I,J$ in $R$:

1. $\sqrt{I}\supset I$

2. $\sqrt{\sqrt{I}} = \sqrt{I}$

3. $\sqrt{I J} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$

4. $\sqrt{I} = R \iff I = R$

5. $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}$

6. If $P\triangleleft R$ is prime then for all $n\ge 1$ we have $\sqrt{P^n} =P$.

Theorem: The radical of an ideal is the intersection of the prime ideals containing it.

Proof: Let $I \triangleleft R$. Then $\sqrt{I} = \phi^{-1}(N_{R/I})$. Recall that $N_{A/I}$ is the intersection of all prime ideals of $R/I$. The result follows since the prime ideals of $R/I$ are precisely those of the form $P/I$ for a prime ideal $P$ of $R$ containing $I$.

Example: Let $R = \mathbb{Z}$, $I = m \mathbb{Z}$. Then if $m = p_1 ^{a_1} ... p_k^{a_k}$ we have $\sqrt{I} = p_1 ... p_k \mathbb{Z}$.

Proposition: Let $I,J\triangleleft R$. If $\sqrt{I}, \sqrt{J}$ are coprime then $I,J$ are coprime.

Proof: $\sqrt{I+J} = \sqrt{\sqrt{I} + \sqrt{J}} = \sqrt{R} = R$, hence $I+J = R$.