Radicals
Let $R$ be a ring. For a subset $X \subset R$ define the radical of $X$ with respect to $R$ by
Clearly $\sqrt{\cup_\alpha X_\alpha} = \cup_\alpha \sqrt{X_\alpha}$ for any family of subsets $X_\alpha$ of $R$.
Proposition: The set of zerodivisors of $R$ is equal to its radical $ \cup_{x \ne 0} \sqrt{\mathrm{Ann} x} $
Proof: Let $D$ be the set of zerodivisors of $R$ so
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 $y(y^{k1}x) =0$. Then if $y^{k1}x \ne 0$ then $y \in D$, otherwise if $y^{k1}x = 0$ we have $y\in D$ by induction.
Thus
Let $I\triangleleft R$. Then
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 $R$:

$\sqrt{I}\supset I$

$\sqrt{\sqrt{I}} = \sqrt{I}$

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

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

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

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 $I+J = R$.