Radical of an ideal

Let $R$ be a commutative ring and let $I\subseteq R$ be an ideal. The radical of $I$ is

Then $\sqrt{I}$ is an ideal containing $I$.

The radical measures “nilpotence modulo $I$”: $r\in \sqrt{I}$ iff the image of $r$ in $R/I$ is nilpotent . The nilradical is the special case $\sqrt{(0)}$, and one has $\sqrt{I}=\bigcap_{\mathfrak p\supseteq I}\mathfrak p$ as an intersection over all prime ideals containing $I$.

Examples:

• In $\mathbb{Z}$, $\sqrt{(12)}=(6)$ since an integer has a power divisible by $12$ iff it is divisible by $2$ and $3$.
• In $k[x]$, $\sqrt{(x^2)}=(x)$.
• In $k[x,y]$, $\sqrt{(xy)}=(x)\cap (y)$.