Radical of an ideal ($\sqrt{I}$)

The radical of an ideal $I$ isAn ideal equal to its radical is a radical ideal.

In a commutative ring, the radical of an ideal is monotone and idempotent: inclusion $I\subseteq J$ gives $\sqrt{I}\subseteq \sqrt{J}$, and $a^n\in \sqrt{I}$ means $a^{nm}\in I$ for some $m$, giving $\sqrt{\sqrt{I}}=\sqrt{I}$. Since $\sqrt{I+J}$ contains both $\sqrt{I}$ and $\sqrt{J}$, it contains their sum of ideals. Taking radicals of ideals gives $\sqrt{\sqrt{I}+\sqrt{J}}\subseteq \sqrt{I+J}$. The reverse inclusion follows from $I+J\subseteq \sqrt{I}+\sqrt{J}$.

In particular, $\sqrt{I}+\sqrt{J}=R$ implies $I+J=R$. One can also see this directly: write $1=a+b$ with $a^m\in I$ and $b^n\in J$, then expand $(a+b{)}^{m+n-1}$ by the binomial theorem. Each term belongs to $I$ or $J$, so $1\in I+J$.

For any ideal $I$ in a commutative ring, every prime ideal containing $I$ contains $\sqrt{I}$, because $a^n\in P$ implies $a\in P$. Conversely, if $a\notin \sqrt{I}$, the multiplicatively closed set $T=\{1,a,a^2,\dots \}$ is disjoint from $I$. By Zorn's lemma, there is an ideal $P\supseteq I$ maximal among those disjoint from $T$.

This $P$ is a prime ideal: if $bc\in P$ but $b,c\notin P$, maximality makes both $P+(b)$ and $P+(c)$ meet $T$. Multiplication of representatives from these two intersections puts an element of $T$ in $P$, a contradiction. Thus there is a prime ideal containing $I$ but avoiding $a$, establishing the identity. For $I=R$, the intersection is empty and means $R$. It follows that an ideal is a radical ideal exactly when it is an intersection of ideals that are all prime ideals.

In a Noetherian ring, every radical of an ideal is an intersection of finitely many prime ideals. A maximal counterexample ideal $J$ cannot be prime. Choose $a,b\notin J$ with $ab\in J$; the strictly larger ideals $J+(a)$ and $J+(b)$ satisfy the result, and $\sqrt{J}=\sqrt{J+(a)}\cap \sqrt{J+(b)}$. This contradicts maximality. The unit ideal uses an empty intersection.

The nilradical of a commutative ring is its ideal of nilpotent elements, equivalently the radical of an ideal applied to its zero ideal.

For any multiplicative subset $S$ of a commutative ring $R$, fractions with nilpotent numerators are nilpotent in $S^{-1}R$. Conversely, if $(a/s{)}^n=0$, the vanishing criterion in a module localization gives $t\in S$ with $t{a}^n=0$. Then $(ta{)}^n=t^{n-1}t{a}^n=0$ and $a/s=ta/(ts)$ has a nilpotent numerator. This proves the formula for the nilradical without any finiteness assumption.

In $R=k[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots )$, the nilradical is generated by all ${\bar{t}}_i$. Every element uses finitely many variables and is nilpotent, while each square-free product ${\bar{t}}_1\cdots {\bar{t}}_j$ is nonzero. Thus this nilradical is not a nilpotent ideal. A Noetherian ring cannot have this behavior: finitely many nilpotent generators give a nilpotent ideal.

For every commutative ring $R$,Every nilpotent element lies in every prime ideal. If $a$ is not nilpotent, an ideal maximal among those disjoint from $\{1,a,a^2,\dots \}$ is a prime ideal avoiding $a$, by the Zorn lemma and the maximality argument.