Radicals and Nilradicals

The radical of an ideal is a contruction in commutative algebra characterizing powers of ring elements within the ideal. Surprisingly, the radical of an ideal and the nilradical (radical of the zero ideal) are closely related to the prime spectrum of the ring, making them important objects in algebraic geometry. Below are a few useful properties and characterizations of radicals and nilradicals. Throughout this section we assume that all rings are commutative with $1$.

Definition 1.
For $R$ a ring (commutative with $1$), the nilradical of $R$ is the set of all nilpotent elements and is notated $$\underline{\nil(R)} = \big\{x\in R\hspace{1mm}\big|\hspace{1mm} \text{$x^n=0$ for some positive $n\in \N$}\big\}$$ Furthermore, $\nil(R)$ is an ideal of $R$ which can be shown via a binomial expansion.

Proposition 1.
One has the equality $\nil(R) = \bigcap_{\p\in \spec(R)}\p$ where $\spec(R)$ is the spectrum of $R$, the set of all prime ideals.

Proof: We show the equality by exhibiting both proper containments:

$\subseteq\colon$ Let $x\in \nil(R)$. There exists some positive integer $n$ such that $x^n=0 \in \p$ for all $\p \in \spec(R)$ (since all ideals contain $0$). However since these ideals are prime, it then follows that $x\in \p$ for all prime ideals and is hence contained in the intersection of all primes.

$\supseteq\colon$ To show the reverse containment, it is enough to show that if there is an element contained in every prime ideal of $R$, that element must be nilpotent. This is equivalent to showing that if an element of $R$ is not nilpotent, then there exists some prime ideal which does not contain it. This puts us in a good position to make use of Zorn's lemma. Let $x\in R$ be a non-nilpotent element and consider the set $$\Sigma = \{I \subset R\hspace{1mm}|\hspace{1mm}\text{$I$ is an ideal and $x^n\notin I$ for all $n>0$}\}$$ Notice that since $x$ is not nilpotent, $(0)\in \Sigma$ and so $\Sigma$ is nonempty. Furthermore $\Sigma$ is partially ordered by inclusion and so by Zorn's lemma, $\Sigma$ has a maximal element $J$ and we claim that $J$ is a prime ideal. Suppose not and there exist elements $y,z\in R$ such that $yz\in J$, but $y\notin J$ and $z\notin J$. Now since neither of these elements are contained in $J$ we have the proper containments $J\subsetneq J+(y)$ and $J\subsetneq J+(z)$. Hence, by the maximality of $J$ in $S$, there must exist positive integers $m$ and $n$ such that $x^n \in J\subsetneq J+(y)$ and $x^m\in J\subsetneq J+(z)$. However, then we see $$x^{n+m}\in \big(J+(y)\big)\big(J+(z)\big) = J^2 +(x)J+(y)J +(yz) \subset J$$ as $yz\in J$. However this is a contradiction as no power of $x$ can be contained in $J$. Hence no such $y$ and $z$ can satisfy the assumptions and so $J$ must be a prime ideal. With this, given any non-nilpotent element $x\in R$ we can find a prime ideal $J$ not containing any power of $x$ and hence not containing $x$ at all. Thus if there is an element contained in every prime ideal, that element must be nilpotent which implies the desired containment.

Definition 2.
For $I$ an ideal of $R$, the radical of $I$ is the set $$\underline{\sqrt{I}} = \big\{x\in R\hspace{1mm}\big|\hspace{1mm} \text{$x^n\in I$ for some positive $n\in \N$}\big\}$$ As this definition appears very similar the previous one, we should expect some sort of relation between the two. Indeed the nilradical of $R$ is actually just the radical of the zero ideal, $\nil(R) = \sqrt{0}$.

Proposition 2.
For $I$ a proper ideal of $R$, one has the equality $\sqrt{I} = \bigcap_{\p\in V(I)}\p$ where $V(I)$ is the subset of $\spec(R)$ consisting of all prime ideals containing $I$..

Proof: We could proceed as before by displaying both proper containments, or rather since we have a relation between radicals and nilradicals and we already know something about the latter, we can save ourselves the trouble of another Zorn's lemma argument. Notice that if $\varphi\colon R\twoheadrightarrow R/I$ is the natural map, then by the definitions above $\sqrt{I} = \varphi^{-1}(\nil(R/I))$. Now since we know that $\nil(R/I)$ is the intersection of all prime ideals of $R/I$, the result immediately follows as all primes $\overline{\p} \in \spec(R/I)$ are of the form $\overline{\p} = \p/I$ for some prime $R$-ideal $\p$ containing $I$ as a consequence of the homomorphism theorem (also an easy exercise).