Let $R$ be a ring and set $N =\{x\in R|x \text{ is nilpotent }\}$. This set is the nilradical of $R$.

Theorem: Let $R$ be a ring and $N$ be the nilradical of $R$. Then $N \triangleleft R$ and $R/N$ has no nonzero nilpotent elements.

Proof: Suppose $x, y \in N$, so $x^n = y^m = 0$ for some positive integers $m,n$. Then for any $a \in R$ we have $(a x)^n = 0$. We also have $(-x)^n = 0$. By considering the binomial expansion of $(x + y)^{m+n-1}$ we see that $(x+y)$ is also nilpotent, thus $N$ is indeed an ideal.

Next, suppose $N+a$ is nilpotent in $R/N$, so for some positive integer $k$ we have $(N+a)^k = N + a^k = N$. But this means $a^k \in N$, which implies $a \in N$ and hence $N + a = N$.

One interpretation of the theorem is that the nilradical of the ring obtained from factoring a ring out by its nilradical is trivial. Another description of the nilradical will be proved later. It turns out that the nilradical of a ring is the intersection of the prime ideals. Before we show this we first introduce some concepts.

Suppose $\{R_i | i\in I]\}$ is a family of rings. Define the direct product $R$ by

$R = \prod_{i\in I}R_i = \{(x_i)_{i\in I} | x_i\in R_i \text {for all } i\}$

This is a ring with coordinatewise operation. It is similar to the direct sum, but direct sums can only have finitely many components.

We say a ring $S$ is a subdirect product of the $R_i$ if there exists an injective ring homomorphism $\psi:S\rightarrow R$ such that $\rho_j \psi : S \rightarrow R_j$ is surjective for all $j$, where $\rho_j$ is the projection mapping $(x_i)_{i\in I} \mapsto x_j$.

Proposition: Let $S$ be a ring and $\{J_i|i\in I\}$ be a family of ideals of $S$. Let $J = \cap_{i\in I} J_i$. Then $J \triangleleft S$ and $S/J$ is a subdirect product of the $S/J_i$.

Proof: Consider the map $\psi:S\rightarrow \prod_{i\in I} S/J_i$ that sends $x\mapsto(J_i +x)_{i\in I}$. $\psi$ is a ring homomorphism with kernel $ker \psi = \cap_{i\in I} J_i = J$, so we have $B/J \cong im \psi$. We also have $\rho_j \psi$ surjective for each $J_i$.

Theorem: Let $R$ be a ring and $N$ be the nilradical of $R$. Then $N$ is the intersection of the prime ideals of $R$.

Proof: Let $N'$ be the intersection of the prime ideals. Let $P$ be some prime ideal, and let $x\in N$. Then we have $x(x^{k-1}) = 0 \in P$ for some positive integer $k$. Since $P$ is prime, this implies $x \in P$ or $x^{k-1}\in P$. By induction we find $x \in P$, hence $N \subset P$, so $N \subset N'$.

Suppose $x\in R\setminus N$, so $x$ is not nilpotent. Set

$\Sigma=\{I\triangleleft R|x^k \notin I \text{ for all } k\ge 1\}$

which is a poset with respect to $\le$. Note $\Sigma$ contains the zero ideal so $\Sigma$ is nonempty. It is easily checked that we may apply Zorn’s lemma, so let $P$ be a maximal element of $\Sigma$.

Now suppose $a b = P$ for some $a, b \in R$. Then if $a,b \notin P$, then $P$ is a proper subset of $P+a R$, and also of $P + b R$. Since $P$ is maximal in $\Sigma$, we must have

$x^k \in P + a R, x^l \in P +b R$

for some positive integers $k, l$. Thus $x^{k+l} \in P+a b R$ so $P + a b R \notin \Sigma$. But $a b \in P$ hence $P + a b R = P$, which is a contradiction. Thus $P$ is prime. Since $x \notin P$, we have that $x \notin N'$ showing that $N' \subset N$.

Corollary: Let $R$ be a ring and $N$ be its nilradical. Then $A/N$ is a subdirect product of integral domains.

Corollary: A ring with a trivial nilradical is a subdirect product of integral domains.