Commutative Algebra - The Nilradical

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

Theorem: Let $R$ be a ring and $N$ be the nilradical of $R$ . Then $N ◃ R$ 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 $R = \prod_{i \in I} R_{i} = {(x_{i})_{i \in I} ∣ x_{i} \in R_{i} for all i}$ 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 $ψ : S \to R$ such that $ρ_{j} ψ : S \to R_{j}$ is surjective for all $j$ , where $ρ_{j}$ 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 $J ◃ S$ and $S / J$ is a subdirect product of the $S / J_{i}$ .

Proof: Consider the map $ψ : S \to \prod_{i \in I} S / J_{i}$ that sends $x \mapsto (J_{i} + x)_{i \in I}$ . $ψ$ is a ring homomorphism with kernel $\ker ψ = \cap_{i \in I} J_{i} = J$ , so we have $B / J ≅ im ψ$ . We also have $ρ_{j} ψ$ 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 $x \in R ∖ N$ , so $x$ is not nilpotent. Set $Σ = {I ◃ R ∣ x^{k} \notin I for all k \geq 1}$ which is a poset with respect to $\leq$ . Note $Σ$ contains the zero ideal so $Σ$ is nonempty. It is easily checked that we may apply Zorn's lemma, so let $P$ be a maximal element of $Σ$ .

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 $Σ$ , 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 $P + a b R \notin Σ$ . 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.