# The Nilradical

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

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

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

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.

*blynn@cs.stanford.edu*💡