Let \(R\) be a commutative ring with identity \(1 \ne 0\). Assume Zorn’s Lemma.
Show that the nilradical \[ N = \{r \in R \mid r^n = 0 \text{ for some } n > 0\} \] is an ideal that is contained in every prime ideal of \(R\).
Let \(r \in R \setminus N\). Let \(\mathcal{S}\) be the collection of all ideals of \(R\) that do not contain any positive power of \(r\). Use Zorn’s Lemma to prove that there is a prime ideal in \(\mathcal{S}\).
Suppose that \(R\) has exactly one prime ideal \(P\). Prove that every element \(r \in R\) is either nilpotent or a unit.
Solution
Goal: Prove that the nilradical is an ideal contained in all prime ideals in (a), use Zorn’s Lemma to construct a prime ideal avoiding powers of a non-nilpotent element in (b), and prove that every element in a ring with a unique prime ideal is nilpotent or a unit in (c).
1. Part (a): \(N\) is an ideal contained in every prime ideal.
1.1. \(N\) is an ideal:- \(0 \in N\) since \(0^1 = 0\).- If \(x, y \in N\), there exist \(m, n \ge 1\) such that \(x^m = 0\) and \(y^n = 0\). By the Binomial Theorem (since \(R\) is commutative):\[(x - y)^{m+n-1} = \sum_{j=0}^{m+n-1} \binom{m+n-1}{j} x^j (-y)^{m+n-1-j}.\]In each term, either \(j \ge m\) (so \(x^j = 0\)) or \((m+n-1-j) \ge n\) (so \(y^{m+n-1-j} = 0\)). Thus \((x - y)^{m+n-1} = 0\), so \(x - y \in N\).- For any \(s \in R\) and \(x \in N\) with \(x^m = 0\): \((s x)^m = s^m x^m = s^m \cdot 0 = 0\), so \(s x \in N\).- Thus \(N\) is an ideal of \(R\).
1.2. \(N \subseteq \mathfrak{p}\) for every prime ideal \(\mathfrak{p}\):- Let \(\mathfrak{p} \subset R\) be a prime ideal and let \(x \in N\).- There exists \(n \ge 1\) such that \(x^n = 0 \in \mathfrak{p}\).- By definition of a prime ideal, \(a b \in \mathfrak{p} \implies a \in \mathfrak{p}\) or \(b \in \mathfrak{p}\).- By induction on \(n\), \(x^n \in \mathfrak{p} \implies x \in \mathfrak{p}\).- Thus \(N \subseteq \mathfrak{p}\).
2. Part (b): Existence of a prime ideal avoiding powers of \(r\).
2.1. Define the multiplicative set \(\Sigma = \{r^k \mid k \in \mathbb{N}_{\ge 1}\}\).
2.2. Define the family of ideals:\[\mathcal{S} = \{I \trianglelefteq R \mid I \cap \Sigma = \emptyset\}.\]
2.3. \(\mathcal{S}\) is non-empty: Since \(r \notin N\), \(r^k \ne 0\) for all \(k \ge 1\), which means \(\{0\} \cap \Sigma = \emptyset\). Thus the zero ideal \(\{0\} \in \mathcal{S}\).
2.4. Partially order \(\mathcal{S}\) by inclusion \(\subseteq\).
2.5. Upper bounds for chains: Let \(\mathcal{C} \subseteq \mathcal{S}\) be a totally ordered chain of ideals. The union \(I_0 = \bigcup_{I \in \mathcal{C}} I\) is an ideal of \(R\). If \(I_0 \cap \Sigma \ne \emptyset\), then \(r^k \in I_0\) for some \(k\), so \(r^k \in I'\) for some \(I' \in \mathcal{C}\), contradicting \(I' \in \mathcal{S}\). Thus \(I_0 \in \mathcal{S}\), and \(I_0\) is an upper bound for \(\mathcal{C}\).
2.6. By Zorn’s Lemma, \(\mathcal{S}\) has a maximal element, denoted \(P \in \mathcal{S}\).
2.7. \(P\) is a proper ideal: Since \(1 = r^0\) or \(r^1 \in \Sigma\), \(1 \notin P\) because \(P \cap \Sigma = \emptyset\).
2.8. \(P\) is prime:- Let \(a, b \in R\) such that \(a b \in P\). Suppose for contradiction that \(a \notin P\) and \(b \notin P\).- The ideals \(P + \langle a \rangle\) and \(P + \langle b \rangle\) strictly contain \(P\).- By the maximality of \(P\) in \(\mathcal{S}\), neither ideal belongs to \(\mathcal{S}\).- Thus there exist powers \(r^m \in P + \langle a \rangle\) and \(r^n \in P + \langle b \rangle\) for some \(m, n \ge 1\).- Write \(r^m = p_1 + u a\) and \(r^n = p_2 + v b\) for some \(p_1, p_2 \in P\) and \(u, v \in R\).- Multiply the two expressions:\[r^{m+n} = (p_1 + u a)(p_2 + v b) = p_1 p_2 + p_1 v b + p_2 u a + u v (a b).\]- Since \(p_1, p_2, a b \in P\), all terms on the right-hand side belong to \(P\), so \(r^{m+n} \in P\).- This contradicts \(P \cap \Sigma = \emptyset\).- Thus \(a \in P\) or \(b \in P\), proving that \(P\) is a prime ideal in \(\mathcal{S}\).
3. Part (c): Rings with a unique prime ideal consist of nilpotents and units.
3.1. Let \(P\) be the unique prime ideal of \(R\).
3.2. Since every maximal ideal of \(R\) is prime, and maximal ideals exist in every non-trivial commutative ring by Krull’s Theorem, \(P\) must also be the unique maximal ideal of \(R\).
3.3. Let \(r \in R\). Suppose \(r\) is not a unit.
3.4. The principal ideal \(\langle r \rangle\) is proper.
3.5. By Krull’s Theorem, \(\langle r \rangle\) is contained in a maximal ideal, so \(\langle r \rangle \subseteq P\), which means \(r \in P\).
3.6. Suppose for contradiction that \(r \notin N\).
3.7. By Part (b), there exists a prime ideal \(Q \subset R\) avoiding all positive powers of \(r\).
3.8. In particular, \(r \notin Q\).
3.9. Since \(P\) is the only prime ideal of \(R\), we must have \(Q = P\).
3.10. But then \(r \notin P\), contradicting 5 where \(r \in P\).
3.11. Thus \(r \in N\), so \(r\) is nilpotent.
3.12. Therefore every element of \(R\) is either a unit or nilpotent.
4. Conclusion:
$N$ is an ideal contained in every prime, Zorn's Lemma yields a prime ideal avoiding powers of non-nilpotents, and every element in a ring with a unique prime is a unit or nilpotent.