----- > [!proposition] Proposition. ([[nilradical equals intersection of all prime ideals]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]]. The [[nilradical of a ring|nilradical]] $\text{Nil}_{R}$ of $R$ equals the intersection of all [[prime ideal|prime ideals]]: $\text{Nil}_{R}=\bigcap_{\mathfrak{p} \text{ prime}}^{}\mathfrak{p}.$ ^proposition > [!proposition] Corollary. > For $I$ any [[ideal]] of $R$, its [[radical of an ideal|radical]] is given by the intersection $\sqrt{ I }=\bigcap_{\mathfrak{p} \supset I \text{ prime }}^{}\mathfrak{p}.$ > [!proof] Proof of Corollary. > [[The correspondence theorem for rings]] gives an inclusion-preserving [[bijection]] $\{ \mathfrak{p} \in \text{Spec }R \text{ containing }I \} \leftrightarrow \text{Spec }R / I.$ > given by sending $\mathfrak{p} \mapsto \mathfrak{p} / I$, the image of $\mathfrak{p}$ under the [[quotient ring|quotient map]] $R \to R / I$. The proposition now says $\text{Nil}_{R / I} = \bigcap_{\overline{\mathfrak{p}} \in \text{Spec } R / I}^{} \overline{\mathfrak{p}}= \bigcap_{\mathfrak{p} \supset I \text{ in } \text{Spec } R}^{} \mathfrak{p} / I.$ > Now lift to $R$: $\text{Nil}_{R / I}$ lifts under the quotient map to $\sqrt{ I }$, while $\bigcap_{\mathfrak{p} \supset I \text{ in } \text{Spec } R}^{} \mathfrak{p} / I$ lifts to $\bigcap_{\mathfrak{p} \supset I \text{ prime }}^{} \mathfrak{p}.$ ^corollary > [!proof]- Proof. ([[nilradical equals intersection of all prime ideals]]) > **$\subset$.** $\text{Nil}_{R}$ is contained in every prime ideal, for the following reason. Let $f \in \text{Nil}_{R}$, $n \in \mathbb{N}$, such that $f^{n}=0$. Let $\mathfrak{p}$ be prime. Then $f^{n}=0 \in \mathfrak{p}$. Hence $f f^{n-1} \in \mathfrak{p}$, and by primality either $f \in \mathfrak{p}$ or $f^{n-1} \in\mathfrak{p}$. Inducting on $n$ implies that $f \in \mathfrak{p}$. > **$\supset$.** Conversely and contrapositively, let $f \in R$ $f \not \in \text{Nil}(R)$. Then $R \neq 0$. Look at the [[localization|localized ring]] $R_{x}$, which is nonzero because $x$ is not [[nilpotent element of a ring|nilpotent]]; let $\mathfrak{p}$ be its [[maximal ideal|maximal]] (we just need [[prime ideal|prime]]) ideal. $\mathfrak{p}$ [[extension and contraction under localization|corresponds to]] a [[prime ideal]] of $R$ disjoint from $\{ x^{n} : n \geq 0\}$. In particular, $x \not \in \mathfrak{p}$. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```