[proofplan] The inclusion $\sqrt{I} \subseteq \bigcap_{I \subset \mathfrak{p}} \mathfrak{p}$ follows directly from the fact that prime ideals are radical: if $x^n \in \mathfrak{p}$ and $\mathfrak{p}$ is prime, then $x \in \mathfrak{p}$. The reverse inclusion is the substantial direction. We show the contrapositive: if $x \notin \sqrt{I}$, then there exists a prime $\mathfrak{p} \supseteq I$ with $x \notin \mathfrak{p}$. The element $x$ is not nilpotent modulo $I$, so we localize the quotient $R/I$ at the multiplicative set generated by the image of $x$, observe the resulting ring is nonzero, extract a prime ideal from it, and pull it back to $R$ via the prime correspondence for localization. [/proofplan] [step:Verify the inclusion $\sqrt{I} \subseteq \bigcap_{I \subset \mathfrak{p}} \mathfrak{p}$] Let $x \in \sqrt{I}$, so $x^n \in I$ for some $n \geq 1$. Let $\mathfrak{p} \in \operatorname{Spec}(R)$ with $I \subseteq \mathfrak{p}$. Then $x^n \in I \subseteq \mathfrak{p}$. Since $\mathfrak{p}$ is prime, $x^n \in \mathfrak{p}$ implies $x \in \mathfrak{p}$ (by induction: $x^n = x \cdot x^{n-1} \in \mathfrak{p}$ gives $x \in \mathfrak{p}$ or $x^{n-1} \in \mathfrak{p}$; iterating, $x \in \mathfrak{p}$). Since this holds for every prime $\mathfrak{p} \supseteq I$, we have $x \in \bigcap_{I \subset \mathfrak{p}} \mathfrak{p}$. [/step] [step:Prove the reverse inclusion by contrapositive: find a prime avoiding $x$ when $x \notin \sqrt{I}$] We prove the contrapositive: if $x \notin \sqrt{I}$, then $x \notin \bigcap_{I \subset \mathfrak{p}} \mathfrak{p}$, i.e., there exists a prime $\mathfrak{p} \supseteq I$ with $x \notin \mathfrak{p}$. Let $\bar{R} = R/I$ and let $\bar{x}$ denote the image of $x$ in $\bar{R}$. Since $x \notin \sqrt{I}$, the element $\bar{x}$ is not nilpotent in $\bar{R}$: if $\bar{x}^n = 0$ for some $n \geq 1$, then $x^n \in I$, giving $x \in \sqrt{I}$, a contradiction. Define the multiplicative subset $S = \{\bar{x}^n : n \geq 0\} \subseteq \bar{R}$. This is multiplicatively closed (since $\bar{x}^m \cdot \bar{x}^n = \bar{x}^{m+n}$) and contains $1 = \bar{x}^0$. Since $\bar{x}$ is not nilpotent, $0 \notin S$, so $S^{-1}\bar{R} \neq 0$ (the localization of a nonzero ring at a multiplicative set not containing $0$ is nonzero). [guided] We prove the harder direction by contrapositive. Suppose $x \notin \sqrt{I}$. We must produce a prime $\mathfrak{p} \supseteq I$ with $x \notin \mathfrak{p}$. The condition $x \notin \sqrt{I}$ means $x^n \notin I$ for all $n \geq 1$. Passing to $\bar{R} = R/I$, this says $\bar{x}^n \neq 0$ for all $n \geq 1$, i.e., $\bar{x}$ is not nilpotent in $\bar{R}$. The idea is to localize away from $\bar{x}$: consider $S = \{\bar{x}^n : n \geq 0\}$, a multiplicative subset of $\bar{R}$. Since $0 \notin S$ (because $\bar{x}$ is not nilpotent), the localization $S^{-1}\bar{R}$ is a nonzero ring. Why is $S^{-1}\bar{R} \neq 0$? In $S^{-1}\bar{R}$, we have $\frac{1}{1} = \frac{0}{1}$ if and only if there exists $\bar{x}^k \in S$ with $\bar{x}^k(1 - 0) = \bar{x}^k = 0$ in $\bar{R}$, which would require $\bar{x}$ to be nilpotent. [/guided] [/step] [step:Extract a prime ideal of $\bar{R}$ disjoint from $S$ and lift to $R$] Since $S^{-1}\bar{R}$ is a nonzero ring, it contains a maximal ideal (by Zorn's lemma applied to the poset of proper ideals), and every maximal ideal is prime. Let $\mathfrak{n}$ be a prime ideal of $S^{-1}\bar{R}$. By the [Prime Ideals Under Localization](/theorems/2927) (part 4), the contraction $\mathfrak{n}^c = \mathfrak{q}$ is a prime ideal of $\bar{R}$ with $\mathfrak{q} \cap S = \varnothing$. Since $\mathfrak{q} \cap S = \varnothing$, in particular $\bar{x} = \bar{x}^1 \notin \mathfrak{q}$. Let $\pi: R \to \bar{R} = R/I$ be the quotient map and set $\mathfrak{p} = \pi^{-1}(\mathfrak{q})$. The preimage of a prime ideal under a ring homomorphism is prime, so $\mathfrak{p} \in \operatorname{Spec}(R)$. Since $\mathfrak{q} \supseteq \{0\}$ in $\bar{R}$, we have $\mathfrak{p} = \pi^{-1}(\mathfrak{q}) \supseteq \pi^{-1}(\{0\}) = \ker \pi = I$, so $\mathfrak{p} \supseteq I$. Finally, $x \notin \mathfrak{p}$: if $x \in \mathfrak{p}$, then $\bar{x} = \pi(x) \in \mathfrak{q}$, contradicting $\bar{x} \notin \mathfrak{q}$. This produces a prime $\mathfrak{p} \supseteq I$ with $x \notin \mathfrak{p}$, completing the proof. [guided] The nonzero ring $S^{-1}\bar{R}$ must contain at least one prime ideal. (Every nonzero ring has a maximal ideal by Zorn's lemma: the set of proper ideals is non-empty --- it contains $(0)$ --- and every chain of proper ideals has an upper bound, namely the union, which is proper since it does not contain $1$. Any maximal ideal is prime in a commutative ring.) Let $\mathfrak{n}$ be a prime ideal of $S^{-1}\bar{R}$. By the prime correspondence for localization ([Prime Ideals Under Localization](/theorems/2927), part 4), there is a bijection between $\operatorname{Spec}(S^{-1}\bar{R})$ and the set $\{\mathfrak{q} \in \operatorname{Spec}(\bar{R}) : \mathfrak{q} \cap S = \varnothing\}$, given by contraction. So $\mathfrak{q} = \mathfrak{n}^c$ is a prime ideal of $\bar{R}$ with $\mathfrak{q} \cap S = \varnothing$. The condition $\mathfrak{q} \cap S = \varnothing$ means no power $\bar{x}^n$ lies in $\mathfrak{q}$. In particular, $\bar{x} \notin \mathfrak{q}$. Now we lift back to $R$. Let $\pi: R \to R/I = \bar{R}$ be the quotient map and set $\mathfrak{p} = \pi^{-1}(\mathfrak{q})$. Since $\mathfrak{q}$ is prime in $\bar{R}$ and $\pi$ is a ring homomorphism, $\mathfrak{p}$ is prime in $R$. The ideal $\mathfrak{p}$ contains $I$ because $\ker \pi = I \subseteq \pi^{-1}(\mathfrak{q}) = \mathfrak{p}$. And $x \notin \mathfrak{p}$ because $\pi(x) = \bar{x} \notin \mathfrak{q}$. So we have found a prime $\mathfrak{p} \supseteq I$ with $x \notin \mathfrak{p}$, proving $x \notin \bigcap_{I \subset \mathfrak{p}} \mathfrak{p}$. This establishes the reverse inclusion $\bigcap_{I \subset \mathfrak{p}} \mathfrak{p} \subseteq \sqrt{I}$. [/guided] [/step]