A radical ideal in a Noetherian ring is the intersection of its minimal primes

If $I=R$, then no prime ideal contains $I$, and the empty intersection is $R$. This is exactly the stated boundary case.

Assume $I\mathrm{\ne }R$. By [L1], choose a minimal primary decomposition $I=Q_1\cap \cdots \cap Q_r\mathrm{.}$ Since the empty intersection is $R$, one has $r\ge 1$. For each $i$, put ${\mathfrak{p}}_i=\sqrt{Q_i}$. Fact [L2] makes every ${\mathfrak{p}}_i$ prime. Since $I$ is radical, $I=\sqrt{I}=\sqrt{Q_1\cap \cdots \cap Q_r}=\sqrt{Q_1}\cap \cdots \cap \sqrt{Q_r}={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_r$ by [L3].

Let $\mathcal{M}$ be the set of inclusion-minimal members among the finite family $\{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r\}$. If one prime in the family contains another, removing the larger one does not change the intersection, so the intersection over $\mathcal{M}$ is still $I$. The product ${\mathfrak{p}}_1\cdots {\mathfrak{p}}_r$ lies in $I$. Thus every prime $\mathfrak{q}$ containing $I$ contains that product, and primality forces ${\mathfrak{p}}_i\subseteq \mathfrak{q}$ for some $i$. If $\mathfrak{q}$ is minimal over $I$, then $\mathfrak{q}={\mathfrak{p}}_i$, so $\mathfrak{q}$ belongs to $\mathcal{M}$. Conversely, if $\mathfrak{p}\in \mathcal{M}$ and a prime $\mathfrak{q}$ satisfies $I\subseteq \mathfrak{q}\subseteq \mathfrak{p}$, the same product argument gives ${\mathfrak{p}}_i\subseteq \mathfrak{q}\subseteq \mathfrak{p}$ for some $i$. Inclusion-minimality in the finite family forces ${\mathfrak{p}}_i=\mathfrak{p}$, hence $\mathfrak{q}=\mathfrak{p}$. Therefore the members of $\mathcal{M}$ are exactly the minimal primes over $I$.

Steps 1.1, 1.2, and 2.1 prove that a radical ideal is the finite intersection of its minimal primes.