Let $A$ be a commutative Noetherian ring, and $C$ a closed subset of $\operatorname{Spec}(A)$.
In some reading, it is an unproven proposition that $C$ is irreducible iff $C=\mathscr{Z}(\mathfrak{p})$ for some prime $\mathfrak{p}$. Here $\mathscr{Z}(\frak{p})$ is the set of zeros of $\frak{p}$, that is the set of primes of $A$ containing $\frak{p}$.
Does anyone have a proof of this proposition I can read? Thank you.
Added: With hints,
If $\mathfrak{p}\in V(\mathfrak{a})$, then $\frak{p}\supset\frak{a}$, so $xy\in\mathfrak{p}$, so say $x\in\mathfrak{p}$. Then $\mathfrak{p}\supset(\mathfrak{a},x)$, and $V(\mathfrak{a})\subseteq V(\mathfrak a, x) \cup V(\mathfrak a, y)$. The converse is clear, so I understand the equality. Since $V(\mathfrak{a})$ is irreducible, $V(\mathfrak{a})=V(\mathfrak{a},x)$ or $V(\mathfrak{a},y)$. Suppose $V(\mathfrak{a})=V(\mathfrak{a},x)$. Then $\text{rad}\mathfrak{a}\supset\text{rad}(\mathfrak{a},x)$, so $x\in\text{rad}\mathfrak{a}=\mathfrak{a}$, and $\mathfrak{a}$ is prime. But why are we allowed to assume $\mathfrak{a}$ is radical?
For the converse, $\mathfrak{p}$ must be in either $V(\mathfrak{a})$ or $V(\mathfrak{b})$ so $\mathfrak{p}$ must contain either $\mathfrak{a}$ or $\mathfrak{b}$. IF $\mathfrak{a}$ and $\mathfrak{b}$ are radical, then this relation implies $\text{rad}\mathfrak{p}\subset\mathfrak{a}$ and $\text{rad}\mathfrak{p}\subset\mathfrak{b}$, so $\mathfrak{a}\supset\mathfrak{p}$ and $\mathfrak{b}\supset\mathfrak{p}$ since $\mathfrak{p}$ is prime. Thus $\mathfrak{p}$ is equal to one of $\mathfrak{a}$ or $\mathfrak{b}$. But again why are we allowed to assume $\mathfrak{a}$ and $\mathfrak{b}$ are radical?