$C$ is irreducible iff $C=\mathscr{Z}(\mathfrak{p})$ for some prime ideal $\mathfrak{p}$?

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.

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?

-
This is a very good exercise, and it probably is an exercise in Atiyah-Macdonald somewhere. Have you given it a shot? I can provide some hints. – Dylan Moreland Feb 17 '12 at 23:39
@DylanMoreland I believe this is related to exercise 20 (iv): If $A$ is a ring and $X = \operatorname{Spec}(A)$, then the irreducible components of $X$ are the closed sets $V(\mathfrak{p})$, where $\mathfrak{p}$ is the minimal prime ideal of $A$. – user38268 Feb 17 '12 at 23:43
Although I don't get how the Noetherian condition comes in here.... – user38268 Feb 17 '12 at 23:44
@Benjamin: If $A$ is Noetherian then $\mathrm{Spec}(A)$ is Noetherian. – Asaf Karagila Feb 17 '12 at 23:47
The Noetherian condition is irrelevant.You may reduce to the case $C=Spec(A)$ (why?) and prove that $Spec(A)$ is irreducible iff $Nil(A)=\mathfrak p$ (the nilpotent radical) is prime. – Georges Elencwajg Feb 18 '12 at 0:33

Let me use $V$ in place of your $\mathscr Z$. I don't think we need the Noetherian hypothesis, although it's useful for some related statements; for example, as Asaf says, it implies that $\operatorname{Spec} A$ is a Noetherian topological space.
Let $V(\mathfrak a)$ be an irreducible closed set in $\operatorname{Spec} A$. I may as well assume that $\mathfrak a$ is radical, and in fact I'd like to prove then that $\mathfrak a$ is prime. For this, let $x, y \in A$ such that $xy \in \mathfrak a$. I claim that $V(\mathfrak a, x) \cup V(\mathfrak a, y) = V(\mathfrak a)$. Why is this true, and what does it imply?
The converse is easier. First, note that $\mathfrak p \in V(\mathfrak p)$. If $V(\mathfrak p) = V(\mathfrak a) \cup V(\mathfrak b)$, then is it possible that $\mathfrak p$ contains neither $\mathfrak a$ nor $\mathfrak b$?
@Buble Not at all. You can assume that $\mathfrak a$ is radical because the radical of an ideal is the intersection of all prime ideals containing it. This is Proposition 2.5 of Milne's notes. For the converse, if I haven't made a mistake then you're overthinking it! If $\mathfrak p \supset \mathfrak a$ then $V(\mathfrak p) \subset V(\mathfrak a)$. – Dylan Moreland Feb 18 '12 at 17:46