How to show $X=Spec(A)$ has maximal components $V(p)$? This is from Atiyah Macdonald, Chapter 1, Execrise 20. 
If $A$ is a ring and $X=Spec(A)$, then the irreducible components of $X$ are the closed sets $V(p)$, where $p$ is a minimal prime ideal of $A$. 
It is not difficult to prove the closed sets $V(p)$ are irreducible. However it isn't clear to me how to prove the second part. It is clear that when closed sets of the form $V(q),q\in X$ then $V(p)$ did provide a maximal irreducible closed set. Neverthelss it is not clear to me why every closed set in $X$ must be of this form(they only provide a basis of closed sets). So theoretically there is the possiblity of some closed set $U$ that contains $V(p)$, and in particular contains some prime ideal $k$ such that $p\not\subset k$. Then what can we do to prove this cannot hold and $k$ must contain $p$?
One way to address this is to take the closure of $U\cap X_{p}=K$. Since the $V(q),q\in X$ form a basis there is some $V(l)\subset \overline K$. Then $V_{l}\cup V_{p}\subset U$, while $V_{l}\cup V_{p}=V_{l\cap p}$. But by definition $l\cap p=p$; so such $V_{l}$ could not exist, and $K$ must be empty. 
I am wondering if this approach is long-winded, as I found almost every other proof used the fact every irreducible component of $X$ must be of the form $V_{r(p)}$, where $r(p)$ denotes the radical of $p$. Why this is true, and how it might help to simplify the above proof?
 A: Let $I$ be an ideal of $A$.
We denote by $rad(I)$ the set $= \{x \in A\colon x^n \in I$ for some integer $n > 0$, where $n$ depends on $x\}$.
It is easy to see that $rad(I)$ is an ideal and $V(I) = V(rad(I))$.
Let $E$ be a subset of $X = Spec(A)$.
We denote by $\mathfrak{I}(E)$ the ideal $\bigcap_{p \in E} p$.
Lemma 0
Let $I$ be an ideal of $A$.
Then $\mathfrak{I}(V(I)) = rad(I)$.
Proof:
This is an immediate consequence of proposition 1.8 of Atiyah-MacDonald(it states that  $\mathfrak{I}(X) = rad(0)$.
Lemma 1
Let $E$ be a subset of $X$.
Then $V(\mathfrak{I}(E))$ is the closure of $E$.
Proof:
Let $\bar E$ be the closure of $E$.
Since $E \subset V(\mathfrak{I}(E))$, $\bar E \subset V(\mathfrak{I}(E))$.
Conversely suppose $E \subset V(I)$ for some ideal $I$.
Then $I \subset \mathfrak{I}(E)$.
Hence $V(\mathfrak{I}(E)) \subset V(I)$.
Hence $V(\mathfrak{I}(E)) \subset \bar E$.
QED
Lemma 2
$X = Spec(A)$ is irreducible if and only if $A/rad(0)$ is an integral domain.
Proof:
Since $X = Spec(A/rad(0))$, we can assume $rad(0) = 0$.
Suppose $X$ is reducible.
There exist proper closed subsets $F_1, F_2$ of $X$ such that $X = F_1 \cup F_2$.
Hence $0 = \mathfrak{I}(X) = \mathfrak{I}(F_1) \cap \mathfrak{I}(F_2)$.
By Lemma 1, $V(\mathfrak{I}(F_1)) = F_1$.
Since $F_1 \neq X$, $\mathfrak{I}(F_1) \neq 0$.
Similarly $\mathfrak{I}(F_2) \neq 0$.
Hence $A$ is not an integral domain.
Conversely suppose $A$ is not an integral domain.
There exist elements $f, g$ of $A$ such that $f \neq 0, g \neq 0$, $fg = 0$.
Then, by Lemma 0, $V(f) \neq X$, $V(g) \neq X$ and $X = V(f) \cup V(g)$.
Hence $X$ is reducible.
QED
Lemma 3
Let $I$ be an ideal of $A$.
Then $V(I)$ is irreducible if and only if $rad(I)$ is a prime ideal.
Proof:
This follows immediately from Lemma 2.
By Lemma 3, an irreducible closed subset of $X$ is of the form $V(q)$, where $q$ is a prime ideal of $A$.
Hence $V(p)$ is a maximal irreducible closed subset of $X$, if $p$ is a minimal prime ideal of $A$.
