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?