Zev's answer is excellent as usual. The following exercises serve to supplement Zev's answer and my comment above.
The following theorem is very important; it is proven in Zev's answer above:
Theorem Let $A$ be a commutative ring. The nilradical of $A$ is equal to the intersection of all prime ideals of $A$.
You should make sure that you understand this theorem and its proof (given in Zev's answer above) before attempting the following exercises.
Exercise 1: Let $p,q$ be prime ideals of a commutative ring $A$. If $p\cap q$ is a prime ideal of $A$, then prove that either $p\subseteq q$ or $q\subseteq p$. More generally, prove that if $p_1,\dots,p_n$ are prime ideals of $A$ and if $p_1\cap\cdots\cap p_n\subseteq P$, then $p_i\subseteq P$ for some $1\leq i\leq n$.
Exercise 2: Let $A$ be a commutative ring. Let $\{p_i\}_{i\in I}$ be a family of prime ideals of $A$ totally ordered by inclusion. Prove that $\bigcap_{i\in I} p_i$ is a prime ideal of $A$.
Exercise 3: Let $A$ be a commutative ring. A prime ideal $p$ of $A$ is said to be a minimal prime ideal of $A$ if there exists no prime ideal $q$ of $A$ strictly contained in $p$. Prove that if $P$ is a prime ideal of $A$, then there exists a minimal prime ideal $Q$ of $A$ contained in $P$. (Hint: use Zorn's lemma and Exercise 2.)
Exercise 4: If $A$ is each of the following commutative rings, then determine the minimal prime ideal(s) of $A$:
(a) $A$ is an integral domain;
(b) $A=\mathbb{Z}/(n\mathbb{Z})$ for some positive integer $n$;
(c) $A=k[x]/(f(x))$ for some non-constant polynomial $f(x)$ where $k$ is a field.
Exercise 5 (Challenge): Let $A$ be a Noetherian ring. Prove that $A$ has finitely many minimal prime ideals. (Hint: it suffices to prove that the nilradical of $A$ is the intersection of finitely many prime ideals of $A$ by Exercise 1. Assume, for a contradiction, that this is not the case and note that the set of ideals $I$ of $A$ that are not the intersection of finitely many prime ideals of $A$ is non-empty. Since $A$ is a Noetherian ring, it follows that there is an ideal $I$ of $A$ maximal with respect to the property of not being an intersection of finitely many prime ideals of $A$. Derive a contradiction.)
I hope this helps!