Nontrivial ideal of a Noetherian domain contains a finite product of nonzero prime ideals If $R$ is a Noetherian domain and $ 0 < U < R$ an nontrivial ideal of $R$. How to prove that there exists nonzero prime ideals $p_1,...,p_n \subset R$ such that the product $ p_1 p_2 ...p_n \subset U$?  
I have no idea on this one. Any help is appreciated.
 A: Theorem: Ideals of Noetherian rings have finite primary decompositions.
Lemma: If $Q$ is a primary ideal in a Noetherian ring whose radical is $P$ (which is prime, of course), then there exists a natural number n such that $P^n\subseteq Q$.
Hint: If $P_1,\dots P_m$ is a complete list of radicals of primary ideals $Q_1,\dots Q_m$ in the decomposition of your ideal $I$, then you can find an $N$ large enough so that $P_1^N\cdot\ldots \cdot P_m^N\subseteq \cap Q_i=I$.
Notice that this doesn't depend on $R$ being a domain or even reduced. (It does of course, depend on foreknowledge of primary decompositions and primary ideals, though.)
A: Consider the collection of nontrivial ideals of $R$ which do not contain a product of nonzero prime ideals. If this collection is nonempty, then it contains a maximal element $I$. I claim that $I$ must be prime (which is absurd). 
Suppose otherwise, so that there exist $x,y \notin I$ such that $xy\in I$. The ideals $I+xR$, $I+yR$ are strictly larger than $I$, and they are not trivial: if, say, $I+xR=R$ then $yI+yxR = yR$ but $yI+yxR$ is contained in $I$, contradicting that $y \notin I$. Therefore by the choice of $I$ each of them must contain a product of prime ideals, but then so does $(I + yR)(I+yR) \subseteq I$, which contradicts the choice of $I$.
A: EDIT: Upon re-reading the question, I realized that my answer worked for Noetherian reduced rings that are not domains. The general case has already been well-covered in the other answers though, so I will leave this as a simple argument in the reduced case.
Every Noetherian ring has finitely many minimal primes $p_1, ..., p_n$, whose intersection is the nilradical, $\text{nil}(R) = \bigcap_{i=1}^n p_i$. Since $R$ is reduced, $\text{nil}(R) = 0$ is contained in every nonzero ideal $I$, but then $p_1...p_n \subseteq p_1 \cap \ldots \cap p_n \subseteq I$.
This gives an explicit set of primes whose product is contained in every nonzero ideal (namely the minimal primes).
