In a noetherian ring there are only finitely many minimal prime ideals containing a given ideal Why in any noetherian ring $N$ there are only finitely many minimal prime ideals containing an ideal $I$ given?
I know that every ideal in $N$ is finitely many generated but I don´t  know how use this fact to get the answer. Thank for any help!
 A: In a Noetherian ring, each ideal is a finite intersection of primary ideals:
$$I=Q_1\cap\cdots\cap Q_k.$$
If $P$ is a prime ideal containing $I$ then $P\supseteq Q_i$
for some $I$. Therefore $P=\sqrt P\supseteq \sqrt{Q_i}$. The ideal
$\sqrt Q_i$ is prime. So the minimal prime ideals containing $I$
are some of the $\sqrt{Q_i}$.
A: Another proof, based on the topological properties of $\operatorname{Spec}N$:
First, observe that, since $N/I$ is noetherian and there's a bijection between the (minimal) prime ideals of $N/I$ and the (minimal) prime ideals of $N$ which contain $I$,  it is enough to prove the finiteness of the minimal prime ideals of $N$.
Second, there's a bijection between the set of minimal prime ideals of $N$ and the set of irreducible components of the topological space $\operatorname{Spec}N$:
$$\mathfrak p\in \operatorname{Min} N\longleftrightarrow V(\mathfrak p)=\{\mathfrak q\in\operatorname{Spec}N\mid \mathfrak p\subset \mathfrak q\}. $$
Last: since $N$ is noetherian ring, $\operatorname{Spec}N$ is a noetherian space, i.e. every non-increasing sequence of closed sets in $\operatorname{Spec}N$ stabilises, and a noetherian space has a finite number of irreducible components.
