Noetherian rings have only finitely many minimal prime ideals.

Question

We say that $p$ is minimal prime if It does not contain any other prime.

Assume that $A$ is Noetherian ring

Question: $A$ has only finitely many minimal primes.

any suggestions please

Answer 1

Here is an approach you might try:

A topological space $X$ is called Noetherian if it satisfies the ascending chain condition on open sets. Equivalently, every collection of open sets has a maximal element. A topological space $Y$ is said to be irreducible if it is not the union of two proper closed sets.

If $R$ is a ring with identity, then the set $\textrm{Spec } R$ of prime ideals of $R$ is a topological space whose closed sets are of the form $$V(I) := \{ \mathfrak p \in \textrm{Spec } R : I \subseteq \mathfrak p \}$$ for $I$ an ideal of $R$.

1 . A maximal irreducible subset of a topological space $X$ is called an irreducible component. They always exist by Zorn's lemma. Show that a Noetherian topological space has only finitely many irreducible components.

2 . Show that the mapping $I \mapsto V(I)$ gives an order reversing bijection between radical ideals of $R$ and closed sets of $\textrm{Spec } R$. Show that under this bijection, prime ideals are in bijective correspondence with irreducible closed sets. In particular, minimal prime ideals correspond to maximal irreducible sets.

3 . Show that if $R$ is a Noetherian ring, then $\textrm{Spec } R$ is a Noetherian topological space.