How to prove a commutative, with unit, Noetherian ring $A$ only has finitely many minimal prime ideals via the following step?
I have proved:
Step. All radical ideals of Noetherian ring $A$ can be expressed as an intersection of finitely many prime ideals.
How to continue?
Best Answer
If you know that radical ideals are finite intersections of primes, in particular $\sqrt{(0)}=\mathfrak p_1\cap\cdots\cap\mathfrak p_n$. Let $\mathfrak p$ be a minimal prime. Since $(0)\subseteq\mathfrak p$ we get $\sqrt{(0)}\subseteq\sqrt{\mathfrak p}=\mathfrak p$, that is, $\mathfrak p_1\cap\cdots\cap\mathfrak p_n\subseteq\mathfrak p$. It follows that there exists $\mathfrak p_i\subseteq\mathfrak p$ and since $\mathfrak p$ is minimal we must have $\mathfrak p_i=\mathfrak p$. (In other words, the minimal primes of $A$ are among the primes $\mathfrak p_1,,\dots,\mathfrak p_n$.)