Noetherian ring under some conditions has at least two minimal prime ideals

Question

Question is :

Suppose $R$ is a noetherian ring. Prove that $R$ is either an integral domain, has nonzero nilpotent elements, or has at least two minimal prime ideals. [Use the previous exercise.]

Previous exercise says :
Let $\mathfrak{p}_1, \mathfrak{p}_2,\cdots \mathfrak{p}_n$ be the associated prime ideals of the ideal $(0)$ in the Noetherian ring $R$.

1. Show that $\bigcap_{i=1}^n\mathfrak{p}_i$ is the collection of nilpotent elements in $R$.
2. Show that $\bigcup_{i=1}^n\mathfrak{p}_i$ is the collection of zero divisors in $R$.

I do not really understand the question clearly... If $R$ is integral domain, there would be no nonzero nilpotent elements. So, what does he mean when he say "either an integral domain, has nonzero nilpotent elements". May be i am missing some point...

Suppose $R$ is not an integral domain and do not have non zero nilpotent elements..

This means that $\bigcap_{i=1}^n\mathfrak{p}_i=\{0\}$ and $\bigcup_{i=1}^n\mathfrak{p}_i\neq \{0\}$

i could not see how to conclude from this that $R$ has at least two minimal prime ideals..

Please give some hints...

This is an exercise from Dummit and Foote's ABSTRACT ALGEBRA, Section $15.2$.

Answer 1

Credits to this answer goes to the user https://math.stackexchange.com/users/127490/hoot

Let $R$ be a noetherian ring so that i can assume that $(0)$ has a primary decomposition.

Let $\mathfrak{p}_1, \mathfrak{p}_2,\cdots \mathfrak{p}_n$ be the associated prime ideals of the ideal $(0)$ in the Noetherian ring $R$. Then

1. $\bigcap_{i=1}^n\mathfrak{p}_i$ is the collection of nilpotent elements in $R$.
2. $\bigcup_{i=1}^n\mathfrak{p}_i$ is the collection of zero divisors in $R$.

Suppose $R$ is not an integral domain and has no nonzero nilpotent elements then we see that $\bigcap_{i=1}^n\mathfrak{p}_i=(0)$...

Suppose $\mathfrak{p}$ is a minimal associated prime of $(0)$ and $\mathfrak{q}$ be a minimal prime ideal in $R$ such that $\mathfrak{q}\subset\mathfrak{p}$.. As $\mathfrak{q}$ is a prime ideal, $\mathfrak{q}$ must contain some minimal associated prime say $\mathfrak{l}$ of $(0)$. So, we have $\mathfrak{l}\subset\mathfrak{q}\subset\mathfrak{p}$. AS $\mathfrak{p}$ and $\mathfrak{l}$ are associated primes of $(0)$ and $\mathfrak{p}$ is minimal we should have $\mathfrak{l}=\mathfrak{q}=\mathfrak{p}$.. Thus $\mathfrak{p}$ is a minimal prime ideal...

Suppose $\mathfrak{p}$ is a minimal prime ideal then, there exist a minimal associated prime $\mathfrak{q}$ of $(0)$ such that $\mathfrak{p}\supset \mathfrak{q}$. As $\mathfrak{q}$ is a prime ideal and $\mathfrak{p}$ is minimal prime ideal we should have $\mathfrak{p}=\mathfrak{q}$... So, $\mathfrak{p}$ is a minimal associated prime.. So, minimal prime ideal is minimal associated prime of $(0)$..

So, minimal primes are precisely the minimal associated primes... So, $$\{0\}=\bigcap_{i=1}^n\mathfrak{p}_i=\bigcap_{\rm{minimal}}\mathfrak{p}_i=\bigcap_{\rm{minimal~ primes ~ of ~ R}}\mathfrak{q}_i$$

So, intersection of all minimal prime ideals is $\{0\}$ and if there is only one minimal prime ideal then that ideal has to be zero ideal... As zero ideal is prime ideal in this $R$, the ring has to be an integral domain which contradicts the assumption.. thus, there exists atleast two minimal prime ideals..

We have $\mathfrak{q}\supset (0)=\mathfrak{q}_1\cap \mathfrak{q}_2\cap\cdots\cap \mathfrak{q}_n$ which implies $r(\mathfrak{q})\supset r(\mathfrak{q}_1)\cap r(\mathfrak{q}_2)\cap\cdots\cap r(\mathfrak{q}_n)$ . As $\mathfrak{q}$ is a prime ideal we have $r(\mathfrak{q})=\mathfrak{q}$... This would mean $\mathfrak{q}\supset r(\mathfrak{q}_1)\cap r(\mathfrak{q}_2)\cap\cdots\cap r(\mathfrak{q}_n)$.. This imply $\mathfrak{q}$ contains some $r(\mathfrak{q}_i)$ for some $i$ and we may even assume $r(\mathfrak{q}_i)$ is minimal primary component... Here i have used a result that "$\mathfrak{a}_1,\mathfrak{a}_2,\cdots, \mathfrak{a}_n$ be ideals of $A$ and $\mathfrak{p}$ is a prime ideal containing $\bigcap_{i=1}^n \mathfrak{a}_i$. Then $\mathfrak{p}\supseteq \mathfrak{a}_i$ for some $i$".

Answer 2

Note that for the given statement to be true, we have to assume that the ring is non-zero. Furthermore, the Noetherian hypothesis is superfluous. I think the following solution is far simpler than the book's hint.

Suppose $R$ is not an integral domain and does not contain non-zero nilpotent elements. Then we can find $a, b \in R - \{0\}$ such that $ab = 0$. Since $a$ is not nilpotent, we can find a minimal prime ideal $\mathfrak p$ such that $a \not \in \mathfrak p$. Similarly, we can find a minimal prime ideal $\mathfrak q$ such that $b \not \in \mathfrak q$. It cannot be the case that $\mathfrak p = \mathfrak q$, for otherwise we would have $ab \in \mathfrak p$ but neither $a \in \mathfrak p$ nor $b \in \mathfrak p$.