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I’d like to show the following theorem.

Theorem

Let A be a ring. If A has infinite minimal prime ideals, then A isn’t noetherian.

I tried as follows.

Let $P_i (i \in N)$ denote minimal prime ideals. The ascending chain of ideals of A

$(0;P_0)\subset (0;P_0•P_1)\subset(0;P_0•P_1•P_3)\subset…$

is strictly increasing.

where (•;•)is ideal quotient and $P_0•P_1$ is a product of ideals.

Hence A isn’t noetherian.

But I can’t prove the chain is strictly increacing.

How can I prove or disprove.

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    $\begingroup$ 1. Please use MathJax to format math - I noticed you've used it in the past, so hopefully it won't be too laborious to add it here. 2. It seems like you have some missing text in your statement of the theorem - you don't mention minimal primes anywhere. $\endgroup$
    – KReiser
    Apr 29 at 3:35
  • $\begingroup$ Note that this is a special case of the Lasker-Noether theorem. This theorem implies that when $A$ is noetherian then the $0$-ideal has a finite primary decomposition $\cap_i {\mathfrak q}_i=(0)$ and the primes minimal among the ${\rm radical}({\mathfrak q}_i)$ are the minimal primes of the ring. See Atiyah-MacDonald, chapters on primary decomposition and on noetherian rings. $\endgroup$
    – Damian Rössler
    Apr 29 at 12:17
  • $\begingroup$ Let $\mathcal F$ be the family of ideals $I$ such that there exist infinitely many minimal primes over $I$. $\mathcal F$ is nonempty, since $(0)\in\mathcal F$ by assumption. If $A$ is Noetherian, $\mathcal F$ has a maximal element $J$. Try to get a contradiction now. $\endgroup$
    – cqfd
    Apr 29 at 15:28

1 Answer 1

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Here is one way method which requires no knowledge of Noetherian rings.

(1) Reduce to the case that $A$ is reduced, since $A$ Noetherian implies $A/\sqrt{0}$ Noetherian and the minimal primes of $A$ and $A/\sqrt{0}$ are in bijection.

(2) For each minimal prime $P_\alpha$, use that $A_{P_\alpha}$ is a field and that $P_\alpha$ is finitely generated (by the Noetherian hypothesis) to produce an $s_\alpha \in A \setminus P_\alpha$ such that $s_\alpha P_\alpha = 0$.

(3) Let $I$ be the ideal generated by the $s_\alpha$. By construction, $I$ is not contained in any minimal prime ideal. Check that this implies that $Ann(I) = 0$, i.e. if $aI = 0$ then $a = 0$.

(4) By the Noetherian hypothesis, $I$ is finitely generated, say by $s_1, \ldots, s_n$. Let $P_i$ be the corresponding minimal prime ideals. Check that $I \prod_{i=1}^n P_i = 0$, and therefore by the previous observation, $\prod_{i=1}^n P_i = 0$.

(5) Check that whenever $A$ is a ring in which some finite product of prime ideals is equal to $0$, $A$ necessarily has only finitely many primes.

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