1
$\begingroup$

I am studying the proof of unique factorisation of ideals into prime ideals in the ring of integers $\mathcal{O}$ of a number field $K$, and the first step is to show that given any proper ideal $I$ of $\mathcal{O}$ there exist non-zero prime ideals $P_1,\dots,P_r$ such that $I \supseteq P_1 \cdots P_r$. The proof uses the fact $\mathcal{O}$ is a noetherian ring.

I was trying to construct an example of a non-noetherian ring such that it has an ideal which does not contain a finite product of non-zero prime ideals, but I have not had any success. I did come across this relevant answer - https://math.stackexchange.com/a/89286/279515 - but I was not able to understand it.

Could someone please help me understand how the example given in that answer works? I did not want to comment directly there because the thread is four years old. My specific questions are :

  1. How do we know that $R$ has any non-trivial prime ideals? I am not able to easily see how to construct any prime ideals in $R$.

  2. If I understood correctly, in the answer it is shown that the zero ideal does not contain a product of finitely many prime ideals. However, the zero ideal is itself prime. In general, every ideal contains the zero ideal, so we have to show that an ideal does not contain a product of non-zero prime ideals, right?

  3. The ring $R$ does not have the identity, but an edit at the end fixes this issue; I am not able to see how that works, either. Some clarification on how the example needs to be modified to take into account this change would be helpful.

Thank you in advance!

EDIT1 : $R$ is not an integral domain, so the zero ideal is not prime. This entirely invalidates whatever I said in point 2.

EDIT2 : I forgot to add this question in my list :

  1. How does the last line $I = \{ 0 \} \neq \bigoplus_{j \neq i_1,\dots,i_k} e_j \mathbb{Z} \subseteq P_1,\dots,P_k$ help us conclude that $I$ does not contain a product of prime ideals?
$\endgroup$
  • 1
    $\begingroup$ Note that the ring in that question is not an integral domain, so the zero ideal is not prime. Every non trivial ring has a proper ideal (e.g the zero ideal). Zorn's lemma allows us to deduce that every ideal is contained in a maximal ideal. $\endgroup$ – Mathmo123 Aug 9 '16 at 9:31
  • $\begingroup$ @Mathmo123 Yes, I only just noticed that it is not an integral domain. $\endgroup$ – user279515 Aug 9 '16 at 9:39
  • 1
    $\begingroup$ To answer point no. 4, If $\{0\} \supseteq P_1, \dots P_k$ for some prime ideals $P_1,\dots,P_k$, then $\{0 \} = P_1,\dots,P_k$. But we have shown that $P_1,\dots,P_k \supsetneq \{ 0 \}$ always. $\endgroup$ – user279515 Aug 9 '16 at 10:37
  • 1
    $\begingroup$ This explains point no. 3 too. $\endgroup$ – user279515 Aug 9 '16 at 10:45
  • $\begingroup$ I'm not sure what's unclear to you, but I've changed that example accordingly to the answerer's edit. In my opinion the answer is pretty clear now. The product $P_1\cdots P_k$ contains an $e_n=e_n\cdots e_n$ ($k$ times) and this is not zero. $\endgroup$ – user26857 Aug 9 '16 at 14:52
1
$\begingroup$

Note: this is an elaboration of the comments above by @Mathmo123 and @user26857.


(1) We use Zorn's lemma to show the existence of prime ideals. Consider the poset (under inclusion) of proper ideals of $R$. It is non-empty because it contains the zero ideal. An application of Zorn's lemma gives us that every ideal is contained in a maximal ideal. Since maximal ideals are prime, this shows the existence of prime ideals in $R$. In fact, this argument is applicable to any ring.

(2) Yes, a ring is Noetherian if every ideal contains a product of non-zero prime ideals. But there is no problem here because the ring $R$ is not an integral domain and so $I = \{0\}$ is not a prime ideal.

(3) The same arguments fall through with either example, namely $$ \begin{align} R &= \{ (a_n)_{n \in \mathbb{N}} : a_n \in \mathbb{Z}, a_n = 0 \text{ for all but finitely many } n \}\\ \text{or} \qquad R' &= \{ (a_n)_{n \in \mathbb{N}} : a_n \in \mathbb{Z}, a_n = a_{n+1} \text{ for all sufficiently large } n \} \end{align} $$ In the second example, the element $(1,1,1,\dots)$, which as the multiplicative identity, belongs to the ring $R'$, whereas it does not belong to the first example $R$.

(4) What we have is that for each prime ideal $P_m$, $1 \leq m \leq k$, there is at most one index, say $i_m \in \mathbb{N}$ such that $e_{i_m} \not\in P_m$. Hence, we can find $j \in \mathbb{N}$ such that $e_j \in P_1,\dots,P_k$. In particular, $$0 \neq e_j = e_j \cdot e_j \cdot \cdots \cdot e_j \in P_1 \cdots P_k$$ which shows that $P_1 \cdots P_k$ is not the zero ideal. But if $P_1 \cdots P_k \subseteq \{0\}$, then $P_1 \cdots P_k = \{0\}$ because $P_1 \cdots P_k \supseteq \{0\}$. So, we have arrived at a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy