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A commutative ring $R$ is called Noetherian if every ascending chain of ideals in $R$ stabilizes, that is, $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ implies the existence of $n\in\mathbb{N}$ such that $I_n=I_{n+1}=I_{n+2}=\cdots$.

My question is the following:

Does there exist a non-Noetherian ring $R$ such that every ascending chain of primary ideals stabilizes?

Remark. Note that there exists non-Noetherian ring $R$ such that every ascending chain of prime ideals stabilizes. This happens exactly when $R$ is non-Noetherian and $\operatorname{Spec}(R)$ is Noetherian topological space. See here and Exercise 12 of Chapter 6 in Introduction to Commutative Algebra by Atiyah & Macdonald.

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I have - at least temporarily - deleted my answer, I'm not sure anymore whether it is indeed so simple, I need to think over it more, sorry. –  Daniel Fischer Jul 8 '13 at 23:22
    
@DanielFischer: No worries! Thanks for giving a thought about the problem :) –  Prism Jul 8 '13 at 23:24

2 Answers 2

up vote 4 down vote accepted

Yes, there do exist rings which aren't Noetherian but which do have ACC on primary ideals. An example is $\prod_{i\in\Bbb N} F_i$ where the $F_i$ are fields.

This is clearly not Noetherian, and because it is commutative and von Neumann regular, all of its primary ideals are maximal.

This is even more dramatic than the ACC really, since you cannot even have a chain of two primary ideals :)

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Excellent! So $\prod_{i\in\Bbb N} F_i$ is not Noetherian because we have the chain of ideals $$(1, 0, 0, 0, ...)\subseteq (1, 1, 0, 0, ...)\subseteq (1,1,1,0,...)\subseteq\cdots$$ that does not terminate, right? And also, can we always conclude that direct product of von Neumann regular rings is always von Neumann regular? Thanks :) –  Prism Jul 9 '13 at 15:31
    
@Prism Yes on both counts. Just think: if all the $F_i$ were VNR, then for each list of $a_i\in F_i$, you can find $x_i$ such that $a_ix_ia_i=a_i$. Then (the obvious product) $axa=a$ for that list $a$ in the product, too. –  rschwieb Jul 9 '13 at 15:33
    
Great! Yes, it would be the component-wise product. –  Prism Jul 9 '13 at 15:36

In Commutative Algebra a suitable construction for counterexamples is given by the idealization of an $A$-module $M$, sometimes denoted by $I_A(M)$. If one considers $R=I_{\mathbb Z}(\mathbb Q)$, then $R$ is not noetherian (since $\mathbb Q$ is not finitely generated as $\mathbb Z$-module) and it satisfies ACC on primary ideals (since these are of the following form: $p^n\mathbb Z\times\mathbb Q$ with $p$ prime and $n\ge 1$, $\{0\}\times\mathbb Q$ or $\{0\}\times\{0\}$).

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Very interesting. I don't know anything about idealization yet, so I will make sure to study the link you mention :) –  Prism Jul 9 '13 at 15:33
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@Prism There is a little cluster of constructions like this that come in handy for making ideals out of modules. The "split null" extension of two rings $R,S$ and an $R,S$ bimodule $M$ is the set of matrices of the form $\begin{bmatrix}r&m\\0&s\end{bmatrix}$ for $r\in R, s\in S, m\in M$. In the link, the "trivial extension" that YACP refers to is a subring of the case when $S=R$ of constant diagonal matrices. A third construction uses a ring $R$ and a rng $M$ and defines $(r,m)+(r',m')=(r+r',m+m')$ and $(r,m)(r',m')=(rr',rm'+r'm+mm')$. In case $M^2=0$, you again get the "trivial extension". –  rschwieb Jul 9 '13 at 16:23
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@prism The third construction is usually called the Dorroh extension. –  rschwieb Jul 9 '13 at 16:25
    
@rschwieb: Thanks for complementing the answer. Oh boy, more cool stuff to learn :) –  Prism Jul 9 '13 at 21:17

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