0
$\begingroup$

Let $R$ be a commutative ring with unity. Let $\mathcal{I} = \{I \subseteq R \mid I \text{ is an ideal of } R\}$. Call $\mathcal{C} \subseteq \mathcal{I}$ a chain if for all $I, J \in \mathcal{C}$ we have either $I \subseteq J$ or $J \subseteq I$. Let $\mathscr{C} = \{\mathcal{C} \subseteq \mathcal{I} \mid \mathcal{C} \text{ is a chain}\}$.

Consider the following property $P$ of a commutative ring with unity $R$: $\forall \mathcal{C} \in \mathscr{C}: |\mathcal{C}| < \infty$. Obviously, we have that

$P(R)$ $\Rightarrow$ $R$ is Noetherian

and

$P(R)$ $\Rightarrow$ $R$ is Artinian.

Is the property $P$ well-known? Does it have a common name? Can you give any reference?

$\endgroup$
3
  • 1
    $\begingroup$ A noetherian/artinian ring is a particular case of a noetherian/artinian module over a ring. An artinian and noetherian $R$-module is said to be of finite length. Jordan-holder theorem holds for this kind of modules (see atiyah-macdonald) and they have some other various properties (see stacks.math.columbia.edu/tag/00IU) $\endgroup$ Apr 2 at 21:16
  • 1
    $\begingroup$ This article of Wikipedia claims that a ring has finite length if and only if it's an Artinian ring. $\endgroup$
    – Crostul
    Apr 2 at 21:53
  • 1
    $\begingroup$ @Crostul Not surprising, considering (right) Artinian rings are (right) Noetherian. $\endgroup$
    – rschwieb
    Apr 3 at 2:44

2 Answers 2

1
$\begingroup$

For completeness, let me present a direct proof that if $R$ is Artinian then it satisfies the property, without using lengths of modules.

Assume by contradiction that an infinite chain $\mathcal{F}$ of ideals is given. Since an Artinian ring is Noetherian, $\mathcal{F}$ has a maximal element $I_1$. Since we are working in a chain, every $J\in\mathcal{F}$ satisfies $J\subseteq I_1$. So if $I_1$ was also a minimal element then we would have $\mathcal{F}=\{I_1\}$, contradiction to $\mathcal{F}$ being infinite. Thus, $I_1$ is not a minimal element.

Again, since $\mathcal{F}$ is Noetherian, there is some $I_2\in\mathcal{F}$ that is maximal with respect to the property that it is strictly contained in $I_1$. In particular $I_2\subsetneq I_1$, and if $J\in\mathcal{F}$ is any ideal then either $J\subseteq I_2$ or $I_2\subseteq J\subseteq I_1$, and the second option implies (using the maximality of $I_2$) that $J=I_2$ or $J=I_1$. So if $I_2$ was a minimal element of $\mathcal{F}$ then we would get $\mathcal{F}=\{I_1, I_2\}$, contradicting the assumption that $\mathcal{F}$ is infinite. Thus, $I_2$ is not minimal. So now let $I_3\in\mathcal{F}$ be maximal between the elements that are strictly contained in $I_2$, and continue this way.

Continuing this way, at the end we get an infinite descending sequence of ideals, a contradiction to $R$ being Artinian.

$\endgroup$
1
$\begingroup$

Edit: I mistook the question as asking what happens when $|\mathscr{C}| < \infty$. The question is really asking when $R$ has finite length as a module over itself, which is equivalent to saying that $R$ is Artinian (see Atiyah-MacDonald Chapters 6, 8 for a reference). I'll leave my original answer below in case anyone finds it helpful.

$|\mathscr{C}| < \infty$ is the same as saying that $R$ only contains finitely many ideals, since any ideal forms a chain by itself. As you noted, $R$ must be Artinian, and in fact we can say more (from a quick google search: https://www.universiteitleiden.nl/binaries/content/assets/science/mi/scripties/zwaanbach.pdf)

$R$ has finitely many ideals if and only if $R$ is the finite direct product of Artinian principal ideal rings and finite rings.

I can roughly sketch the argument presented in this paper for you here: first, an Artinian ring is a principal ideal ring iff $\dim_k{\mathfrak{m}/\mathfrak{m}^2} \leq 1$, and in this case every ideal is a power of the maximal ideal (see Atiyah-MacDonald 8.8). Thus, if $R$ is of the above form, it can only have finitely many ideals. Conversely, since every Artin ring is the finite direct product of Artin local rings (Atiyah-MacDonald 8.7) it suffices to show that an Artinian local ring with only finitely many ideals must be either principal or finite. Indeed, if it was neither, then the residue field $k$ would have to be infinite, with $\mathfrak{m}/\mathfrak{m}^2$ a $k$-vector space of dimension $\geq 2$, thus having infinitely many distinct $k$-submodules. Pulling back to $R$ gives you infinitely many ideals.

$\endgroup$
3
  • 1
    $\begingroup$ Why is it equivalent to $R$ having finitely many ideals? I don't see it. If there are infinitely many, how do you construct an infinite chain from them? $\endgroup$
    – Mark
    Apr 2 at 21:32
  • 1
    $\begingroup$ Oh, sorry! I thought you meant $|\mathscr{C}| < \infty$. The condition mentioned is equivalent to $R$ being Artinian. The question is really asking when $R$ has finite length as a module over itself; see Atiyah-MacDonald Ch6 and Ch8 for a reference. $\endgroup$
    – Emory Sun
    Apr 2 at 21:37
  • 2
    $\begingroup$ By the way, I'm not OP. But you are right, this is equivalent to being Artinian. $\endgroup$
    – Mark
    Apr 2 at 21:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .