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It is well known that for every exact sequence $$0 → M' → M → M'' → 0$$ of modules over some ring, $M$ is Noetherian / Artinian if and only if both $M'$ and $M''$ are.

If the arrows in such a sequence are given by $α \colon M' → M$ and $β \colon M → M''$, then the submodule lattices of $M'$ and $M''$ imbed to the submodule lattice of $M$ by $α$ and $β^{-1}$. These two sublattices are the lattices given by all submodules between $0$ and $\operatorname{img} α$ and all submodules between $\ker β$ and $M$ respectively. By the exactness of the sequence, these two lattices touch at $\operatorname{img} α = \ker β$.

This suggests that we might prove the result by abstracting it to lattices. However, the naïve generalisation is not true, which would be:

Let $L = (L, ≤, ∧, ∨, ⊥, ⊤)$ be a lattice with top $⊤$ and bottom $⊥$ and let $x ∈ L$. Then $L$ is Noetherian / Artinian if and only if both sublattices $$↓x = \{z ∈ L;~z ≤ x\} \quad\text{and}\quad ↑x = \{z ∈ L;~z ≥ x\}$$ are Noetherian / Artinian.

As a counterexample, I imagine adjoining to $ℤ$, with the usual order, elements $-∞, \star, ∞$ with $-∞ < ℤ < ∞$ and $-∞ < \star < ∞$. This should yield a lattice $L$ for which $↓\star$ and $↑\star$ are both Noetherian and Artinian (being finite), but $L$, containing $ℤ$ as a sublattice, is neither.

So my question is:

Can we demand some extra structure or properties of $L$ or $x$ such that the statement becomes true?

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  • $\begingroup$ Infectiousness...? $\endgroup$ – Giuseppe Negro Nov 21 '17 at 16:51
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    $\begingroup$ @GiuseppeNegro : ) I like figurative language. How would you say that? Feel free to improve the rather lengthy title if you can manage to do so without losing its meaning. $\endgroup$ – k.stm Nov 21 '17 at 16:56
  • $\begingroup$ In the module situation, if $N$ is a submodule of $M$, then $N$ is between an element of $\downarrow M'$ and $\uparrow M'$, precisely $M'\cap N$ and $M'+N$. This doesn't happen in your counterexample $\endgroup$ – egreg Nov 21 '17 at 17:03
  • $\begingroup$ @egreg What to you mean by precisely? In my counterexample, for $n ∈ ℤ$, $-∞ < n < ∞$ and $-∞ ∈↓n$ and $∞ ∈↑n$. $\endgroup$ – k.stm Nov 21 '17 at 17:06
  • $\begingroup$ @k.stm At least one among $M'\cap N$ and $M'+N$ is not top or bottom $\endgroup$ – egreg Nov 21 '17 at 17:13
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What makes the module example work is that any lattice of submodules is modular. If you assume that $L$ is a modular lattice, $x\in L$, and both $\downarrow x$ and $\uparrow x$ satisfy the ACC/DCC, then $L$ also satisfies ACC/DCC.

In fact, it can be shown that every equationally definable class of lattices is either modular or contains your counterxample. Thus, the class of modular lattices is the largest equationally definable class $K$ of lattices where, whenever $L\in K$, one has that for each $x\in L$, if $\downarrow x$ and $\uparrow x$ satisfy the ACC/DCC, then $L$ also satisfies ACC/DCC.


Can you elaborate on why modularity suffices or give a reference for this?

I will explain why any counterexample must be nonmodular.

Let $L$ be a lattice, and suppose that $x\in L$ has the properties that $\downarrow x$ and $\uparrow x$ satisfy the ACC. Suppose also that somewhere in $L$ there is an infinite strictly increasing chain $y_1<y_2<\cdots$. Then $$ x\wedge y_1\leq x\wedge y_2\leq \cdots$$ is a weakly increasing chain (a $\leq$-chain) below $x$, and $$ x\vee y_1\leq x\vee y_2\leq \cdots$$ is a weakly increasing chain above $x$. The intervals below and above $x$ have ACC, so for some $n_0$ it is the case that for every $n\geq n_0$ we have $x\wedge y_n=x\wedge y_{n+1}$ and $x\vee y_n=x\vee y_{n+1}$. But now the elements $x, y_{n_0}, y_{n_0+1}$ witness a failure of the modular law. Specifically $y_{n_0} < y_{n_0+1}$, but $y_{n_0+1}\wedge (x\vee y_{n_0})\neq (y_{n_0+1}\wedge x)\vee (y_{n_0+1}\wedge y_{n_0})$, since the left hand side is $y_{n_0+1}$ and the right hand side is $y_{n_0}$.

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  • $\begingroup$ Can you elaborate on why modularity suffices or give a reference for this? $\endgroup$ – k.stm Nov 21 '17 at 21:48

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