Algebras with supremum-founded subalgebra lattice I am interested in algebras whose subalgebra lattice is supremum-founded. Let us call those algebras small.
A complete lattice $(L, \leq)$ is called supremum-founded, if for any two elements $x < y$ from $L$ there is an element $s \in L$ which is minimal with respect to $s \leq y$, $s \not\leq x$.
Of course, every finite lattice is supremum-founded whence every finite algebra is small. Moreover, it is not very difficult to show that every $1$-locally finite algebra (that is, an algebra whose $1$-generated subalgebras are finite) is small.
Does anybody know some more/bigger classes of examples? Is there a suitable reference?
 A: I just noticed this old question.
I have a sufficient condition for smallness,
which covers all cases mentioned so far.
When the subalgebra lattice is distributive, the
condition I give is both necessary and sufficient.

Call a subalgebra $S$ of an algebra $A$
completely join irreducible or CJI
if it is a completely join irreducible element
of the lattice $L=\textrm{Sub}(A)$.
This means that $S$ has a largest proper subalgebra, $S_*$.
Lemma.
A subalgebra $S\leq A$ is CJI iff
whenever a nonempty subset $X\subseteq S$ generates $S$, then
some single $x\in X$ already generates $S$. (in particular, any CJI
subalgebra is $1$-generated.)
Reasoning.
If $S$ is CJI with lower cover $S^*$, and $X\subseteq S$
generates $S$, then $X\not\subseteq S_*$. Any $x\in X-S_*$
will generate $S$.
Conversely, if $S$ has the generation property
of the lemma statement, let $S_*$
be the subalgebra of $S$ generated by the set of
all singleton nongenerators of $S$. $S_*$ will be
the largest proper subalgebra of $S$.
\\\\\
Now consider:
Condition I. Every element of $L$ is a join
of CJI elements. (I.e., every subalgebra of $A$
is generated by the union of its CJI subalgebras.)
Condition II. $L$ satisfies DCC on its CJI elements.
($A$ satisfies DCC on its CJI subalgebras.)
Lemmas.
(1) If Conditions I and II hold, then $L=\textrm{Sub}(A)$
is supremum founded.
(2) If $L$
is supremum founded, then Condition I holds.
(3) If $L$ is distributive and 
is supremum founded, then Conditions I and II both hold.
Hence Conditions I and II characterize the property
of supremum-foundedness when $L$ is distributive.
(4) If $A$ is 1-locally finite or if all elements of $\textrm{Sub}(A)$
are dually compact, then Conditons I and II hold.
Reasoning.
For (1); If $X<Y$ in $L$, Condition I guarantees that there are some
CJI subalgebras satisfying $S\leq Y$ and $S\not\leq X$. By Condition II,
there is a minimal such CJI subalgebra, $S_0$. This subalgebra
must be minimal for the conditions $S_0\leq Y$ and $S_0\not\leq X$,
showing that $L$ is supremum founded.
For (2): Suppose that $L$ is supremum founded. Choose $Y\in L$
arbitrarily. We must show that $Y$ is generated by CJI
subalgebras. If not, let $X$ be the join of the union of the
CJI subalgebras contained in $Y$. We must have $X<Y$. Since $L$ is
supremum founded, there is a subalgebra $S\leq Y$
minimal for $S\not\leq X$. But this minimality forces
$S_*:=S\cap X$ to be the largest proper subalgebra of $S$.
This implies $S$ is CJI,
contradicting the definition of $X$.
For (3): (sketch only) Assume that $L=\textrm{Sub}(A)$
is supremum founded and distributive. By (2),
Condition I holds.
Now let $P$ be the subposet of CJI
elements of $L$ ordered as in $L$ (by inclusion).
By assuming that $L$ is algebraic we get that
CJI elements are compact.
By assuming that $L$ is distributive,
we then derive that
CJI elements are completely join prime in $L$.
By Condition I, the mapping from $L$ to the 
lattice of order ideals of $P$ which sends $X\in L$ to the set of
CJI elements below it is an isomorphism between $L$
and the lattice of order ideals of $P$.
I claim that Condition II must hold.
This claim is the assertion that $P$ satisfies DCC.
If this is not the case, let $S_1>S_2>\cdots$ be a
strictly descending chain in $P$. Let $F$ be the order filter
generated by $\{S_i\}$, and let $I$ be the complementary
order ideal.
(It is the largest order ideal containing none of the
$S_i$.) Let $X=\bigvee I$ be the element of $L$ that
corresponds to the order ideal $I$. $X$
is the largest element of $L$
containing none of the $S_i$'s. Let $Y=A$ = the top of $L$.
We have $X<Y$, so since $L$ is supremum founded there should
be some minimal $S\leq Y$ such that $S\not\leq X$. This $S$
must contain some $S_i$, hence properly contain $S_{i+1}$,
hence $S$ is not minimal after all. This contradiction shows that
Conditions I and II determine supremum-foundedness for distributive
$L$.
For (4): Just check. \\\\\

To reiterate an interesting point from the proof of (3); An algebraic lattice is supremum founded and distributive iff it is isomorphic to the ideal lattice of a poset with DCC.

Example.
Let $R$ be an infinite left Artinian ring, and let $M$
be the $\omega$-generated free left $R$-module.
Then $M$ is not 1-locally finite (since $R$ is infinite)
and $\textrm{Sub}(M)$ does not have the property
that every element is dually compact (since this lattice
does not satisfy DCC). But $M$ does satisfy Conditions I and II,
so $\textrm{Sub}(M)$ is supremum founded.
