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\begin{document}
\title[Two Statements about Infinite Products that Are Not Quite True]
{Two Statements about Infinite Products\\
that Are Not Quite True}
\keywords{infinite product module, inverse limit module, dichotomy of
finite or uncountable generation, homomorphism to infinite direct sum,
left perfect ring; infinite symmetric group, finite generation of
power or ultrapower of an algebra over its diagonal subalgebra.}
\author{George M. Bergman}
\dedicatory{To Don Passman, on his 65th birthday}
\address{Dept. of Mathematics\\
University of California\\
Berkeley, CA 94720-3840, USA}
\email{gbergman@math.berkeley.edu}
\subjclass[2000]{Primary: 08B25, 16D70.
% *P&& mdls+ids:str+cl
Secondary: 03C20, 16P70, 16S50, 20B30, 20M20, 22A05}
% ultrapr CC.other end_rgs symgps tfnsmgps topgps
\date{}
% \begin{abstract}
% \end{abstract}
% \thanks{{\em URLs of this preprint.} %%
% http://math.berkeley.edu/{$\!\sim$}gbergman%
% /papers/P\_\P}
There is no surjective left $\!R\!$-module homomorphism
$\bigoplus_\omega R\rightarrow R\<^\omega.$
\end{xlist}
%
\begin{xlist}\item\label{x.P!->cP}
There is no surjective left $\!R\!$-module homomorphism
$R\<^\omega\to\bigoplus_\omega R.$
\end{xlist}
In \S\ref{S.cegs} we will note classes of rings $R$ for
which each of these statements fails.
In \S\S\ref{S.diag}-\ref{S.top}, however, we will see
that~(\ref{x.cP!->P}) holds, i.e., $R\<^\omega$ requires
uncountably many generators as a left $\!R\!$-module, unless
$R\<^\omega$ is {\em finitely} generated, and that~(\ref{x.P!->cP})
holds unless $R$ has descending chain condition on finitely
generated right ideals.
From the above assertion regarding~(\ref{x.cP!->P}),
and the statement of~(\ref{x.P!->cP}), it is
easy to see that for every $R,$ at least one
of~(\ref{x.cP!->P}),~(\ref{x.P!->cP}) holds.
We shall also see that the above restriction on
rings for which~(\ref{x.P!->cP}) fails implies that for every
$R,$ either~(\ref{x.P!->cP}) or the statement
%
\begin{xlist}\item\label{x.P!c>cP}
There is no embedding of left $\!R\!$-modules
$R\<^\omega \rightarrow \bigoplus_\omega R.$
\end{xlist}
%
holds.
(I did not count~(\ref{x.P!c>cP}) among the ``not quite true''
statements of the title, because it does not appear
that it is ``nearly'' true; i.e.,
that its failure implies strong restrictions on $R.)$
The result asserted above in connection with~(\ref{x.cP!->P})
will in fact be proved with $R\<^\omega$ replaced by $M^\omega$
for $M$ any $\!R\!$-module, while
the result on~(\ref{x.P!->cP}) will be obtained with $R\<^\omega$
generalized to the inverse limit of any countable inverse system
of finitely generated $\!R\!$-modules and surjective homomorphisms.
In~\S\ref{S.gen} we change gears: We will note that our proof of the
result about~(\ref{x.cP!->P}) generalizes to the context of general
algebra (a.k.a.\ `universal algebra'), and in \S\ref{S.Sym}, we
deduce from this that if $A$ is an algebra such
that $A^\omega$ is countably generated over the diagonal
image of $A,$ then it is finitely generated over that image.
We will then show that the monoid (respectively, the group) of all maps
(respectively, invertible maps) of an infinite set $\Omega$
into itself has this finite generation property, and will obtain
general results on algebras with this and related properties.
I am grateful to P.\,M.\,Cohn, T.\,Y.\,Lam, A.\,W.\,Miller,
B.\,Osofsky and G.\,Sabbagh for pointing me to relevant literature,
to the referee and Zak Mesyan for helpful proofreading,
to H.\,W.\,Lenstra,~Jr., A.\,Blass, and T.\,Scanlon for contributions
that will be noted below, and to Hans Lewy (1904--1988) for assigning,
in an undergraduate problems-course I took around 1961, a problem
equivalent to showing that the partially ordered set of sequences
of natural numbers has uncountable cofinality, which was the source
of the method of Lemma~\ref{L.Pinfg} below.
\section{Counterexamples.}\label{S.cegs}
It is easy to give rings for which statement~(\ref{x.P!->cP}) fails:
If $R$ is a division ring, then $R\<^\omega$ is
infinite-dimensional as a left $\!R\!$-vector-space, hence admits a
homomorphism onto $\bigoplus_\omega R.$
More generally, if $R$ is a quasi-Frobenius ring, then
the submodule $\bigoplus_\omega R \subset R\<^\omega,$ being free,
is injective~\cite[first paragraph]{CF}, so
$R\<^\omega$ can be retracted onto it, and again~(\ref{x.P!->cP}) fails.
The result to be proved in \S\ref{S.top}, that for~(\ref{x.P!->cP})
to fail $R$ must have descending chain condition on finitely
generated right ideals, shows that examples of failure
of~(\ref{x.P!->cP}) are all fairly close to these.
Counterexamples to~(\ref{x.cP!->P}) are less evident.
To construct these, we begin by noting that if $M$
is a left module over any ring $K$ and $(N_i)_{i\in I}$ any family
of such modules, then under the standard convention
(cf.~\cite{GMB.rl}) that homomorphisms of left modules are written
on the right of their arguments and composed accordingly, we have
%
\begin{xlist}\item\label{x.HomO+,}
${\rm Hom}_K(\bigoplus_I N_i,\,M)\,\cong\,\prod_I {\rm Hom}_K(N_i,\,M)$
as {\em right} $\!{\rm End}_K(M)\!$-modules,
\end{xlist}
%
and
%
\begin{xlist}\item\label{x.Hom,*P}
${\rm Hom}_K(M,\ \prod_I N_i)\,\cong\,\prod_I {\rm Hom}_K(M,\,N_i)$
as {\em left} $\!{\rm End}_K(M)\!$-modules.
\end{xlist}
%
Indeed, the bijective correspondences follow from the universal
properties of the direct sum in~(\ref{x.HomO+,}) and the
direct product in~(\ref{x.Hom,*P}); it remains only to note that
$M$ is a right $\!{\rm End}_K(M)\!$-module, and that this module
structure carries over to the hom-sets of~(\ref{x.HomO+,}), while it is
turned by contravariance into left module structures on the
hom-sets of~(\ref{x.Hom,*P}).
We can now get our examples.
\begin{lemma}\label{L.R=R^*k}
Let $K$ be a ring and $\kappa$ a cardinal.
Then if $M$ is either a {\em right} $\!K\!$-module which satisfies
%
\begin{xlist}\item\label{x.M=(+)M}
$M~\cong~\bigoplus_\kappa M$
\end{xlist}
{\rm (}for instance, if $\kappa$ is infinite and $M$ a right
$\!K\!$-module of the form $\bigoplus_\kappa N),$
or a {\em left} $\!K\!$-module which satisfies
%
\begin{xlist}\item\label{x.M=M^}
$M~\cong~M^\kappa$
\end{xlist}
%
{\rm (}for instance, if $\kappa$ is infinite and $M$ a left
$\!K\!$-module of the form $N^\kappa),$
and if, in either case, we take $R~=~{\rm End}_K(M),$ then
%
\begin{xlist}\item\label{x.R=R^*k}
$R~\cong~R^\kappa$ as {\em left} $\!R\!$-modules.
\end{xlist}
%
\end{lemma}\begin{proof}
In~(\ref{x.HomO+,}) (with left and right interchanged),
respectively~(\ref{x.Hom,*P}) (as it stands), take $I=\kappa,$ put
the given module $M$ in the role of both the $M$ and all the~$N_i,$
and simplify the left hand side using~(\ref{x.M=(+)M}),
respectively~(\ref{x.M=M^}).
\end{proof}
A statement which embraces both cases of Lemma~\ref{L.R=R^*k}
is that if $\mathcal C$ is an $\!\mathcal{A}b\<\!$-cat\-e\-gory
with an object $M$ that is a $\!\kappa\!$-fold coproduct or
product of itself, then its endomorphism ring in $\mathcal C$ (or the
opposite of that ring, depending on which of these cases one is in,
and one's choice of how morphisms are composed in $\mathcal C)$
satisfies~(\ref{x.R=R^*k}).\vspace{6pt}
Clearly, when~(\ref{x.R=R^*k}) holds,~(\ref{x.cP!->P})
and~(\ref{x.P!c>cP}) fail.\vspace{6pt}
Let me say here how this subject came up.
H.\,W.\,Lenstra~Jr., in connection with a course he was teaching
in Leiden, e-mailed me asking whether, for a nontrivial ring $R,$
one could have
%
\begin{xlist}\item\label{x.hwl=?}
$R\<^\omega~\cong~\bigoplus_\omega R$
\end{xlist}
%
as left $\!R\!$-modules.
I replied that this was impossible because $R\<^\omega$ could
not be countably generated.
He pointed out the first case of Lemma~\ref{L.R=R^*k} (for $M$ an
abelian group, i.e., $K=\mathbb Z),$ which shows the contrary.
I then thought further and found the argument of the next section,
showing that~(\ref{x.hwl=?}) nonetheless cannot occur.
I have not been able to find in the literature any occurrence of
the idea of Lemma~\ref{L.R=R^*k} for infinite $\kappa,$
so it appears that the result belongs to Lenstra.
However, two less elementary examples of similar phenomena
were proven earlier.
J.\,D.\,O'Neill \cite{JO'Nrg} constructs for any $\kappa>1$ a ring $R$
without zero-divisors such that $R^\kappa\cong R^2$ as left modules,
and as noted in the MR review of that paper, W.\,Stephenson
showed in~\cite{WSt} that for any non-right-Ore ring $S$
without zero divisors, the right quotient ring $R$ of $S$
satisfies~(\ref{x.R=R^*k}) for any cardinal $\kappa$ such that
$S$ has $\geq\kappa$ right linearly independent elements.
That result is, in fact, an instance of the generalization
noted immediately after the proof of Lemma~\ref{L.R=R^*k} above.
For the right quotient ring of $S$ is the endomorphism
ring of $S$ in the $\!\mathcal{A}b\<\!$-category
$\mathcal C$ whose objects are right $\!S\!$-modules, but where
$\mathcal C(M,N)$ is the set of morphisms from essential
submodules of $M$ into $N,$ modulo the equivalence relation that
identifies morphisms which agree on essential submodules; and for
a ring $S$ without zero-divisors and having
$\geq\kappa$ right linearly independent elements,
the free right $\!S\!$-module on one generator has an essential
submodule consisting of a direct sum of $\geq\kappa$ copies of~$S,$
leading to an isomorphism~(\ref{x.M=(+)M}) in that category.
As suggested in the introduction, further counterexamples
to~(\ref{x.P!c>cP}) are not hard to come by.
For instance, if, starting with any ring $K,$
we form the noncommuting formal power series ring $R$
in a $\!\kappa\!$-tuple of indeterminates, with no
restriction on the number of monomials allowed in each degree, then
its ideal of elements with constant term zero is isomorphic as a
left module to $R^\kappa,$ so $R^\kappa$ embeds in $R.$
Since examples where~(\ref{x.cP!->P}) fails are harder to
find, I will record here a slight extension of the class of
examples given by Lemma~\ref{L.R=R^*k}.
First note that any ring $R$ having a module
isomorphism~(\ref{x.R=R^*k}), e.g., a ring as in that lemma,
will contain elements $f_i$ $(i\in\kappa)$
such that the isomorphism is given by
%
\begin{xlist}\item\label{x.R->R^*k}
$r~\mapsto~(r f_i)_{i\in\kappa}.$
\end{xlist}
%
(For example, in a case arising from a right module
isomorphism~(\ref{x.M=(+)M}), $(f_i)_{i\in\kappa}$
can be any family of one-to-one endomorphisms of
$M$ such that $M=\bigoplus_\kappa f_i(M).$
In general, the $f_i$ will be the components of the image
under~(\ref{x.R=R^*k}) of $1\in R.)$
But the surjectivity of~(\ref{x.R->R^*k}) is preserved
on replacing $R$ by any homomorphic image; so such a homomorphic
image will again be a ring for which~(\ref{x.cP!->P}) fails.
To get rings $R$ as in Lemma~\ref{L.R=R^*k} which have ideals $I$ such
that $R/I$ does not (so far as I know) itself satisfy~(\ref{x.R=R^*k}),
one can (a)~let $K$ be a field, $R$ the endomorphism ring of an
infinite-dimensional $\!K\!$-vector-space $V,$ and $I$ the ideal of
finite-rank endomorphisms of $V,$ or (b)~let $K$ be a ring having
a non-finitely-generated $\!2\!$-sided ideal $I_0,$ let $R=
{\rm End}_K(\bigoplus_\kappa K),$ and let $I$ be the ideal of $R$
generated by the diagonal image of $I_0.$
Because $I_0$ is non-finitely-generated, $I$ is generally smaller than
the kernel of the natural map ${\rm End}_K(\bigoplus_\kappa K)
\rightarrow {\rm End}_{K/I_0}(\bigoplus_\kappa K/I_0)$
(it does not contain endomorphisms whose components all lie in
$I_0,$ but do not all lie in any finitely generated subideal of $I_0),$
so $R/I$ is not simply the latter ring, which would just be
another example of the construction of Lemma~\ref{L.R=R^*k}.
It is interesting to note that a ring $R$ can
satisfy~(\ref{x.R=R^*k}) simultaneously as a right and as a left
$\!R\!$-module (though necessarily by different bijections).
Namely, if $K$ is a division ring and $\kappa$ an infinite
cardinal, then the left vector-space $M=K^\kappa$ not
only satisfies $M^\kappa\cong M,$ but also,
being a $\!K\!$-vector-space of dimension at least $\kappa,$ satisfies
$\bigoplus_\kappa M\cong M;$ hence combining Lemma~\ref{L.R=R^*k}
and its dual, we get the desired right and left module isomorphisms.
We now turn to results showing that despite these instances
of~(\ref{x.R=R^*k}), no nontrivial ring satisfies~(\ref{x.hwl=?}).
\section{A diagonal argument.}\label{S.diag}
In this section, we shall restrict ourselves, for simplicity of
presentation, to the case $\kappa=\omega.$
(In~\S\ref{S.gen}, in addition to passing from module theory
to general algebra, we will give the corresponding results
for a general infinite cardinal $\kappa.)$
The hypothesis of the next lemma may seem irrelevant to the
question at hand, but appearances are deceiving.
%
\begin{lemma}\label{L.Pinfg}
Let $R$ be a ring and $(M_i)_{i\in\omega}$ a family of
non-finitely-generated left $\!R\!$-modules.
Then $\prod_{i\in\omega}M_i$ is not countably generated.
\end{lemma}\begin{proof}
It will suffice to show that for any countable family of elements
$x_j\in\prod_{i\in\omega} M_i$ $(j\in\omega),$ we can construct
an element $y$ not in the submodule generated by the $x_j.$
We do this by a diagonal construction: For each $i\in\omega,$
the assumption that $M_i$ is not finitely generated allows us
to take as the $\!i\!$-component of $y$ an element of $M_i$ not
in the span of the $\!i\!$-components of $x_0,\dots,x_i.$
If $y$ were in the span of all the $x$'s, it would be in the
span of finitely many of them, say $x_0,\dots,x_i.$
But looking at its $\!i\!$-component, we get a contradiction.
\end{proof}
The relevance of that lemma is seen in the proof of
\begin{theorem}\label{T.fgorunc}
Let $M$ be a left module over a ring $R.$
Then the left $\!R\!$-module $M^\omega$ is either finitely generated,
or not countably generated.
\end{theorem}\begin{proof}
If $M^\omega$ is not finitely generated, then the above
lemma shows that $(M^\omega)^\omega$ is not countably generated.
But $(M^\omega)^\omega\cong M^{\omega\times \omega}\cong M^\omega,$
so $M^\omega$ is non-countably-generated, as claimed.
\end{proof}
So Lenstra's question is answered:
\begin{corollary}\label{C.hwl_ans1}
No nonzero ring $R$ can satisfy $R\<^\omega\cong\,\bigoplus_\omega R$
as left $\!R\!$-modules.
\end{corollary}\begin{proof}
$\bigoplus_\omega R$ is neither finitely generated nor
non-countably-generated, while we have seen that $R\<^\omega$
must have one of these properties.
\end{proof}
In the context of Theorem~\ref{T.fgorunc}, if $M^\omega$ is not
finitely generated, it is natural to ask whether the least cardinality
of a generating set must be $\geq 2^{\aleph_0}.$
Under some conditions, this is indeed true.
Trivially, it is so if we assume the Continuum Hypothesis.
Almost as trivially, writing $|X|$ for the cardinality of a
set $X,$ we see that if $|R\<|<|M|^{\aleph_0},$
in particular if $|R\<|<2^{\aleph_0},$ then, even without an explicit
assumption that $M^\omega$ is not finitely generated, as long
as $M$ is nonzero, a generating set for $M^\omega$ must have cardinality
$|M|^{\aleph_0},$ since a module of infinite cardinality cannot be
generated by a subset of smaller cardinality
over a ring also having smaller cardinality.
Without any restriction on $|R\<|,$ the module $M^\omega$
will {\em contain} a direct sum of $2^{\aleph_0}$ copies of $M.$
This can be seen either by applying
\cite[Corollary~3.5]{ZB} to the discrete topology
on $\omega,$ or, alternatively, as follows:
Note that $M^\omega\cong M^\mathbb Q$ where $\mathbb Q$ is the set
of rational numbers.
Associate to every real number
$r$ the map $f_r: M\rightarrow M^\mathbb Q$ carrying each
$x\in M$ to the function on $\mathbb Q$ which has the
value $x$ at all rational numbers $qP})
and~(\ref{x.P!->cP}) cannot fail simultaneously.
He then raised the question of whether for some $R$
the module $R\<^\omega$ could admit both injective and surjective
maps to $\bigoplus_\omega R;$ i.e., whether~(\ref{x.P!->cP})
and~(\ref{x.P!c>cP}) could both fail.
This is also impossible.
I know now that this fact can be obtained without too much work
from a result of S.\,U.\,Chase on homomorphisms from direct product
modules to direct sums~\cite[Theorem~1.2]{SUC3}.
% because (DCC on principal => DCC on fg) MR 40\#1419 (2)
However, I will prove some stronger statements, including a
generalization of Chase's result, Theorem~\ref{T.MNB-Chase} below
-- essentially, the convex hull of his theorem and the similar
result that I had obtained before learning of~\cite{SUC3}.
Before leaping into the proof, let us note that a general tool
in proving restrictions on homomorphisms $f$ out of an infinite
product module $M=\prod_i M_i$
is to assume those restrictions fail, and
construct an element $x\in M$ by specifying its values
on successive coordinates, in such a way that the properties of $f(x)$
would lead to a contradiction.
To ``control'' the effects of these successive specifications, one
generally takes the $\!i\!$th coordinate to lie in an additive subgroup
$I_i\,M_i\subseteq M_i,$ using smaller and smaller right ideals
$I_i\subseteq R,$ and calling (explicitly or implicitly) on the fact
that if a right ideal $I$ is finitely generated, and we modify an
arbitrary set of coordinates $x_i$ of $x$ by elements of $I\k_{i-1}$ giving a proper inclusion
%
\begin{xlist}\item\label{x.smaller}
$I_i\,q_i(N_{k_i})~\subset\ I_{i-1}\,q_i(N_{k_{i-1}}).$
\end{xlist}
Once these choices have been made for all $i,$ Lemma~\ref{L.IN} tells us
that $M$ is complete in the $\!(I_i N_{k_i})\!$-adic topology, hence by
Lemma~\ref{L.Baire}, there is some $j\in\omega$ such that the closure of
$B_j$ in that topology is also open.
Thus for some $i_0\in\omega,$ that closure contains
$I_{i_0} N_{k_{i_0}},$ which is easily seen to imply that
%
\begin{xlist}\item\label{x.allIN}
$B_j+I_i\,N_{k_i}~\supseteq\ I_{i_0} N_{k_{i_0}}$ for all $i\in\omega.$
\end{xlist}
%
This means that when $i\geq i_0,$ taking larger values of $i$
(and hence smaller subgroups $I_i\,N_{k_i})$
does not decrease the left-hand side of~(\ref{x.allIN}).
Also, note that~(\ref{x.allIN}), and hence
that conclusion, is preserved on increasing $j.$
We easily deduce
%
\begin{xlist}\item[]
$B_{j'} + I_i\,N_k~=\ B_{j'} + I_{i'}\,N_{k'}$
for all $i,i'\geq i_0,\ j'\geq j,\ k,k'\geq k_{i_0}$
\end{xlist}
%
or as an equation in $M/B_{j'},$
%
\begin{xlist}\item[]
$I_i\,q_{j'}(N_k)~=\ I_{i'}\,q_{j'}(N_{k'})$
for all $i,i'\geq i_0,\ j'\geq j,\ k,k'\geq k_{i_0}.$
\end{xlist}
%
But the strict inclusion~(\ref{x.smaller}), for any
$i\geq\mathrm{max}(j,i_0{+}1),$ contradicts this equality,
completing the proof.
\end{proof}
The conclusion of the above theorem is rather complicated, with
quantification over
$\mathcal F,$ $(B_j)_{j\in\omega}$ and $(N_k)_{k\in\omega}.$
One can get a statement involving only the first two of these
if one assumes that $M$
is an inverse limit of {\em finitely generated} $\!R\!$-modules $M_i.$
%
\begin{corollary}\label{C.fg}
Suppose in the situation of Theorem~\ref{T.MNB} that the $\!R\!$-modules
$M_i=M/N_i$ of~{\rm (\ref{x.invsys})} are finitely generated.
Then there exists $j^*\in\omega$ such that
%
\begin{xlist}\item\label{x.IM/B}
the set of additive subgroups $\{I\,(M/B_{j^*})\ |\ I\in\mathcal F\}$
of $M/B_{j^*}$ has a least member.
\end{xlist}
{\rm (}This will also be the least member of the larger family
of additive subgroups described as in~{\rm (\ref{x.IN/B}).)}
\end{corollary}\begin{proof}
Let $j^*,$ $I^*,$ $k^*$ be as in Theorem~\ref{T.MNB}, and
note that the property asserted by that theorem is preserved under
increasing $j^*$ while leaving $I^*$ and $k^*$ unchanged.
Let $M/N_{k^*}$ be generated by the image
of a finite set $T\subseteq M.$
Since $\{B_j\}$ has union $M,$ we can assume
by increasing $j^*$ if necessary that $T\subseteq B_{j^*}.$
This says that the image of $B_{j^*}$ in $M/N_{k^*}$ is all of
$M/N_{k^*},$ equivalently, that $M=B_{j^*} + N_{k^*},$ equivalently,
that the image of $N_{k^*}$ in $M/B_{j^*}$ is all of $M/B_{j^*}.$
Substituting $M/B_{j*}$ for $q_{j*}(N_{k*})$ in the final assertion
of Theorem~\ref{T.MNB} (with $k=k*)$ gives the desired conclusion.
\end{proof}
We can now prove our earlier
assertion about maps onto $\bigoplus_\omega R.$
\begin{corollary}\label{C.DCC}
Suppose $R$ is a ring such that the free left $\!R\!$-module
$\bigoplus_\omega R$ can be written as a homomorphic image of the
inverse limit $M$ of an inversely directed system~{\rm (\ref{x.invsys})}
of finitely generated left $\!R\!$-modules and surjective homomorphisms.
Then $R$ is left perfect, i.e., has descending chain condition
on finitely generated right ideals.
In particular, any ring $R$ for which
there exists a surjective left $\!R\!$-module
homomorphism $R\<^\omega\rightarrow\bigoplus_\omega R,$
i.e., for which {\rm (\ref{x.P!->cP})} fails, is left perfect.
\end{corollary}\begin{proof}
Given $M$ as above
and a surjective homomorphism $f:M\rightarrow \bigoplus_\omega R,$
let $B_i=f^{-1}(\bigoplus_0 ^{i-1}\,R)$ $(i\in\omega).$
Then $M=\bigcup B_i;$ moreover, each factor module $M/B_i$ is isomorphic
to $\bigoplus_{j\geq i} R,$ again a free left module of countable
(hence nonzero) rank,
so for every $i\in\omega,$ distinct right ideals
$I$ yield distinct abelian groups $I(M/B_i).$
The conclusion of Corollary~\ref{C.fg} therefore tells us
that every downward directed system of finitely generated right ideals
of $R$ has a least member.
Applied to chains of ideals, this says that $R$ has
DCC on finitely generated right ideals, as claimed.
The final assertion is clear.
\end{proof}
We noted at the beginning of \S\ref{S.cegs} that~(\ref{x.P!->cP})
always failed when $R$ was a quasi-Frobenius ring;
so the class of rings for which it fails lies between the
classes of quasi-Frobenius and left perfect rings.
It would be of interest to characterize this class more precisely.
H.\,Lenzing \cite[Proposition~2, in left-right dual form]{Lenzing}
shows that if $\bigoplus_{i\in\omega} R\subseteq R^\omega$
is a direct summand as left modules, then $R$ is semiprimary
with ascending chain condition on left annihilator ideals.
(Whether the converse holds is left open
\cite[sentence beginning at bottom of p.687]{Lenzing}.)
The failure of our condition~(\ref{x.P!->cP}) is equivalent to the
statement that $\bigoplus_{i\in\omega} R$ can be embedded in {\em some}
way as a direct summand in $R^\omega;$ so it is natural to ask
\begin{question}\label{Q.summand}
If $\bigoplus_{i\in\omega} R$ has some embedding as a direct summand
in $R^\omega,$ will the {\em canonical} copy
of $\bigoplus_{i\in\omega} R$ in $R^\omega$ be a direct summand?
\end{question}
Getting back to our original
goal, we can now combine Corollary~\ref{C.DCC} with known results to
obtain the result Lenstra asked about.
\begin{corollary}\label{C.hwl_ans2}
For a nonzero ring $R,$ statements~{\rm (\ref{x.P!->cP})}
and~{\rm (\ref{x.P!c>cP})} cannot fail simultaneously;
i.e., $R\<^\omega$ cannot be both mappable onto and
embeddable in $\bigoplus_\omega R.$
\end{corollary}\begin{proof}
By the observations in the third paragraph
following Corollary~\ref{C.hwl_ans1}, a power module
$R\<^\omega$ always contains a direct sum of $2^{\aleph_0}$ copies
of $R,$ hence if $R\<^\omega$ is embeddable in $\bigoplus_\omega R,$
that module will contain
a set $X$ of $2^{\aleph_0}$ left linearly independent elements.
Writing $\bigoplus_\omega R$ as the union of the countable
ascending chain of submodules $\bigoplus_0 ^{i-1} R,$ we conclude that
for some $i,$ the latter submodule will contain uncountably many
% had $2^{\aleph_0}.$ "PlanetMath": cof(c) uncountable, so "OK" but...
members of $X;$ so in particular it contains $i\,{+}\,1$ members
of $X;$ in other words, a free left $\!R\!$-module
of rank $i$ contains a free submodule of rank $i\,{+}\,1.$
This leads to an $i{+}1\times i$ matrix over $R$ which is left
regular (has zero left annihilator), hence
to an $i{+}1\times i{+}1$ matrix $A$ with left linearly
independent rows, but with a zero column.
Thus, in the $i{+}1\times i{+}1$ matrix ring
over $R,$ the element $A$ is left but not right regular.
On the other hand, if $R\<^\omega$ is mappable onto
$\bigoplus_\omega R,$ then by the preceding corollary, $R$ is left
perfect, hence so is its $i{+}1\times i{+}1$ matrix ring.
A generalization by Lam \cite[Exercise~21.23]{Lam} of a result of
Asano's says that the left regular elements, the right regular elements
and the invertible elements in such a ring coincide, so an element $A$
cannot be left but not right regular.
Hence the two conditions on $R$ are incompatible.
\end{proof}
Returning to Theorem~\ref{T.MNB}, let me now obtain from it a version
with the conclusion in a form closer to Chase's formulation.
Note that the hypothesis below is as in Theorem~\ref{T.MNB}, except that
instead of a countable ascending chain of submodules
$(B_i)_{i\in\omega},$ we assume given a
homomorphism from $M$ into a module $\bigoplus_{\alpha\in J} C_\alpha,$
with no restriction on the cardinality of the index set~$J.$
\begin{theorem}
[{\rm after Chase \cite[Theorem~1.2]{SUC3}}\textbf{}]\label{T.MNB-Chase}
Let $R$ be a ring and $M$ a left $\!R\!$-module which has
a descending chain of submodules~{\rm (\ref{x.Ni})}; equivalently,
which is the inverse limit of a system~{\rm (\ref{x.invsys})} of
modules and surjective homomorphisms, and let $\mathcal F$ be
a downward directed system of right ideals of $R.$
Suppose we are given a family $(C_\alpha)_{\alpha\in J}$ of left
$\!R\!$-modules, and
a homomorphism $f: M\rightarrow \bigoplus_{\alpha\in J}\,C_\alpha.$
For each $\beta\in J$ let $\pi_\beta: \bigoplus_{\alpha\in J}\,C_\alpha
\rightarrow C_\beta$ denote the $\!\beta\!$th projection map.
Then there exist $k^*\in\omega,$ a finite subset $J_0\subseteq J,$
and an $I^*\in\mathcal F$ such that
%
\begin{xlist}\item\label{x.pfN=cap}
$I^*\,\pi_\beta(f(N_{k^*}))~\subseteq\ \bigcap_{I\in\mathcal F}\,
I\,C_\beta$ for all $\beta\in J-J_0.$
\end{xlist}
\end{theorem}\begin{proof}
Assuming the contrary, let us construct recursively
a chain of ideals $I_0\supseteq I_1\supseteq\dots$ in $\mathcal F$ and
a sequence of indices $\alpha_0,\,\alpha_1,\dots\in J,$ starting
with arbitrary $I_0\in\mathcal F$ and $\alpha_0\in J.$
Say $I_0\supseteq\dots\supseteq I_{i-1}$
and $\alpha_0,\dots,\alpha_{i-1}$ have been chosen for some $i>0.$
By assumption, the choices
$I^*=I_{i-1},\ k^*=i,\ J_0=\{\alpha_0,\dots,\alpha_{i-1}\}$ do
not satisfy~(\ref{x.pfN=cap}), hence we can choose some $\alpha_i\not\in
\{\alpha_0,\dots,\alpha_{i-1}\}$
such that $I_{i-1} \,\pi_{\alpha_i}(f(N_i))\not\subseteq
\bigcap_{I\in\mathcal F} I\,C_{\alpha_i}.$
This in turn says that we can choose some $I_i\in\mathcal F$ such that
%
\begin{xlist}\item\label{x.nonstop}
$I_{i-1}\,\pi_{\alpha_i}(f(N_i))~\not\subseteq\ I_i\,C_{\alpha_i}.$
\end{xlist}
%
Since this property is preserved under replacing $I_i$ by
a smaller ideal, and since $\mathcal F$ is downward directed,
we may assume that $I_i\subseteq I_{i-1}.$
After constructing these ideals and indices for all $i,$
we define for each $j\in\omega$
%
\begin{xlist}\item[]
$B_j~=\ \{x\in M\ |\ \forall i\geq j,
\ \pi_{\alpha_i}(f(x))\in\bigcap_{I\in\mathcal F} I\,C_{\alpha_i}\}.$
\end{xlist}
%
Note that for any $x\in M,$ the finite support of $f(x)$ contains
$\alpha_i$ for only finitely many $i,$ so taking $j\in\omega$ greater
than these finitely many values, we see that $x\in B_j.$
Thus, the ascending chain of submodules $B_j$ has union $M.$
Applying Theorem~\ref{T.MNB}, with $\{I_i\ |\ i\in\omega\}$ for
$\mathcal F,$ we get $i^*,\,j^*,\,k^*$ such that in $q_{j^*}(M)=
M/B_{j^*},$
%
\begin{xlist}\item[]
$I_i(q_{j^*}(N_k))~=
\ I_{i^*}(q_{j^*}(N_{k^*}))$ for all $i\geq i^*,\ k\geq k^*.$
\end{xlist}
%
From the definition of $B_{j^*}$ and $q_{j^*},$ this implies that
%
\begin{xlist}\item[]
$I_i\,\pi_{\alpha_j}(f(N_k))\,+\,
\bigcap_{I\in\mathcal F} I\,C_{\alpha_j}
~=\ I_{i^*}\,\pi_{\alpha_j}(f(N_{k^*}))\,+\,
\bigcap_{I\in\mathcal F} I\,C_{\alpha_j}$
for all $i\geq i^*,\ j\geq j^*,\ k\geq k^*.$
\end{xlist}
%
Hence if we fix $j\geq j^*$ and $k\geq k^*$ and vary $i\geq i^*,$
the left hand side above does not depend on the latter value.
In particular, for $i={\rm max}(i^*{+}1,j^*,k^*)$ we have
%
\begin{xlist}\item[]
$I_{i-1}\,\pi_{\alpha_i}(f(N_i))+
\bigcap_{I\in\mathcal F} I\,C_{\alpha_i}
~=\ I_i\,\pi_{\alpha_i}(f(N_i))+
\bigcap_{I\in\mathcal F} I\,C_{\alpha_i,}$
\end{xlist}
%
so
%
\begin{xlist}\item[]
$I_{i-1}\,\pi_{\alpha_i}(f(N_i))
\subseteq\ I_i\,C_{\alpha_i}+\bigcap_{I\in\mathcal F}
I\,C_{\alpha_i}\ =~I_i\,C_{\alpha_i},$
\end{xlist}
%
contradicting~(\ref{x.nonstop}) and completing the proof.
\end{proof}
As with Theorem~\ref{T.MNB}, if one assumes the modules $M_i=
M/N_i$ finitely generated, one gets a simplified conclusion, with
$I^*\,\pi_\beta(f(M))$ in place of $I^*\,\pi_\beta(f(N_{k^*})).$
\vspace{6pt}
Is my development above an improvement on Chase's
proof of \cite[Theorem~1.2]{SUC3}?
It improves the result by
generalizing direct products to inverse limits, and principal
right ideals to finitely generated right ideals, but these changes
could have been made without significantly altering his argument.
% key: p.848_12: a_{n+1}y_n replaced by member of I_{n+1}A; is member
% because I_{n+1} fg and all coords are members
The above development, with the auxiliary lemmas, and the derivation of
Theorem~\ref{T.MNB-Chase} via Theorem~\ref{T.MNB}, is
longer than Chase's.
The best I can say is that it provides alternative perspectives on
what underlies these results, complementing those provided by the
original proof.
For additional results of Chase and
others on maps from direct product modules
to direct sums, see \cite{SUC1}, \cite{SUC2}, \cite{JO'N},
\cite{BZH}, and papers referred to in the latter two works.
The more recent works obtain results on maps from products
of not necessarily countable families of modules.
It would be interesting to know whether similar results can be
obtained for maps on inverse limits of general not
necessarily countable inverse systems of modules.
Incidentally, reading Chase~\cite{SUC3} led me to strengthen my
own results by replacing an original descending chain of right ideals
with a downward directed system $\mathcal F,$ and to explicitly state
Theorem~\ref{T.MNB} rather than passing directly to the case where
the $M_i$ are finitely generated, i.e., Corollary~\ref{C.fg}.
In~\cite{BZH}, the right ideals occurring in these results are
generalized still further, the operation of multiplying by such an
ideal being replaced
with any subfunctor of the forgetful functor from left
$\!R\!$-modules to abelian groups that commutes with arbitrary direct
products, and it is noted that these include not only functors of
multiplication by finitely generated right ideals, but also functors
obtainable from those by transfinite iteration.
Where above we have examined homomorphisms from direct products and
related constructions to direct sums,~\cite{EK} investigates
homomorphisms from direct products and
related constructions to an arbitrary fixed module, and~\cite{EKN}
homomorphisms from an arbitrary fixed module to direct sums and related
constructions; though in exchange for this greater generality,
those papers study a more restricted set of questions.
It is amusing that where, above, I took an argument by successive
approximation and replaced it with an application of the
Baire Category Theorem, the authors of~\cite{EK} take a similar
argument, presented in~\cite{Lady}
in terms of the Baire Category Theorem, and translate it
back into a construction by successive approximation.
{\em Plus \c{c}a change~\dots\,.}
Some similar results with nonabelian groups in place of
modules are proved in \cite{GH}, \cite{KE}, \cite{KE+SSh}.
\section{Generating products of general algebras.}\label{S.gen}
The diagonal argument by which we proved Theorem~\ref{T.fgorunc}
used nothing specific to modules.
Also, as noted at the beginning
of~\S\ref{S.diag}, the focus there on countable products was
for the sake of presentational simplicity.
We shall give the result here in its natural generality.
(However, for consistency of notation, having begun this paper
in the context of module theory, I will not follow the
general algebra convention of using different symbols for
algebras and their underlying sets.)
As noted in the statement of the next theorem, though the hypothesis
there restricts the {\em arities} of the operations, there is
no restriction on the cardinality of the {\em set} of operations.
That cardinality corresponds to the cardinality of the ring $R$ in
Theorem~\ref{T.fgorunc}.
The latter theorem is trivial for $R$ of
cardinality $<2^{\aleph_0},$ and this one is likewise trivial
if there are $<2^\kappa$ operations.
The second conclusion of the theorem, concerning ultraproducts,
was pointed out to me by T.\,Scanlon.
(That conclusion formally subsumes the first conclusion, and of course
implies the corresponding intermediate statement with $U$ replaced by
any filter containing no set of cardinality $<\kappa,$ since any
such filter extends to an ultrafilter with the same property.)
%
\begin{theorem}\label{T.>*kgens}
Let $\kappa$ be an infinite cardinal, and $T$ a type of
algebras such that all operations of $T$ have arities $<\kappa,$
and if $\kappa$ is singular, all have arities $<$ some common
cardinal $<\kappa;$ but where no assumption is made on the cardinality
of the set of those operations.
Let $(M_i)_{i\in\kappa}$ be a $\!\kappa\!$-tuple of algebras
of type $T,$ each of which requires at least $\kappa$ generators.
Then $\prod_{i\in\kappa}M_i$ requires $>\kappa$ generators.
In fact, if $U$ is any ultrafilter on $\kappa$ which
contains no set of cardinality $<\kappa,$ then the ultraproduct
$(\prod_{i\in\kappa}M_i)/U$ requires $>\kappa$ generators.
\end{theorem}\begin{proof}
The restriction on arities of operations of $T$
insures that if an algebra of type $T$
is generated by a set $X,$ then each element of that algebra
belongs to a subalgebra generated by $<\kappa$ elements of $X,$
and hence that any family of $<\kappa$ elements is contained
in such a subalgebra.
Given this observation, the proof of the first
assertion is exactly analogous to that of Lemma~\ref{L.Pinfg}.
For $X\subseteq \prod M_i$ of cardinality $\leq \kappa,$ $y$ an element
of $\prod M_i$ constructed, as in that proof, to avoid
the subalgebra generated by $X,$ and $z$ any element of the
latter subalgebra, note that $y$ and $z$ will in fact agree
at fewer than $\kappa$ coordinates.
Hence for an ultrafilter $U$ containing no set of cardinality $<\kappa,$
the image of $y$ in $(\prod M_i)/U$ will not equal the image of $z;$ so
$(\prod M_i)/U$ is not generated by the image of such a set $X;$
i.e., it, too, requires $>\kappa$ generators.
\end{proof}
We can now generalize Theorem~\ref{T.fgorunc}.
The value of $\lambda$ of greatest interest in the next result is, of
course, $\aleph_0.$
\begin{theorem}\label{T.bigorsmall}
Let $\lambda\leq\kappa$ be infinite cardinals, with $\lambda$
regular, let $T$ be an algebra type such that all operations of
$T$ have arities $<\lambda,$ and let
$M$ be any algebra of type $T.$
Then the algebra $M^\kappa$ either
requires $<\lambda$ or $>\kappa$ generators.
\end{theorem}\begin{proof}
Let $\mu$ be the least cardinality of a generating set for $M^\kappa.$
If $\mu$ were neither $<\lambda$ nor $>\kappa,$ then noting that
$M^\kappa\cong(M^\kappa)^\mu,$ we could apply Theorem~\ref{T.>*kgens},
with $\mu$ in place of $\kappa$ and $M^\kappa$ in place of each
$M_i,$ to conclude that $M^\kappa$ requires $>\mu$ generators,
a contradiction.
\end{proof}
Unlike Theorem~\ref{T.>*kgens}, Theorem~\ref{T.bigorsmall} contains
no statement about ultrapowers, so we ask
\begin{question}\label{Q.ultra}
Is the analog of Theorem~\ref{T.bigorsmall} true with $M^\kappa$
replaced by the ultrapower $M^\kappa/U,$ for $U$ any
ultrafilter on $\kappa$ containing no set of cardinality $<\kappa$?
If so, do all such ultrapowers of $M$ agree with one another as to
whether the number of generators required is $<\lambda$ or $>\kappa$?
If so, do they agree with $M^\kappa$ in this respect?
\end{question}
Something one {\em can} show is that if there is
an ultrafilter $U$ on $\kappa$ such that
$M^\kappa/U$ can be generated by $\mu$ but no fewer generators,
where $\lambda\leq\mu\leq\kappa,$ then there is another
ultrafilter $U'$ such that $M^\kappa/U'$ requires $>\mu$ generators.
To see this, take any ultrafilter $V$ on $\mu$
containing no subset with $<\mu$ elements.
Then Theorem~\ref{T.>*kgens} applied with $\mu$ in the role
of $\kappa$ shows that $(M^\kappa/U)^\mu/V$ requires $>\mu$ generators.
But there exists an ultrafilter $UV$ on $\kappa\times\mu$ such that
$(M^\kappa/U)^\mu/V\cong M^{\kappa\times\mu}/UV;$
and since $\mu\leq\kappa,$ we may choose an a bijection
$\kappa\times\mu\rightarrow\kappa,$ and this will take $UV$ to the
desired ultrafilter $U'$ on $\kappa.$
I have no reason to expect a positive answer to the next
question; the present wording is just the shortest way of asking for
a counterexample that should exist, but that I don't know how to find.
\begin{question}\label{Q.inf}
In the final conclusion of Theorem~\ref{T.bigorsmall},
can one strengthen ``$\!\<<\lambda\!$'' to ``$\!\<<\aleph_0\!$''?
\end{question}
Let us note a ``difficulty'' with Theorem~\ref{T.bigorsmall}
as a generalization of Theorem~\ref{T.fgorunc}: Classes of algebras with
infinitely many operations are not commonly considered in most fields
other than ring- and module-theory.
We will now note one close analog of the module-theoretic situation;
then, in the next section, introduce a class of examples of a quite
different sort, and obtain some results on these.
The nonlinear analog of an $\!R\!$-module is
a $\!G\!$-set where $G$ is a group or monoid, and the next lemma, which
is proved just like Lemma~\ref{L.R=R^*k}, shows
that in that context, the ``$\!<\lambda\!$'' case
of Theorem~\ref{T.bigorsmall} again occurs.
In the category $\mathcal C$ referred to in the statement,
morphisms are assumed to
compose like functions written on the left of their arguments,
i.e., the composite of morphisms $f:X\rightarrow Y$
and $g: Y\rightarrow Z$ is written $gf: X\rightarrow Z.$
\begin{lemma}\label{L.M=M^*k}
Let $\mathcal C$ be a category, $\kappa$ a cardinal, and
$M$ an object of $\mathcal C$ which is either a $\!\kappa\!$-fold
coproduct of copies of itself
%
\begin{xlist}\item\label{x.M=(+)Mgen}
$M~\cong~\coprod_\kappa M$
\end{xlist}
%
{\rm (}for instance, if $\kappa$ is infinite and $M$ any object
of the form $\coprod_\kappa N),$ or
a $\!\kappa\!$-fold product of copies of itself
%
\begin{xlist}\item\label{x.M=M^gen}
$M~\cong~\prod_\kappa M$
\end{xlist}
%
{\rm (}for instance, if $\kappa$ is infinite and $M$ any object
of the form $\prod_\kappa N).$
In the former case, if we let $G$ be the monoid ${\rm End}(M),$ or in
the latter, if we let $G~=~{\rm End}(M)^{\rm op},$ then
%
\begin{xlist}\item\label{x.G=G^*k}
$G~\cong~G^\kappa$ as left $\!G\!$-sets.
\end{xlist}
\vspace{-10pt}\qed
%
\end{lemma}
The situation is strikingly different when $G$ is a group, no matter
how constructed.
If $M$ is any left $\!G\!$-set of more than one element, then for
every proper nonempty subset $S\subseteq\kappa,$ the set $A_S$ of
elements of $M^\kappa$ having exactly two values, one at all
indices in $S$ and the other at all indices in $\kappa-S,$ is a union
of $\!G\!$-orbits, and sets $A_{S_1}$ and $A_{S_2}$ are disjoint unless
$S_1$ and $S_2$ are equal or complements of one another.
So for $\kappa$ infinite, $M^\kappa$ consists of $\geq 2^\kappa$
orbits, and so cannot be generated by fewer elements.
The concept of a $\!G\!$-set, for $G$ a monoid or a group, is
generalized by that of an algebra $M$ of any type on which
$G$ acts by endomorphisms, respectively by automorphisms.
$\!R\!$-modules for $R$ a ring are a particular class of examples;
there may be others for which Theorem~\ref{T.bigorsmall}
would be of interest, and in particular, where the exotic
phenomenon of generation of $M^\kappa$ by $<\lambda$ elements occurs.
\section{Generation of power algebras over their diagonal
subalgebras.}\label{S.Sym}
A different way to get interesting examples of algebras with
operation-sets of arbitrarily large cardinalities is to
start with algebras of arbitrary type $T,$ fix one such
algebra $X$ whose underlying set has large cardinality, and consider
algebras $Y$ given with a homomorphism $i$ of $X$ into them, formally
treating the image of each element of $X$ as a zeroary operation.
The simplest case is that in which $Y=X$ and $i$ is the identity map.
For that case, Theorem~\ref{T.bigorsmall} (with $\lambda$ taken
to be $\aleph_0$ for simplicity) says
\begin{corollary}\label{C.overdiag}
Let $\kappa$ be a cardinal, and $M$ an algebra with
operations all of finite arity.
Then the algebra $M^\kappa$ is either generated by the
diagonal $\Delta(M)$ and finitely many additional elements,
or requires $>\kappa$ additional elements.\qed
\end{corollary}
The next result gives, for any infinite
cardinal $\kappa,$ nontrivial algebras of two
familiar sorts having only finitely many operations, whose
$\!\kappa\!$th direct powers are finitely generated
over their diagonal subalgebras.
In the proof, we continue to write maps on the left of their
arguments and compose them accordingly, though this reverses the
convention in the material from \cite{Sym_Omega:1} cited
in the proof of the second statement.
\begin{theorem}\label{T.Sym}
Let $\kappa$ be an infinite cardinal and $\Omega$ a set
of cardinality $\geq\kappa.$\\
{\rm (i)} If $M$ is the monoid of all maps $\Omega\rightarrow\Omega,$
then $M^\kappa$ can be generated over $\Delta(M)$ by two elements.\\
{\rm (ii)} If $S$ is the group of all permutations of $\Omega,$
then $S^\kappa$ can be generated over $\Delta(S)$ by one element.
\end{theorem}\begin{proof}
(i): Since $|\Omega|=\kappa\cdot|\Omega|,$ we can write $\Omega$ as the
union of $\kappa$ disjoint sets of the same cardinality
as $\Omega,$ $\Omega=\bigcup_{i\in\kappa}\Sigma_i.$
For each $i\in\kappa,$ let $a_i\in M$ be a one-to-one map
with image $\Sigma_i,$ and $b_i\in M$ a left inverse to $a_i.$
I claim $M^\kappa$ is generated by $\Delta(M)$ and the
two elements $a=(a_i)_{i\in\kappa}$ and $b=(b_i)_{i\in\kappa}.$
For given any $x=(x_i)_{i\in\kappa}\in M^\kappa,$ let us ``encode''
$x$ in a single element $x'\in M,$ defined to act on each
subset $\Sigma_i$ by $a_i\,x_i\,b_i.$
Then we see that $b\,\Delta(x')\,a=(b_i\,x'\,a_i)_{i\in\omega}=
(b_i\,a_i\,x_i\,b_i\,a_i)_{i\in\omega}=(x_i)_{i\in\omega}=x,$
as required.
(ii): Again the idea will be to encode elements of $S^\kappa$ in single
elements of $S.$
The trouble is that our structure no longer contains elements
$a_i$ and $b_i$ giving bijections between $\Omega$
and the subsets of $\Omega$ on which we will encode the
components of our $\!\kappa\!$-tuple.
It will, however, contain permutations carrying each of these subsets
bijectively to a common set $\Sigma_1\subset\Omega.$
Unfortunately, if we take an element whose actions on
these subsets ``encode'' the coordinates of a member of $S^\kappa,$
and conjugate it by each of these permutations to get permutations
having those same actions on
$\Sigma_1,$ the information encoded in its other components
will not disappear; it will move to other parts of $\Omega,$ where it
will constitute ``garbage'' that we must get rid of.
We will do this using commutator operations, in which those
parts of our maps are commuted with identity maps.
Finally, we will call on a result from \cite{Sym_Omega:1} to go from
permutations of $\Sigma_1$ to permutations of $\Omega.$
So let $\Omega=\Sigma_1\cup\Sigma_2,$ where $\Sigma_1$ and $\Sigma_2$
are disjoint sets of the same cardinality as $\Omega,$ and
let $S_1$ be the subgroup of $S$ consisting of elements which
fix all members of $\Sigma_2$ and act arbitrarily on $\Sigma_1.$
We shall first show that $S_1^\kappa$ is contained in the subgroup of
$S^\kappa$ generated over $\Delta(S)$ by a single
element $f=(f_i)_{i\in\kappa}.$
Let us identify $\Sigma_2$ with $\Sigma_1\times\kappa\times\{0,1\},$
as we may since $|\Omega|\geq\kappa.$
For each $i\in\kappa,$ we define an element $f_i\in S$ which cyclically
permutes the three sets $\Sigma_1,$ $\Sigma_1\times\{i\}\times\{0\}$
and $\Sigma_1\times\{i\}\times\{1\},$
and for each $j\in\kappa-\{i\}$ interchanges the two sets
$\Sigma_1\times\{j\}\times\{0\}$ and $\Sigma_1\times\{j\}\times\{1\}.$
Precisely, we let
%
\begin{xlist}\item\label{x.fdef}
$f_i:\ \alpha\mapsto(\alpha,i,0)\mapsto(\alpha,i,1)\mapsto\alpha,$~
$(\alpha,j,0)\leftrightarrow(\alpha,j,1),$\\
\indent for $\alpha\in\Sigma_1,\ j\neq i$ in $\kappa.$
\end{xlist}
Now in the group of permutations of an infinite
set, every element is a commutator by \cite{OO} or
\cite[Lemma~14]{Sym_Omega:1}, so given $x=(x_i)_{i\in\omega}$ in
$S_1^\kappa,$ let us, for each $i\in\kappa,$ regard
$x_i$ as a permutation of $\Sigma_1$ and write it as the commutator of
permutations $u_i,\,v_i$ of that set.
Let us now define elements $y,z\in S$ which act trivially
on $\Sigma_1$ and on $\Sigma_1\times\kappa\times\{1\},$
while on each set $\Sigma_1\times\{i\}\times\{0\},$
$y$ behaves like $u_i$ and $z$ like $v_i,$ i.e.,
$y(\alpha,i,0)=(u_i(\alpha),i,0),\ z(\alpha,i,0)=(v_i(\alpha),i,0).$
I claim that $x$ is the commutator of $f^{-1}\Delta(y)f$ and
$f^2\Delta(z)f^{-2}.$
Indeed, looking at the $\!i\!$th component of $f^{-1}\Delta(y)f$
for any $i,$ we see that it acts trivially except on $\Sigma_1$ and on
the sets $\Sigma_1\times\{j\}\times\{1\}$ with $j\neq i,$
while $f^2\Delta(z)f^{-2}$ is trivial except on $\Sigma_1$ and on
the sets $\Sigma_1\times\{j\}\times\{0\}$ with $j\neq i.$
Hence their commutator is trivial except on $\Sigma_1,$ where it
behaves as the commutator of $u_i$ and $v_i,$ i.e., as $x_i.$
So the subgroup of $S^\kappa$ generated by $\Delta(S)\cup\{f\}$
contains every $x\in S_1^\kappa,$ as claimed.
For the remainder of the proof we put aside the description of
$\Omega$ as $\Sigma_1\cup(\Sigma_1\times\kappa\times
\{0,1\})$ used to get this fact, and take a decomposition of
a simpler sort, keeping $\Sigma_1$ as above, but
choosing a set $\Sigma_3$ such that
$\Sigma_1\cup\Sigma_3=\Omega,$ and such that
$\Sigma_1\cap\Sigma_3,$ $\Sigma_1-\Sigma_3$ and $\Sigma_3-\Sigma_1$
all have cardinality $|\Omega|.$
From the proof of~\cite[Lemma~2]{Sym_Omega:1} one finds that in this
situation, if we denote by $S_3$ the group of permutations of $\Omega$
that fix all elements not in $\Sigma_3,$ then every
permutation of $\Omega$ can be written either as a member of the
product-set $S_1\,S_3\,S_1$ or of the product-set
$S_3\,S_1\,S_3;$ hence we have $S=S_1\,S_3\,S_1\,S_3.$
(See last three paragraphs proof of~\cite[Lemma~2]{Sym_Omega:1}.
If one doesn't want to read between the lines of
that proof, one can use the statement of that lemma,
which, for subsets $U$ and $V$ of $S$ satisfying certain weaker
conditions than being equal to the above subgroups, shows --
after adjusting to left-action notation -- that every element of $S$
belongs either to $V (V\,U)^4$ or to $U (U\,V)^4.$
Hence if $U$ and $V$ contain $1,$ every element will belong
to $(U\,V)^6,$ hence in our present
situation, to $(S_1\,S_3)^6,$ which one may use
in place of $S_1\,S_3\,S_1\,S_3$ in the reasoning of the next sentence.)
Now taking $t\in S$ which interchanges $\Sigma_1$ and $\Sigma_3,$
we get $S_3=t^{-1}S_1\,t,$
so $S=S_1\,(t^{-1}S_1\,t)\,S_1\,(t^{-1}S_1\,t),$
hence $S^\kappa = S_1^\kappa\,\Delta(t^{-1})S_1^\kappa\,
\Delta(t)\,S_1^\kappa\,\Delta(t^{-1})S_1^\kappa\,\Delta(t)$.
Since the subgroup of $S^\kappa$ generated by $\Delta(S)\cup\{f\}$
contains $S_1^\kappa,$ the above equation shows that it is
all of $S^\kappa.$
\end{proof}
% could we have avoided commutator trick, merely getting $\Sigma_1$
% as "full moiety"? Not clear; the way proof from \cite{Sym_Omega:1}
% gets rid of garbage can depend on element ...
The result from \cite{Sym_Omega:1} called on in the last
paragraph of the above proof was used there in proving
two other properties of infinite symmetric
groups, essentially~(\ref{x.unccof}) and~(\ref{x.bddgen}) below.
One may ask whether there is a direct relation between
the conclusion of the above lemma and those properties.
An implication in one direction is proved, under simplifying
restrictions on the value of $\kappa$ and the algebra-type, in
the next result.
\begin{theorem}\label{T.fg/=>}
Let $S$ be an algebra with finitely many primitive operations,
all of finite arity, which satisfies
%
\begin{xlist}\item\label{x.fg/*D}
The countable power algebra $S\<^\omega$ is finitely generated
over the diagonal subalgebra $\Delta(S).$
\end{xlist}
%
Then $S$ also satisfies both of
%
\begin{xlist}\item\label{x.unccof}
$S$ cannot be written as the union of a countable chain of
proper subalgebras.
\end{xlist}
%
\begin{xlist}\item\label{x.bddgen}
For every subset $X\subseteq S$ which generates $S,$ there exists a
positive integer $n$ such that all elements of $S$ can be
represented by words of length $\leq n$ in the elements of $X.$
\end{xlist}
%
\end{theorem}\begin{proof}
It is not hard to show that the conjunction of~(\ref{x.unccof})
and~(\ref{x.bddgen}) is equivalent to the statement
%
\begin{xlist}\item\label{x.unc+bdd}
Whenever $X_0\subseteq X_1\subseteq\dots$ is an $\!\omega\!$-indexed
chain of subsets of $S$ with $\bigcup_{i\in\omega} X_i=S,$
such that for each primitive operation
$f$ of $S$ and each $i\in\omega,$
one has $f(X_i,\dots,X_i)\subseteq X_{i+1},$
then $X_i=S$ for some $i\in\omega.$
\end{xlist}
%
So it suffices to show that~(\ref{x.fg/*D}) implies~(\ref{x.unc+bdd}).
(This reduction only needs the fact that~(\ref{x.unc+bdd}) implies
both~(\ref{x.unccof}) and~(\ref{x.bddgen}).
These implications are seen by taking for $X_i,$ in the first case,
the $\!i\!$th member of an $\!\omega\!$-indexed ascending chain of
subalgebras, and in the second, the set of elements of $S$ that can be
represented by words of depth $\leq i$ in the elements of $X.)$
We will prove this in contrapositive form, showing that if
$(X_i)_{i\in\omega}$ is a family which satisfies the
hypotheses of~(\ref{x.unc+bdd}) but not the conclusion, and
$Y\subseteq S\<^\omega$ is finite, then the subalgebra
of $S\<^\omega$ generated by $\Delta(S)\cup Y$ must be proper.
Indeed, define the {\em rank} of an element $s\in S$ to be the
least $r$ such that $s\in X_r.$
Then our assumption that the conclusion
of~(\ref{x.unc+bdd}) fails says that the rank function is unbounded,
so for each $i\in\omega$ let us choose an element $x_i$ of rank
at least $i+{\rm max}_{y\in Y}\,{\rm rank}(y_i)$ (where
each $y\in Y\subseteq S\<^\omega$ is written $(y_i)_{i\in\omega}).$
We claim that the element $x=(x_i)_{i\in\omega}$ does not lie in the
subalgebra of $S\<^\omega$ generated by $\Delta(S)\cup Y.$
For if it did, it would lie in the subalgebra
generated by $\Delta(Z)\cup Y$ for some finite
subset $Z\subseteq S,$ and be represented by a word of
some depth $d$ in these elements; thus for each
$i,$ $x_i$ would be expressed as a word of depth $d$ in the
elements of $Z$ and the $\!i\!$th components of the elements of $Y.$
We now get a contradiction on taking any
$i>d+{\rm max}_{z\in Z}\,{\rm rank}(z),$ since then by choice of $x_i,$
%
\begin{xlist}\item[]
${\rm rank}(x_i)\ \geq\ i+{\rm max}_{y\in Y}\,{\rm rank}(y_i)\\
\indent
>\ d+{\rm max}_{z\in Z}\,{\rm rank}(z)+
{\rm max}_{y\in Y}\,{\rm rank}(y_i),$
\end{xlist}
%
contradicting the existence of such a depth-$\!d\!$ expression
for $x_i.$
\end{proof}
Conditions~(\ref{x.unccof}) and, more recently,~(\ref{x.bddgen}) have
been proved for several sorts of groups that arise in ways similar to
infinite symmetric groups (see \cite[\S3]{Sym_Omega:1} for references),
and also for the endomorphism ring of a direct sum
or product of infinitely many copies of any nontrivial module \cite{ZM}.
It seems likely that some or all of those proofs can be
modified to establish~{\rm (\ref{x.fg/*D})} in these cases as well.
The endomorphism-ring case, indeed, follows easily from
Lemma~\ref{L.R=R^*k} above and the observation that for any ring $R,$
%
\begin{xlist}\item\label{mod>bi>rg}
$R\<^\omega$ fin.gen.\ as left $\!R\!$-module $\implies$
$R\<^\omega$ fin.gen.\ as $\!(R,R)\!$-bimodule\\
\indent
$\implies$ $R\<^\omega$ fin.gen.\ over $\Delta(R)$ as ring,
\end{xlist}
%
where the last step uses the fact that in
$R\<^\omega,$ left or right module multiplication by an element of $R$
is equivalent to ring multiplication by an element of $\Delta(R).$
In fact, since Lemma~\ref{L.R=R^*k} shows that
$R^\omega$ is cyclic as a left $\!R\!$-module, we have not only
``finitely generated'' but ``generated by one element'' in this case.
Likewise, replacing rings and modules by monoids and
$\!M\!$-sets, and the application of Lemma~\ref{L.R=R^*k}
with an application of Lemma~\ref{L.M=M^*k}, with
$\mathcal C=\mathcal{S}et,$ we get Theorem~\ref{T.Sym}(i)
with ``two elements'' strengthened to ``one element''.
(I stated Theorem~\ref{T.Sym}(i) as I did because the proof
given seemed the best way to lead up to the more
difficult proof of part~(ii).
In proving directly the ``one element'' statement, one can take
that one element to be what we called $a$ in the earlier proof.
Given $x=(x_i),$ if we define $x''\in M$ to have restriction to
each $\Sigma_i$ given by $x_i b_i,$ we see that $\Delta(x'')a=x.)$
A nontrivial {\em finite} algebra clearly satisfies~(\ref{x.unccof})
and~(\ref{x.bddgen}), but will not satisfy~(\ref{x.fg/*D}),
since $\Delta(S)$ is finite while $S^\omega$ is uncountable;
so the converse of Theorem~\ref{T.fg/=>} is false.
On the other hand (still considering only algebras
with finitely many operations, all of finite arities), we see that
a finitely generated infinite algebra
cannot satisfy~(\ref{x.bddgen}), and a countably generated but not
finitely generated algebra cannot satisfy~(\ref{x.unccof}),
so any infinite algebra satisfying both~(\ref{x.unccof})
and~(\ref{x.bddgen}) must be uncountable, raising the hope
that all such algebras might satisfy~(\ref{x.fg/*D}).
However, this is not so,
for it is shown in \cite{YdC} and in \cite[Corollaire~18]{AK} (both
generalizing a result in \cite{SK+JT})
that an infinite direct power of a finite perfect group
satisfies~(\ref{x.unccof}) and~(\ref{x.bddgen}); but such a group cannot
satisfy~(\ref{x.fg/*D}), since it admits a homomorphism
onto a nontrivial finite group, and~(\ref{x.fg/*D}) clearly
carries over to homomorphic images.
Since the proof of Theorem~\ref{T.fg/=>} constructs
(when~(\ref{x.unccof}) or~(\ref{x.bddgen}) fails) an element $x$
which disagrees at all but finitely many coordinates with
every element of the subalgebra generated by $\Delta(S)\cup Y,$
one can in fact say (using again the idea of the last
sentence of Theorem~\ref{T.>*kgens})
that for any nonprincipal ultrafilter $U,$ the image of such
an element $x$ in $S^\kappa/U$ fails to lie in the image
of the subalgebra generated by $\Delta(S)\cup Y.$
Thus, for $U$ a nonprincipal ultrafilter on $\omega,$
one can insert in Theorem~\ref{T.fg/=>} the intermediate condition
%
\begin{xlist}\item\label{x.fg//U}
The ultrapower $S\<^\omega/U$ is finitely generated
over the image of the diagonal subalgebra $\Delta(S),$
\end{xlist}
%
i.e., strengthen that theorem
to say that (\ref{x.fg/*D})$\Rightarrow$(\ref{x.fg//U})$\Rightarrow$%
(\ref{x.unccof})$\wedge$(\ref{x.bddgen}).
Unlike~(\ref{x.fg/*D}), but like~(\ref{x.unccof}) and~(\ref{x.bddgen}),
condition~(\ref{x.fg//U}) is clearly satisfied by finite algebras.
For groups, it is also preserved by the operation of pairwise direct
product; this follows from the fact that if groups $G_1$ and $G_2$ are
generated by subsets $X_1$ and $X_2,$ each containing
$1,$ then $G_1\times G_2$ will be generated by $X_1\times X_2.$
Hence the direct product of the permutation
group on an infinite set with any nontrivial finite group is an
example of an infinite algebra that separates~(\ref{x.fg/*D})
from~(\ref{x.fg//U}).
The example mentioned earlier of an infinite direct power of a
finite perfect group turns out to separate~(\ref{x.fg//U}) from
(\ref{x.unccof})$\wedge$(\ref{x.bddgen}), in view of the next result.
\begin{proposition}\label{P.xPfin}
If an algebra $S$ satisfies~{\rm (\ref{x.fg//U})} for some
nonprincipal ultrafilter $U$ on $\omega,$ then there is
a {\rm (}finite{\rm )} upper bound on the cardinalities of
finite homomorphic images of $S.$
Equivalently, $S$ has a least congruence $C$ such that $S/C$ is finite.
\end{proposition}\begin{proof}
Let $S$ be an algebra which does not have a least congruence $C$
making $S/C$ finite.
We shall show that $S$ does not satisfy~(\ref{x.fg//U}).
Denote by $\mathcal F$ the class of all congruences $C$ on $S$ with
finite quotient $S/C,$ and let
$C_{\rm res.fin.}=\bigcap_{C\in\mathcal F} C.$
Since~(\ref{x.fg//U}) is preserved under passing to homomorphic
images, we may divide out by $C_{\rm res.fin.}$ and
assume $S$ residually finite, but still infinite.
I claim now that we can find a chain of congruences
$C_0\supset C_1\supset\dots\in\mathcal F$ and a sequence of
elements $u_0,\,u_1,\,\dots\in S$ such that
$(u_i,\,u_{i+1})\in C_i-C_{i+1}$ for all $i.$
Indeed, let $C_0\in\mathcal F$ be arbitrary.
Since $S/C_0$ is finite, some congruence class $c_0$ with respect to
$C_0$ is infinite.
Choose a congruence $C_1\subset C_0$ such that $c_0$
decomposes into more than one congruence class under $C_1.$
At least one of these must be infinite; choose such an infinite
class $c_1\subset c_0,$ and repeat the process.
For each $i,$ choose $u_i\in c_i-c_{i+1};$
then the relations $(u_i,\,u_{i+1})\in C_i-C_{i+1}$ clearly hold.
Writing $p_i$ for the canonical map $S\rightarrow S/C_i,$
these relations say that for all~$i,$
%
\begin{xlist}\item\label{x.cong+not}
$p_{i+1}(u_i)~\neq\ p_{i+1}(u_{i+1})~=\ p_{i+1}(u_{i+2})~
=\ p_{i+1}(u_{i+3})~=\ \dots\,.$
\end{xlist}
To show that~(\ref{x.fg//U}) fails, let
$Y\subseteq S\<^\omega$ be any finite subset;
we must show that the image in $S\<^\omega/U$ of the subalgebra
of $S\<^\omega$ generated by $\Delta(S)\cup Y$ is a proper subalgebra.
Since $S/C_0$ and $Y$ are finite, and $U$ is an
ultrafilter, there exists $R_0\in U$ such that for each
$y=(y_j)_{j\in\omega}\in Y,$
the $\!\omega\!$-tuple $(p_0(y_j))_{j\in\omega}\in (S/C_0)^\omega$
becomes constant when $j$ is restricted to $R_0\subseteq\omega.$
Likewise, we can find $R_1\subseteq R_0$ in $U$ such that
for each $y\in Y,$ $(p_1(y_j))_{j\in\omega}$ becomes constant when
restricted to $j\in R_1,$ and so on,
getting a chain $R_0\supseteq R_1\supseteq\dots$ of members of
$U$ such that for each $i\in\omega$ and each $y\in Y,$
the elements $p_i(y_j)\in S/C_i$ are the same for all $j\in R_i.$
Also, since $U$ is nonprincipal, we can along the way make sure, for
each $i,$ that $i\notin R_i,$ so that $\bigcap_i R_i=\emptyset.$
Now for any $i,$ $j,$ and $j'$ such that the $\!j\!$- and
$\!j'\!$-components of all elements of $Y$ have equal images under
$p_i,$ the same will be true for all elements of
the subalgebra generated by $\Delta(S)\cup Y;$ hence
the above conditions on the $R_i$ yield
%
\begin{xlist}\item\label{x.ij=ij'}
For all $z=(z_j)$ in the subalgebra of $S\<^\omega$ generated by
$\Delta(S)\cup Y,$ all $i\in\omega,$ and all $j,j'\in R_i,$
we have $p_i(z_j)=p_i(z_{j'}).$
\end{xlist}
We shall now construct an element $x\in S\<^\omega$ which does not
agree modulo $U$ with any element $z$ satisfying~(\ref{x.ij=ij'}),
in other words, such that for any such $z,$ and any $R\in U,$
%
\begin{xlist}\item\label{x.notinR}
$\{j\in\omega~|\ x_j=z_j\}\ \not\supseteq\ R.$
\end{xlist}
%
For each $j\in\omega,$ let $i(j)$ be the
greatest integer such that $j\in R_{i(j)}$ if $j\in R_0,$ and $-1$
otherwise; and let $x_j$ be $u_{i(j)-1}$ if $i(j)\geq 1,$
arbitrary otherwise.
To prove~(\ref{x.notinR}) given $R\in U,$ take any $j\in R_1\cap R,$
and then any $j'\in R_{i(j)+1}\cap R,$ noting that $1\leq i(j)*}, for each $i$ let $x_i$ have rank
$\geq i+{\rm max}_{y\in Y_i}\,{\rm rank}(y_i),$ using the
finite set $Y_i$ rather than all of $Y.$
Likewise, in the proof of Proposition~\ref{P.xPfin}, construct the
sets $R_i$ so that the elements $p_i(y_j)\in S/C_i$ $(j\in R_i)$ are
merely the same for each $y\in Y_i.)$
It is also natural to ask
\begin{question}\label{Q.U,U'}
For a given algebra $S,$ if~{\rm(\ref{x.fg//U})} holds for
some nonprincipal ultrafilter $U,$ will it hold for all
nonprincipal ultrafilters?
If not, what implications can be obtained among these conditions for
different $U$?
\end{question}
\section{Still more conditions.}\label{S.more}
Khelif~\cite[D\'efinition~5]{AK} introduces a property having an
interesting similarity to~(\ref{x.fg/*D}).
Given an algebra-type $T$ and a natural number $n,$ let $W(n,T)$ denote
the set of all $\!n\!$-ary words
% don't change to "derived ops": reference to "word length" below!
in the operations of $T.$
Note that if $S$ is an algebra of type $T,$
then an element of $W(n,T)^\omega$ induces an $\!n\!$-ary
operation on $S\<^\omega.$
Khelif's definition (reformulated to bring out the parallelism
with~(\ref{x.fg/*D})) says that an algebra $S$ ``has property P*\,'' if
there exists a natural number $n$ and an element
$w=(w_i)_{i\in\omega}\in W(n,T)^\omega$ such that every element
$x\in S\<^\omega$ is the value of $w$ at some $\!n\!$-tuple of
elements of $\Delta(S).$
In~\cite[D\'efinition~4]{AK} he defines a weaker condition P, which
says that there exists a natural
number $n,$ and a function $\eta$ from the
natural numbers to the natural numbers, such that for every
$x\in S\<^\omega$ there exists $w\in W(n,T)^\omega$ whose $\!i\!$th
component is a word of length $\leq\eta(i)$ for each $i,$ such that,
again, $x$ is the value of $w$ at an $\!n\!$-tuple of
elements of $\Delta(S);$ and he shows that P (and hence
also P*) implies (\ref{x.unccof})$\wedge$(\ref{x.bddgen}).
Since there are only finitely many words of a given length, and
there is a bound on the lengths of any finite set of words, Khelif's
property P is equivalent to the statement that for some $n$ one can
associate to every natural number $i$ a finite subset
of $W_i\subseteq W(T,n)$ such that every element
$x\in S\<^\omega$ is the value of some $w\in\prod_i W_i$
at some $\!n\!$-tuple of elements of $\Delta(S).$
This suggests a parallel generalization of~(\ref{x.fg/*D}),
where the finite generating set is replaced by one that is
merely ``locally finite'':
%
\begin{xlist}\item\label{x.pfg/*D}
There exists a sequence $(X_i)_{i\in\omega}$ of finite
subsets $X_i\subseteq S$ such that $S\<^\omega$ is generated
by $\Delta(S)\,\cup\,\prod_i\,X_i.$
\end{xlist}
And indeed, one sees that the proof of Theorem~\ref{T.fg/=>}
(and its strengthening by the insertion of conditions~(\ref{x.fg//U})
and~(\ref{x.cg//U}))
works equally well under this weaker hypothesis, since the construction
there of the sequence $(x_i)$ called on the finiteness of the
generating set $Y$ only via the finiteness of the
subsets $\{y_i~|~y\in Y\}$ $(i\in\omega).$
One can also combine features of~(\ref{x.fg/*D}) or~(\ref{x.pfg/*D}) and
Khelif's conditions, allowing, for instance, finitely
many elements of $W(T,n)^\omega$ and finitely many
elements of $S\<^\omega$ to be used together in generating
$S\<^\omega$ over $\Delta(S\<^\omega).$
This plethora of conditions leads one to wonder whether there is
some small number of ``core'' conditions to which most or all of
those we have mentioned are related in simple ways; e.g., by adding
conditions such as ``$\!S\!$ has no finite homomorphic images'',
adding cases such as ``all finite algebras'', and/or varying
some natural parameters in the conditions.
It is noted in \cite[paragraph following Question~8]{Sym_Omega:1}
that no non-finitely-generated abelian group satisfies
either~(\ref{x.unccof}) or~(\ref{x.bddgen}), so
Theorem~\ref{T.fg/=>} shows (in two ways) that
no nontrivial abelian group satisfies~(\ref{x.fg/*D}).
One may ask generally
\begin{question}\label{Q.vars}
What can be said about varieties of algebras containing
non-finitely-generated algebras $S$ that satisfy
{\rm (\ref{x.fg/*D})}?
{\rm (\ref{x.unccof})}?
{\rm (\ref{x.bddgen})}?
{\rm (\ref{x.fg//U})}?
{\rm (\ref{x.cg//U})}?
Khelif's P*?
P?
Further variants of these?
\end{question}
Another condition, weaker than both~(\ref{x.bddgen}) and
Khelif's property P, and of a more elementary nature,
which might nevertheless be of interest in the study
of these conditions, is the statement that for every generating
set $X$ for $S,$ there exist an integer $n$ such that every
element of $S$ belongs to a subalgebra generated by $\leq n$
elements of $X.$
This is satisfied by the abelian group $Z_{p^\infty},$
with $n=1$ independent of $X.$
Turning in a different direction, observe that if an algebra $S$
satisfies~(\ref{x.fg/*D}), then so will $S\<^\omega.$
For if we write $S\<^\omega=S',$ and identify $(S')^\omega$ with
$S^{\omega\times\omega}\cong S\<^\omega,$ then the diagonal
subalgebra of $(S')^\omega$ contains the diagonal subalgebra of
$S^{\omega\times\omega},$ over which~(\ref{x.fg/*D}) shows
that it is finitely generated.
In particular, $S\<^\omega$ satisfies~(\ref{x.bddgen}); so given
a finite set $Y$ such that $\Delta(S)\cup Y$ generates $S\<^\omega$
as in~(\ref{x.fg/*D}), there will be a bound on the lengths of
words needed to so express every element of $S\<^\omega.$
It would be interesting to know whether one can in general
get all elements of $S\<^\omega$ using a
single word, with specified positions in each of which a
specified element of $Y$ occurs, while arbitrary elements of $\Delta(S)$
are put in the other positions.
The proof of Theorem~\ref{T.Sym}
shows that this is so in the case of the monoid or group of all
maps or invertible maps of an infinite set into itself.
To an algebra $S$ satisfying~(\ref{x.fg/*D}), we can associate several
natural-number-valued invariants witnessing that condition:
The least cardinality of
a set $Y$ such that $S\<^\omega$ is generated by $\Delta(S)\cup Y;$
the least $n$ such that for some finite $Y\in S\<^\omega$ every element
of $S\<^\omega$ can be expressed using words of length (or depth)
$\leq n$ in terms of elements of $\Delta(S)\cup Y,$ etc..
Note that if we have a sequence $S_0,\ S_1,\,\dots$ of algebras of the
same type which each satisfy~(\ref{x.fg/*D}), but for which
the sequence of values of
such an invariant is unbounded, then the product algebra
$\prod_{i\in\omega} S_i$ cannot satisfy~(\ref{x.fg/*D}),
in contrast to the observation of the preceding
paragraph on a direct power of one such $S.$
A confession about Theorem~\ref{T.fg/=>}: The hypothesis that $S$ had
only finitely many primitive operations was not used in the proof.
I made that assumption because it would be needed in places in our
subsequent discussion, and because without it, the
conclusion~(\ref{x.bddgen}) of that theorem is {\em weaker}
than optimal.
The reader will not find it hard to verify that if $S$ has
{\em countably} many primitive operations, then
for any generating set $X,$ all elements of $S$ may be obtained
not merely using words of bounded length as in~(\ref{x.bddgen}),
but using such words in finitely many of these operations,
and more generally, that if $S$ has an arbitrarily large set of
primitive operations, and this set is
partitioned in any way into countably many subsets, then $S$
can be obtained as in~(\ref{x.bddgen}) using the operations
in the union of finitely many of these subsets.
(Key idea: In~(\ref{x.unc+bdd}), replace the
relation $f(X_i,\dots,X_i)\subseteq X_{i+1}$
by $f(X_i,\dots,X_i)\subseteq X_{i+a(f)},$ where $a$ is a function
from the set of primitive operations to the positive integers giving the
partition into countably many subsets, and in place of ``depth'' in the
proof, use a ``weighted depth'' that takes account of this function.
Incidentally, this result is analogous to~(\ref{x.unccof}), since the
latter can be translated
as saying that whenever a generating set for $S$ is partitioned into
countably many subsets, the union of some finite number of these
subsets is still a generating set.)
For an example of an algebra $S$ satisfying~(\ref{x.fg/*D})
with uncountably many primitive operations, such that no
{\em finite} set of these suffices, let $S=\omega,$ and let the set of
primitive operations be $\omega^\omega,$ i.e., the set of
{\em all} unary operations on $S.$
Then $S\<^\omega$ is generated under these
operations by a single element, the identity
function, but one cannot obtain all of $S$ from the generating
set $\{0\}$ using finitely many operations.
We saw in Theorem~\ref{T.>*kgens} that large direct products led
to algebras requiring increased numbers of generators; but
Proposition~\ref{P.any} showed that this increase could be
arbitrarily small.
The next result shows that when this growth is
``too'' small, this can have restrictive consequences.
Recall that $\mathfrak{d}$ denotes the cofinality of the partially
ordered set $\omega^\omega$ of all sequences of natural numbers under
componentwise comparison, a cardinal somewhere
between $\aleph_1$ and $2^{\aleph_0}.$
(It called the ``dominating number'', and generally described
slightly differently \cite{JEV}, but the definitions are
equivalent \cite[Remark 2.3]{AB}.)
\begin{proposition}\label{P.cof}
Assume the cofinality $\mathfrak{d}$ of the partially
ordered set $\omega^\omega$ is a regular cardinal.
Let $T$ be a type of finitary algebras \textup{(}with no
restriction on the cardinality of its set of operations\textup{)} and
$(M_i)_{i\in I}$ an infinite family of nonempty algebras of that type,
such that $\prod_{i\in I}M_i$ can be generated by $<\mathfrak{d}$
elements \textup{(}or more generally, such that $\prod_{i\in I}M_i$
cannot be written as the union of a well-ordered chain
of $\mathfrak{d}$ proper subalgebras\textup{)}.
Then all but finitely many of the $M_i$
satisfy~\textup{(\ref{x.unccof})} and~\textup{(\ref{x.bddgen})}.
\end{proposition}\begin{proof}
It follows from the assumption that $\mathfrak{d}$ is regular
that if $\prod_{i\in I}M_i$ can be generated
by $<\mathfrak{d}$ elements, then it cannot be written as the union of a
well-ordered chain of $\mathfrak{d}$ proper subalgebras; so the
latter conditions does indeed generalize the former.
To complete the proof of the proposition, let us assume given a
family of algebras $(M_i)_{i\in I}$ having
an infinite subfamily, which we can clearly assume
countable, $(M_{i(n)})_{n\in\omega},$ such that none of
the $M_{i(n)}$ satisfy both~(\ref{x.unccof}) and~(\ref{x.bddgen}),
and show that $\prod_{i\in I}M_i$ {\em can} be written as such a union.
As noted in the proof of Theorem~\ref{T.fg/=>},
the assumption that no $M_{i(n)}$ satisfies
both~(\ref{x.unccof}) and~(\ref{x.bddgen}) is equivalent
to the statement that each $M_{i(n)}$ may be written
$M_{i(n)} = \bigcup_{m\in\omega} X_{n,m}$ for some chain of proper
subsets $\dots\subseteq X_{n,m}\subseteq X_{n,m+1}\subseteq\dots,$ such
that for each primitive operation $f$ of $T,$ and each $m\in\omega,$
one has $f(X_{n,m},\dots,X_{n,m})\subseteq X_{n,m+1}.$
Choosing such a chain for each $n,$ let us define
for each $x\in\prod_{i\in I}M_i$ (where $x=(x_i)_I)$
the element $\mathrm{rank}(x)\in\omega^\omega$ to have
as $\!n\!$th component, for each $n\in\omega,$ the
least $m$ such that $x_{i(n)}\in X_{n,m}.$
Note that for $f$ a primitive operation of $T,$ say of arity $r,$
and $x^{(1)},\dots,x^{(r)}\in\prod_{i\in I}M_i,$ the condition
just stated relating $f$ and the sets $X_{m,n}$ shows that
an upper bound for $\mathrm{rank}(f(x^{(1)},\dots,x^{(r)}))$ will be
$(\mathrm{rank}(x^{(1)})\vee\dots\vee\mathrm{rank}(x^{(r)}))+\Delta(1),$
where $\vee$ denotes componentwise supremum, and
$\Delta:\omega\rightarrow\omega^\omega$ is the diagonal map.
Now let $\{s_\alpha\mid\alpha\in\mathfrak{d}\}$ be a cofinal subset
of $\omega^\omega;$ and for each $\alpha\in\mathfrak{d},$
let $N_\alpha\subseteq\prod_{i\in I}M_i$ be the set of
elements $x$ such that $\mathrm{rank}(x)$ is majorized
by some element $(s_{\beta_1}\vee\dots\vee s_{\beta_r})+\Delta(k)$
with $\beta_1,\dots,\beta_r\in\alpha$ and $k\in\omega.$
From our bound on $\mathrm{rank}(f(x^{(1)},\dots,x^{(r)})),$
we see that each $N_\alpha$ is a subalgebra; and
the map $\alpha\mapsto N_\alpha$ is clearly isotone.
Since $\{s_\alpha\mid\alpha\in\mathfrak{d}\}$ is cofinal
in $\omega^\omega,$ the union of these subalgebras
is all of $\prod_{i\in I}M_i.$
Finally, given $\alpha,$ to see that the subalgebra $N_\alpha$ is
proper, note that the cardinality of $\{s_\beta\mid\beta\in\alpha\}$
is $<\mathfrak{d},$ hence so is the cardinality of the set of
elements $(s_{\beta_1}\vee\dots\vee s_{\beta_r})+\Delta(k)$
$(\beta_1,\dots,\beta_r\in\alpha,\ k\in\omega),$
hence that set is not cofinal in $\omega^\omega,$ hence we can
construct an $x\in\prod_{i\in I}M_i$ whose rank is not majorized
by any member of that set, i.e., which is not in $N_\alpha.$
\end{proof}
Some final miscellaneous remarks:
An easily described class of algebras that satisfy~(\ref{x.bddgen})
but not~(\ref{x.unccof}) are the infinite algebras in which
the value of each operation is always one of its arguments; for
instance, an infinite chain regarded as a lattice, or
an infinite set with no operations.
These examples also show that if we generalize
\cite[Question~8]{Sym_Omega:1},
which asked whether a countably infinite group can
satisfy~(\ref{x.bddgen}), to algebras of arbitrary type,
the answer is affirmative.
Although~(\ref{x.fg/*D}) and~(\ref{x.unccof}) are preserved under
taking homomorphic images, and although the same is true
of~(\ref{x.bddgen}) in any algebra whose structure
includes a structure of group, it is not true
of~(\ref{x.bddgen}) in general, for if $S'$ is a homomorphic
image of $S,$ the inverse image of a generating set for $S'$
may not be a generating set for $S.$
For example, let $S'$ be the semilattice of finite subsets
of $\omega$ under union, and $S$ the semilattice obtained by
hanging from each $x\in S',$ a new element
%
\begin{xlist}\item\label{x.x0\kappa$''.)
And, of course, the diagonal subalgebra $\Delta(S)$
of a power $S^\kappa$ was merely the most obvious example of the
context introduced in the first paragraph of \S\ref{S.Sym},
which would be worth considering in greater generality.
Another kind of algebra structure involving an arbitrarily large
set of operations, all of finite arity, is a set on which
a Boolean ring $B$ ``acts'', in the sense of \cite{bool_act}.
So Theorem~\ref{T.bigorsmall} also applies to these structures,
and it might be of interest to examine that case,
possibly combining the Boolean operations with others.
\begin{thebibliography}{00}
\bibitem{GMB.rl} George M. Bergman,
{\it Notes on composition of maps,} 6pp,
http://math.berkeley.edu/\linebreak[0]{$\!\sim$}gbergman/%
grad.hndts/left$+$right.ps\,, 1997.
\bibitem{ZB} \bysame,
{\it Boolean rings of projection maps,}
J. London Math. Soc. (2) {\bf 4} (1972), 593-598.
MR~{\bf 47}\#93.
\bibitem{bool_act} \bysame,
{\it Actions of Boolean rings on sets,}
Algebra universalis {\bf 28} (1991), 153--187.
MR~{\bf 92g}:06021.
\bibitem{Sym_Omega:1} \bysame,
{\it Generating infinite symmetric groups,}
to appear, J.\ London Math.\ Soc..
Preprint 8\,pp., at http://math.berkeley.edu/{$\!\sim$}gbergman/%
papers/Sym\<\_\,Omega:1.\{tex,dvi\} and
arXiv:math.GR/0401304\,.
\bibitem{AB} Andreas Blass,
{\it Combinatorial cardinal characteristics of the continuum,}
readable at http://www.math.lsa.umich.edu/{$\!\sim$}ablass/set.html\,,
to appear in {\it Handbook of Set Theory,} ed. M. Foreman, M. Magidor,
and A. Kanamori.
\bibitem{SUC1} Stephen U. Chase, % QA1 A53
{\it Direct products of modules,}
Trans. Amer. Math. Soc. {\bf 97} (1960), 457--473.
MR~{\bf 22}\#11017.
\bibitem{SUC2} \bysame, % QA1 A56
{\it A remark on direct products of modules,}
Proc. Amer. Math. Soc. {\bf 13} (1962), 214--216.
MR~{\bf 25}\#1197.
\bibitem{SUC3} \bysame, % QA1 P18
{\it On direct sums and products of modules,}
Pacific J. Math. {\bf 12} (1962), 847--854.
MR~{\bf 26}\#2472.
\bibitem{YdC} Yves de Cornulier, {\it Strongly bounded groups and
infinite powers of finite groups}, to appear, Communications in Algebra.
Preprint 8\,pp.\ at arXiv/math.GR/0411466.
\bibitem{KE} Katsuya Eda,
{\it Free $\!\sigma\!$-products and non-commutatively slender groups,}
J. Algebra {\bf 148} (1992), 243--263.
MR~{\bf 94a}:20040.
\bibitem{KE+SSh} \bysame\ and Saharon Shelah,
{\it The non-commutative Specker phenomenon in the uncountable case,}
J. Algebra {\bf 252} (2002), 22--26.
MR~{\bf 2003g}:20041.
% case of measurable cardinal
\bibitem{EK} Robert El Bashir and Tom\'a\v{s} Kepka,
{\it Modules commuting (via Hom) with some limits,}
Fund. Math. {\bf 155} (1998), 271--292.
MR~{\bf 99b}:16002.
\bibitem{EKN} \bysame, \bysame\ and Petr N\v{e}mec,
{\it Modules commuting (via Hom) with some colimits,}
Czechoslovak Math. J. {\bf 53} (128) (2003), 891--905.
MR~{\bf 2004i}:16002.
\bibitem{CF} Carl Faith,
{\it Rings with ascending condition on annihilators,}
Nagoya Math. J. {\bf 27} (1966), 179--191.
MR~{\bf 33}\#1328.
% qF <=> oA projs are inj <=> right self-inj w ACC on rt annulets
\bibitem{GH} Graham Higman,
{\it Unrestricted free products, and varieties of topological groups,}
J. London Math. Soc. {\bf 27} (1952), 73--81.
MR~{\bf 13},623d.
\bibitem{NJ} Nathan Jacobson, % QA162 .J3 1975 v.2
{\it Lectures in abstract algebra. Volume II: Linear algebra,}
Van Nostrand, 1953, and Springer GTM, No. 31, 1975.
MR~{\bf 14},837e\, and\, MR~{\bf 51}\#5614.
\bibitem{AK.CR} Anatole Khelif, {\it A propos de la
propri\'et\'e de Bergman,}
C.R. Acad. Sci. Paris, Ser.I {\bf 342} (2006), 377--380.
\bibitem{AK} \bysame, {\it A propos de la
propri\'et\'e de Bergman,}
preprint, 16\,p., July 2005.
(Author's e-mail address:
khelif@logique.jussieu.fr\,.)
\bibitem{SK+JT} Sabine Koppelberg and Jacques Tits,
{\it Une propri\'et\'e des produits directs
infinis de groupes finis isomorphes,}
C. R. Acad. Sci. Paris S\'er. A {\bf 279} (1974), 583--585.
MR~{\bf 51}\#13058.
%inf/ dir power of a nonab' simple f' gp [has unctbl cofinlty]
\bibitem{Lam} T. Y. Lam,
{\it Exercises in classical ring theory,}
Springer-Verlag Problem Books in Mathematics.
1st ed. 1995, 2nd ed. 2003.
MR~{\bf 95m}:16001 and MR~{\bf 2004g}:16001.
\bibitem{Lady} E. L. Lady,
{\it Slender rings and modules,}
Pacific J. Math. {\bf 49} (1973), 397--406.
MR~{\bf 50}\#7262.
\bibitem{Lenzing} Helmut Lenzing,
{\it Direct sums of projective modules as direct summands of their
direct product,}
Comm. Algebra {\bf 4} (1976), 681--691.
MR~{\bf 53}\#8137.
\bibitem{ZM} Zachary Mesyan,
{\it Generating endomorphism rings of infinite direct sums and
products of modules,} J.\,Alg. {\bf 283} (2005), 364--366.
MR~{\bf 2005g}:16057.
\bibitem{JO'N} John D. O'Neill,
{\it On direct products of modules,}
Comm. Algebra {\bf 12} (1984), 1327--1342.
MR~{\bf 85f}:20046.
\bibitem{JO'Nrg} \bysame,
{\it An unusual ring,}
J. London Math. Soc. (2) {\bf 44} (1991), 95--101.
MR~{\bf 92i}:16005.
\bibitem{OO} Oystein Ore,
{\it Some remarks on commutators,}
Proc. Amer. Math. Soc. {\bf 2} (1951), 307--314.
MR~{\bf 12},\,671e.
\bibitem{WSt} W. Stephenson,
{\it Characterizations of rings and modules by means of
lattices}, Ph.D.\ thesis, University of London, 1967, unpublished.
\bibitem{JEV} Jerry E. Vaughan,
{\it Small uncountable cardinals and
topology. With an appendix by S. Shelah,}
http://www1.elsevier.com/homepage/sac/opit/11/article.pdf,
MR1078647, pp.195--218 in
{\it Open problems in topology,}
ed. Jan van Mill and George M. Reed,
North-Holland, Amsterdam, 1990.
MR~{\bf 92c}:54001.
\bibitem{BZH} Birge Zimmermann-Huisgen,
{\it Direct products of torsion modules,}
Arch. Math. (Basel) {\bf 38} (1982), 426--431.
MR~{\bf 84a}:13011.
\bibitem{ZH+Z} \bysame\ and Wolfgang Zimmermann,
{\it Algebraically compact rings and modules,}
Math. Z. {\bf 161} (1978), 81--93.
MR~{\bf 58}\#16792.
\end{thebibliography}
\end{document}
*