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\begin{document}
\begin{center}
\texttt{Comments, corrections, and related references welcomed,
as always!}\\[.5em]
{\TeX}ed \today
\vspace{2em}
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\title%
{Homomorphic images of pro-nilpotent algebras}
\thanks{This preprint is accessible online at
\url{/~gbergman/papers/}
and at \url{http://arXiv.org/abs/arXiv:0912.0020}\,.
The former version is likely to be updated more frequently than
the latter.
}
\subjclass[2010]{Primary: 16N20, 16N40, 17A01, 18A30.
% J(R), nilrad, nonass lim&co
Secondary: 13C13, 17B30.} % 16S50, 17A65
% other_mdls slv&np.Lie End() rad.th
\keywords{Homomorphic image of an inverse limit of nilpotent algebras,
multiplication algebra of a nonassociative algebra,
Jacobson radical associative algebra, Hopfian module,
solvable Lie algebra.
}
\author{George M. Bergman}
\address{University of California\\
Berkeley, CA 94720-3840, USA}
\email{gbergman@math.berkeley.edu}
\begin{abstract}
It is shown that any finite-dimensional homomorphic image of an
inverse limit of nilpotent not-necessarily-associative algebras
over a field is nilpotent.
More generally, this is true of algebras over a general commutative
ring $k,$ with ``finite-dimensional'' replaced by ``of
finite length as a $\!k\!$-module''.
These results are obtained by considering the multiplication algebra
$M(A)$ of an algebra $A$ (the associative algebra of $\!k\!$-linear maps
$A\to A$ generated by left and right multiplications by elements
of $A),$ and its behavior with respect
to nilpotence, inverse limits, and homomorphic images.
As a corollary, it is shown that a finite-dimensional homomorphic
image of an inverse limit of finite-dimensional {\em solvable}
Lie algebras over a field of characteristic~$0$ is solvable.
It is also shown by example that {\em infinite}-dimensional
homomorphic images of pro-nilpotent algebras can have
properties far from those of nilpotent algebras; in particular,
properties that imply that they are not residually nilpotent.
Several open questions and directions for further
investigation are noted.
\end{abstract}
\maketitle
% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
\section{General definitions.}\label{S.defs}
Throughout this note, $k$ will be a commutative associative unital ring,
and an ``algebra'' will mean a $\!k\!$-algebra;
i.e., a $\!k\!$-module $A$ given with a $\!k\!$-bilinear
multiplication $A\times A\to A,$ not necessarily associative or unital.
A right, left, or $\!2\!$-sided ideal of a $\!k\!$-algebra $A$ will
mean a $\!k\!$-submodule closed under left, respectively right
multiplication by elements of $A,$ respectively both.
Recall that if $A$ is a {\em nonunital associative} algebra contained
in a {\em unital} associative algebra $A',$ then the identity
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.(1+x)(1+y)}
$(1+x)(1+y)\ =\ 1+(x+y+xy)\quad (x,y\in A)$
\end{minipage}\end{equation}
%
which holds in $A'$ motivates
one to define, on $A,$ the operation of {\em quasimultiplication},
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.quasimult}
$x*y\ =\ x+y+xy.$
\end{minipage}\end{equation}
%
This is again associative, and has $0$ as identity element;
an element $x\in A$ is called {\em quasiinvertible} if
there exists $y\in A$ such that $x*y=y*x=0;$ equivalently,
if $1+x$ is invertible in the submonoid
$\{1+u\mid u\in A\}$ of $A'$ with respect to ordinary multiplication.
In particular, every nilpotent element $x\in A$ is
quasiinvertible, with quasiinverse $-x+x^2-\dots+(-x)^n+\cdots\,.$
The Jacobson radical of $A$ is the largest ideal consisting of
quasiinvertible elements; so an associative algebra is
Jacobson radical if and only if every element is quasiinvertible.
We shall write ``Jacobson radical'' and ``radical'' interchangeably
in this note, using the former mainly in statements of results.
We shall only use these terms in reference to associative algebras.
If $A$ is a not-necessarily-associative algebra,
let us write $\En(A)$ for the associative unital $\!k\!$-algebra
of all endomorphisms of $A$ as a $\!k\!$-module.
(Since $\mathrm{End}(A)$ generally denotes the monoid of algebra
endomorphisms of $A,$ we use this slightly different symbol
for its algebra of module endomorphisms.)
For every $x\in A,$ we define the left and right multiplication maps
$l_x,\ r_x\in\En(A)$ by
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.rl}
$l_x(y)\ =\ x\,y,\quad r_x(y)\ = \ y\,x,$
\end{minipage}\end{equation}
%
and denote by $M(A)$ the generally {\em nonunital} subalgebra
of $\En(A)$ generated by these maps, as $x$ runs over~$A;$ this is
called the {\em multiplication algebra} of $A$ \cite[p.14]{Schafer}.
An algebra $A$ is called {\em nilpotent} if for some $n>0,$ all
length-$\!n\!$
products of elements of $A,$ no matter how bracketed, are zero.
We shall see that $M(A)$ is nilpotent if and only if $A$ is
nilpotent (not hard to prove, but not quite trivial).
\section{Preview of the proof of our main result, and of a counterexample.}\label{S.preview}
If $A$ is a nilpotent algebra, then the associative algebra
$M(A),$ being nilpotent, will in particular be radical.
Now though the property of being nilpotent is not preserved by inverse
limits, that of being radical is, and is
likewise preserved under surjective homomorphisms.
To use these facts, we have to know how $M(A)$
behaves with respect to homomorphisms and inverse limits.
In general, a homomorphism of algebras $h:A\to B$ does not induce a
homomorphism $M(h):M(A)\to M(B);$ but we shall see that
it does if $h$ is surjective, and that $M(h)$ is then also surjective.
The need for $h$ to be surjective will not be a problem for us, because
if an algebra $A$ is an inverse limit of nilpotent algebras $A_i,$
then by replacing the $A_i$ with appropriate subalgebras, we can
get a new system having the same inverse limit $A,$ and such
that the new projection maps $A\to A_i$ and connecting
maps $A_i\to A_j$ are surjective.
Once these conditions hold, we shall find that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.Mlm}
$M(\lm A_i)\ \subseteq\ \lm M(A_i)\ \subseteq\ \En(\lm A_i).$
\end{minipage}\end{equation}
%
Hence, if the $A_i$ are all nilpotent, the elements
of $M(\lm A_i)$ will all have quasiinverses in the radical
algebra $\lm M(A_i),$ and hence in $\En(\lm A_i).$
From\ this we shall be able to deduce that if $B$ is a homomorphic
image of $A=\lm A_i,$ then for all $u\in M(B),$ the
linear map $1+u\in\En(B)$ is surjective.
If, moreover, $B$ has finite length as a $\!k\!$-module,
this surjectivity makes these
maps $1+u$ $(u\in M(B))$ invertible; i.e., it makes
the elements $u$ quasiinvertible in $\En(B).$
If we could say that they were quasiinvertible in $M(B),$
this would make $M(B)$ radical.
We can't initially say that; but we shall find that the
quasiinvertibility of these images in $\En(B)$
allows us to extend the domain of our map $M(A)\to\En(B)$ to a radical
subalgebra of $\En(A)$ containing $M(A).$
Since a homomorphic image of a radical algebra is radical,
we get a radical subalgebra of $\En(B)$ containing $M(B).$
Using once more the finite length assumption on $B,$ we will conclude
that that subalgebra of $\En(B)$ is nilpotent, hence so is $M(B);$
hence so is $B,$ yielding our main result (first paragraph of abstract).
It is curious that in the above development, before
assuming that $B$ had finite length, we could conclude that the maps
$1+u$ $(u\in M(B))$ were surjective, but not that they were injective.
Let me sketch a concrete example (to be given in detail in \S\ref{S.eg})
showing how injectivity can fail, and why surjectivity must
nonetheless hold.
Suppose one takes an inverse limit $A$ of nilpotent associative
algebras $A_i,$ and divides out by the two-sided ideal $(w)$
generated by an element of the form
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.w=y-xyz}
$w\ =\ y\,-\,x\,y\,z\ =\ (1-l_x r_z)(y),$
\end{minipage}\end{equation}
%
where $x,y,z\in A.$
In the resulting algebra $A/(w),$ let us, by abuse of notation, use
the same symbols
$x,y,z$ for the images of the corresponding elements of $A.$
Thus, in that algebra we have $y=x\,y\,z;$ equivalently, $y$
is annihilated by the operator $1-l_x r_z.$
Hence the latter operator will not be injective if
$y\neq 0$ in $A/(w),$ in other words, if $y\notin(w)$ in $A.$
Can $y$ in fact fail to lie in $(w)$?
Note that we can formally solve~(\ref{d.w=y-xyz}) for $y,$ getting
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.y=w+}
$y\ =\ w+xwz+x^2wz^2+\dots\,.$
\end{minipage}\end{equation}
%
In each of the nilpotent algebras $A_i$ of which
$A$ is the inverse limit,~(\ref{d.y=w+}) is literally true,
since the images of $x$ and $z$ are nilpotent; so in each of those
algebras, the image of $y$ does lie in the image of $(w).$
But as we pass to larger and larger algebras $A_i,$
the number of nonzero terms of~(\ref{d.y=w+}) can grow without bound,
so there is no evident way to express $y\in A$ as a member
of the ideal $(w);$ and indeed,
we shall see in \S\ref{S.eg} that for appropriate choice
of these algebras and elements, $y$ does not belong to
that ideal, so that on $A/(w),$ $1-l_x r_z$ is non-injective.
By the above considerations,
this makes $A/(w)$ non-residually-nilpotent; a quicker way
to see this is to note that the equations $y=x\,y\,x=x^2y\,x^2=\dots$
show that $y\in\bigcap_n (A/(w))^n,$ whence the image of
$y$ in any nilpotent homomorphic image of $A/(w)$ must be zero.
On the other hand, I claim that whenever $x$ and $z$ are elements
of a homomorphic image $A/U$ of an inverse limit $A$ of nilpotent
associative
algebras $A_i,$ the operator $1-l_x r_z$ will be {\em surjective}.
Given an element $w\in A/U$ which we want to show is in the range
of this operator, let us lift $x,\,w,z$ to elements of $A,$
which we will denote by the same symbols.
Seeking an element $y\in A$ mapped by $1-l_x r_z$ to $w,$
we get the same formal expression as before, $y=w+xwz+x^2wz^2+\dots\,.$
Again, this sum cannot be evaluated using the
algebra operations of $A;$ but it can in each of the $A_i,$
and we find that the resulting elements of the $A_i$ yield, in
the inverse limit $A,$ an element $y$ satisfying $w=y-xyz,$ as desired.
\section{Acknowledgements, and some related literature.}\label{S.ackn}
I am indebted to Nazih Nahlus for conjecturing the main result of this
note in the case where the $A_i$ are finite-dimensional nilpotent
Lie algebras over an algebraically closed field of characteristic~$0.$
I am also grateful to Georgia Benkart, Christian Jensen,
Karl H. Hofmann, Greg Marks, Nazih Nahlus, and the referee
of this note for helpful comments on earlier drafts.
For related results, by Nahlus and the present author,
on homomorphic images of {\em direct products}
of algebras, see~\cite{prod_Lie1},~\cite{prod_Lie2}.
Inverse limits of finite-dimensional Lie groups and Lie algebras are
also studied in \cite{KHH+SAM},
though with a different emphasis from this note, focusing on
homomorphisms continuous in the inverse limit topology.
\section{Basic properties of nilpotence.}\label{S.np}
The condition of nilpotence on a nonassociative algebra $A$ can
be characterized in several ways.
In what follows, whenever $B$ and $C$ are $\!k\!$-submodules
of $A,$ we understand $BC$ to mean the $\!k\!$-submodule of $A$
spanned by all products $bc$ $(b\in B,\,c\in C).$
Let us define recursively $\!k\!$-submodules
$A_{[n]}$ and $A_{(n)}$ $(n=1,2,\dots)$ of any algebra $A$ by
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.weak_series}
$A_{[1]}\ =\ A,\qquad A_{[n+1]}\ =\ A\,A_{[n]}+A_{[n]}\,A,$
\end{minipage}\end{equation}
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.strong_series}
$A_{(1)}\,\ =\ A,\qquad A_{(n+1)}\ =
\ \sum_{00,$\quad $A_{[n]}\ \supseteq\ A_{[n+1]}$\quad
and\quad $A_{(n)}\ \supseteq\ A_{(n+1)},$
\end{minipage}\end{equation}
%
and also that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.[]<()}
for all $n,\quad A_{[n]}\ \subseteq\ A_{(n)}.$
\end{minipage}\end{equation}
If $A$ is associative, then $A_{[n]}$ and $A_{(n)}$ clearly
coincide, their common value being the submodule of $A$ spanned
by all $\!n\!$-fold products, which we shall write $A^n.$
In the next lemma, for an arbitrary algebra $A,$
we apply the latter notation to the associative
algebra $M(A)\subseteq\En(A),$ defined in~\S\ref{S.defs}.
\begin{lemma}\label{L.nilp}
If $A$ is an algebra, then the following conditions are equivalent:
\vspace{.2em}
\strt\ \textup{(i)}~ There exists a positive integer $n_1$ such
that $A_{[n_1]}=\{0\}.$
\vspace{.2em}
\strt\,\textup{(ii)}~ There exists a positive integer $n_2$ such
that $A_{(n_2)}=\{0\}.$
\vspace{.2em}
\textup{(iii)}~ There exists a positive integer $n_3$ such
that $M(A)^{n_3}=\{0\}.$
\vspace{.2em}
Moreover, if the above equivalent conditions hold, then
letting $N_1,\ N_2,\ N_3$ be the least $n_1,\ n_2,\ n_3$
as in those statements, we have
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.N1N2N3}
$N_3\ =\ \max(1,\,N_1-1),\qquad N_1\ \leq\ N_2\ \leq\ 2^{N_1-2}+1.$
\end{minipage}\end{equation}
%
\end{lemma}
\begin{proof}
We will first establish the stated relations between
conditions~(i) and~(iii), and between $N_1$ and $N_3.$
Let $l_A\subseteq M(A)$ denote the $\!k\!$-submodule of all
left-multiplication operators $l_x$ $(x\in A)$ and
$r_A$ the $\!k\!$-submodule of all right-multiplication operators $r_x.$
We claim that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.M(A)^n+1}
for all $n>0,$\quad $M(A)^{n+1}\ =\ (l_A+r_A)\,M(A)^n.$
\end{minipage}\end{equation}
%
Here ``$\supseteq$'' is clear.
To see ``$\subseteq$'', note that $M(A)$ consists of all sums
of products of one or more elements of $l_A + r_A,$
hence $M(A)^{n+1}$ consists of all sums of products
of $n+1$ or more such elements.
If such a product has more than $n+1$ such factors, we can, by
associativity, group these
into $n+1$ subproducts, of which the first is a single factor.
Moreover, the assumption $n>0$ assures us that the first of $n+1$
factors is not the only one.
So written, our product clearly belongs to $(l_A+r_A)M(A)^n,$
giving~``$\subseteq$''.
Now the recursive step
of~(\ref{d.weak_series}) says that $A_{[n+1]}=(l_A+r_A)A_{[n]},$
so using~(\ref{d.M(A)^n+1}), and induction from the case $n=1,$
one concludes that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.M^nA}
for all $n>0,$\quad $A_{[n+1]}\ =\ M(A)^n(A)$
\end{minipage}\end{equation}
%
This gives the equivalence of~(i) and~(iii) on the
one hand, and the initial equality of~(\ref{d.N1N2N3}) on the other.
Turning to the submodules $A_{(n)},$ the inclusion~(\ref{d.[]<()})
yields the implication (ii)$\!\implies\!$(i) and the first
inequality of~(\ref{d.N1N2N3}).
To get the reverse implication and the final
inequality of~(\ref{d.N1N2N3}), we first note that both
hold trivially if $A=\{0\},$ in which case $N_1=N_2=1.$
To prove them for nonzero $A,$ in which case any $n_1$ as
in~(i), or $n_2$ as in~(ii), must be $\geq 2,$ it suffices to show that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.2^n-2+1}
for $n\geq 2,$\quad $A_{(2^{n-2}+1)}\ \subseteq\ A_{[n]}.$
\end{minipage}\end{equation}
For $n=2,$ we have equality.
Assuming we know~(\ref{d.2^n-2+1}) for some $n\geq 2,$ we look at
the definition of $A_{(2^{n-1}+1)}$
as in~(\ref{d.strong_series}), and note that in each of
the summands $A_{(m)}A_{((2^{n-1}+1)-m)},$ one of the indices
$m$ or $(2^{n-1}+1)-m$ will be $\geq 2^{n-2}+1;$
hence the summand will be contained
in $A_{(2^{n-2}+1)}A+A\,A_{(2^{n-2}+1)}.$
By inductive hypothesis,
this is $\subseteq A_{[n]}A+A\,A_{[n]}=A_{[n+1]},$ as required.
\end{proof}
(If we think of an arbitrarily parenthesized nonassociative
product as representing
a binary tree of multiplications, the last part of the above proof
is essentially a calculation
showing that a binary tree with $2^{n-2}+1$ leaves $(n\geq 2)$
must contain a chain with $n$ nodes.)
\begin{definition}\label{D.nilp}
An algebra $A$ satisfying the equivalent conditions \textup{(i)-(iii)}
of Lemma~\ref{L.nilp} will be called {\em nilpotent}.
\end{definition}
Lemma~\ref{L.nilp} now gives
\begin{corollary}\label{C.nilpiff}
For any algebra $A,$ $M(A)$ is nilpotent if and only if
$A$ is nilpotent.\qed
\end{corollary}
(In the sketch in the preceding section, we defined
nilpotence in terms of condition~(ii) of Lemma~\ref{L.nilp},
as is often done.
The verification that this is equivalent to~(iii) required
the ``$2^{n-2}+1$'' part of the proof of that
lemma, which is why we described it as not quite trivial.)
\vspace{.5em}
We end this section with some further observations on nilpotence
that will not be needed for our main results.
In the inequality $N_1\leq N_2$ of~(\ref{d.N1N2N3}),
we have equality whenever $A$ is associative,
by the sentence following~(\ref{d.[]<()}).
For examples where the upper bound $N_2\leq 2^{N_1-2}+1$ is
achieved, take any positive integer $n,$
and consider the (nonassociative) $\!k\!$-algebra $A$ such that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.xixi}
$A$ is free as a $\!k\!$-module on a basis $\{x_1,\dots,x_{n-1}\},$
with multiplication given by $x_m\,x_m\ =\ x_{m+1}$ for
$1\leq m\leq n-2,$ and all other products of basis
elements equal to zero (including $x_{n-1}\,x_{n-1}).$
\end{minipage}\end{equation}
It is easy to verify by induction that for every $i\leq n,$
$A_{[i]}$ is the submodule spanned by $\{x_i,\dots,x_{n-1}\}.$
In particular, $A_{[i]}$ becomes $\{0\}$ starting with $i=n,$
so the $N_1$ of Lemma~\ref{L.nilp} is $n$ for this algebra.
Less obvious, but no harder to verify, is the statement that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.A()=}
for every $i\leq n,$ and $j$ with $2^{i-2}1,$ and $j$ lies in
the above range, then $j$ can be written as the sum of two
integers $\leq 2^{i-2},$ but not as the sum of two
integers $\leq 2^{i-3}.$
Using this fact, and the definitions~(\ref{d.strong_series})
and~(\ref{d.xixi}), one gets~(\ref{d.A()=}) by induction on $i.$
So for this algebra, $N_2=2^{n-2}+1=2^{N_1-2}+1.$
The next lemma shows that Lie algebras
behave like associative algebras in this respect.
\begin{lemma}\label{L.[]=()}
\textup{(i)}~ If $A$ is an associative or Lie algebra, then
for all positive integers $p$ and $q,$
$A_{[p]}A_{[q]}\subseteq A_{[p+q]}.$
\textup{(ii)}~ If $A$ is any algebra
for which the conclusion of\textup{~(i)} holds, then
for every positive integer $n,$ $A_{[n]}=\nolinebreak A_{(n)}.$
\end{lemma}
\begin{proof}
For associative algebras,~(i) is a weakening of the familiar
identity $A^p A^q = A^{p+q}.$
For Lie algebras, let us switch to bracket notation, and note
that by anticommutativity, the recursive step of
our definition~(\ref{d.weak_series})
can be written $A_{[n+1]}=[A,\,A_{[n]}].$
This immediately gives~(i) for $p=1$ and arbitrary $q.$
So let $p>1,$ and assume inductively that the result is
true for all smaller $p.$
Using the Jacobi identity at the second step below, and
our inductive assumption at the third and fourth steps, we compute
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.jacobi}
$
[A_{[p]},\,A_{[q]}]\ =
\ [\,[A,\,A_{[p-1]}],\,A_{[q]}]\ \subseteq
\ [A,\,[A_{[p-1]},\,A_{[q]}]\,]\ +
\ [A_{[p-1]},\,[A,\,A_{[q]}]\,]\\[.3em]
\strt\qquad\subseteq\ [A,\,A_{[p+q-1]}]\ +
\ [\,A_{[p-1]},\,A_{[q+1]}]\ \subseteq
\ A_{[p+q]}\ + \ A_{[p+q]}\ =
\ A_{[p+q]}.$
\end{minipage}\end{equation}
To get~(ii), recall from~(\ref{d.[]<()}) that
$A_{[n]}\subseteq A_{(n)}$ for arbitrary algebras, and note that
by definition we have equality when $n=1.$
Thus, it suffices to prove the inclusion
$A_{[n]}\supseteq A_{(n)}$ when $n>1,$
inductively assuming this inclusion for smaller $n.$
The inclusion we are to
prove is clearly equivalent to the statement that each
summand $A_{(m)}A_{(n-m)}$ in the definition of $A_{(n)}$
is contained in $A_{[n]}.$
By our inductive
hypothesis, $A_{(m)}A_{(n-m)}$ is contained in $A_{[m]}A_{[n-m]},$
and since we are assuming the conclusion of~(i), this is
contained in~$A_{[n]}.$
\end{proof}
\section{Properties of $M(A).$}\label{S.MA}
As recalled above, the multiplication algebra $M(A)$ of any
algebra $A$ is the (generally nonunital) subalgebra
of the associative algebra $\En(A)$ generated by the
left and right multiplication operators $l_x$ and $r_x,$
as $x$ ranges over $A.$
For a general homomorphism of algebras $h:A\to B,$
there is no natural way to map $M(A)$ to $M(B).$
For instance, if $h$ is the inclusion of a subalgebra $A$ in
an algebra $B,$ such that a central element $x\in A$
becomes noncentral in $B,$ then $l_x=r_x$ in $M(A),$ but the
corresponding members of $M(B)$ are distinct.
For surjective homomorphisms, however, this problem goes away:
\begin{lemma}\label{L.surj}
If $h:A\to B$ is a surjective homomorphism of algebras, then
there exists a unique homomorphism $M(h):M(A)\to M(B)$ such that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.Mfx}
for all $x\in A,\quad M(h)(l_x)\ =\ l_{h(x)}$\quad
and \quad $M(h)(r_x)\ =\ r_{h(x)},$
\end{minipage}\end{equation}
%
equivalently, such that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.Mfu}
for all $u\in M(A)$ and $a\in A,$\quad $(M(h)(u))(h(a))\ =\ h(u(a)).$
\end{minipage}\end{equation}
Moreover, $M(h)$ is surjective.
\end{lemma}
\begin{proof}
$\mathrm{Ker}(h)$ is an ideal of $A,$ hence it is
carried into itself by every map $l_x$ and every map $r_x,$ and thus
by every element $u$ of the algebra $M(A)$ generated by such maps.
Hence if two elements $a,\ a'\in A$ differ by an element
of $\mathrm{Ker}(h),$ so do $u(a)$ and $u(a');$ that is,
if $h(a)=h(a'),$ then $h(u(a))=h(u(a'));$ so as $B=h(A),$ we get
a well-defined linear map $M(h)(u):B\to B$ satisfying~(\ref{d.Mfu}).
It is immediate that $M(h)$ is an algebra homomorphism, and
acts by~(\ref{d.Mfx}) on elements $l_x$ and $r_x.$
It is surjective because it carries the generating
set $\{l_x,\,r_x\mid x\in A\}$ of $M(A)$ to the corresponding
generating set of $M(B).$
\end{proof}
It is also immediate that for a composable pair of surjective
algebra homomorphisms $h,\ g,$ we have $M(hg)=M(h)M(g);$
and that if we write $\mathrm{id}_A$ for
the identity homomorphism $A\to A,$ then
$M(\mathrm{id}_A)=\mathrm{id}_{M(A)}.$
Thus, $M$ is a functor from the category
whose objects are $\!k\!$-algebras, and whose morphisms
are {\em surjective} algebra homomorphisms, to the category of
associative $\!k\!$-algebras.
Now suppose we are given an inverse system of $\!k\!$-algebras; i.e.,
that for some inversely directed partially ordered set $I,$ we are
given a family of algebras $(A_i)_{i\in I}$ and algebra
homomorphisms $f_{ji}: A_i\to A_j$ $(i\leq j),$ such that
$f_{ii}=\mathrm{id}_{A_i}$ for $i\in I,$ and
$f_{kj}f_{ji}=f_{ki}$ for $i\leq j\leq k.$
Recall that the {\em inverse limit} $\lm_I A_i$
of this system can be constructed (or alternatively, may be defined)
as the subalgebra $A\subseteq \prod_I A_i$ consisting
of those elements $(a_i)_{i\in I}$ such that
$f_{ji}(a_i)=a_j$ for all $i\leq j.$
Thus, the projection maps $p_j: A\to A_j$ carrying
$(a_i)_{i\in I}$ to $a_j\in A_j$ satisfy
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.fjipi}
$f_{ji}\,p_i\ =\ p_j$ $(i\leq j).$
\end{minipage}\end{equation}
%
The algebra $A,$ with these maps, is universal for~(\ref{d.fjipi})
(see~\cite[\S\S7.4-7.5]{245} for motivation and details).
For a general inverse system of algebras $A_i,$ we cannot talk
of applying $M$ to the $f_{ji}$ and $p_i,$
since these may not be surjections.
(Even if all the $f_{ji}$ are surjective, the resulting
$p_i$ may fail to be: \cite{LH} \cite{GH+AHS}~\cite{WW}.)
However, given any inverse system of
algebras $(A_i)_{i\in I},$ and writing $A$ for
its inverse limit, if we
replace each $A_i$ with its subalgebra $p_i(A),$ the result
will be an inverse system still having inverse limit $A,$ but where
the restricted maps $f_{ji}$ and $p_i$ are all surjective.
(Actually, surjectivity of the $p_i$ implies
surjectivity of the $f_{ji},$ in view of~(\ref{d.fjipi}).)
Also, of course, if the original algebras $A_i$ were nilpotent,
the subalgebras with which we have replaced them will still be.
Hence in what follows, we shall often restrict attention
to inverse systems where all $f_{ji}$ and $p_i$ are surjective.
\begin{lemma}\label{L.M+lm}
Let $(A_i, f_{ji})_{i,j\in I}$ be an inverse system of $\!k\!$-algebras,
and $A=\lm A_i$ its inverse limit, with projection maps $p_i:A\to A_i;$
and suppose the $p_i$ \textup{(}and hence the $f_{ji})$ are all
surjective.
Then $\lm_I M(A_i)$ may be identified with a subalgebra of $\En(A)$
containing $M(A),$ by letting each $(u_i)_{i\in I}\in\lm_I M(A_i)$
act on $A$ by sending $(a_i)_{i\in I}\in A$ to
$(u_i(a_i))_{i\in I}\in A.$
\end{lemma}
\begin{proof}
The condition for $(a_i)_{i\in I}$ to belong to $A=\lm_I A_i$ says that
each $f_{ji}$ takes $a_i$ to $a_j,$ and the condition for
$(u_i)_{i\in I}$ to belong to $\lm_I M(A_i)$ says that
each $M(f_{ji})$ takes $u_i$ to $u_j.$
By~(\ref{d.Mfu}), with $f_{ji}$ for $h,$ the latter condition
tells us that $u_j(f_{ji}(a_i))=f_{ji}(u_i(a_i)),$ and by the former,
the left-hand side of this relation equals $u_j(a_j).$
This shows that the $\!I\!$-tuple $(u_i(a_i))_{i\in I}$
again belongs to $A=\lm_I A_i.$
Thus, each $u\in\lm_I M(A_i)$ induces a map $A\to A$
acting as described in the last phrase of the lemma.
It is routine to verify that these maps $A\to A$ are module
endomorphisms, that this action of $\lm_I M(A_i)$ on $A$ respects the
ring operations of $\lm_I M(A_i),$ and that it is faithful; so we get
an identification of $\lm_I M(A_i)$ with a subalgebra of $\En(A).$
Finally, for any $x=(x_i)_{i\in I}\in A,$ one easily verifies
that $(l_{x_i})_{i\in I}$
is an element of $\lm_I M(A_i)$ that acts on $A$ as $l_x;$
so as a subalgebra of $\En(A),$ $\lm_I M(A_i)$ contains each
operator $l_x.$
It similarly contains each $r_x,$ hence it contains $M(A).$
In fact, one easily checks that each $u\in M(A)$ agrees
with the element $(M(p_i)(u))_{i\in I}\in\lm_I M(A_i).$
\end{proof}
In general, $\lm_I M(A_i)$ will be larger than $M(A).$
Indeed, as noted in~\S\ref{S.preview}, if all $A_i$ are nilpotent,
then the algebras $M(A_i)$ are nilpotent, hence are radical, hence
$\lm_I M(A_i)$ is radical.
But in the example we sketched there (to be given in
detail in~\S\ref{S.eg}), $M(A)$ was
not radical (since the image of $1-\,l_x\,r_z$ under the
map $M(A)\to M(B)$ was not invertible, so
that element could not have been invertible in $M(A)).$
Thus, in such an example, $M(A)$ cannot coincide with $\lm_I M(A_i),$
and, indeed, must fail to be closed therein under quasiinverses.
\section{Hopfian modules, and modules of finite length.}\label{S.Hopf+}
As also noted in~\S\ref{S.preview}, the operator $1-l_x\,r_z$
of the example referred to above will nevertheless
be surjective on any homomorphic image $B$ of $A.$
A key to the proof of our main result will be to restrict
attention to image algebras $B$ whose $\!k\!$-module structure
is such that every surjective module endomorphism is injective.
In getting our main conclusion, we will need the
stronger assumption that $B$ has finite length as a $\!k\!$-module;
but let us take a look at the weaker condition just stated,
under which we will be able to carry our proof part of the way.
An algebraic structure is said to be {\em Hopfian} if it has no
surjective endomorphisms other than automorphisms \cite{VAH} \cite{KV}.
Here are some quick examples of Hopfian modules:
A vector space is Hopfian if and only if it is finite-dimensional.
A Noetherian module $M$ over any ring is Hopfian,
for if $h:M\to M$ were surjective but not injective, then the chain
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.strict_chain}
$\{0\}\ \subsetneq\ h^{-1}(\{0\})\ \subsetneq
\ h^{-1}(h^{-1}(\{0\}))\ \subsetneq\ \dots$
\end{minipage}\end{equation}
%
would contradict the Noetherian condition \cite[Prop.~IV.5.3(i)]{HB}
\cite[Prop.~6(i)]{VAH} \cite[Prop.~1.14]{TYL}.
In particular, any module of finite length is Hopfian.
Over a commutative ring, every finitely generated module
is Hopfian \cite[Prop.~IV.5.3(ii)]{HB}, and over
a commutative integral domain $k$ with field of fractions $F,$
any $\!k\!$-submodule of a finite-dimensional
$\!F\!$-vector-space is Hopfian (cf.~\cite[Prop.~11]{VAH}).
So, for instance, $\mathbb{Q}$ is a Hopfian $\!\mathbb{Z}\!$-module --
though its homomorphic image $\mathbb{Q/Z}$ is an example
of a non-Hopfian module.
(The classes of Hopfian modules listed above are all closed under
finite direct sums; however, examples are known of non-Hopfian finite
direct sums of Hopfian modules; indeed, of a Hopfian abelian group
$A$ such that $A\oplus A$ is not Hopfian~\cite{ALSC}.)
The next result only assumes $A$ and $B$ are modules over a ring,
and does not require that ring to be commutative.
In view of our convention that $k$ denotes a commutative
ring, we shall call the ring there $K.$
(In our application of the result, however, $K$
will be our commutative ring $k.)$
\begin{proposition}\label{P.Hopfian}
Suppose $A$ and $B$ are right modules over an associative
unital ring $K,$ let $h:A\to B$ be a surjective module
homomorphism, and let $\En(A;\ker(h))$ be the
subring of the endomorphism ring $\En(A)$ consisting
of the endomorphisms that carry $\ker(h)$ into itself
\textup{(}and hence induce endomorphisms of~$B).$
Suppose $R$ is a radical subring
of $\En(A),$ and $B$ is Hopfian as a $\!K\!$-module.
Then $R\cap\nolinebreak\En(A;\ker(h))$ is also a radical ring; hence its
image in $\En(B)$ is a radical subring of $\En(B).$
\end{proposition}
\begin{proof}
To show that the ring $R\cap\En(A;\ker(h))$ is radical, it suffices
to verify that it is closed under quasi\-inverses in $R.$
Let $r\in R\cap\En(A;\ker(h)),$ and $s$ be its quasiinverse in $R.$
Thus, $1+r$ and $1+s$ are mutually inverse elements of $\En(A).$
Since $1+r$ is invertible as an endomorphism of $A,$ it is
in particular surjective, from which it is easy to see that the
endomorphism of $B$ it induces is surjective.
Since $B$ is Hopfian, that endomorphism is also injective, and
this says that back in $\En(A),$
$1+r$ carries no element from outside $\ker(h)$ into $\ker(h).$
Thus, the inverse map $1+s\in\En(A)$ carries no element of $\ker(h)$
out of $\ker(h),$ i.e., $1+s\in\En(A;\ker(h));$
hence $s\in\En(A;\ker(h)),$ proving the latter ring radical.
The final assertion follows immediately.
\end{proof}
We shall use the above result in conjunction with
part~(iii) of the next lemma.
Note that in that lemma, we return to the general hypothesis
of a commutative base field $k;$ and that
parts~(i) and~(ii), but not part~(iii), assume $B$ an algebra.
(Even in part~(iii), it will be an algebra in our application.)
\begin{lemma}\label{L.idpt}
\textup{(i)}~ In a radical associative algebra $B,$ a finite set of
elements $X\subseteq B$
which are not all zero cannot satisfy $X\subseteq BX.$
\textup{(ii)}~ A radical associative algebra $B$ cannot have
a nonzero subalgebra which is both idempotent $(S=\nolinebreak S^2)$
and finitely generated as an algebra.
\textup{(iii)}~ If $B$ is a $\!k\!$-module of finite length, then
any radical subalgebra $R\subseteq\En(B)$ is nilpotent.
\end{lemma}
\begin{proof}
(i): Writing $U=(k+B)X$ for the left ideal of $B$ generated by $X,$
the condition $X\subseteq BX$ implies that $U=BU;$
so by Nakayama's Lemma \cite[Lemma~4.22(2)]{TYL1}
\cite[Exercise XVII.7.4, p.661]{SL.Alg},
$U=\{0\},$ hence $X\subseteq\{0\},$ contradicting our hypothesis.
(The references cited state Nakayama's Lemma for unital rings.
In our present context, we can apply that version of the
lemma to the left module $U$ over the unital ring $k+B,$ using
the fact that $B$ is an ideal contained in the radical thereof.)
(ii): Suppose $S$ were an idempotent subalgebra of
$B$ generated as a $\!k\!$-algebra by a finite set $X.$
The fact that $S$ is generated by $X$ implies that
$S\subseteq(k+S)X,$ giving the third inclusion of
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.X<...}
$X\ \subseteq\ S\ =\ S^2\ \subseteq B\,S\ \subseteq
\ B(k+S)X\ \subseteq\ B\,X,$
\end{minipage}\end{equation}
%
which contradicts~(i).
(iii): Since $B$ has finite length as a $\!k\!$-module, the chain of
submodules $B\supseteq RB\supseteq R^2B\supseteq\dots$ stabilizes;
say $R^{n+1}B=R^nB.$
Again using finite length of $B,$ we see that $R^nB$ is finitely
generated as a $\!k\!$-module, hence as a $\!(k+R)\!$-module;
hence, since it is carried to itself by the radical ideal $R$
of $k+R,$ Nakayama's lemma shows that it is zero.
Hence $R^n=\{0\}.$
\end{proof}
\section{The main theorem.}\label{S.main}
\begin{definition}\label{D.pro-np}
A $\!k\!$-algebra $A$ which can be written as
an inverse limit of nilpotent $\!k\!$-algebras is called
{\em pro-nilpotent}.
\end{definition}
Part~(iii) of the next result is what we have been aiming at.
The first two parts note what can be said under weaker assumptions.
\begin{theorem}\label{T.main}
Let $B=h(A)$ be a surjective homomorphic image of a pro-nilpotent
$\!k\!$-algebra $A.$
Then
\vspace{.2em}
\textup{(i)}~ For every $r\in M(B),$ the operator $1+r\in\En(B)$
is surjective.
\textup{(}More generally, for every $n>0$ and
$r\in\mathrm{Mat}_n(M(B)),$ $1+r$ acts surjectively on the
direct sum of $n$ copies of $B.)$
\vspace{.2em}
\textup{(ii)}~ If $B$ is Hopfian as a $\!k\!$-module, $M(B)$ is
contained in a Jacobson radical subalgebra of $\En(B).$
\vspace{.2em}
\textup{(iii)}~ If $B$ is of finite length as a $\!k\!$-module,
then it is nilpotent as an algebra.
\end{theorem}
\begin{proof}
Let $A=\lm_I A_i,$ where $(A_i, f_{ji})_{i,j\in I}$ is an inverse
system of nilpotent $\!k\!$-algebras.
As noted earlier, if we replace each $A_i$ by the image $p_i(A)$
therein, and restrict the $f_{ji}$ to these subalgebras,
we get a new inverse system having the same inverse limit $A,$
and such that the restricted maps $p_i$ and $f_{ji}$ are surjective;
moreover, the new $A_i,$ being subalgebras of the given algebras,
are still nilpotent.
Hence without loss of generality, let us assume all
the $p_i$ and $f_{ji}$ surjective.
By Corollary~\ref{C.nilpiff}, the multiplication algebras $M(A_i)$
are nilpotent, hence are radical, and an inverse
limit of radical rings is radical; so under the identification
of Lemma~\ref{L.M+lm}, $\lm_I M(A_i)$ is a radical subalgebra
of $\En(A)$ containing $M(A)\subseteq\En(A;\ker(h)).$
With no additional assumptions, we see that the radicality
of $\lm_I M(A_i)$ implies that for every $u\in M(A),$ the
operator $1+u$ is invertible on $A,$ hence in particular,
acts surjectively, hence that its image in $M(B)$ acts surjectively
on $B,$ giving the first statement of~(i).
Since a full matrix algebra $\mathrm{Mat}_n(R)$ over a radical algebra
$R$ is radical,
this argument applies, more generally, to $\mathrm{Mat}_n(M(A))$ and
$\mathrm{Mat}_n(M(B)),$ acting on a direct sum of copies of
$A,$ respectively $B,$ yielding the parenthetical generalization.
If $B$ is Hopfian as a $\!k\!$-module, then
by Proposition~\ref{P.Hopfian},
$(\lm_I M(A_i))\cap\En(A;\ker(h))$ is a radical $\!k\!$-algebra.
Since $\ker(h)$ is an ideal of $A,$ it is carried into itself by the
operators $l_x$ and $r_x$ $(x\in A),$ so
$M(A)\subseteq (\lm_I M(A_i))\cap\En(A;\ker(h)),$ hence
its image $M(B)=M(h)(M(A))\subseteq\En(B)$ is contained in the radical
subalgebra $M(h)((\lm_I M(A_i))\cap\En(A;\ker(h))),$ giving~(ii).
Finally, if $B$ has finite length, Lemma~\ref{L.idpt}(iii) shows
that the above radical subalgebra of $\En(B)$ is nilpotent, hence its
subalgebra $M(B)$ is nilpotent,
hence by Corollary~\ref{C.nilpiff} again, $B$ is nilpotent.
\end{proof}
In the next section we will give counterexamples to the
conclusions of Theorem~\ref{T.main} and Lemma~\ref{L.idpt} in
the absence of some of the hypotheses; in particular, the
finite length hypothesis of Theorem~\ref{T.main}(iii).
On the other hand, in~\S\ref{S.Nak-like} (after some general
observations in~\S\ref{S.chain}), we will get a few
additional positive results.
In~\S\ref{S.solv} we note a consequence of our main theorem
for solvable Lie algebras,
and in the final~\S\S\ref{S.variants}-\ref{S.questions},
some questions and topics for further study.
\section{Counterexamples.}\label{S.eg}
The first example below is the promised case of a
homomorphic image $B$ of a pro-nilpotent algebra containing elements
$x,\,z$ such that the map $1-l_x r_z\in M(B)$ is not one-to-one.
In constructing that example and the next,
we shall make use of unital free associative
algebras $k\<\langle X\rangle$ on finite sets $X$ of
noncommuting indeterminates (e.g., $X=\{x,y,z\})$ over a field $k,$
their completions, which are noncommuting formal power series algebras
$k\<\langle\langle X\rangle\rangle,$
and the nonunital versions of these two
constructions (their ``augmentation ideals'', i.e., the kernels of the
unital homomorphisms to $k$ sending the indeterminates to zero),
which we will denote $[k\<]\langle X\rangle,$ respectively
$[k\<]\langle\langle X\rangle\rangle.$
Within these algebras, we shall write
$(a,b,\dots)$ for the $\!2\!$-sided ideal generated by
elements $a,b,\dots\,.$
In the completed algebras, we shall also
write $((a,b,\dots))$ for the {\em closure} of such an ideal in the
inverse limit topology.
These examples will start by taking
a set $T$ of monomials in the given free generators,
which does not contain the monomial $1,$ and
forming the factor algebra $k\<\langle X\rangle/(T).$
Note that this has a $\!k\!$-basis consisting of all monomials not
containing any subword belonging to $T.$
We shall then form the
completion $k\<\langle\langle X\rangle\rangle/((T))$
and take for our $A$ the subalgebra
$[k\<]\langle\langle X\rangle\rangle/((T)).$
It is not hard to see that $k\<\langle\langle X\rangle\rangle/((T))$
is the inverse limit of the factor-algebras
$k\<\langle X\rangle/(T\cup X^i)$
where $X^i$ denotes the set of monomials of length $i$ in the
given generators, so that
$[k\<]\langle\langle X\rangle\rangle/((T))$ is the inverse limit of
the nilpotent algebras $A_i = [k\<]\langle X\rangle/(T\cup X^i).$
This inverse limit consists of all formal infinite
$\!k\!$-linear combinations of monomials having no subword in $T.$
By abuse of notation, we use the same symbols
$x,\,\dots$ for our original generators and
for their images in our various factor-algebras.
\begin{example}\label{E.y-xyz}
For any field $k$
there exists a pro-nilpotent associative $\!k\!$-algebra $A$
having elements $x,y,z$ such that $y\notin (y-xyz).$
Thus, in the algebra $B=A/(y-xyz),$
the operator $1-l_x r_z$ annihilates the nonzero element $y.$
In particular, $0\neq y\in B\,y\,B,$ so $B$
cannot be residually nilpotent.
Hence also, though the algebras $A$ and $B$ are Jacobson radical,
their multiplication algebras $M(A)$ and $M(B)$ are not: in each,
the element $-l_x r_z$ is not quasiinvertible \textup{(}though
it is the product of the quasiinvertible
elements $-l_x$ and $r_z).$
\end{example}
\begin{proof}[Construction and proof]
Since it is easier to study the ideal
of an algebra $[k\<]\langle\langle X\rangle\rangle/((T))$
generated by one of the indeterminates than the ideal
generated by a more complicated expression, we shall take for $A$ an
algebra of the form $[k\<]\langle\langle x,w,z\rangle\rangle/((T)),$
find a $y\in A$ such that $w=y-xyz,$
and then obtain $B$ by dividing $A$
by the ideal generated by the indeterminate $w.$
Let the set of monomials $T$ be chosen so that the only nonzero
monomials in $[k\<]\langle x,w,z\rangle/(T)$ are the words
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.xiwzj}
$x^iw\,z^j$ $(i,j\geq 0),$\quad and subwords of such words.
\end{minipage}\end{equation}
%
Thus, we take
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.T=1}
$T\ =\ \{xz,\,wx,\,ww,\,zw,\,zx\}.$
\end{minipage}\end{equation}
%
and let
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.A=1}
$A\ =\ [k\<]\langle\langle x,w,z\rangle\rangle/((T)).$
\end{minipage}\end{equation}
%
For convenient calculation with ideals, we also introduce the notation
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.k+A=1}
$k+A\ =\ k\<\langle\langle x,w,z\rangle\rangle/((T)).$
\end{minipage}\end{equation}
On $A,$ which by our preceding
discussion is pro-nilpotent, consider the operator
$-\,l_x r_z\in M(A)\subseteq\lm M(A_i).$
Since the latter algebra is Jacobson radical, $-\,l_x r_z$ is
quasiinvertible in $\En(A);$ so let $y=(1-l_x r_z)^{-1}(w).$
Clearly,
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.y=1}
$y\ =\ w+xwz+x^2wz^2+\dots+x^nwz^n+\dots\ .$
\end{minipage}\end{equation}
%
(Indeed, one can see immediately that this series satisfies $w=y-xyz.)$
We claim that $y\notin (w).$
To see this, note that every element of $(w)$ is a {\em finite} sum
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.sumaiwbi}
$\sum_{i=1}^n\,a_i\,w\,b_i\quad(a_1,\dots,a_n,\,b_1,\dots,b_n\in k+A).$
\end{minipage}\end{equation}
%
Now an element $a\in A$
such that every monomial occurring in $a$ contains
a factor $w$ or $z$ will annihilate $w$ on the
left, and elements in which all monomials occurring contain
factors $w$ or $x$ likewise annihilate $w$
on the right (see~(\ref{d.xiwzj}),~(\ref{d.T=1})); so let us write
each $a_i$ in~(\ref{d.sumaiwbi}) as $a_i'+a_i'',$
where $a_i'\in k[[x]],$ while the monomials occurring in
$a_i''$ all have factors $w$ or $z,$ and each
$b_i$ as $b_i'+b_i'',$ where $b_i'\in k[[z]]$ while the monomials
occurring in $b_i''$ all have factors $w$ or $x.$
Then~(\ref{d.sumaiwbi}) becomes
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.sumai'wbi'}
$\sum_{i=1}^n\,a_i'\,w\,b_i'\quad(a_1',\dots,a_n'\in k[[x]],
\ b_1',\dots,b_n'\in k[[z]]).$
\end{minipage}\end{equation}
We now see that if in~(\ref{d.sumai'wbi'}) we take the right
coefficient, in $k[[z]],$ of $x^j w$ for any $j\geq 0,$ this will be
a $\!k\!$-linear combination of $b_1',\dots,b_n'.$
In particular,
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.f.dim}
the $\!k\!$-vector-subspace of $k[[z]]$ spanned by the
right coefficients in that algebra of the words $x^j w$
$(j=0,1,\dots)$ in any element~(\ref{d.sumai'wbi'}) of $(w)$
is finite-dimensional over $k.$
\end{minipage}\end{equation}
However, by~(\ref{d.y=1}),
the right coefficient of $x^j w$ in $y$ is $z^j.$
The elements $z^j$ span an infinite-dimensional subspace of
$k[[z]],$ so $y\notin (w)=(y-xyz),$ proving our first
assertion about this example.
The second assertion is a restatement thereof.
Since $y=xyz$ in $B,$ we have $y\in ByB\subseteq B(ByB)B\subseteq\dots,$
hence $y$ maps to $0$ in every nilpotent homomorphic image
of $B,$ so $B$ is not residually nilpotent.
$A$ is radical because
it is an inverse limit of radical algebras, and $B$ is radical
because it is a homomorphic image of $A.$
We have shown that $l_x r_z\in M(B)$ is not
quasiinvertible, hence the same is true of the element of
$M(A)$ (denoted by the same symbol) which maps to it;
so neither $M(A)$ nor $M(B)$ is radical.
Finally, the maps $A\to M(A)$ given by $a\mapsto l_a$
and $a\mapsto r_a$ are a homomorphism and an
antihomomorphism, so the quasiinvertibility
of $x$ and $z$ in $A$ implies quasiinvertibility of $l_x$ and $r_z$
in $M(A),$ hence also in $M(B).$
\end{proof}
We remark that if we write
$A'=[k\<]\langle\langle x,w,z\rangle\rangle,$
so that $A=A'/((T)),$ and define $y=(1-l_x l_z)^{-1}(w)$
in the pro-nilpotent algebra $A',$ then the fact that
$y\notin(w)$ in $A$ implies that the same holds in $A';$
so the conclusions proved above for $A$ and $B$
also hold for $A'$ and $B'=A'/(w).$
Dividing out by $((T))$ just made it easier
to see what we were doing.
In a different direction, suppose that instead of dividing
a pro-nilpotent algebra $A$ by the ideal generated
by an element of the
form $r=y-xyz=(1-l_x r_z)(y),$ we had divided such an algebra
by the ideal generated by an element of the
form $s=y-xy=(1-l_x)(y).$
Using the fact that $-x,$ and hence $l_{-x},$ are quasiinvertible,
we find that in this case, $y$ {\em does} belong to the ideal $(s),$
so it goes to zero in our factor-ring.
Thus, this simpler
construction does not give an example of non-injectivity.
The same, of course, happens if we divide out by an element
of the form $t=y-yz=(1-r_z)(y).$
Thus, the two-sided nature of the operator $l_x r_z$ was needed
to make Example~\ref{E.y-xyz} work.
However, the fact we just called on,
that $a\mapsto l_a$ is a homomorphism $A\to M(A),$
holds only for associative algebras $A.$
In the next example, we shall see that on an inverse
limit of nilpotent Lie algebras, an operator of the
form $1-l_x$ can fail to be quasiinvertible.
(Our example will in fact be ``one-sided'' from the Lie point of
view, but ``two-sided'' from the associative point of view.)
The construction will be formally a little simpler than the preceding
example, but the verification will be a bit more complicated.
\begin{example}\label{E.y-xy+yx}
There exists a pro-nilpotent associative algebra $A$ over any field $k$
having elements $x,y$ such that $y\notin (y-xy+yx).$
Thus, under commutator brackets, $A$ is a pro-nilpotent
{\em Lie algebra}
with elements $x,y$ such that $y\notin (y-[x,y])_\mathrm{Lie}$
\textup{(}where $(\ )_{\mathrm{Lie}}$ denotes
``Lie ideal generated by''\textup{)}.
Hence, in the Lie algebra $B=A/(y-[x,y])_\mathrm{Lie},$
$1-\mathrm{ad}_x$ annihilates the nonzero element~$y.$
In particular,
$0\neq y\in [B,\,y],$ so $B$ is not residually nilpotent.
As in the previous example, the associative algebras $A$ and
$A/(y-[x,y])$ are Jacobson radical, while their multiplier algebras
are not: $r_x-l_x$ is not quasiinvertible
\textup{(}though $r_x$ and $l_x$ are\textup{)}.
\end{example}
\begin{proof}[Construction and proof]
This time, let us start with the associative
algebra $[k\<]\langle\langle x,w\rangle\rangle/((T)),$
with $T$ chosen so that the only nonzero monomials are the words
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.xiwxj}
$x^iw\,x^j$ $(i,j\geq 0),$\quad and their subwords.
\end{minipage}\end{equation}
%
Thus, we take
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.T=2}
$T\ =\ \{w x^i w\mid i\geq 0\},$
\end{minipage}\end{equation}
%
and let
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.A=2}
$A\ =\ [k\<]\langle\langle x,w\rangle\rangle/((T)).$
\end{minipage}\end{equation}
We now define
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.y=2}
$y\ =\ (1-l_x+r_x)^{-1}(w)\ \in\ A.$
\end{minipage}\end{equation}
Though the obvious way to begin the calculation of this
element would be to write
$(1-l_x+r_x)^{-1}=\sum_{i=0}^\infty\ (l_x-r_x)^i,$ we can
get the right coefficient of $x^i w$ in~(\ref{d.y=2}) more
quickly if we instead use the formula
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.1-ad-1}
$(1-l_x+r_x)^{-1}\ =\ \sum_{i=0}^\infty\ l_x^i\,(1+r_x)^{-1-i},$
\end{minipage}\end{equation}
%
which is valid because $l_x$ and $1+r_x$ commute.
This gives
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.y=3}
$y\ =\ \sum_{i=0}^\infty\ x^i\,w\,(1+x)^{-1-i}$
\end{minipage}\end{equation}
%
in the formal power series algebra
$A=[k\<]\langle\langle x,w\rangle\rangle/((T)).$
Again, if this lay in $(w),$ it would follow that the right
factors $(1+x)^{-1-i}$ $(i=0,1,\dots)$ of the monomials $x^i\,w$
would lie in a finite-dimensional $\!k\!$-subspace of $[k\<][[x]].$
But they do not: the positive and negative
powers of $1+x$ are $\!k\!$-linearly independent in the
field $k(x),$ so they are $\!k\!$-linearly independent in the
larger formal Laurent series field
$k((x)),$ hence in the smaller
formal power series algebra $k[[x]]\subseteq A.$
Hence $y\notin (w)=((1-l_x+r_x)(y))=(y-xy+yx),$ and since the
Lie ideal generated by $y-xy+yx=y-[x,y]$ is contained in the
associative ideal generated by that element, we likewise
have $y\notin (y-[x,y])_\mathrm{Lie}.$
The other assertions follow as before.
\end{proof}
The above example may seem suspicious:
In $B,$ $y=[x,y],$ so $x$ and $y$ span a $\!2\!$-dimensional
sub-Lie-algebra $B^*\subseteq B.$
Suppose we let $A^*$ be the inverse image of this algebra in $A,$
and replace each member of the family of algebras $A_i$
of which $A$ is the inverse limit by the image $A_i^*$ of $A^*$ therein.
Won't the resulting inverse system have $A^*$ as inverse limit,
giving a description of the finite-dimensional non-nilpotent
Lie algebra $B^*$ as a homomorphic image of an inverse limit $A^*$ of
nilpotent Lie algebras?
What is wrong with this argument is the assumption
that the inverse limit of the $A_i^*$ will be $A^*.$
Rather, one finds that that inverse limit will be the closure
of $A^*$ in the inverse limit topology on $A.$
Since the map $A\to B$ is not continuous in that topology
on $A$ and the discrete topology on $B$
(since its kernel is $(y-[x,y])_\mathrm{Lie},$ not
$((y-[x,y]))_\mathrm{Lie}),$ the closure of $A^*$ may
have a much larger image than $B^*.$
Another thought:
Looking intuitively at Examples~\ref{E.y-xyz} and~\ref{E.y-xy+yx},
we can say that in each, we took a pro-nilpotent algebra $A,$
and were able to arrange for an element
$y\in A$ to ``survive'' under a homomorphism $A\to B$
that made it fall together with a member of $AyA$ or $Ay+yA.$
In these cases, $y$ survived ``with the help of'' other
elements $(x,$ and possibly $z)$ which did not themselves fall together
with higher-degree expressions.
We may ask whether a family $X$ of elements can
all ``help one another'' to survive under a homomorphism that makes
each fall together with a linear combination of higher degree monomials
in it and the others.
One way of posing this question is: Can a homomorphic image $B$
of a pro-nilpotent algebra $A$ contain a nonzero
subalgebra $S$ that is idempotent, i.e., satisfies $S=S^2$?
If our algebras are associative, and the set $X$ generating
$S$ is finite, the answer is no.
Indeed, $A,$ and hence $B,$ will be
Jacobson radical, and Lemma~\ref{L.idpt}(ii) says that such an algebra
cannot have a nonzero finitely generated idempotent subalgebra.
Lemma~\ref{L.idpt}(i) describes a more general restriction.
However, these restrictions fail for nonassociative algebras.
Indeed, in Example~\ref{E.y-xy+yx} we had $y\in [B,y],$
contradicting the analog of Lemma~\ref{L.idpt}(i).
We record next a much simpler (though non-Lie) example with the same
property (which
we will want to call on, for a different property, later), then
an example with a finite-dimensional simple subalgebra.
\begin{example}\label{E.y-xy}
Another pro-nilpotent algebra $A$ over any field $k$
having elements $x,y$ such that $y\notin (y-xy),$
hence such that on $B=A/(y-xy),$
$1-l_x$ annihilates the nonzero element $y;$ hence such
that $0\neq y\in B\,y$
\textup{(}in contrast to Lemma~\ref{L.idpt}\textup{(i))}.
\end{example}
\begin{proof}[Construction and proof]
A natural approach, paralleling our earlier constructions, would
be to start with a nonassociative $\!k\!$-algebra on
generators $x,\,w,$ in which all monomials are set to zero
except for $x,$ $w,$ $xw,$ $x(xw),$ $x(x(xw)),\dots\,.$
But rather than dealing with a free nonassociative algebra, and the
resulting proliferation of parentheses, let us simply name the
resulting basis
of our algebra, and say how the multiplication acts.
(The ``free associative algebra modulo monomials''
approach of our previous examples insured that the algebra described was
associative; but no such condition is needed here.)
So let us start with an algebra having a basis
$\{x,\,w_0,\,w_1,\,w_2,\dots\},$ and multiplication given by
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.xwi...}
$x\,w_i\ =\ w_{i+1}$ $(i=0,1,\dots),$\quad
and all other products of basis elements zero.
\end{minipage}\end{equation}
Clearly, for each $i\geq 0,$ this algebra has a homomorphic image
$A_i$ in which all $w_j$ with $j\geq i$ are set to zero, and these
images form an inverse system of nilpotent
algebras, whose inverse limit $A$ consists of all
formal infinite sums $\alpha\,x + \sum_{i=0}^\infty\ \beta_i\,w_i$
$(\alpha,\,\beta_i\in k).$
The ideal $(w_0)$ of $A$ is easily shown to consist of the
finite sums $\beta_0\,w_0+\dots+\beta_n\,w_n.$
In particular, it does not contain the element
$y=\sum_{i=0}^\infty w_i=(1-l_x)^{-1} w_0,$
which satisfies $y-xy=w_0.$
This gives the first assertion;
the remaining assertions follow immediately.
\end{proof}
Still more striking is
\begin{example}\label{E.y-yy}
There exists a pro-nilpotent algebra $A$ over any field $k$
having an element $y$ such that $y\notin (y-y^2).$
Hence in $B=A/(y-y^2),$ the element $y$ spans
an idempotent $\!1\!$-dimensional \textup{(}associative!\textup{)}
subalgebra, in contrast to Lemma~\ref{L.idpt}\textup{(ii)}.
\end{example}
\begin{proof}[Construction and proof]
This time, we start with an algebra having basis
$\{w_0,\,w_1,\dots\},$ and multiplication given by
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.xwizi}
$w_i\,w_i\ =\ w_{i+1}$ $(i=0,1,\dots),$\quad
and all other products of basis elements zero.
\end{minipage}\end{equation}
We again get nilpotent homomorphic images $A_i$ on
setting $w_j$ equal to zero for all $j\geq i.$
The inverse limit
$A$ of these algebras consists of all formal infinite sums
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.sum}
$\sum_{i=0}^n\,\alpha_i\,w_i\quad (\alpha_i\in k).$
\end{minipage}\end{equation}
%
Again, it is not hard to see that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.w0z0}
the ideal $(w_0)$ of $A$ consists of all
finite sums $\alpha_0\,w_0+\dots+\alpha_n\,w_n.$
\end{minipage}\end{equation}
Again let $y=\sum_{i=0}^\infty\ w_i.$
We find that $y-y^2=w_0,$
so $(y-y^2)$ is $(w_0),$ described in~(\ref{d.w0z0}),
which clearly does not contain $y.$
This proves the main assertion; the final statement again follows.
\end{proof}
Even for associative algebras, Lemma~\ref{L.idpt}(ii)
only excludes nonzero {\em finitely generated} idempotent subalgebras.
An easy example of a radical associative algebra $B$ with a
non-finitely-generated idempotent subalgebra $S$
is gotten by taking for both $B$ and $S$ the maximal ideal of any
nondiscrete valuation ring.
It is harder to get an example with $B$ a homomorphic image of a
pro-nilpotent algebra, but the following celebrated construction of
S\c{a}siada and Cohn~\cite{ES+PMC} has that property.
(For parallelism with the other examples of this section, I have
interchanged below the use of the symbols $x$ and $y$
in~\cite{ES+PMC}.)
Note that the ideal $(y)$ in the statement, though generated by
a single element as a $\!2\!$-sided ideal of $B,$ must,
by Lemma~\ref{L.idpt}(i)-(ii), require infinitely many elements to
generate it as a left ideal or as a subalgebra of $B.$
\begin{example}[{S\c{a}siada and Cohn \cite{ES+PMC}}]\label{E.ES+PMC}
If $k$ is a field, then in the pro-nilpotent
associative algebra $A=[k\<]\langle\langle x,\,y\rangle\rangle,$
one has $y\notin (y-x\,y^2\,x).$
Thus, in $B=A/(y-x\,y^2\,x),$ we have $0\neq y=xy^2x,$
so the ideal $(y)$ is a nonzero left ideal satisfying $B(y)=(y)$ and a
nonzero subalgebra satisfying $(y)^2=(y).$
Moreover, if we take any ideal $U$ of $B$ maximal for the property
of not containing $y,$ then in $B'=B/U,$ the subalgebra $(y)$ is simple.
Thus, $(y)\subseteq B'$ is a simple Jacobson radical algebra.
\end{example}
\begin{proof}[Sketch of proof.]
The proof that $y\notin (y-x\,y^2\,x)$ occupies most of the
five pages of~\cite{ES+PMC}, and I will not discuss it here.
% If we try [k]<> and look for y with w=y-xyyx, find
% = sum of set of expressions obtained recursively, starting with w,
% replacing any w by xwwx. Not clear how to show \notin (w).
The relation $y=xy\cdot yx$ in $B$ yields
the asserted equalities, $B(y)=(y)$
(which one sees also holds in Example~\ref{E.y-xyz}),
and $(y)^2=(y)$ (which does not hold in our previous
associative examples).
Let us now verify the simplicity as a ring of the
ideal $(y)$ of $B'=B/U$ (though this is also done in~\cite{ES+PMC}).
By maximality of $U,$
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.nosubid}
$(y)$ contains no proper nonzero $\!B'\!$-ideal.
\end{minipage}\end{equation}
%
Suppose, however, by way of contradiction, that it contained
a proper nonzero $\!(y)\!$-ideal $V.$
If $(y)V=\{0\},$ then the right annihilator
of $(y)$ in $(y)$ is a nonzero ideal of $B'$ contained
in $(y),$ hence, by~(\ref{d.nosubid}), is all of $(y).$
But this implies that in $B',$ $y\,y=0,$ hence
$0=xyyx=y,$ a contradiction.
So $(y)V\neq\{0\}.$
Now from this it follows in
turn that if $(y)V(y)=\{0\},$ the left annihilator
of $(y)$ in $(y)$ gives the same contradiction.
So $(y)V(y)\neq\{0\}.$
But this is an ideal of $B'$ contained in $V,$
hence properly contained in
$(y),$ a final contradiction that completes the proof.
In the final assertion of the lemma (the goal of \cite{ES+PMC}),
radicality holds because any ideal of a radical ring is radical.
\end{proof}
Each of Examples~\ref{E.y-xyz}-\ref{E.ES+PMC} above shows,
among other things,
that the conclusion of Theorem~\ref{T.main}(ii) can fail if one
deletes the condition that the underlying $\!k\!$-module of $B$
be Hopfian.
We end this section with a much easier example showing that even if
that module is Hopfian, this is not enough to give the full assertion
of part~(iii).
\begin{example}\label{E.hopf}
There exist a commutative ring $k$
and a pro-nilpotent commutative associative $\!k\!$-algebra
$A$ which is Hopfian as a $\!k\!$-module, but not nilpotent
as an algebra.
\end{example}
\begin{proof}[Construction and proof]
Let $k$ be a complete discrete valuation ring, with
maximal ideal $(p),$ and consider the inverse system of nilpotent
algebras $A_i=(p)/(p^i)$ $(i\geq 1)$ with the obvious surjective
connecting homomorphisms.
Because $k$ is complete, the inverse limit $A$ of this
system is isomorphic as a $\!k\!$-algebra
to the ideal $(p)\subseteq k,$ which is free of rank~$1$ as a
$\!k\!$-module, hence is Hopfian, but is not nilpotent as an algebra.
\end{proof}
(If we had left out the assumption that $k$ was complete, our $A$
would have been the maximal ideal $p\,\hat{k}$ of the completion
$\hat{k}$ of $k.$
In that situation, $\hat{k},$ and hence that ideal, are again Hopfian,
not only as $\!\hat{k}\!$-modules but
as $\!k\!$-modules, though for less obvious reasons.)
\section{A chain of conditions.}\label{S.chain}
The proof of Theorem~\ref{T.main} involves a chain of
conditions on an algebra $A:$
%
\vspace{.5em}
\begin{equation}\label{d.nilp}
\mbox{$A$ is nilpotent; equivalently, $M(A)$ is nilpotent.}
\end{equation}
$$\Downarrow$$
\begin{equation}\begin{minipage}[c]{18pc}\label{d.rad}
$M(A)$ is Jacobson radical; equivalently, for every
$u\in M(A),$ $1+u$ is invertible in $1+M(A).$
\end{minipage}\end{equation}
$$\Downarrow$$
\begin{equation}\label{d.M_in_rad}
\mbox{$M(A)$ is contained in a Jacobson radical subalgebra of $\En(A).$}
\end{equation}
$$\Downarrow$$
%
\begin{equation}\begin{minipage}[c]{20pc}\label{d.mxsurj}
For every $n>0$ and $u\in\mathrm{Mat}_n(M(A)),$ $1+u$
is \mbox{surjective} as a map on the direct sum of $n$ copies of~$A.$
\end{minipage}\end{equation}
%
$$\Downarrow$$
\begin{equation}\label{d.surj}
\mbox{For every $u\in M(A),$ $1+u$ is surjective as a map $A\to A.$}
\end{equation}
\vspace{.5em}
We note some quick examples showing that the first four of these
conditions are distinct:
In any commutative local integral domain $R$ which is not
a field, the maximal ideal $A$
is a radical subalgebra, and satisfies $M(A)\cong A,$ hence
$A$ satisfies~(\ref{d.rad}), but not~(\ref{d.nilp}).
Amplifying the first sentence of the
paragraph preceding Example~\ref{E.hopf}, we note that
for the algebras $A$ of Examples~\ref{E.y-xyz}-\ref{E.ES+PMC} the
inclusion $M(A)\subseteq\lm_I M(A_i)$
yields~(\ref{d.M_in_rad}), but that these
algebras cannot satisfy~(\ref{d.rad}), since they have
homomorphic images $B$ on which certain operators $1+u$ $(u\in M(B))$
are non-invertible.
The algebras $B$ of those same examples
satisfy~(\ref{d.mxsurj}) by Theorem~\ref{T.main}(i),
but they have elements $u\in M(B)$ such that $1+u$ is not injective,
hence not invertible, so they do not satisfy~(\ref{d.M_in_rad}).
I do not know whether there are algebras satisfying~(\ref{d.surj})
but not~(\ref{d.mxsurj}).
\vspace{.5em}
Conditions~(\ref{d.nilp}), (\ref{d.rad}), (\ref{d.mxsurj})
and~(\ref{d.surj}) clearly carry over to homomorphic images; but
Examples~\ref{E.y-xyz}-\ref{E.ES+PMC} show
that~(\ref{d.M_in_rad}) does not;
though Proposition~\ref{P.Hopfian} shows that it does
when the image algebra is Hopfian as a $\!k\!$-module.
Condition~(\ref{d.nilp}) carries over to subalgebras,
but none of the others do.
E.g., in a discrete
valuation ring, such as the localization $\mathbb{Z}_{(p)}$
of $\mathbb{Z}$ at a prime $p$
(notation unrelated to the $A_{(n)}$ of \S\ref{S.np}!),
or a formal power series algebra $k[[t]]$ over a field
$k,$ the maximal ideal
(in these cases, $p\,\mathbb{Z}_{(p)},$ respectively $[k\<][[t]]),$
regarded as an algebra, satisfies~(\ref{d.rad}), and
hence~(\ref{d.M_in_rad})-(\ref{d.surj});
but in these two examples, the $\!\mathbb{Z}\!$-subalgebra
$p\,\mathbb{Z}\subseteq p\,\mathbb{Z}_{(p)},$ respectively
the $\!k[t]\!$-subalgebra $[k\<][t]\subseteq [k\<][[t]],$ fail to
satisfy~(\ref{d.surj}), hence likewise~(\ref{d.rad})-(\ref{d.mxsurj}).
What about inverse limits; say with respect to inverse systems
where the $p_i$ are surjective?
Examples~\ref{E.y-xyz}-\ref{E.ES+PMC} show that~(\ref{d.nilp})
and~(\ref{d.rad}) fail to carry over to these.
Probably~(\ref{d.mxsurj}) and~(\ref{d.surj}) do not carry over either
-- those conditions make the maps $1+u$ $(u\in M(A))$ surjective,
but do not make inverse images of elements under those maps
unique, and this leads to no way of lifting such inverse images to the
inverse limit algebra.
However,~(\ref{d.M_in_rad}) does carry over.
Indeed, for each $i$ let $N(A_i)$ denote the least radical
subalgebra of $\En(A_i)$ containing $M(A_i),$ i.e., the closure
of $M(A_i)$ under quasiinverses and the algebra operations.
It is not hard to verify that
the $f_{ji}$ induce (surjective) homomorphisms $N(A_i)\to N(A_j).$
The inverse limit of this system of homomorphisms
will be a radical subalgebra of $\En(A)$ containing $M(A).$
\vspace{.5em}
We remark that variants of~(\ref{d.mxsurj}) and~(\ref{d.surj})
in which ``surjective'' is replaced by ``injective'' or by
``bijective'' might also be of interest.
\section{A Nakayama-like property.}\label{S.Nak-like}
In the proof of Theorem~\ref{T.main}, we obtained statement~(iii)
from statement~(ii) essentially by showing that
for an algebra of finite length as a
module, condition~(\ref{d.M_in_rad}) implies~(\ref{d.nilp}).
On the other hand, we did not obtain~(ii) directly from~(i)
-- I do not know whether for algebras that are Hopfian as
modules,~(\ref{d.mxsurj}) or~(\ref{d.surj}) implies~(\ref{d.M_in_rad}).
(If the implication requires the stronger statement~(\ref{d.mxsurj}),
then it probably needs not only the hypothesis that $A$ is
Hopfian, but that all finite direct sums $\bigoplus_n A$ are Hopfian.)
If $A$ is Hopfian and satisfies~(\ref{d.surj}),
the maps $1+u$ $(u\in M(A))$ are invertible, hence all
$u\in M(A)$ are quasiinvertible in $\En(A);$ the difficulty is that if
try to close $M(A)$ within $\En(A)$ under quasiinverses and the
algebra operations, some of the quasiinverses we introduce
may have sums or products that fail to be
quasiinvertible (like the $l_x r_z$ and $l_x+r_x$
of Examples~\ref{E.y-xyz} and~\ref{E.y-xy+yx}),
leading to the failure of~(\ref{d.M_in_rad}).
However, whether or not we can deduce~(\ref{d.M_in_rad}),
if~(\ref{d.mxsurj}) holds and the
modules $\bigoplus_n A$ are Hopfian, then
$M(A)$ will behave somewhat like a radical algebra,
in that it will satisfy a version of Nakayama's lemma.
Let us formulate this result with $M(A)$ and $A$ generalized
to an arbitrary ring and module.
\begin{lemma}\label{L.Nak-like}
Let $R$ be a nonunital associative ring,
$A$ a left $\!R\!$-module, and $n$ a positive integer.
Suppose that for every $u\in\mathrm{Mat}_n(R),$ the
element $1+u$ acts in a one-to-one fashion on $\bigoplus_n A.$
Then for any $\!n\!$-generator $\!R\!$-submodule
$C$ of $A$ \textup{(}or more generally,
for any $\!n\!$-generator $\!R\!$-submodule $C$ of a direct
product of copies of $A),$ one has $R\,C=C\implies C=\{0\}.$
\end{lemma}
\begin{proof}
The case where $C$ is a submodule of a direct product
reduces immediately to the case $C\subseteq A$ by projecting
onto any coordinate where some member of $C$ has nonzero component.
So assume $C$ is a submodule of $A,$ generated by $x_1,\dots,x_n.$
The condition $R\,C=R$ means that each $x_m$ can be written
as an $\!R\!$-linear combination of itself and the others;
which says that if we let $x=(x_1,\dots,x_n)\in \bigoplus_n A,$
then for some $u\in\mathrm{Mat}_n(R),$ we have $x=ux.$
But this says that $1-u$ annihilates $x,$ contradicting our hypothesis.
\end{proof}
We can now get part~(iii) of Theorem~\ref{T.main} directly from
part~(i), via the following consequence of the above lemma.
\begin{corollary}\label{C.i>iii}
For $A$ an algebra of finite length as a
$\!k\!$-module,\textup{~(\ref{d.mxsurj})}
implies\textup{~(\ref{d.nilp})}.
\textup{(}So for such
$A,$\textup{~(\ref{d.nilp})-(\ref{d.mxsurj})} are equivalent.\textup{)}
\end{corollary}
\begin{proof}
If~(\ref{d.nilp}) fails, then the decreasing chain of submodules
$M(A)^d(A)$ of $A$ $(d=0,1,\dots)$ never becomes zero;
but by the finite length assumption, it must stabilize.
Thus, say $\{0\}\neq C=M(A)^d(A)$ satisfies $C\ =\ M(A)(C).$
By our finite length hypothesis, $C$
is finitely generated, say by $n$ elements.
Since a module of finite length is
Hopfian, condition~(\ref{d.mxsurj}) says
that all elements $1+u$ $(u\in\mathrm{Mat}_n(M(A)))$
act invertibly on $\bigoplus_n A,$ hence in a one-to-one fashion;
so Lemma~\ref{L.Nak-like} says that $C=\{0\},$ a contradiction.
\end{proof}
The next corollary to Lemma~\ref{L.Nak-like} shows that
Theorem~\ref{T.main}(ii) has concrete consequences for the
structure of Hopfian homomorphic images of pro-nilpotent algebras.
(Note that the hypothesis of finite generation as an ideal
is weaker than finite generation as a one-sided ideal, which is
weaker in turn than finite generation as an algebra.)
\begin{corollary}\label{C.not_idpt}
A nonzero $\!k\!$-algebra $A$ satisfying\textup{~(\ref{d.mxsurj})}
\textup{(}in particular, a nonzero homomorphic image of
a pro-nilpotent algebra\textup{),} which
has the property that for all $n$
the $\!k\!$-module $\bigoplus_n A$ is Hopfian, cannot be both
idempotent as a $\!k\!$-algebra and finitely generated as an ideal.
\end{corollary}
\begin{proof}
For any algebra $A,$ the ideals of $A$ are
its $\!M(A)\!$-submodules, and we see that the conditions
of idempotence as an algebra and finite generation as an ideal say
that $M(A)A=A$ and $A$ is finitely generated as an $\!M(A)\!$-module.
However, the Hopfian condition together
with~(\ref{d.mxsurj}) yield the hypothesis of
Lemma~\ref{L.Nak-like}, implying that $A=\{0\}.$
\end{proof}
\section{Solvable Lie algebras.}\label{S.solv}
The {\em derived series} of an algebra $A$ is the
sequence of subalgebras $A^{(n)}$ $(n=0,1,\dots)$ defined by
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.der_series}
$A^{(0)}=A,\qquad A^{(n+1)}= A^{(n)}\,A^{(n)}.$
\end{minipage}\end{equation}
%
This concept is standard in the theory of Lie algebras (where
the $A^{(n)}$ are in fact ideals).
It is less so for general nonassociative algebras,
but is introduced in that context in \cite[p.17]{Schafer}.
An algebra $A$ is called {\em solvable} if $A^{(n)}=\{0\}$ for
some $n\geq 0.$
It is easy to see that $A^{(n)}\subseteq A_{(2^n)}$ $(n=0,1,\dots\,),$
so every nilpotent algebra is solvable; but the converse is
not true, as shown by the $\!2\!$-dimensional Lie algebra
with basis $\{x,\,y\}$ and multiplication $[x,\,y]=y.$
There is, however, in the classical context, a characterization
of solvable Lie algebras in terms of nilpotent Lie algebras:
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.solv}
\cite[Corollary~1 to Theorem~13, p.51.]{NJ_Lie}
If $A$ is a finite-dimensional Lie algebra over a field
of characteristic~$0,$ then $A$ is solvable
if and only if its commutator ideal $A^{(1)}=[A,\,A]$ is nilpotent.
\end{minipage}\end{equation}
Nazih Nahlus has pointed out to me that using
this fact, one gets as a consequence of Theorem~\ref{T.main} the
following result, which he conjectured some years ago.
\begin{corollary}[{to Theorem~\ref{T.main}(iii).~ N.\,Nahlus (personal communication)}]\label{C.solv}
Let $A$ be an inverse limit of finite-dimensional solvable Lie
algebras $A_i$ over a field $k$ of characteristic~$0.$
Then any finite-dimensional homomorphic image $B$ of $A$ is solvable.
\end{corollary}
\begin{proof}
By~(\ref{d.solv}), the commutator ideals of the $A_i$
form an inverse system of nilpotent algebras.
The inverse limit $A^*\subseteq A$ of this system contains
all brackets of elements of $A;$ so
when we map $A$ homomorphically onto the finite-dimensional
algebra $B,$
the image of $A^*$ contains all brackets of elements of $B.$
Theorem~\ref{T.main}(iii) tells us that that image
is nilpotent, so $B$ is solvable.
\end{proof}
On the other hand, there are both infinite-dimensional Lie algebras $A$
in characteristic $0,$ and finite-dimensional Lie algebras $A$
in positive characteristic, which are solvable,
but for which $A^{(1)}$ is not nilpotent.
An example of the former is given by the vector space $A$ of operators
on $\mathbb{R}[x]$ spanned by the operators $X^n$ of multiplication
by $x^n$ $(n=0,1,\dots),$ the operator $D=d/dx,$
and the composite operator $XD=x\,d/dx.$
Indeed, one verifies that $A$
is closed under commutator brackets, hence forms a Lie algebra
(the semidirect product of the $\!2\!$-dimensional
Lie algebra $L$ spanned by $\{D,\,XD\},$
and the $\!L\!$-module $\mathbb{R}[x]).$
One finds that the subalgebra $A^{(1)}=[A,A]$ is spanned
by all the above operators except $XD.$
(In particular, $[D,\,XD]=D$ does appear.)
This subalgebra is not nilpotent, since $[D,X^n]=nX^{n-1},$
so that there are elements which can be bracketed with $D$
arbitrarily many times before going to zero.
At the next step, however,
$A^{(2)}=[A^{(1)},A^{(1)}]$ is spanned by the operators $X^n$ only,
and hence has zero bracket operation, so $A^{(3)}=\{0\},$
showing that $A$ is solvable, even though $A^{(1)}$ is not nilpotent.
To get a finite-dimensional example in positive characteristic, let
us first note a variant of the above characteristic~$0$ example.
Consider the ring of functions $\mathbb{R}[x,\,e^x],$ and the space
of operators on that ring spanned by $D$ and $XD$ as above, together
with (rather than the operators $X^n)$ the operators
$X^n\,Y$ $(n\geq 0),$
where $Y$ is the operator of multiplication by $e^x.$
Again, one verifies that this is closed under commutator
brackets, and so gives a Lie algebra $A$
(the semidirect product of $L$ as above and
the $\!L\!$-module $\mathbb{R}[x]\,e^x).$
Where in the preceding example, the infinite-dimensionality of
$\{X^0,\,X^1,\,X^2,\dots\}$ was involved in establishing the
non-nilpotence of $A^{(1)},$ here non-nilpotence follows from the
single relation
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.DX0Y}
$[D,\,X^0\,Y]=X^0\,Y.$
\end{minipage}\end{equation}
%
This does not allow us to pass to a finite-dimensional
subalgebra with the desired properties, because the iterated
action of $XD$ on $X^0\,Y$ brings in all the $X^n\,Y.$
However, one finds that the structure constants of this
Lie algebra with respect to our basis are integers, and
that when one reduces them modulo a prime $p,$ then the span
of $\{X^p\,Y,\,X^{p+1}Y,\,\dots\}$ becomes an ideal.
(Key calculation: in the original algebra,
$[D,\,X^pY]=pX^{p-1}Y+X^pY,$ and modulo $p,$
the first term of that expression vanishes.)
The factor-algebra by that ideal
is thus a $\!(p+2)\!$-dimensional Lie algebra $B,$ and the
relation~(\ref{d.DX0Y}) shows that $B^{(1)}$ is still not nilpotent.
However, we find that $B^{(1)}$ again loses the operator $XD,$ and
$B^{(2)}$ likewise loses $D,$ hence has zero bracket operation,
so that again $B^{(3)}=\{0\}$ and $B$ is solvable.
Further examples in prime characteristic may
be found in~\cite{KB+DAT}.
So if Corollary~\ref{C.solv} is to be extended
to positive characteristic, or to inverse limits of not
necessarily finite-dimensional Lie algebras (or, indeed,
to non-Lie algebras), a very different proof will be needed.
One can, of course, generalize that corollary and its
present proof by strengthening
the hypothesis to {\em assume} $A$ is an inverse limit
of Lie algebras for which $A^{(1)}$ is nilpotent.
Indeed, one can generalize the resulting statement to arbitrary
algebras, replacing solvability by any condition specifying that
the values of a given family of algebra terms should generate
a nilpotent subalgebra.
\section{Possible variants of our main theorem.}\label{S.variants}
Let us look at a few ways Theorem~\ref{T.main}
can, or might, be generalized.
We start with one which as a generalization proves disappointing
-- but does so by showing that our present
Theorem~\ref{T.main} is stronger than we realized.
\subsection{General limits.}\label{S2.genlim}
Recall that the concept of the inverse limit of an inversely directed
system of algebraic structures is a case of the more general
category-theoretic notion of the ``limit'' of a functor
\cite[\S\S7.6]{245} \cite[\S III.4]{CW}, other important examples
of which are the fixed-point algebra of a group acting
on an algebra, and the equalizer of a pair of algebra homomorphisms.
If one examines the proof of Theorem~\ref{T.main}, one sees no
reason why it should not work for limits in this general sense.
It does -- but that extension gives nothing new:
\begin{lemma}\label{L.gen_lims}
For a $\!k\!$-algebra $A,$ the following conditions are equivalent.
\textup{(i)}~ $A$ can be written as the limit of a system of
nilpotent $\!k\!$-algebras indexed by a small category.
\textup{(ii)}~ $A$ is pro-nilpotent, i.e.,
can be written as an inverse limit of an
inversely directed system of nilpotent $\!k\!$-algebras.
\end{lemma}
\begin{proof}[Sketch of proof]
Clearly, (ii)$\implies$(i).
Conversely, suppose $F:\mathbf{C}\to\mathbf{Alg}_k$ is a functor
from a small category $\mathbf{C}$ to the category $\mathbf{Alg}_k$
of not-necessarily-associative $\!k\!$-algebras, such
that for all $X\in\mathrm{Ob}(\mathbf{C}),$ $F(X)$ is nilpotent.
Let $I$ be the partially ordered set of finite subsets of
$\mathrm{Ob}(\mathbf{C}),$ ordered by reverse inclusion;
clearly, $I$ is inversely directed.
For each $i\in I,$ let $\mathbf{C}_i$ be the full subcategory
of $\mathbf{C}$ with object-set $i,$ and let
$A_i=\lm\,(F\,|\,\mathbf{C}_i),$ where $F\,|\,\mathbf{C}_i$ denotes
the restriction of $F$ to $\mathbf{C}_i.$
Each $A_i$ is a subalgebra of the finite product
$\prod_{X\in i} F(X),$ and the class of nilpotent
algebras is closed under finite products and
subalgebras, hence each $A_i$ is nilpotent.
Given $i\leq j\in I,$ which by our ordering of $I$
means $i\supseteq j,$ the inclusion
$\mathbf{C}_j\subseteq\mathbf{C}_i$ induces a restriction
homomorphism $A_i\to A_j.$
It is straightforward to verify that $\lm F=\lm_I A_i,$ yielding~(ii).
\end{proof}
\subsection{Variant sorts of nilpotence.}\label{S2.varnilp}
Within the multiplier algebra $M(A)$ of an algebra $A,$ we may
look at the subalgebra $M_l(A)$ generated by the left multiplication
operators $l_x,$ and the subalgebra $M_r(A)$ generated by
the right multiplication operators $r_x.$
If $A$ is associative, these give nothing very new:
$M_l(A)$ is isomorphic to the factor-algebra of $A$
by its left annihilator ideal $\{x\in A\mid xA=\{0\}\},$
and $M_r(A)$ is antiisomorphic to the factor-algebra
of $A$ by the analogous right annihilator ideal;
in particular, each is nilpotent if and only if $A$ is.
If, rather, $A$ is anticommutative (e.g., is a Lie algebra)
or is commutative (e.g., is a Jordan algebra), then $M_l(A)$
and $M_r(A)$ coincide with $M(A).$
But for a general nonassociative algebra $A,$ these two
subalgebras of $M(A)$ can look very different.
For instance, for the algebra with
multiplication~(\ref{d.xwi...}), it is easy to see that $(AA)A=\{0\},$
so that $M_r(A)^2=\{0\},$ but that $M_l(A)^n\neq \{0\}$ for all $n.$
The conditions $(\exists\,n)\,M_l(A)^n=\{0\}$
and $(\exists\,n)\,M_r(A)^n=\{0\}$ are known
as {\em left} nilpotence and {\em right} nilpotence \cite{AMSl}.
An algebra can be both left and right nilpotent
without being nilpotent, as shown by the algebra with
basis $x, w_0, w_1,\dots\,,$ and multiplication
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.xw2i2i+1}
$x\,w_{2i}\ =\ w_{2i+1},\quad w_{2i+1}\,x\ =\ w_{2i+2},$\quad
all other products of basis elements being zero.
\end{minipage}\end{equation}
The development of Theorem~\ref{T.main} goes over,
with no change, if the condition of nilpotence
is replaced by that of left nilpotence or of right nilpotence!
It is not clear to me what the most useful common generalization
of these various sorts of nilpotence is, so I leave it to the experts
in nonassociative rings to develop that observation further.
Let us record a few other versions of nilpotence,
corresponding to still other subalgebras of $M(A).$
Given any $\alpha,\beta\in k,$
one can define a new multiplication on any $\!k\!$-algebra $A$ by
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.x*y}
$x*y\ =\ \alpha\,x\,y+\beta\,y\,x,$
\end{minipage}\end{equation}
%
(from which the original multiplication is recoverable by a
transformation of the same form if $\alpha^2-\beta^2$
is invertible in $k).$
Left nilpotence of this operation is a property of $A$ that is not,
in general,
equivalent to either nilpotence, left nilpotence, or right nilpotence
of the original operation; but any results on
left nilpotence of a general algebra will necessarily apply to
left nilpotence of this operation.
Recall next that for any algebra $A$ one can define the family of
{\em associator} operations,
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.associator}
$a_{x,z}(y)\ =\ x(yz)-(xy)z\qquad (x,y,z\in A).$
\end{minipage}\end{equation}
%
Hence we may consider the subalgebra $M_a(A)\subseteq M(A)$ generated by
all these maps, and study algebras $A$ for which $M_a(A)$ is nilpotent.
Does the fact that the generating set of maps $\{a_{x,z}\mid x,z\in A\}$
is not a linear image of $A$ but a bilinear image of $A\times A$
affect the usefulness of this construction?
I don't know.
Finally, note that to every finite binary tree with $n$ leaves, one
can associate a way of bracketing $n$ symbols, and hence a way of
associating to every algebra $A$ a derived $\!n\!$-ary operation.
Various nilpotence-like conditions can be expressed conveniently in
terms of this formalism.
Thus, an algebra $A$ is left nilpotent if and only if for some $n,$ the
$\!n+1\!$-ary operation induced by the length-$\!n\!$ right-branching
chain is zero on $A;$ right nilpotent,
likewise, if and only if for some $n,$ the
operation induced by the length-$\!n\!$ left-branching chain is zero.
(Here we call a tree a ``chain'' if after pruning all leaves, it has
the form usually called a chain.)
An algebra $A$ is nilpotent if and only if for some $n,$ the operations
induced by all length-$\!n\!$ chains are zero; equivalently, if
and only if for
some $n'$ the operations induced by all trees with $n'$ leaves are zero.
An algebra is solvable if and only if for some $n$ the operation
induced by the full depth-$\!n\!$ binary tree (with $2^{n+1}-1$
nodes) is zero.
One might develop a general framework for studying such
conditions, associating a nilpotence-like condition on algebras
to every filter of subsets of the set of finite binary trees.
\subsection{What about restricted Lie algebras?}\label{S2.p-Lie}
Over a field $k$ of characteristic $p>0,$ a more useful concept than
that of a Lie algebra is that of a {\em restricted} Lie
algebra or {\em $\!p$-Lie algebra}\/: a Lie algebra given with
an additional operation, $x\mapsto x^{(p)},$ satisfying certain
identities which, in associative $\!k\!$-algebras,
relate the $\!p$-th power map with the
$\!k\!$-module structure and commutator brackets.
Though $\!p$-Lie algebras are
not algebras in the sense of this note, it would, of course,
be of interest to know whether versions of our
results hold for these objects.
\subsection{What about nilpotent groups?}\label{S2.groups}
The relation between nilpotence of Lie algebras over
$\mathbb{R}$ and $\mathbb{C},$ and
nilpotence, in the sense of group theory, of the corresponding
Lie groups makes it natural to ask whether the methods and results
of this note have analogs for groups $G$ (not necessarily Lie).
In view of the way the brackets of a Lie algebra are related
to the group operation, the natural analogs of the maps
$r_x$ and $l_x$ in the above development would seem to be
the commutator maps $c_x(y)=x^{-1}y^{-1}x\,y.$
The analog of ``quasiinvertibility'' for a map $u:G\to G$ might
be invertibility of the set-map $g\mapsto g\,u(g)$ of $G$ to itself.
But it is not clear under what operations it would be natural to close
the set of commutator maps to form the analog of $M(A),$ and whether
this (or any method) will lead to an analog of Theorem~\ref{T.main}.
\subsection{A different connection with groups}\label{S2.pro-p}
Yiftach Barnea has pointed out a lemma in group
theory which is similar to our main theorem in a different way.
It says that a finite homomorphic image of a pro-$\!p\!$ group
is a $\!p\!$-group \cite[Lemma~1.18]{pro-p}.
The proof can, in fact, be put in a form
strikingly similar to that of our Theorem~\ref{T.main}: Note that in
a $\!p\!$-group, for every integer $q$ relatively prime
to $p,$ the operation $(~)^q$ of exponentiation by $q$ is invertible.
Hence this remains true in an inverse limit of such groups.
Hence in any homomorphic image $G$ of such an inverse limit,
$(~)^q$ is surjective.
In that situation, $(~)^q$ need not be
injective; but if $G$ is finite, surjectivity implies injectivity,
so all the maps $(~)^q$ are invertible,
forcing $G$ to be a $\!p\!$-group.
That lemma is a key step in Serre's proof that any homomorphism
from a finitely generated (in the topological sense)
pro-$\!p\!$ group to a finite group is continuous in the
natural topology \cite[Theorem~1.17]{pro-p},
\cite[\S I.4.2, Exercise~5-6, p.32]{Serre}.
Following the parallism noted above, and using
Theorem~\ref{T.main} of this note in place of \cite[Lemma~1.18]{pro-p},
we obtain in \cite{pro-np2} an analog of that result
for pro-nilpotent algebras in a large class of varieties of algebras
over a field $k,$ including those of associative, Lie and Jordan
algebras.
\section{Questions.}\label{S.questions}
Several topics for further investigation were noted above.
Here are some more specific questions.
Regarding the chain of conditions in~\S\ref{S.chain}, we ask
\begin{question}\label{Q.chain}
\textup{(i)}~ For $A$ an algebra, is the implication
\textup{(\ref{d.mxsurj})$\!\implies\!$(\ref{d.surj})} reversible?
More generally, if $A$ is a module over an associative nonunital
ring $R$ such that for each $r\in R,$ the operator
$1+r$ is surjective on $A,$ does the action of
$\mathrm{Mat}_n(R)$ on the direct sum of $n$
copies of $A$ have the same property?
\textup{(ii)}~ For $A$ an algebra which
is Hopfian as a $\!k\!$-module, is any of the the implications
\textup{(\ref{d.rad})$\!\implies\!$(\ref{d.M_in_rad})$\!\implies\!$%
(\ref{d.mxsurj})$\!\implies\!$(\ref{d.surj})} reversible?
\end{question}
Examples~\ref{E.y-yy} and~\ref{E.ES+PMC} show that a homomorphic image
of a pro-nilpotent algebra over a field can contain a simple
subalgebra, and so, in particular, an idempotent subalgebra.
This leaves open the question
\begin{question}\label{Q.idpt}
Can a nonzero homomorphic image $B$ of a pro-nilpotent algebra $A$
over a field $k$ itself be idempotent?
Simple?
If so, can this happen when our algebras are associative?
\end{question}
Of course, by Theorem~\ref{T.main}(iii), such a $B$ cannot
be finite-dimensional and by Lemma~\ref{L.idpt}(i)-(ii), if our
algebras are associative, $B$ cannot be finitely generated.
For the case where $k$ is not, as assumed above, a field,
Corollary~\ref{C.not_idpt} gives a weaker restriction of the same sort.
We have seen ways in which Lie algebras behave like
associative algebras (Lemma~\ref{L.[]=()}), and ways
in which they differ (the contrast between
Lemma~\ref{L.idpt}(i) and Example~\ref{E.y-xy+yx}).
The next question concerns some cases where it isn't
clear on which side of the fence Lie algebras will fall.
\begin{question}\label{Q.Lie_idpt}
Can a homomorphic image $B$ of a pro-nilpotent Lie algebra
have a nonzero finitely generated idempotent subalgebra?
If so, can it have a nonzero finitely generated simple subalgebra?
If so, can such a subalgebra be finite-dimensional?
\end{question}
(A curious difference between the behaviors of Lie and
associative algebras is noted in \cite[Example~25.49]{coalg}, where
it is observed that a topological Lie algebra (over a field) with
a linearly compact topology need not be an inverse limit of
finite-dimensional Lie algebras.
That example is the Lie algebra spanned by the operator denoted
$D$ in the discussion following Corollary~\ref{C.solv}, together
with formal power series (rather than just polynomials)
in the operator there denoted $X.$
Under the duality between vector spaces and linearly compact
vector spaces, this shows that the ``Fundamental
theorem on coalgebras'', a result on coassociative coalgebras,
is not valid for co-Lie-algebras.)
In~\S\ref{S.solv}, where we considered solvable Lie algebras,
we raised
\begin{question}\label{Q.solv}
In Corollary~\ref{C.solv}, is it possible to remove or weaken
\textup{(i)}~the condition that the $A_i$ be finite-dimensional, or
\textup{(ii)}~the condition of characteristic~$0,$ or
\textup{(iii)}~the condition that the algebras be Lie?
\end{question}
In~\cite{prod_Lie1} and~\cite{prod_Lie2}, N.\,Nahlus
and the present author study homomorphic
images of {\em direct product} algebras $\prod_I A_i.$
The form of the results obtained there suggests some possible
strengthenings of Theorem~\ref{T.main}(iii):
\begin{question}\label{Q.cf_prod_Lie}
In Theorem~\ref{T.main}\textup{(iii)}, if $k$ is a field
\textup{(}or perhaps, more restrictively, an infinite field\textup{)},
can the hypothesis that $B$ is finite-dimensional
\textup{(}the form that the finite-length hypothesis takes for vector
spaces\textup{)} be weakened to {\em countable}-dimensional?
\textup{(}Cf.~\cite[Theorem~11]{prod_Lie1},
\cite[Theorem~8]{prod_Lie2}.\textup{)}
For any algebra $B,$ let us
write $Z(B)$ for the ideal $\{b\in B\mid b\,B=B\,b=\{0\}\,\}.$
Then if $k$ is infinite and $\mathrm{card}(I)$ is less than
any uncountable measurable cardinal, can the conclusion of
Theorem~\ref{T.main}\textup{(iii)} be strengthened to say that
the composite map $A\to B\to B/Z(B)$ factors through
one of the projections $p_i:A\to A_i$ \textup{(}equivalently,
is continuous in the pro-discrete topology\textup{)}?
Without those cardinality hypotheses, can we say that
$B/Z(B)$ is a homomorphic image of one of the $A_i$?
\textup{(}Cf.~\cite[Proposition~16]{prod_Lie1}.\textup{)}
\end{question}
The concept of measurable cardinal is reviewed
in~\cite[\S15]{prod_Lie1}.
The need, in the second paragraph of the above question, for
the cardinality conditions and for the denominator ``$Z(B)$''
arises from the need for these same restrictions in~\cite{prod_Lie1}
and~\cite{prod_Lie2}.
Indeed, an infinite direct product of algebras is an
inverse limit of finite subproducts, so counterexamples
to statements for infinite products
are also counterexamples for inverse limits.
On the other hand, in \cite{pro-np2} we obtain positive results
without the cardinality restriction on $I,$ or passing to $B/Z(B),$
under the assumption that $A$ has a finitely generated subalgebra
dense in the inverse limit topology, and satisfies identities
with certain properties (which hold, inter alia, for associative,
Lie, and Jordan algebras).
Thinking about the counterexamples in \S\ref{S.eg},
and the differences between the kinds of examples that can exist
for associative and for nonassociative algebras, suggests
\begin{question}\label{Q.assoc_im}
If an associative algebra $B$ can be written as a homomorphic
image of a pro-nilpotent algebra, can it be written as
a homomorphic image of an associative pro-nilpotent algebra?
Same question, with associativity replaced by an arbitrary
family of identities.
\end{question}
% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
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MR~{\bf 92f}:16001.
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MR~{\bf 99i}:16001.
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MR~{\bf 50}\#7275.
% \bibitem{ES+PMC} E.\,S\polhk{a}siada and P.\,M.\,Cohn,
% \bibitem{ES+PMC} E.\,S\lasp{a}siada and P.\,M.\,Cohn,
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MR~{\bf 35}\#1643, MR~{\bf 96j}:17001.
% 17_I derived series, "solvable"; 18M "nilpotent"
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\end{thebibliography}
\end{document}