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\begin{document}
\vspace{-19em}
\begin{center}
\texttt{
This is the final preprint version of a paper which appeared at \\[.3em]
Journal of Algebra and its Applications,
{\bf 11} (2012) No.5, 1250090\\[.3em]
The published version is accessible to subscribers at\\[.3em]
\url{http://dx.doi.org/10.1142/S0219498812500909} .}
\end{center}
\vspace{12em}
\title%
{Bilinear maps on Artinian modules}
\thanks{\url{http://arXiv.org/abs/1108.2520}\,.
After publication of this note, updates, errata, related references
etc., if found, will be recorded at
\url{http://math.berkeley.edu/~gbergman/papers/}.
}
\subjclass[2010]{Primary: 13E10, 16P20;
% Artin_rgs+mdls ditto
Secondary: 13C05, 13E05, 15A63, 16D20, 16P40.}
% mdl-str Noeth quadr+bil bimod Noeth
\keywords{Nondegenerate bilinear map, Artinian module,
Noetherian module, bimodule.
}
\author{George M. Bergman}
\address{University of California,
Berkeley, CA 94720-3840, USA}
\email{gbergman@math.berkeley.edu}
\begin{abstract}
It is shown that if a bilinear map $f:A\times B\to C$ of modules
over a commutative ring $k$ is nondegenerate (i.e.,
if no nonzero element of $A$ annihilates all of $B,$ and vice
versa), and $A$ and $B$ are Artinian, then
$A$ and $B$ are of finite length.
Some consequences are noted.
Counterexamples are given to some attempts to generalize the above
result to balanced bilinear maps
of bimodules over noncommutative rings, while the
question is raised whether other such generalizations are true.
\end{abstract}
\maketitle
% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Below, rings and algebras are associative and
unital, except where the contrary is stated.
Examples of modules over a commutative
ring $k$ that are Artinian but not Noetherian are
well known; for example, the $\!\Z\!$-module $Z_{p^\infty}.$
However, such modules do not show up as underlying modules of
$\!k\!$-algebras.
(We shall see in \S\ref{S.HL} that this can be deduced, though
not trivially, from the Hopkins-Levitzki Theorem, which says
that left Artinian rings are also left Noetherian.)
The result of the next section can be thought of as a generalization
of this fact.
I am grateful to J.\,Krempa for pointing out two misstatements
in the first version of this note, to K.\,Goodearl, D.\,Herbera,
T.\,Y.\,Lam and L.\,W.\,Small for references to related literature,
and to the referee of a previous version of this note for an
alternative proof of the main theorem, noted at the end
of~\S\ref{S.main}.
\section{Our main result}\label{S.main}
In the proof of the following theorem, it is
interesting that everything before the next-to-last sentence works,
mutatis mutandis, if $A$ and $B$ are both
assumed Noetherian rather than Artinian, though the
final conclusion is clearly false in that case.
(On the other hand, the argument does not work at all if
one of $A$ and $B$ is assumed Artinian, and the other Noetherian.)
\begin{theorem}\label{T.main}
Suppose $k$ is a commutative ring, and
$f:A\times B\to C$ a bilinear map of $\!k\!$-modules
which is nondegenerate, in the sense that for every nonzero
$a\in A,$ the induced map $f(a,-):B\to C$ is nonzero, and for
every nonzero $b\in B,$ the induced map $f(-,b):A\to C$ is nonzero.
Then if $A$ and $B$ are Artinian, they both have finite length.
\end{theorem}
\begin{proof}
If elements $a\in A,\ b\in B$ satisfy $f(a,b)=0,$ we shall say
they \emph{annihilate} one another.
(The concept of an element $c$ of the base
ring $k$ annihilating an element
$x$ of $A,$ $B,$ or $C$ will retain its usual meaning, $c\,x=0.)$
For subsets $X\subseteq A,$ respectively $Y\subseteq B,$ we
define the annihilator sets
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.perp}
$X^\perp\ =\ \{b\in B\mid(\forall\,x\in X)\ f(x,b)=0\}
\ \subseteq\ B,$\\[0.3em]
\strut$Y^\perp\ =\ \{a\in A\mid(\forall\,y\in Y)\ f(a,y)=0\}
\ \subseteq\ A.$
\end{minipage}\end{equation}
We see that these are submodules of $B$ and $A$ respectively,
that the set of annihilator submodules in $B$ (respectively, in $A)$
% "in" rather than "of" to avoid reading as "annihilator of"
forms a lattice (in the order-theoretic sense) under inclusion,
and that these two lattices of annihilator submodules
are antiisomorphic to one another, via the maps $U\mapsto U^\perp.$
(This situation is an example of a ``Galois connection''
\cite[\S5.5]{245}, but I will not assume familiarity with
that formalism.)
When $A$ and $B$ are Artinian, these lattices of submodules
of $A$ and $B$ both have descending chain condition; so since
they are antiisomorphic, they also have ascending chain condition.
Hence all their chains have finite length.
Let us choose a maximal
(i.e., unrefinable) chain of annihilator submodules of $A,$
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.A_i}
$\{0\}=A_0\ \subseteq\ A_1\ \subseteq\ \dots\ \subseteq A_n=A.$
\end{minipage}\end{equation}
%
This yields a maximal chain of annihilator submodules of $B,$
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.B_i}
$B=B_0\ \supseteq\ B_1\ \supseteq\ \dots\ \supseteq B_n=\{0\},$
\end{minipage}\end{equation}
%
where
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.*}
$B_i\ =\ A_i^\perp,\qquad A_i\ =\ B_i^\perp.$
\end{minipage}\end{equation}
It is easy to see that for each $i