m$ elements, at least one lies in none of these spaces, giving the inductive step. To get the second statement, recall that maximal subspaces of $Y$ are kernels of elements of the dual space $Y^*=\r{Hom}(Y,{_D}D),$ a right $\!D\!$-vector-space. Hence we can apply to that space the left-right dual of the first statement, and get the desired result. \end{proof} The next corollary applies the two preceding results to the modules and map of~\eqref{d.setup}. Note that the statement refers to both {\em minimal} idempotents and {\em minimal central} idempotents. \begin{corollary}\label{C.N-strong} Suppose $h:{_T}A_S \times {_S}B\to {_T}C$ is as in~\eqref{d.setup}, and satisfies~\eqref{d.nondeg},~\eqref{d.b}, and~\eqref{d.c}. Let $N_T$ be the minimum of the cardinalities of the division rings $f\,Tf$ as $f$ ranges over the minimal idempotents of $T,$ if that minimum is finite, or the symbol $\infty$ if all those division rings are infinite. Let $d_T$ be the maximum, as $f$ ranges over the minimal {\em central} idempotents of $T,$ of the number of minimal central idempotents $e$ of $S$ such that $fA\,e\neq 0.$ Then for each minimal idempotent $f$ of $T,$ there is at least one minimal idempotent $e$ of $S$ such that $fA\,e$ is left $\!N_T/d_T\!$-strong as a set of maps $e\,B\to f\,C.$ Likewise, let $N_S$ be the minimum of the cardinalities of the division rings $eSe$ as $e$ ranges over the minimal idempotents of $S$ if this is finite, or the symbol $\infty$ if that minimum is infinite, and $d_S$ the maximum, as $e$ ranges over the minimal {\em central} idempotents of $S,$ of the number of minimal central idempotents $f$ of $T$ such that $fA\,e\neq 0.$ Then for each minimal idempotent $e$ of $S,$ there is at least one minimal idempotent $f$ of $T$ such that $fA\,e$ is right $\!N_S/d_S\!$-strong as a set of maps $e\,B\to f\,C.$ \end{corollary} \begin{proof} In the situation of the first assertion, we see from condition~\eqref{d.c} that for each minimal central idempotent $f$ of $T,$ the set $f\,A$ of morphisms $f\,a: B\to f\,C$ $(a\in A)$ has the property that for every minimal $\!f\,Tf\!$-submodule $C_0\subseteq f\,C,$ there is at least one nonzero $f\,a\in f\,A$ which takes $B$ into $C_0.$ Now the minimal central idempotents $e\in S$ sum to $1,$ so there is at least one such $e\in S$ such that $f\,a\,e\neq 0.$ This map clearly still carries $B$ into $C_0;$ so we conclude that for every minimal $C_0\subseteq f\,C,$ some nonzero element of $\bigcup_e (fA\,e-\{0\})$ has image in $C_0,$ where the union is over the minimal central idempotents $e$ of $S.$ By Lemma~\ref{L.not_union}, this shows that if we write $D$ for the division ring (unique up to isomorphism) given by $f'Tf'$ where $f'$ is any minimal idempotent of $f\,Tf,$ then $\bigcup_e (fA\,e-\{0\})$ is left $\!\card(D)\!$-strong; hence left $\!N_T\!$-strong. Now by definition of $d_T,$ there are at most $d_T$ minimal central idempotents $e$ such that $fA\,e\neq 0;$ so $\bigcup_e (fA\,e-\{0\})$ involves at most $d_T$ nonempty sets $fA\,e-\{0\}.$ Hence by Lemma~\ref{L.strong_union}, at least one of these sets is left $\!N_T/d_T\!$-strong, as claimed. The second assertion is proved in the analogous fashion. \end{proof} The above corollary will enable us to prove our desired generalizations of~\eqref{d.cm_dim} and~\eqref{d.cm_local_dim} {\em unless} one or more of the division rings $eSe$ and $f\,Tf$ are finite fields of small cardinality. (If such a field occurs, $N_T/d_T$ or $N_S/d_S$ may be too small for our arguments to work.) It is curious that a similar condition in my original proof of~\eqref{d.cm_dim} and~\eqref{d.cm_local_dim} was eliminated by de~Seguins Pazzis's argument; but we shall see by example, in \S\ref{S.misc}, that the corresponding condition in the present situation cannot be dropped. While these cardinality conditions are not very restrictive, and are needed for the result to hold, we come now to an embarrassingly restrictive condition, needed for the proofs of the results of the next section, though I have no example showing that those results fail without it. The condition is awkward to state in maximum generality. A fairly natural special case is the hypothesis that in~\eqref{d.setup}, % \begin{equation}\begin{minipage}[c]{35pc}\label{d.small_algcl} The semisimple Artinian rings $S$ and $T$ are finite-dimensional algebras over a common algebraically closed field $k,$ and the induced actions of $k$ on the two sides of the bimodule $_T A_S$ are the same. \end{minipage}\end{equation} % Actually, we need the assumption that $k$ is algebraically closed only to make the simple factors of $S$ and $T$ full matrix algebras over $k;$ so we can instead put that assumption on the table, as the condition % \begin{equation}\begin{minipage}[c]{35pc}\label{d.small_mxs} For some field $k,$ each of the semisimple Artinian rings $S$ and $T$ is a direct product of full matrix algebras over $k,$ and the induced actions of $k$ on the two sides of the bimodule $_T A_S$ are the same. \end{minipage}\end{equation} % A condition that is still more general (as we shall show in the next lemma), and will suffice for our purposes, is % \begin{equation}\begin{minipage}[c]{35pc}\label{d.small_both} If $a\in A$ is a nonzero element whose image $a\,B$ is contained in a simple submodule of $C,$ then there exists nonzero $a'\in TaS$ whose kernel contains a maximal submodule of $B;$ and likewise if $a\in A$ is a nonzero element whose kernel contains a maximal submodule of $B,$ then there exists nonzero $a'\in TaS$ whose image is contained in a simple submodule of $C.$ \end{minipage}\end{equation} % In fact, we can make do with the following still weaker (though still wordier) condition. % \begin{equation}\begin{minipage}[c]{35pc}\label{d.small_either} For each minimal central idempotent $e\in S$ and minimal central idempotent $f\in T$ such that $f\,A\,e\neq\{0\},$ it is {\em either} true that for every $a\in fA\,e$ such that the induced map $e\,B\to f\,C$ has image in a simple submodule of $f\,C,$ some nonzero $a'\in TaS$ has kernel containing a maximal submodule of $B,$ {\em or} that for every $a\in fA\,e$ such that the induced map $e\,B\to f\,C$ has kernel containing a maximal submodule of $B,$ some nonzero $a'\in TaS$ has image in a simple submodule of $f\,C.$ (But which of these is true may vary with the pair $(f,e).)$ \end{minipage}\end{equation} Let us note the implications among these conditions. \begin{lemma}\label{L.conditions} For $h:{_T}A_S \times {_S}B\to {_T}C$ a bilinear map as in~\eqref{d.setup} which satisfies~\eqref{d.nondeg}, one has the implications \eqref{d.small_algcl}$\!\implies\!$% \eqref{d.small_mxs}$\!\implies\!$% \eqref{d.small_both}$\!\implies\!$\eqref{d.small_either}. \end{lemma} \begin{proof} \eqref{d.small_algcl}$\implies$\eqref{d.small_mxs} follows from the standard description of the structures of semisimple Artin rings, and \eqref{d.small_both}$\!\implies\!$\eqref{d.small_either} is clear. Let us prove \eqref{d.small_mxs}$\implies$\eqref{d.small_both}. Suppose as in~\eqref{d.small_both} that $a\neq 0,$ and $a\,B$ is contained in a simple submodule $C_0\subseteq C.$ Since the identity elements of $S$ and $T$ are sums of minimal central idempotents, we can find such idempotents $e\in S$ and $f\in T$ such that $f\,a\,e\neq 0;$ and we will have $f\,a\,e\,B\subseteq f\,a\,B\subseteq f\,C_0\subseteq C_0.$ Hence, replacing $a$ by some $a'=f\,a\,e$ if necessary, we may assume without loss of generality that $a\in fA\,e$ for such a pair of idempotents. Now by assumption, $eSe$ and $f\,Tf$ have the forms $\r{Matr}_{m,m}(k)$ and $\r{Matr}_{n,n}(k)$ for some positive integers $m$ and $n.$ Identifying them with these matrix rings, it is easy to verify that there exist finite-dimensional $\!k\!$-vector-spaces $B',$ $C'$ such that we can identify $eB$ and $fC$ with the spaces $B'^m$ and $C'^n$ of column vectors of elements of $B'$ and $C',$ made into modules over $eSe=\r{Matr}_{m,m}(k)$ and $fTf=\r{Matr}_{n,n}(k)$ in the natural way; that $fAe$ can then be identified with $\r{Matr}_{n,m}(A')$ where $A'$ is a $\!k\!$-vector-space of $\!k\!$-linear maps $B'\to C',$ and finally, that the simple submodule $C_0\subseteq f\,C$ will have the form $C_0'^n$ for some $\!1\!$-dimensional subspace $C'_0\subseteq C'.$ (Explicitly, letting $e'$ and $f'$ denote minimal idempotents of $S$ and $T,$ say those given by the matrix units $e_{11}$ of their matrix representations, we can take $B'=e'B,$ $C'=f'C,$ $C'_0=f'C_0,$ and $A'=f'A\,e'.)$ Thus, our element $a\in fA\,e$ will be an $n\times m$ matrix of linear maps $e\,B'\to f\,C',$ each having range in $C'_0.$ We can now choose $\bar{a}\in TaS-\{0\}$ to be nonzero and have all components in some $\!1\!$-dimensional subspace $k\,a'\subseteq A'.$ (E.g., we can let $\bar{a}$ be a product $e_{1,i}\,a\,e_{j,1}$ such that the $(i,j)$ component of $a$ is nonzero.) Since $a': B'\to C'$ has range in the $\!1\!$-dimensional subspace $C'_0$ of $C',$ it is a rank-$\!1\!$ $\!k\!$-linear map, hence has kernel $B'_0\subseteq B'$ of codimension $1.$ Hence $\bar{a},$ having all components in $k\,a',$ will have kernel containing the maximal proper submodule $(B'_0)^m\subseteq B,$ as required. The second assertion of~\eqref{d.small_both} is proved similarly. \end{proof} \section{Minimal faithful modules over left Artin rings}\label{S.lt_art} Recall that if $G$ is a finite graph, its {\em Euler characteristic} $\chi(G)$ is the number of vertices of $G$ minus the number of edges, an integer which may be positive, negative or zero. Recall also that a {\em bipartite} graph is a graph whose vertex-set is given as the disjoint union of two specified sets, such that every edge connects a member of one set with a member of the other. We shall call those sets (nonstandardly) the {\em left} and {\em right} vertex-sets of~$G.$ The hard work of this section comes right at the beginning: proving the following noncommutative analog of Proposition~\ref{P.cm_bilin}, or more precisely, of Corollary~\ref{C.cm}. \begin{proposition}\label{P.main} Suppose $h:{_T}A_S \times {_S}B\to {_T}C$ is a bilinear map as in~\eqref{d.setup}, which satisfies~\eqref{d.nondeg}, \eqref{d.b}, \eqref{d.c} and~\eqref{d.small_either}. For notational convenience we shall assume $S$ and $T$ disjoint. Let $N_S,$ $d_S,$ $N_T$ and $d_T$ be defined as in Corollary~\ref{C.N-strong}. Further, let $l_S$ denote the maximum of the values $\lt_{eSe}(e\,B)$ as $e$ ranges over the minimal central idempotents $e\in S,$ and $l_T$ the maximum of $\lt_{f\,Tf}(f\,C)$ as $f$ ranges over the minimal central idempotents $f\in T;$ and assume that % \begin{equation}\begin{minipage}[c]{35pc}\label{d.cardD} $N_T\ \geq\ d_T\,l_S,$\quad and\quad $N_S\ \geq\ d_S\,l_T.$ \end{minipage}\end{equation} Finally, let $G$ be the bipartite graph whose right vertex-set is the set of minimal central idempotents $e\in S$ satisfying $e\,B\neq 0$ \textup{(}equivalently, $A\,e\neq 0),$ whose left vertex-set is the set of minimal central idempotents $f\in T$ satisfying $f\,C\neq 0$ \textup{(}equivalently, $f\,A\neq 0),$ and such that two such vertices $e,$ $f$ are connected by an edge $(f,e)$ if and only if $fA\,e\neq\{0\}.$ Then % \begin{equation}\begin{minipage}[c]{35pc}\label{d.P.main} $\lt({_S}B)\ +\ \lt({_T}C)\ \leq\ \lt({_T}A_S)\ +\ \chi(G).$ \end{minipage}\end{equation} % \end{proposition} \begin{proof} The parenthetical equivalences in the definition of $G$ follow from~\eqref{d.nondeg},~\eqref{d.b} and~\eqref{d.c}. Combining Corollary~\ref{C.N-strong} with our hypothesis~\eqref{d.cardD}, we find that % \begin{equation}\begin{minipage}[c]{35pc}\label{d.strong} For each minimal idempotent $f$ of $T,$ there is at least one minimal idempotent $e$ of $S$ such that $fA\,e$ is left $\!l_S\!$-strong as a set of maps $e\,B\to f\,C;$ and for each minimal idempotent $e$ of $S,$ there is at least one minimal idempotent $f$ of $T$ such that $fA\,e$ is right $\!l_T\!$-strong as a set of maps $e\,B\to f\,C.$ \end{minipage}\end{equation} We shall now perform a series of reductions and decompositions on our system ${_T}A_S \times {_S}B\to {_T}C,$ verifying at each stage that if the inequality corresponding to~\eqref{d.P.main} holds for our simplified system(s), then it also holds for the original system; and, finally, we shall establish that inequality for the very simple sorts of system we end up with. In preparation, let us harness~\eqref{d.strong} by choosing, arbitrarily, for each minimal central idempotent $f\in T,$ {\em one} minimal central idempotent $e\in S$ such that $fA\,e$ is left $\!l_S\!$-strong, and call $(f,e)$ the {\em left-marked} edge of the graph $G$ associated with the vertex $f;$ and similarly, for each minimal central idempotent $e\in S,$ let us choose a minimal central idempotent $f\in T$ such that $fA\,e$ is right $\!l_T\!$-strong, and call $(f,e)$ the {\em right-marked} edge of $G$ associated with the vertex $e.$ Some edges may be both right- and left-marked (for their respective right and left vertices). We begin our reductions by considering any edge $(f,e)\in G$ which is neither right- nor left-marked, and seeing what happens if we drop the summand $fA\,e$ from $A;$ i.e., replace $A$ with $(1-f)A + A(1-e);$ and thus drop the edge $(f,e)$ from $G,$ leaving the rest of our system unchanged. Because $(f,e)$ is neither right nor left marked, condition~\eqref{d.strong} has not been lost. (The constant $d_S$ and/or $d_T$ may have decreased by $1,$ but we don't have to think about this, because our use of these constants was only to obtain~\eqref{d.strong}, which has been preserved.) The removal of $fA\,e$ has no effect on the left-hand side of~\eqref{d.P.main}, while on the right-hand side, it decreases $\lt({_T}A_S)$ by $\lt({_T}(fA\,e)_S)\geq 1,$ and increases $\chi(G)$ by $1.$ Hence if the new system satisfies~\eqref{d.P.main}, then the original system, whose right-hand side is $\geq$ that of the new system, also did. Hence by induction, the task of proving~\eqref{d.P.main} is reduced to the case where {\em every} edge of $G$ is left and/or right marked. Suppose, next, that there is some left vertex $f$ of $G$ such that the only edge adjacent to $f$ is its associated left-marked edge, say $(f,e),$ and such that this is {\em not} also right-marked, and consider what happens if we remove both the vertex $f$ and the edge $(f,e);$ i.e., replace $C$ by $(1-f)C,$ and $A$ by $(1-f)A.$ Clearly, the {\em remaining} vertices and edges continue to witness condition~\eqref{d.strong}. Our new system also has the same Euler characteristic as the old one, since just one vertex and one edge have been removed from the graph. To see how~\eqref{d.P.main} is affected, let $d=\lt({_T}f\,C).$ I claim that we can find elements $a_1,\dots,a_d\in fA\,e$ such that the submodules $T\,a_i\,B\subseteq f\,C$ are simple and their sum is direct. Indeed, assuming we have constructed $a_1,\dots,a_j$ with $j2,$ or a graph having just two vertices, and a single edge which is marked for both of these. It follows that our system ${_T}A_S \times {_S}B\to {_T}C$ decomposes into a direct sum of subsystems corresponding to those connected components, and that~\eqref{d.P.main} will be the sum of the corresponding inequalities for those components. Hence, it will suffice to prove~\eqref{d.P.main} in the two cases where $G$ is a loop, and where $G$ has just a single edge. The $l_S$ and $l_T$ for each such system are $\leq$ the $l_S$ and $l_T$ for our original system, so~\eqref{d.strong} will hold for these systems, because it held for the original system. If $G$ is a loop, then each edge is marked for only one vertex, and the proof is quick: The Euler characteristic $\chi(G)$ is zero, while for each vertex, the bimodule corresponding to its marked edge has at least the length of the module corresponding to that vertex, by the same ``$\sum_{i\leq j} T\,a_i\,S$'' arguments used in the preceding reduction. Summing these inequalities over the vertices, we have~\eqref{d.P.main}. We are left with the case where $G$ has just one right vertex, $e,$ one left vertex, $f,$ and the single edge $(f,e).$ Thus, $B=eB$ and $C=fC.$ Let % \begin{equation}\begin{minipage}[c]{35pc}\label{d.m,n} $m\ =\ \lt({_S}B),\qquad n\ =\ \lt({_T}C).$ \end{minipage}\end{equation} % Note that the fact that $(f,e)$ is both left- and right-marked tells us that $A$ is both left $\!l_S\!$-strong and right $\!l_T\!$-strong. It is now that we will use our hypothesis~\eqref{d.small_either}. Let us begin by assuming the second of the two alternatives it offers, which in this situation says that whenever an $a\in A$ has kernel containing a maximal submodule of $B,$ then some nonzero element of $TaS$ has image in a simple submodule of $C.$ Under this assumption, we begin by constructing, for $m$ as in~\eqref{d.m,n}, elements $a_1,\dots,a_m\in A$ such that % \begin{equation}\begin{minipage}[c]{35pc}\label{d.a1...am} $T\,a_1\,S,\dots,T\,a_m\,S$ have for kernels maximal submodules of $B,$ none of which contains the intersection of the kernels of the others, and each $T\,a_i\,S$ has for image a simple submodule of~$C.$ \end{minipage}\end{equation} % To see that we can do this, suppose inductively that we have constructed $i N$ submodules of length $r$ between $Y''$ and $Y'$ relative to which $W$ is left $\!N\!$-strong. We will simply say $W$ is left $\!N\!$-strong (in our new sense) if it is left $\!N\!$-strong relative to its codomain $Y.$ It is now easy to verify that if, for every simple submodule of $Y_0\subseteq Y,$ there is a nonzero element of $W$ with image in $Y_0,$ and if the division ring over which $V$ is a full matrix ring has cardinality $\geq N,$ then $W$ is left $\!N\!$-strong under our new definition. Moreover, an easy induction shows that if a union $W=W_1\cup\dots\cup W_d$ is left $\!N\!$-strong, then at least one of the $W_i$ is left $\!N/d\!$-strong. The definition of right $\!N\!$-strong would be modified analogously. In the proof of Proposition~\ref{P.cm_bilin}, the hypotheses~\eqref{d.cm_b} and~\eqref{d.cm_c} could then be weakened to say that (up to the change of notation appropriate to a subspace $A\subseteq B\otimes_k C$ rather than a map $A\times B\to C),$ $A$ is both left and right $\!1\!$-strong in our modified sense. I suspect that the same method could be adapted to the last part of the proof of Proposition~\ref{P.main}, and would in fact allow us to weaken the hypothesis~\eqref{d.cardD} by dropping the factors $l_S$ and~$l_T.$ I haven't worked out the details, because they do not get at the serious restrictions in our results. In particular, the suggested change in the end of the proof of Proposition~\ref{P.main} would not get rid of the need for condition~\eqref{d.small_either}. Let us now look at what we wish we could do. \subsection{Some questions}\label{SS.question} Here is an innocent-sounding generalization of our first main result that we might ask for. \begin{question}\label{Q.socle_central} Does Theorem~\ref{T.TG} remain true if the assumption that $\soc(R)$ is central in $R$ is deleted? \end{question} To prove such a result, we would want a version of Proposition~\ref{P.cm_bilin} involving a division ring rather than a field. As in the development of our results on non-local rings, I don't see a convenient way of ``symmetrizing'' the general statement we need, so in the next question, I will ask, not for the analog of that proposition, but of its corollary. Moreover, although the result needed just to get a positive answer to Question~\ref{Q.socle_central} would have for $B$ and $C$ vector spaces over the same division ring, I expect it would be no more difficult to prove such a result without that restriction; so let us pose the question as follows. \begin{question}\label{Q.loc_bilin} Let $S$ and $T$ be division rings, ${_S}B$ and ${_T}C$ be nonzero finite-dimensional vector spaces, and ${_T}A_S$ be a subbimodule of the $\!(T,S)\!$-bimodule of all additive group homomorphisms ${_S}B\to {_T}C.$ Suppose moreover that % \begin{equation}\begin{minipage}[c]{35pc}\label{d.loc_b} For every proper subspace $B'\subsetneq B,$ there is at least one $a\in A$ whose restriction to $B'$ is zero. \end{minipage}\end{equation} % and % \begin{equation}\begin{minipage}[c]{35pc}\label{d.loc_c} For every proper homomorphic image $C/C_0$ of $C,$ there is at least one $a\in A$ whose composite with the factor map $C\to C/C_0$ is zero. \end{minipage}\end{equation} % Then must it be true that % \begin{equation}\begin{minipage}[c]{35pc}\label{d.loc_lt} $\lt({_T}A_S)\ \geq\ \dim_S(B)\ +\ \dim_T(C)\ -\ 1$ \end{minipage}\end{equation} % \textup{(}where $\lt({_T}A_S)$ denotes the length of $A$ as a bimodule\textup{)}? \end{question} I have posed this question in a relatively easy-to-state form, but my hope is that if a positive answer can be proved, then one will be able push the proof further, and get the same result with the hypotheses~\eqref{d.loc_b} and~\eqref{d.loc_c} weakened to say that $A$ is left and right $\!N\!$-strong for appropriate $N,$ in the modified sense sketched in \S\ref{SS.better_N-str} above (or something like it), and that this could be used to prove generalizations of Proposition~\ref{P.main} and Theorem~\ref{T.main}. Incidentally, it would be harmless to throw into our hoped-for variant of Proposition~\ref{P.main} the added assumption that $A$ is semisimple as a left module, a right module, and a bimodule, since those conditions are true of the socle of a left Artin ring, the situation to which we apply that result in Theorem~\ref{T.main}. \subsection{A waffle about wording}\label{SS.ubiquitous} I would have preferred a more suggestive term for what I have called left and right $\!N\!$-strong families of maps. I was tempted to replace ``strong'' with ``ubiquitous''; but this might suggest too much -- a property like~\eqref{d.b} and~\eqref{d.c}, rather than the weaker property actually defined. Another thought was ``prevalent''; but this seemed a bit vague. \section{Getting minimal faithful modules from module decompositions}\label{S.sub_or_hom} This section is essentially independent of the rest of this note, though the final assertion proved will complement the results proved above. Namely, Proposition~\ref{P.socle}(iii) below, though it has a weaker conclusion than Theorems~\ref{T.TG} and~\ref{T.main}, is applicable when the hypotheses of those theorems are not satisfied. Aside from that final result, the focus will be on modules with only one of the two minimality properties considered in preceding sections. (The one other exception to independence from the rest of this note is that at one point below, we will refer to an example from the preceding section.) The two preliminary lemmas below may be of interest in their own right. The final statement of each describes how, under certain conditions, a module can be decomposed into ``small pieces'': in the first, as a sum of submodules $N$ such that $N/J(R)N$ is simple; in the second, as a subdirect product of modules $N$ with $\soc(N)$ simple. \begin{lemma}\label{L.M_as_sum} Let $R$ be a left Artinian ring, and $M$ a left $\!R\!$-module \textup{(}not necessarily Artinian\textup{)}. Then \textup{(i)} If $L$ is a simple submodule of $M/J(R)M,$ then $M$ has a submodule $N$ such that the inclusion $N\subseteq M$ induces an isomorphism $N/J(R)N\cong L\subseteq M/J(R)M.$ Hence \textup{(ii)} Given a decomposition of $M/J(R)M$ as a sum of simple modules $\sum_{i\in I} L_i,$ one can write $M$ as the sum of a family of submodules $N_i$ $(i\in I),$ such that for each $i,$ $N_i/J(R)N_i\cong L_i,$ and $L_i$ is the image of $N_i$ in $M/J(R)M.$ \end{lemma} \begin{proof} In the situation of~(i), let $x$ be any element of $M$ whose image in $M/J(R)M$ is a nonzero member of $L.$ Thus, the image of $R\,x$ in $M/J(R)M$ is $L.$ Since $R$ is left Artinian, so is $R\,x,$ hence we can find a submodule $N\subseteq R\,x$ minimal for having $L$ as its image in $M/J(R)M.$ Now since $N/J(R)N$ is semisimple, it has a submodule $L'$ which maps isomorphically to $L$ in $M/J(R)M.$ If $L'$ were a proper submodule of $N/J(R)N,$ then its inverse image in $N$ would be a proper submodule of $N$ which still mapped surjectively to $L,$ contradicting the minimality of $N.$ Hence $N/J(R)N=L'\cong L,$ completing the proof of~(i). In the situation of~(ii), choose for each $L_i$ a submodule $N_i\subseteq M$ as in~(i). Then we see that $M=J(R)M+\sum_i N_i,$ hence since $R$ is Artinian, $M=\sum_i N_i$ \cite[Theorem~23.16\,$(1)\!{\implies}\!(2')$]{TYL1}, as required. \end{proof} The next result is of a dual sort, but the arguments can be carried out in a much more general context, so that the result we are aiming for (the final sentence) looks like an afterthought. \begin{lemma}\label{L.M_as_subdir} Let $R$ be a ring and $M$ a left $\!R\!$-module. Then \textup{(i)} If $L$ is any submodule of $M,$ then $M$ has a homomorphic image $M/N$ such that the composite map $L\hookrightarrow M\to M/N$ is an embedding, and the embedded image of $L$ is essential in $M/N$ \textup{(}i.e., has nonzero intersection with every nonzero submodule of $M/N).$ Hence \textup{(ii)} If $E$ is an essential submodule of $M,$ and $f: E\to\prod_I L_i$ a subdirect decomposition of $E,$ then there exists a subdirect decomposition $g:M\to\prod_i M_i$ of $M,$ such that each $M_i$ is an overmodule of $L_i$ in which $L_i$ is essential, and $f$ is the restriction of $g$ to $E\subseteq M.$ In particular, every locally Artinian module can be written as a subdirect product of locally Artinian modules with simple socles. \end{lemma} \begin{proof} In the situation of~(i), let $N$ be maximal among submodules of $M$ having trivial intersection with $L.$ The triviality of this intersection means that $L$ embeds in $M/N,$ while the maximality condition makes the image of $L$ essential therein. (If it were not essential, $M/N$ would have a nonzero submodule $M'$ disjoint from the image of $L,$ and the inverse image of $M'$ in $M$ would contradict the maximality of $N.)$ In the situation of~(ii), for each $j\in I$ let $K_j$ be the kernel of the composite $E\to\prod_I L_i\to L_j.$ Applying statement~(i) with $M/K_j$ in the role of $M,$ and $E/K_j\cong L_j$ in the role of $L,$ we get an image $M_j$ of $M/K_j,$ and hence of $M,$ in which $L_j$ is embedded and is essential. Now since $E$ is essential in $M,$ every nonzero submodule $M'\subseteq M$ has nonzero intersection with $E,$ and that intersection has nonzero projection to $L_i$ for some $i;$ so in particular, for that $i,$ $M'$ has nonzero image in $M_i.$ Since this is true for every $M',$ the map $M\to\prod_I M_i$ is one-to-one, and gives the desired subdirect decomposition. To get the final assertion, note that the socle of a locally Artinian module is essential, and, being semisimple, can be written as a subdirect product (indeed, as a direct sum) of simple modules; so we can apply~(ii) with $E=\soc(M)$ and the $L_i$ simple. Each of the $M_i$ in the resulting decomposition will have a simple essential submodule $L_i,$ so that submodule must be its socle. \end{proof} We can now get the following result, showing that given a faithful module $M$ over an Artinian ring, we can carve out of $M$ a ``small'' faithful submodule, factor-module, or subfactor. Note that in the statement, the length of $\soc(R)$ as a bimodule may be less than its length as a left or right module. (For example, the full $n\times n$ matrix ring over a division ring is its own socle, and has length $n$ as a right and as a left module, but length $1$ as a bimodule. Similarly, in the $R$ of \S\ref{SS.small}, each of the direct summands $x\,F$ and $F\,y$ of $\soc(R)$ has length $2$ as left and as right $\!R\!$-module, but length $1$ as an $\!R\!$-bimodule; so $\soc(R)$ has length $4$ on each side, but $2$ as a bimodule.) \begin{proposition}\label{P.socle} Let $R$ be a left Artinian ring, let $n$ be the length of $\soc(R)$ as a bimodule \textup{(}equivalently, as a $\!2\!$-sided ideal\textup{)}, and let $M$ be a faithful $\!R\!$-module. Then \textup{(i)} $M$ has a submodule $M'$ which is again faithful over $R,$ and satisfies $\lt(M'/J(R)\,M')\leq n.$ \textup{(}In particular, $M'$ is generated by $\leq n$ elements.\textup{)} \textup{(ii)} $M$ has a homomorphic image $M''$ which is faithful over $R,$ and satisfies $\lt(\soc(M''))\leq\nolinebreak n.$ \textup{(iii)} $M$ has a subfactor faithful over $R$ satisfying both these inequalities. \end{proposition} \begin{proof} To get~(i), note that since $R/J(R)$ is semisimple Artin, $M/J(R)M$ can be written as a direct sum of simple modules $L_i$ $(i\in I)$ over that ring, and hence over $R,$ so we can construct a generating family of submodules $N_i\subseteq M$ related to these as in Lemma~\ref{L.M_as_sum}(ii). Since $M=\sum_i N_i$ is faithful, and $\soc(R)$ has length $n$ as a $\!2\!$-sided ideal, the sum of some family of $\leq n$ of these submodules, say % \begin{equation}\begin{minipage}[c]{35pc}\label{d.M'} $M'\ =\ N_{i_1}+\dots+N_{i_m}$\quad where $m\leq n,$ \end{minipage}\end{equation} % must have the property that $M'$ is annihilated by no nonzero subideal of $\soc(R).$ (Details: one chooses the $N_{i_j}$ recursively; as long as $N_{i_1}+\dots+N_{i_j}$ is annihilated by a nonzero subideal $I\subseteq\soc(R),$ one can choose an $N_{i_{j+1}}$ which fails to be annihilated by $I.$ The annihilators in $\soc(R)$ of successive sums $M'=N_{i_1}+\dots+N_{i_j}$ $(j=0,1,\dots)$ form a strictly decreasing chain, so this chain must terminate after $\leq n$ steps.) Since an ideal of an Artinian ring having zero intersection with the socle is zero, $M'$ has zero annihilator, i.e., is faithful. Since each $N_i$ satisfies $\lt(N_i/J(R)N_i)=1,$ we have $\lt(M'/J(R)M')\leq m\leq n.$ Statement (ii) is proved in the analogous way from the final statement of Lemma~\ref{L.M_as_subdir}, using images of $M$ in products of finite subfamilies of the $M_i$ in place of submodules of $M$ generated by finite subfamilies of the $N_i.$ Statement (iii) follows by combining~(i) and~(ii). \end{proof} \section{Acknowledgements}\label{S.ackn} I am indebted to Luchezar Avramov for pointing me to Gulliksen's \cite[Lemma~2]{TG}, to Cl\'{e}ment de Seguins Pazzis for providing, at \cite{overflow}, the proof of the present version of~Proposition~\ref{P.cm_bilin}, and to Pace Nielsen for many helpful comments on an earlier draft of this note. \begin{thebibliography}{00} \bibitem{cmtg_mxs} George M. Bergman, {\em Commuting matrices, and modules over Artinian local rings,} unpublished note, 14 pp., readable at \url{/~gbergman/papers/unpub} and at \url{http://arxiv.org/abs/1309.0053}~. \bibitem{DE} David Eisenbud, {\em Linear sections of determinantal varieties}, Amer. J. Math. {\bf 110} (1988) 541--575. MR\ {\bf 89h}:14041. \bibitem{TG} Tor H.\,Gulliksen, {\em On the length of faithful modules over Artinian local rings}, Math. Scand. {\bf 31} (1972) 78-82. % http://www.mscand.dk/article.php?id=2084 MR\,{\bf 47}\,\#3370. \bibitem{TYL1} T.\,Y.\,Lam, {\em A first course in noncommutative rings}, Springer GTM, v.131, 1991. MR~{\bf 92f}:16001. \bibitem{O+C+V} Kevin C.\,O'Meara, John Clark and Charles I.\,Vinsonhaler {\em Advanced topics in linear algebra. Weaving matrix problems through the Weyr form}, Oxford University Press, 2011. xxii$+$400~pp. % ISBN: 978-0-19-979373-0. MR\,2849857. \bibitem{overflow} \url{http://mathoverflow.net/questions/141378/is-this-lemma-in-elementary-linear-algebra-new}\,. \end{thebibliography} \end{document}