Finite matrix and module preliminaries: right inverses, rank invariance, finite length and nilpotent trace

Write the unique coordinate expansion $v_j={\sum }_{i<n}{a}_{ij}{e}_i$ of each $v_j$ in the basis $(e_i)$; this is the coordinate matrix $A=(a_{ij})$, and identifying $M$ with $R^n$ through coordinate expansions of [F1] carries $v_j$ to the $j$-th column $a_j$ of $A$ and is an $R$-module isomorphism. Since the $v_j$ span $M$, the columns $a_j$ span $R^n$, so for every $i<n$ the set $\{b\in R^n:Ab=e_i\}$ is nonempty; by [F6] applied to the function $i\mapsto \{b\in R^n:Ab=e_i\}$ on $n$ there are $b_0,\dots ,b_{n-1}\in R^n$ with $A{b}_i=e_i$, and $B\in M_n(R)$ denotes the matrix with $i$-th column $b_i$.

Let $U\subseteq F^n$ be the $F$-row space of $A$ and $(u_0,\dots ,u_{r-1})$ an $F$-basis of it, so that $r={\mathrm{rank}}_F(A)$ by [F7]; by [F12] the field extension makes $E^n$ an $F$-vector space containing $F^n$, and viewing rows and the $u_i$ in $E^n$, every row of $A$ is an $F$-combination of the $u_i$ and hence an $E$-combination, so every $E$-combination of the rows is an $E$-combination of the $u_i$: each $u_i$ is itself an $F$-combination of the rows and therefore lies in $U_E$, so the list $(u_0,\dots ,u_{r-1})$ $E$-spans the $E$-row space $U_E$ of the matrix $A$ read over $E$.

A submodule of $N$ is closed under $k$-scalars through the structure map of the $k$-algebra $B$ of [F8], hence is a $k$-linear subspace of $N$; consequently, if $L⊊L^{\mathrm{\prime }}\subseteq N$ are submodules then ${\dim }_{k}L<{\dim }_k{L}^{\mathrm{\prime }}$ by [F9].

Claim 4 is proved by induction on $m$. For $m=0$ the direct sum is $0$, so $N=0$ is the empty direct sum with complement $0$; for $m=1$ the only submodules of the simple module $S_1$ are $0$ and $S_1$, the direct sums over the empty and the full subfamily, with complements $S_1$ and $0$; for $m\ge 2$, write $S:=S_1$ and $M:=S_2\oplus \cdots \oplus S_m$, a direct sum of $m-1$ simple modules, so that $S_1\oplus \cdots \oplus S_m=S\oplus M$ with the coordinate projections as in [F14], and let $N\subseteq S\oplus M$ be a submodule.

If $T:N\to N$ is $k$-linear with $T^r=0$, put $K_j:=\ker (T^j)$ for $j\ge 0$, so that $K_0=0$ and $0=K_0\subseteq K_1\subseteq \cdots \subseteq K_r=N$ is a chain of $k$-subspaces of the finite-dimensional space $N$, with $K_j$ finite-dimensional for every $j$ by [F9]; extend a basis of $K_j$ successively to a basis of $K_{j+1}$ using the extension clause of [F9] and concatenate the successive blocks to an ordered basis $(w_1,\dots ,w_d)$ of $N$ in the sense of [F10].

By [F2] the $i$-th column of $AB$ is $A{b}_i=e_i$, so $AB=I_n$; by [F4] $\det (A)\det (B)=\det (AB)=\det (I_n)$, and the Leibniz expansion of $\det (I_n)$ has vanishing products ${\prod }_{i<n}(I_n{)}_{\sigma (i),i}$ unless $\sigma =\mathrm{id}$, so $\det (I_n)=1$ for $n\ge 1$ and $\det (A)\det (B)=1$: the determinant $\det (A)$ is a unit of $R$ with inverse $\det (B)$.

Suppose ${\sum }_{i<r}{\lambda }_i{u}_i=0$ in $E^n$ with all ${\lambda }_i\in E$. The coefficients span an $F$-subspace $L\subseteq E$ that is spanned by the finitely many ${\lambda }_i$; discarding from that finite spanning list each vector lying in the span of those retained leaves an $F$-basis $({\mu }_1,\dots ,{\mu }_t)$ of $L$, and ${\lambda }_i={\sum }_t{c}_{it}{\mu }_t$ with $c_{it}\in F$. For each coordinate $k<n$ the equality gives $0={\sum }_{i<r}{\lambda }_i{u}_i(k)={\sum }_t{\mu }_t({\sum }_{i<r}{c}_{it}{u}_i(k))$ in $E$; the ${\mu }_t$ are $F$-independent in $E$ and the coefficients ${\sum }_{i<r}{c}_{it}{u}_i(k)$ lie in $F$, so they all vanish, that is ${\sum }_{i<r}{c}_{it}{u}_i=0$ in $F^n$ for every $t$; the $u_i$ are $F$-independent, so every $c_{it}=0$ and hence every ${\lambda }_i=0$: the list $(u_0,\dots ,u_{r-1})$ is $E$-independent.

Let $N\mathrm{\ne }0$. The $k$-dimensions ${\dim }_{k}L$ of the proper submodules $L⊊N$ form a nonempty set of natural numbers, since the zero submodule is proper and has dimension $0$, bounded above by ${\dim }_{k}N-1$ by [F9]; let $d$ be its largest element and let $L$ be a proper submodule with ${\dim }_{k}L=d$ (one selection from a nonempty set of submodules). Then $L$ is maximal proper: if $L⊊L^{\mathrm{\prime }}\subseteq N$, step 1.3 gives ${\dim }_k{L}^{\mathrm{\prime }}>{\dim }_{k}L=d$, so $L^{\mathrm{\prime }}$ is not a proper submodule with dimension at most $d$, whence $L^{\mathrm{\prime }}=N$.

Case $N\cap S\mathrm{\ne }0$ of claim 4. Since $S$ is simple and $N\cap S$ is a nonzero submodule of it, $N\cap S=S$, and for $n=s+u\in N$ with $s\in S$, $u\in M$ one has $u=n-s\in N\cap M$; hence $N=S\oplus (N\cap M)$ with the sum direct because $S\cap M=0$. By the induction hypothesis applied to $N\cap M\subseteq M$ there are a subfamily $(S_i{)}_{i\in J}$, $J\subseteq \{2,\dots ,m\}$, with $N\cap M\cong {⨁}_{i\in J}{S}_i$ and a submodule $K^{\mathrm{\prime }}\subseteq M$ with $(N\cap M)\cap K^{\mathrm{\prime }}=0$ and $(N\cap M)+K^{\mathrm{\prime }}=M$; then $N\cong S\oplus {⨁}_{i\in J}{S}_i$, and $K^{\mathrm{\prime }}$ is a complement of $N$ in $S\oplus M$, because $N+K^{\mathrm{\prime }}=S+(N\cap M)+K^{\mathrm{\prime }}=S\oplus M$ and an element of $N\cap K^{\mathrm{\prime }}$ lies in $M$, hence has zero $S$-component and lies in $(N\cap M)\cap K^{\mathrm{\prime }}=0$.

By [F5] the unit determinant of step 2.1 makes $A$ invertible, so [F3] gives $C$ with $CA=AC=I_n$; if ${\sum }_{j<n}{\lambda }_j{a}_j=0$ in $R^n$, then $0=C(0)=CA\lambda =\lambda$, so the columns $a_j$ of $A$ are linearly independent and, being spanning, they form a basis of $R^n$; applying the coordinate isomorphism to the preimages shows $(v_0,\dots ,v_{n-1})$ is a basis of $M$, which proves claim 1 for $n\ge 1$; for $n=0$ the module is $0$ with empty basis by [F1] and there is no determinant clause.

The list $(u_0,\dots ,u_{r-1})$ is $E$-independent by step 2.2 and $E$-spans $U_E$ by step 1.2, so it is an $E$-basis of $U_E$; therefore ${\mathrm{rank}}_E(A)={\dim }_E{U}_E=r={\dim }_{F}U={\mathrm{rank}}_F(A)$ by [F7], which is claim 2.

Iteration for claim 3: starting from $N$ and repeatedly replacing a nonzero module by a maximal proper submodule, which exists by step 2.3, produces a strictly decreasing chain $N=N_0⊋N_1⊋\cdots ⊋N_t=0$ of finite-dimensional submodules, because each $N_i$ is finite-dimensional by [F9] and each step strictly lowers the $k$-dimension, so the process terminates after at most ${\dim }_{k}N$ steps. Reading the chain upwards gives $0=N_t⊊\cdots ⊊N_0=N$, and each factor $N_{i-1}\mathrm{/}{N}_i$ is simple: if it had a nonzero proper submodule $Q$, the preimage $\{x\in N_{i-1}:x+N_i\in Q\}$ would be a submodule strictly between $N_i$ and $N_{i-1}$, contradicting maximality of $N_i$ in $N_{i-1}$; this is a finite composition series of $N$ in the sense of [F13], so $N$ has finite length.

Case $N\cap S=0$ of claim 4. The projection $\pi :N\to M$ is then injective, and $\pi (N)\subseteq M$ is a submodule to which the induction hypothesis applies: there are a subfamily $(S_i{)}_{i\in J}$, $J\subseteq \{2,\dots ,m\}$, with $\pi (N)\cong {⨁}_{i\in J}{S}_i$ and a submodule $K\subseteq M$ with $\pi (N)\cap K=0$ and $\pi (N)+K=M$. For $u\in \pi (N)$ let $f(u)\in S$ be the $S$-component of the unique $n\in N$ with $\pi (n)=u$, so that this $n$ is $f(u)+u$ and $N=\{f(u)+u:u\in \pi (N)\}$; the projection and the $S$-component map are $B$-module homomorphisms; the inverse of the restricted projection respects addition and $B$-scalars, so $f$ is a $B$-module homomorphism. Every $x=s+u\in S\oplus M$ with $u=u_1+u_2$, $u_1\in \pi (N)$, $u_2\in K$, equals $(f(u_1)+u_1)+((s-f(u_1))+u_2)\in N+(S\oplus K)$, and if $n=f(u)+u\in N\cap (S\oplus K)$ with $n=s^{\mathrm{\prime }}+k$, $k\in K$, then projecting to $M$ gives $u=k$, so $u\in \pi (N)\cap K=0$, $u=0$ and $n=f(0)=0$ by linearity of $f$, whence $N\cap (S\oplus K)=0$; therefore $S\oplus K$ is a complement of $N$ in $S\oplus M$, and $N\cong \pi (N)\cong {⨁}_{i\in J}{S}_i$ through the injective projection.

Cases 2.4 and 3.4 cover every submodule $N\subseteq S\oplus M$ according to whether $N\cap S$ is zero, and in both cases $N$ is isomorphic to a direct sum of a subfamily of $S_1,\dots ,S_m$ and has a complement in $S_1\oplus \cdots \oplus S_m$; with the case $m\le 1$ of step 1.4 this proves claim 4 by induction on $m$.

In the ordered basis of step 1.5, a basis vector $w\in K_j$ coming from the $j$-th block satisfies $T(w)\in K_{j-1}$, which is the span of the blocks preceding $w$; the matrix $[T]$ of $T$ in this ordered basis of [F10] therefore has zeros on its diagonal, and $\mathrm{tr}(T)$ is the trace of that matrix by [F11] and thus the sum of its diagonal entries, which is $0$. This proves claim 5, and with steps 3.1, 3.2, 3.3, 4.1 all five claims are proved.