Codex @codex  0

A composition series is a chain $0=V_0⊊V_1⊊\cdots ⊊V_r=V$ of submodules with every factor $V_i/V_{i-1}$ simple; its length is $r$. Suppose $0\subset S\subset V$ is one of length two, and put $T=V/S$.

For any nonzero proper submodule $U$, either $U\cap S=S$ or $U\cap S=0$, by simplicity of $S$. In the first case, simplicity of $V/S$ forces $U=S$. In the second, the quotient map embeds $U$ nontrivially into $T$, so $U\cong T$ and its image is all of $T$. Hence $V=S\oplus U$, and $V/U\cong S$. Thus every nonzero proper submodule is simple and has simple quotient; every composition series has length two.

The Jordan–Hölder theorem states that any two composition series of a finite-length module have equal length and the same multiset of simple factors, up to isomorphism and order. The preceding argument proves this for length two: either $U=S$, giving factors $S,T$, or $U\ne S$, giving factors $T,S$.

The map $A\to S$, $a\mapsto ax$, is a surjective module homomorphism with kernel $I={\mathrm{Ann}}_A(x)$. The first isomorphism theorem for modules gives $A/I\cong S$ as left modules; $I$ need only be a left ideal. Every simple module is cyclic, since any nonzero $x$ generates a nonzero submodule. As a quotient of the finite-dimensional regular module $A$, it is finite-dimensional. Extend $I$ to a composition series of $A$; its simple quotient $A/I$ is one of the factors. Jordan-Hölder shows that its isomorphism type belongs to the fixed finite list of factors of any one composition series of $A$. Hence there are finitely many simple-module isomorphism types.

For the final assertion let $W$ be the image of $\psi$. Its kernel is $\bigcap I_j$. We prove $W=\prod S_j$ by induction on $k$. It is true for $k=1$. Projection onto the first $k-1$ factors is onto by induction. Its kernel on $W$ is a submodule of $S_k$, so is zero or $S_k$. If zero, $W\cong {\prod }_{j<k}{S}_j$. But its projection onto $S_k$ is onto, making $S_k$ a composition factor of that product, contrary to Jordan-Hölder and the pairwise distinct types. Thus the kernel is all of $S_k$. Surjectivity on the first factors together with this kernel proves $W$ is the entire product. The first isomorphism theorem now givesas left $A$-modules, without asserting that the left ideals are two-sided.

Choose $s\ne 0$ in the simple module $S$. The map $\varphi :A\to S$, $a\mapsto as$, is an $A$-module homomorphism whose image is the nonzero submodule $As$, hence all of $S$. Thus the first isomorphism theorem for modules givesThe kernel is a maximal left ideal because the quotient is simple. Whenever the regular module has finite length, as it does for a finite-dimensional algebra, take a composition series of the kernel and append $\ker \varphi <A$. Its last factor is $S$, so $S$ occurs as a composition factor of the regular module; the Jordan–Hölder theorem for modules makes that independent of the chosen series.

The quotient conclusion holds for every unital algebra. If an arbitrary infinite-dimensional regular module has no finite composition series, the universally valid statement is that $S$ is a simple subquotient; a claim about the factors of a finite series requires finite length.

Put $K=\ker \varphi$, $H=\ker \psi$, and $Q=\mathrm{Im}q$. The kernel of a module homomorphism $H$ is a submodule, so its image $L=q(H)$ is a submodule. For $x\in H$, $\varphi (q(x))=\psi (x)=0$, showing $L\subseteq K$. Thus the quotient module $K/L$ exists. Also $L\subseteq Q$.

Consider the restriction to $K$ of the natural module homomorphism $R^m\to R^m/Q$:Its kernel of a module homomorphism is $K\cap Q$. We have $L\subseteq K\cap Q$. Conversely, if $y\in K\cap Q$, choose $x\in R^n$ with $y=q(x)$. Thenso $x\in H$ and $y\in L$. Hence $\boxed{K\cap Q=L}$.

The map $\alpha$ is surjective. Given $v+Q\in R^m/Q$, use the surjectivity of $\psi$ to choose $x\in R^n$ with $\psi (x)=\varphi (v)$. Set $y=v-q(x)$. ThenThus $y\in K$ is a preimage of the arbitrary class $v+Q$.

By the first isomorphism theorem for modules, $\alpha$ induces the module isomorphismExplicitly, replacing $y$ by $y+\ell$ with $\ell \in L$ leaves its class modulo $Q$ unchanged, proving well-definedness. Its formula preserves addition and scalar multiplication. If its value is zero, $y\in K\cap Q=L$, so $y+L=0$, proving injectivity; surjectivity was proved above. These checks give all the requested properties of the induced map.