Chain Conditions, Semisimple Modules and the Wedderburn–Artin Theorem

For submodules $L,N\le M$, map $L\to (L+N)\mathrm{/}N$.

Its kernel is $L\cap N$, and it is surjective by the definition of $L+N$; the first isomorphism theorem gives $L\mathrm{/}(L\cap N)\cong (L+N)\mathrm{/}N$.

The coincident and zero cases are admitted and give equalities rather than exceptions. For $L=0$ both sides are $0$, since $0\mathrm{/}(0\cap N)=0$ and $(0+N)\mathrm{/}N=N\mathrm{/}N=0$; for $N=0$ both sides are $L$; and for $L=N$ both sides are $0$. This proves the stated claim.

For $N\le L\le M$, send $m+N$ to $m+L$.

If $m+N=m^{\mathrm{\prime }}+N$, then $m-m^{\mathrm{\prime }}\in N\le L$, so the images modulo $L$ agree. The map is surjective, and its kernel consists exactly of the cosets $m+N$ with $m\in L$, namely $L\mathrm{/}N$; the first isomorphism theorem gives the displayed isomorphism.

The two coincident cases are admitted by $N\le L\le M$ and hold. For $N=L$ the submodule $L\mathrm{/}N$ is zero and the isomorphism reads $(M\mathrm{/}N)\mathrm{/}0\cong M\mathrm{/}N=M\mathrm{/}L$; for $L=M$ it is $L\mathrm{/}N=M\mathrm{/}N$ and the isomorphism reads $(M\mathrm{/}N)\mathrm{/}(M\mathrm{/}N)=0\cong M\mathrm{/}M=0$. This proves the stated claim.

Inverse image and quotient give mutually inverse inclusion-preserving bijections between submodules of $M\mathrm{/}N$ and submodules of $M$ containing $N$.

If $P\le M\mathrm{/}N$, then $\pi ({\pi }^{-1}P)=P$ by surjectivity of $\pi$; if $L\le M$ contains $N$, then ${\pi }^{-1}(L\mathrm{/}N)=L$. Direct calculation with inverse images gives preservation of sums and intersections, while [L3] identifies successive quotients. This proves the stated claim.

Assume condition 1 and let $(N_n)$ be ascending. The union $N:={\bigcup }_n{N}_n$ contains zero. Any two of its elements lie in a common stage, so their combination $ru+v$ lies there too. Thus $N$ is a submodule by [L2]. By condition 1, $N$ has a finite generating set $S$. A finite subset of this ascending union lies in one stage: start with stage zero for the empty set, and when adjoining one element take the larger of the previous stage and a stage containing that element. Hence $S\subseteq N_k$ for some $k$, so $N=⟨S{⟩}_R\subseteq N_k\subseteq N$. For $n\ge k$, $N_k\subseteq N_n\subseteq N=N_k$. This proves condition 2 without countably many choices.

Assume condition 2 and DC. Let $\mathcal{F}$ be a nonempty set of submodules. If it has no maximal member, the relation $ARB$ defined by $A⊊B$ is entire on $\mathcal{F}$. Fix $A_0\in \mathcal{F}$. By [L3] there is a sequence $(A_n)$ in $\mathcal{F}$ with $A_n⊊A_{n+1}$ for every $n$, contrary to condition 2. Thus $\mathcal{F}$ has a maximal member, proving condition 3.

Assume condition 3 and fix a submodule $N\le M$. Its finitely generated submodules form a set $\mathcal{G}$ containing the zero submodule, generated by the empty set. Condition 3 gives a maximal $L\in \mathcal{G}$. Fix a finite generating set $S$ of $L$. If $L\mathrm{\ne }N$, an element $x\in N\setminus L$ makes $⟨S\cup \{x\}{⟩}_R$ a finitely generated submodule of $N$ strictly containing $L$, a contradiction. Hence $L=N$, proving condition 1 without DC.

Steps 1.1, 1.2 and 1.3 prove the stated implications. Only step 1.2 uses DC.

DCC gives a minimal member of every nonempty family by contradiction: under dependent choice, absence of a minimal member yields a strict descending chain.

Conversely, a nonstabilizing descending chain has no minimal member.

Repeated terms are what make step 2.1 correct rather than an equivocation: minimality of $M_k$ in the family $\{M_n\}$ means no member is properly contained in it, so $M_n=M_k$ for all $n\ge k$, which is stabilization. For the zero module the only submodule is $0$, every nonempty family is $\{0\}$ with minimal member $0$, and every descending chain is constant, so both conditions hold. This proves the stated claim.

Identify $N$ with $i(N)\subseteq M$ using the injective homomorphism in [F3]. The generated submodule of a finite set is precisely its finite linear combinations: these combinations form a submodule containing the set, and every submodule containing it contains all such combinations. This follows from [F2] and the module laws. Thus homomorphisms carry finite generating sets to generating sets of their images.

Suppose $M$ is Noetherian. Every submodule of $N$ is a submodule of $M$, hence finitely generated. If $V\le Q$, its preimage $p^{-1}(V)$ is a submodule of $M$ and has finite generators; their images generate $V$ because $p$ is surjective. Consequently $N$ and $Q$ are Noetherian.

Conversely suppose $N$ and $Q$ are Noetherian, and let $L\le M$. Choose finite generators $a_1,\dots ,a_r$ of $L\cap N$ and $q_1,\dots ,q_s$ of $p(L)$. By [L1], lift the finitely many $q_j$ to $b_j\in L$. For $x\in L$, write $p(x)={\sum }_j{c}_j{q}_j$. Then $x-{\sum }_j{c}_j{b}_j$ belongs to $L\cap \ker p=L\cap N$, so is a linear combination of the $a_i$. Thus the $a_i$ and $b_j$ generate $L$. Empty generating lists cause no change to this argument. Since $L$ was arbitrary, $M$ is Noetherian, without any countable or dependent choice.

Suppose $M$ is Artinian. A descending chain of submodules of $N$ is one in $M$, hence stabilizes. A descending chain $(V_j)$ in $Q$ lifts to the descending chain $(p^{-1}(V_j))$ in $M$; after it stabilizes, its images $V_j$ stabilize by surjectivity. Thus $N$ and $Q$ are Artinian.

Conversely suppose $N$ and $Q$ are Artinian and $L_0\supseteq L_1\supseteq \cdots$ is a descending chain in $M$. The chains $L_j\cap N$ and $p(L_j)$ stabilize. Take an index $j_0$ beyond both stabilization indices. For $j\ge j_0$ and $x\in L_{j_0}$, equality of images gives $y\in L_j$ with $p(y)=p(x)$. Then $x-y\in L_{j_0}\cap N=L_j\cap N$, so $x\in L_j$. Hence $L_j=L_{j_0}$ for all $j\ge j_0$, and $M$ is Artinian.

Steps 2.1 and 2.2 prove the Noetherian equivalence, and steps 2.3 and 2.4 prove the Artinian equivalence.

Induct on the number of summands using the split short exact sequence for a binary direct sum.

Reflection follows because every summand is a submodule and a quotient.

Both extremes of the induction are admitted. The empty direct sum is the zero module, whose only chains of submodules are constant, so it satisfies both conditions while the "every summand" side is vacuously true; a single summand makes the direct sum that summand, so the equivalence is an identity and supplies the base of the induction in step 1.1. This proves the stated claim.

A module generated by $n$ elements is a quotient of $R^n$.

The left regular module is Noetherian, [L3] makes $R^n$ Noetherian, and [L5] makes its quotient $M$ Noetherian.

The case $n=0$ is admitted: a module generated by the empty set is $0$, and $R^0$ is the zero module, so step 1.1 presents $M=0$ as a quotient of $0$. The zero module has only the constant chain of submodules and is Noetherian, so the argument needs no separate base case. This proves the stated claim.

For a surjective endomorphism $f$, the ascending chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes.

Choose $n$ with $\ker f^n=\ker f^{n+1}$. For $x\in \ker f$, surjectivity of $f^n$ gives $y$ with $f^n(y)=x$. Then $f^{n+1}(y)=0$, so $y\in \ker f^{n+1}=\ker f^n$ and $x=f^n(y)=0$.

The zero module is admitted: its only endomorphism is the identity, which is injective, so the conclusion holds there and the argument of step 2.1 is not vacuous by accident. This proves the stated claim.

Let $0=M_0<\cdots <M_n=M$ be a composition series [L1] and induct on $i$ that $M_i$ is Noetherian and Artinian. The zero module $M_0$ satisfies both conditions vacuously. A simple factor has only the submodules $0$ and itself; any one nonzero element generates it, since its cyclic submodule is nonzero. Thus all its submodules are finitely generated and all descending chains stabilize. Applying [L4] to $0\to M_{i-1}\to M_i\to M_i\mathrm{/}{M}_{i-1}\to 0$ carries both conditions to $M_i$. At $i=n$ this gives the forward implication, without any choice principle.

Conversely, assume dependent choice and let $M$ be Noetherian and Artinian. Every submodule of $M$ is Noetherian by [L4], so a nonzero submodule $N$ has a maximal proper submodule by the nonempty maximal condition of [L2], using DC. Define a relation on the submodules of $M$ by taking a maximal proper submodule at each nonzero term and letting zero be its own successor. This relation is serial. DC, available from the assumed AC, supplies a sequence starting at $M$; it strictly decreases until it reaches zero and is then constant. These are the precise choice uses in the converse.

The chain of step 1.2 is strictly descending while its terms are nonzero, so the descending chain condition forces some $N_r=0$. Since $N_{k+1}$ is maximal proper in $N_k$, the quotient $N_k\mathrm{/}{N}_{k+1}$ is nonzero and has no proper nonzero submodule, hence is simple. Reversing the chain gives $0=N_r<\cdots <N_0=M$, a composition series. For $M=0$ the empty chain is already the required series, so no choice is consumed in that case. This proves the stated claim.

Fix a composition series $0=M_0<\cdots <M_n=M$ and prove by induction on $n$ that it has the asserted comparison with every other composition series of $M$. We use simultaneously the elementary consequence that, for any $C\le M$, intersecting the fixed series with $C$ and deleting repetitions gives a composition series of $C$: each remaining factor embeds in the corresponding simple factor $M_i\mathrm{/}{M}_{i-1}$ and is therefore simple.

The case $n=0$ is $M=0$. For $n>0$, let $A=M_{n-1}$ and let $B$ be the penultimate term of a second series. If $A=B$, the induction hypothesis applied in $A$ matches all lower factors, and the common top factor finishes.

Suppose $A\mathrm{\ne }B$. Since $A$ and $B$ are maximal proper submodules, $A+B=M$. Put $C=A\cap B$. The second isomorphism theorem gives $A\mathrm{/}C\cong M\mathrm{/}B,B\mathrm{/}C\cong M\mathrm{/}A,$ so both quotients are simple.

By step 1.1, $C$ has a composition series. Appending $A$ gives a composition series of $A$ ending in $A\mathrm{/}C$. Compare it with $0=M_0<\cdots <M_{n-1}=A$ using the induction hypothesis, whose fixed first series has length $n-1$. It follows that the series of $C$ has length $n-2$ and that its factors together with $A\mathrm{/}C$ are exactly the factors below $M\mathrm{/}A$ in the fixed series.

Appending $B$ to the same series of $C$ gives a composition series of $B$ of length $n-1$. Using this as the fixed first series, the induction hypothesis compares it with the lower part of the second series. The isomorphisms in step 3.1 exchange the two top simple factors $A\mathrm{/}C$ and $M\mathrm{/}A$ with $M\mathrm{/}B$ and $B\mathrm{/}C$. Hence the two original series have length $n$ and the same factors up to permutation. This also covers $n=1$, when $C=0$.

If $N$ and $Q$ have composition series, lift the series of $Q$ along $M\to Q$ and splice it above the series of $N$. Correspondence identifies all lifted factors, so this is a composition series of $M$ with ${\mathrm{\ell }}_R(N)+{\mathrm{\ell }}_R(Q)$ factors.

Conversely, let $0=M_0<\cdots <M_n=M$ be a composition series. Put $N_i=M_i\cap N$ and let $Q_i$ be the image of $M_i$ in $Q$. For each $i$, the simple factor $M_i\mathrm{/}{M}_{i-1}$ has submodule $N_i\mathrm{/}{N}_{i-1}$ and corresponding quotient $Q_i\mathrm{/}{Q}_{i-1}$; exactly one is that simple factor and the other is zero. Deleting repetitions therefore gives composition series of $N$ and $Q$, and their numbers of factors add to $n$.

Jordan–Hölder makes all three lengths independent of the chosen series, so steps 1.1 and 1.2 prove both directions and the formula. If $N=0$, $Q=0$, or $M=0$, the relevant series is empty and the same count applies.

If $f,g\in {\mathrm{End}}_R(M)$ then $f\circ g$ is again a module homomorphism, since $(f\circ g)(m+n)=f(g(m)+g(n))=(f\circ g)(m)+(f\circ g)(n)$ and $(f\circ g)(rm)=f(rg(m))=r(f\circ g)(m)$. So composition is a binary operation on ${\mathrm{End}}_R(M)$.

Taking $N=M$ in [L2] makes $({\mathrm{End}}_R(M),+)$ an abelian group, with the zero homomorphism as neutral element and $(-f)(m)=-f(m)$ as the inverse of $f$.

Composition is associative: for all $m$, $(f\circ (g\circ h))(m)=f(g(h(m)))=((f\circ g)\circ h)(m)$.

Both distributive laws hold. For all $m$, $(f\circ (g+h))(m)=f(g(m)+h(m))=f(g(m))+f(h(m))=(f\circ g+f\circ h)(m)$, where the middle equality is additivity of $f$; and $((g+h)\circ f)(m)=g(f(m))+h(f(m))=(g\circ f+h\circ f)(m)$ directly from pointwise addition.

The identity map ${\mathrm{id}}_M$ satisfies ${\mathrm{id}}_M(m+n)=m+n$ and ${\mathrm{id}}_M(rm)=rm$, so it lies in ${\mathrm{End}}_R(M)$, and $f\circ {\mathrm{id}}_M=f={\mathrm{id}}_M\circ f$ for every $f$.

Steps 1.1 through 1.5 are exactly the axioms of a unital ring for $({\mathrm{End}}_R(M),+,\circ ,{\mathrm{id}}_M)$. For the zero module the only map $0\to 0$ is ${\mathrm{id}}_0$, so ${\mathrm{End}}_R(0)$ has one element and is the one-element ring, in which the identity coincides with the zero element. This proves the stated claim.

Evaluate an endomorphism at $1$.

Left linearity gives $f(r)=rf(1)$, so endomorphisms are right multiplications; composition reverses the order of their defining elements.

The map $a\mapsto {\rho }_a$ with ${\rho }_a(x)=xa$ is inverse to $f\mapsto f(1)$: it is a left $R$-module homomorphism because ${\rho }_a(rx)=rxa=r{\rho }_a(x)$, it satisfies ${\rho }_{f(1)}=f$ by step 2.1, and ${\rho }_a(1)=a$. It is additive, and ${\rho }_a\circ {\rho }_b(x)=xba={\rho }_{ba}(x)={\rho }_{a\star b}(x)$ with $\star$ the multiplication of [L1], so it is a ring isomorphism onto ${\mathrm{End}}_R({}_{R}R)$. For the zero ring $1=0$, both ${\mathrm{End}}_R({}_{R}R)$ and $R^{\mathrm{o}\mathrm{p}}$ are the one-element ring and the correspondence is the unique map between them. This proves the stated claim.

We use inclusions and projections to send $f$ to entries $f_{ij}={\pi }_{i}f{\iota }_j$, and reconstruct $f$ by finite sums.

Composition becomes matrix multiplication because ${\sum }_k{\iota }_k{\pi }_k=\mathrm{id}$ on a finite direct sum, so the $(i,j)$ entry of $f\circ g$ is ${\pi }_i(f\circ g){\iota }_j={\pi }_{i}f\left({\sum }_k{\iota }_k{\pi }_k\right)g{\iota }_j={\sum }_k({\pi }_{i}f{\iota }_k)({\pi }_{k}g{\iota }_j)={\sum }_k{f}_{ik}\circ g_{kj},$ which is the matrix product with composition in the entries; the sum is finite because $n$ is.

For $n=0$ the direct sum is the zero module, ${\mathrm{End}}_R(0)$ has one element by [L1], and the set of $0\times 0$ matrices also has exactly one element, so both sides are the one-element zero ring as the Statement records. For $n=1$ the matrix is the single entry $f_{11}={\pi }_{1}f{\iota }_1=f$, and the correspondence is the identity on ${\mathrm{End}}_R(M_1)$. This proves the stated claim.