Chain Conditions, Semisimple Modules and the Wedderburn–Artin Theorem — Examples

Given: The hypotheses and objects in the Example.

A left $R$-module $M$ is Noetherian when every submodule of $M$ is finitely generated (def-generated-cyclic-finitely-generated-and-free-modules). This finite-generation definition is the convention; its equivalence with ACC and the maximal condition is proved in thm-equivalent-characterizations-of-noetherian-modules. (Noetherian modules: every submodule is finitely generated).

A left $R$-module $M$ is Artinian when every descending chain $M_0\supseteq M_1\supseteq \cdots$ of submodules stabilizes: there is $N$ such that $M_n=M_N$ for all $n\ge N$. This is the descending chain condition. (Artinian modules by the descending chain condition).

Every subgroup $H\le (\mathbb{Z},+)$ equals $n\mathbb{Z}=⟨n⟩$ for exactly one $n\in \mathbb{N}$; in particular every subgroup is cyclic. (Every subgroup of $(\mathbb{Z},+)$ is $⟨n⟩=n\mathbb{Z}$ for exactly one natural number $n$).

Given: The hypotheses and objects in the Example.

A left $R$-module $M$ is Noetherian when every submodule of $M$ is finitely generated (def-generated-cyclic-finitely-generated-and-free-modules). This finite-generation definition is the convention; its equivalence with ACC and the maximal condition is proved in thm-equivalent-characterizations-of-noetherian-modules. (Noetherian modules: every submodule is finitely generated).

A left $R$-module $M$ is Artinian when every descending chain $M_0\supseteq M_1\supseteq \cdots$ of submodules stabilizes: there is $N$ such that $M_n=M_N$ for all $n\ge N$. This is the descending chain condition. (Artinian modules by the descending chain condition).

$(\mathbb{Q},+,\cdot ,0,1)$ with the operations of def-rat-operations is a field: a commutative ring with $1\mathrm{\ne }0$ in which every nonzero element has a multiplicative inverse. (The rationals form a field).

The map $j(k)=[(k,1)]$ is injective and preserves addition, multiplication, and order. Composing with lem-nat-embeds-int embeds $\mathbb{N}$ in $\mathbb{Q}$; we write $k$ for $j(k)$ throughout. (The integers embed in the rationals).

For $N\le M$, the additive cosets $m+N$ form the quotient module $M\mathrm{/}N$ under the well-defined scalar action $r(m+N):=rm+N\mathrm{.}$ (Quotient module $M\mathrm{/}N$ with scalar multiplication on additive cosets).

Given: The hypotheses and objects in the Example.

A unital ring $R$ is left Noetherian when its left regular module ${}_{R}R$ is Noetherian, and right Noetherian when the right regular module $R_R$ is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur. (Left and right Noetherian rings).

Let $R$ be a unital ring and $(M_i{)}_{i\in I}$ a family of left $R$-modules (def-left-and-right-modules). Their direct product is the module ${\prod }_{i\in I}{M}_i$ with coordinatewise operations. The support of $m=(m_i)$ is $\{i\in I:m_i\mathrm{\ne }0\}$, and the direct sum is the submodule ${⨁}_{i\in I}{M}_i=\{m\in {\prod }_{i\in I}{M}_i:\mathrm{supp}(m)\text{ is finite}\}$ (def-submodule). (The direct sum of an indexed family of modules).

For every $n\in \mathbb{N}$, with addition and multiplication as in def-addition-and-multiplication-modulo-n: 1. $(\mathbb{Z}\mathrm{/}n,+,[0{]}_n)$ is an abelian group (def-group), with $-[a{]}_n=[-a{]}_n$; 2. $(\mathbb{Z}\mathrm{/}n,\cdot ,[1{]}_n)$ is a commutative monoid (def-semigroup-and-monoid); 3. multiplication distributes over addition on both sides. (For every natural $n$, $(\mathbb{Z}\mathrm{/}n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Given: The hypotheses and objects in the Example.

For a division ring $D$ and $n\ge 1$, the left regular module of $M_n(D)$ is the direct sum of its simple column ideals $M_n(D)e_{jj}\cong D^n$. (Matrix rings over division rings are semisimple).

For $r\ge 1$, $n_i\ge 1$, and division rings $D_i$, every simple left module over ${\prod }_i{M}_{n_i}(D_i)$ is supported on exactly one factor and is isomorphic to that factor's column module $D_i^{n_i}$. (Simple modules over a product of matrix rings over division rings).

The matrix unit $E_{ij}\in M_{m\times n}(F)$ has entries $(E_{ij}{)}_{rs}={\delta }_{ri}{\delta }_{sj},$ so it has entry $1$ in position $(i,j)$ and $0$ everywhere else. (Matrix units $E_{ij}$ and the Kronecker delta).

Given: The hypotheses and objects in the Example.

Let $R$ be a nonzero unital ring. Then $R$ is semisimple if and only if $R\cong {\prod }_{i=1}^r{M}_{n_i}(D_i)$ for positive integers $r,n_i$ and division rings $D_i$. (Wedderburn–Artin theorem for semisimple rings).

For $r\ge 1$, $n_i\ge 1$, and division rings $D_i$, every simple left module over ${\prod }_i{M}_{n_i}(D_i)$ is supported on exactly one factor and is isomorphic to that factor's column module $D_i^{n_i}$; these give all isomorphism classes. (Simple modules over a product of matrix rings over division rings).

Given: The hypotheses and objects in the Example.

A composition series of a left $R$-module $M$ is a finite chain $0=M_0<M_1<\cdots <M_n=M$ whose factors $M_i\mathrm{/}{M}_{i-1}$ are simple. If such a series exists, the length ${\mathrm{\ell }}_R(M)$ is its number $n$ of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length $0$. (Composition series and length of a module).

Any two composition series of a module have the same length, and their simple factors agree up to permutation and isomorphism. (Jordan–Hölder theorem for modules).

For a short exact sequence $0\to N\to M\to Q\to 0$, the module $M$ has finite length if and only if $N$ and $Q$ do, and then ${\mathrm{\ell }}_R(M)={\mathrm{\ell }}_R(N)+{\mathrm{\ell }}_R(Q)\mathrm{.}$. (Module length is additive in short exact sequences).

Let $G$ be a finite group such that the positive integer $\mathrm{\mid }G\mathrm{\mid }$ is prime. Then every $g\mathrm{\ne }e$ has order $\mathrm{\mid }G\mathrm{\mid }$, satisfies $⟨g⟩=G$, and hence generates $G$. In particular, $G$ is cyclic. (A finite group of prime order is cyclic and every nonidentity element generates it).

Given: The hypotheses and objects in the Example.

Let $A\to B$ be a homomorphism of commutative rings. An element $b\in B$ is integral over $A$ when it is a root of a monic polynomial in $A[X]$. The extension is integral when every element is integral. An algebraic integer is a complex number integral over $\mathbb{Z}$. (Integral elements over a commutative ring and algebraic integers).

A rational number is an algebraic integer if and only if it is an integer. (The rational algebraic integers are exactly the integers).

Let $F$ be a complete ordered field (def-complete-ordered-field). Then every $a\in F$ with $a\ge 0$ has a unique $s\in F$ with $s\ge 0$ and $s^2=a$; we write $s=\sqrt{a}$. Consequently the positive elements of $F$ are exactly the nonzero squares: $x>0$ if and only if $x=y^2$ for some $y\mathrm{\ne }0$. (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$).

Given: The hypotheses and objects in the false claim.

For every prime $p$, the Prüfer group $\mathbb{Z}(p^{\mathrm{\infty }})\le \mathbb{Q}\mathrm{/}\mathbb{Z}$ is Artinian but not Noetherian as a $\mathbb{Z}$-module. (The Prüfer $p$-group is Artinian but not Noetherian).

Given: The hypotheses and objects in the false claim.

A module with a composition series is both Noetherian and Artinian. The zero module has the empty composition series. (A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice).

The regular $\mathbb{Z}$-module $\mathbb{Z}$ is Noetherian but not Artinian. ($\mathbb{Z}$ is Noetherian but not Artinian as a module over itself).

Given: The hypotheses and objects in the false claim.

For every field $F$ and $n\ge 1$, the left regular module of $M_n(F)$ is the internal direct sum $M_n(F)={⨁}_{j=1}^n{M}_n(F)e_{jj}$ of simple left ideals. ($M_n(F)$ as a direct sum of minimal left ideals).

For every field $F$ and every natural $n\ge 2$, the ring $M_n(F)$ is not commutative. ($M_n(F)$ is noncommutative for every $n\ge 2$).

Given: The hypotheses and objects in the false claim.

A unital ring $R$ is left Noetherian when its left regular module ${}_{R}R$ is Noetherian, and right Noetherian when the right regular module $R_R$ is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur. (Left and right Noetherian rings).

If $D$ is an integral domain, then $D\setminus \{0\}$ is multiplicative. Its localisation $\mathrm{Frac}(D)=(D\setminus \{0\}{)}^{-1}D$ is the field of fractions of $D$. Thus its elements are fractions $a\mathrm{/}b$ with $a,b\in D$ and $b\mathrm{\ne }0$, modulo the localisation equivalence relation. (The field of fractions $\mathrm{Frac}(D)=(D\setminus \{0\}{)}^{-1}D$ of an integral domain).

For every integral domain $D$, the localisation $\mathrm{Frac}(D)$ is a field. Its canonical map $D⟶\mathrm{Frac}(D),d⟼d\mathrm{/}1,$ is an injective unital ring homomorphism. ($\mathrm{Frac}(D)$ is a field and $d\mapsto d\mathrm{/}1$ embeds the integral domain $D$).

Given: The hypotheses and objects in the false claim.

A unital ring $R$ is left Noetherian when its left regular module ${}_{R}R$ is Noetherian, and right Noetherian when the right regular module $R_R$ is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur. (Left and right Noetherian rings).

Let $R$ be a commutative ring. Matrices of one shape are added and scaled entrywise. For $A\in M_{m\times n}(R)$ and $B\in M_{n\times p}(R)$, their product is $AB\in M_{m\times p}(R)$ with $(AB{)}_{ik}:={\sum }_{j<n}{a}_{ij}{b}_{jk}\mathrm{.}$ The sum is the finite sum in the additive commutative monoid of $R$, and is $0$ when $n=0$. (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

$(\mathbb{Q},+,\cdot ,0,1)$ with the operations of def-rat-operations is a field: a commutative ring with $1\mathrm{\ne }0$ in which every nonzero element has a multiplicative inverse. (The rationals form a field).