Semisimple Artinian Rings and Algebras

Semisimple

Definition

A ring $R$ with identity is called semisimple if ${}_{R}R$ is semisimple as a left $R$-modules.

Remark. ${}_{R}R$ is called (left) regular module.

Theorem

TFAE for an Artinian ring $R$.

• $R$ is a semisimple Artinian ring.
• Every $R$-module $V$ is semisimple.

\begin{proof} i)→ii) Let ${}_{R}R={\sum }_{\lambda }{I}_{\lambda }$ be a sum of left simple $R$-submodules. Then $V={\sum }_{v\in V}{\sum }_{\lambda }{I}_{\lambda }v$. Note that $f:I_{\lambda }\to I_{\lambda }v,i\mapsto iv$ is an $R$-homomorphism, then $\ker f=0$ or $I_{\lambda }$, that is, $I_{\lambda }v$ is $0$ or simple.

ii)→i) is trivial. \end{proof}

Some Isomorphisms

In this section, we discuss some statements that will be useful later.

Lemma

Let $M$ be a left $R$-module. Then $M\simeq {\mathrm{Hom}}_R({}_R{R}_R,{}_{R}M)$.

\begin{proof} Define

$${\mathrm{Hom}}_R(R,M)\to M,f\mapsto f(1)$$

and define

$$M\to {\mathrm{Hom}}_R(R,M),m\mapsto {\phi }_m:r\mapsto rm.$$

It is easy to check they are homomorphism and are inverse each other. \end{proof}

Corollary

For any ring $R$ with identity, one has ${\mathrm{End}}_R({}_{R}R)\simeq R^{\mathrm{op}}$ as rings.

\begin{proof} Define

$${\mathrm{End}}_R({}_{R}R)\to R^{\mathrm{op}},f\mapsto f(1).$$

Note that $f_1{f}_2(1)=f_1(f_2(1))=f_1(f_2(1)\cdot 1)=f_2(1)f_1(1)$. \end{proof}

Lemma

Let $R$ be a ring, and let $M_n(R)$ be the set of all square matrices of degree $n$ over $R$. Then ${\mathrm{M}}_n(R{)}^{\mathrm{op}}\simeq {\mathrm{M}}_n(R^{\mathrm{op}})$.

\begin{proof} Consider

$$\phi :{\mathrm{M}}_n(R{)}^{\mathrm{op}}\simeq {\mathrm{M}}_n(R^{\mathrm{op}}),A\mapsto A^T.$$

Assume that $({\mathrm{M}}_n(R),\cdot )$, $({\mathrm{M}}_n(R{)}^{\mathrm{op}},\ast )$ and $({\mathrm{M}}_n(R{)}^{\mathrm{op}},\circ )$. Then $\phi (A\ast B)=\phi (B\cdot A)=(B\cdot A{)}^T$ and $\phi (A)\circ \phi (B)=A^T\circ B^T$, and we can prove that $(BA{)}^T=A^T\circ B^T$. \end{proof}

Schur Lemma and Artin-Wedderburn Theorem

Schur lemma

Let $S_1$ and $S_2$ be simple $R$-modules. Then ${\mathrm{Hom}}_R(S_1,S_2)=0$ unless $S_1\simeq S_2$ in which case the endomorphism ring ${\mathrm{End}}_R(S_1)$ is a division ring.

\begin{proof} See Schur lemma. \end{proof}

Lemma

Let $R$ be a ring with identity $1$. Suppose that $U=S_1+\cdots +S_n$ is an $R$-module that can be expressed as the sum of finitely many simple submodule. If $V$ is any submodule of $U$, then there is a subset $I\subseteq \{1,\cdots ,n\}$ such that $U=V\oplus ({\oplus }_{i\in I}{S}_i)$.

In particular, $V$ is a direct summand of $U$, and $U$ is a direct sum of some subset of the $S_i$‘s and hence necessarily semisimple.

\begin{proof} Let $\mathcal{F}$ be the collection of all subsets $J\subseteq \{1,\dots ,n\}$ such that $V\cap \left({⨁}_{j\in J}{S}_j\right)=\{0\}$. Note that $\mathcal{F}$ is non-empty since $\varnothing \in \mathcal{F}$, then $\mathcal{F}$ contains a maximal element by $\{1,\cdots ,n\}$ finite. Let $I$ be such a maximal subset, and let $W=V\oplus \left({⨁}_{i\in I}{S}_i\right)$. Now we claim that $W=U$. Otherwise, since $U=S_1+\cdots +S_n$, there exists some $k\in \{1,\dots ,n\}$ such that $S_k⊈W$. Then $S_k\cap W=\{0\}$ by $S_k$ simple. Define $I^{\prime }=I\cup \{k\}$, then $I^{\prime }\in \mathcal{F}$, leading to a contradiction by the maximality of $I$. Now we finish the proof. \end{proof}

Theorem

Let $U$ be an $R$-module. TFAE:

• $U$ is semisimple;
• $U$ is a sum of simple $R$-modules;
• any $R$-submodule of $U$ is a direct summand of $U$.

\begin{proof} See here. \end{proof}

Corollary

Let $U$ be the $R$-module with composition series of finite composition length.

• The sum of all the simple submodule of $U$ is a semisimple module that is the unique largest semisimple submodule of $U$, which is called socle.
• The sum of all submodules of $U$ isomorphic to some given simple module $S$ is a submodule isomorphic to a direct sum of copies of $S$. It is the unique largest submodule with this property.

Corollary

Let $U=S_1^{a_1}\oplus \cdots \oplus S_r^{a_r}$ be a semisimple submodule where the $S_i$ are non-isomorphic simple $R$-modules. Then each submodule $S_i^{a_i}$ is uniquely determined and characterized as unique largest submodule of $U$ expressed as a direct sum of copies of $S_i$.

\begin{proof} It suffices to show that $S_i^{a_i}$ contains every submodule of $U$ isomorphic to $S_i$. Let $T$ be a non-zero simple submodule of $U$ with $T⊈S_i^{a_i}$. Define a projective ${\pi }_i:U\to S_i^{a_i}$, then ${\pi }_j(T)\ne 0$ for some $j\ne i$ and $T\simeq S_j\not\simeq S_i$. Now we finish the proof. \end{proof}

Theorem

Let $R$ be a ring with $1$. Let ${}_{R}V=V_1\oplus V_2\oplus \cdots \oplus V_n$ be a direct sum of simple $R$-modules $V_i$ such that $V_i\simeq V_j$ for any $i,j$. Then ${\mathrm{End}}_R(V)$ is isomorphic to the full matrix ring of degree $n$ over the division ring ${\mathrm{End}}_R(V_1)$.

\begin{proof} Let ${\pi }_i:V\to V_i$ be the projection, and let ${\theta }_i:V_1\stackrel{\sim }{\to }{V}_i$ be an isomorphism. For $\sigma \in {\mathrm{End}}_R(V)$, define the composition

$$V_1\stackrel{{\sigma }_j}{\to }{V}_j\stackrel{\sigma }{\to }V\stackrel{{\pi }_i}{\to }{V}_i\stackrel{{\theta }_i^{-1}}{\to }{V}_1$$

as ${\phi }_{ij}(\sigma ):={\theta }_i^{-1}{\pi }_i\sigma {\theta }_j\in {\mathrm{End}}_R(V_1)$. Then we have map

$$\phi :{\mathrm{End}}_R(V)\to {\mathrm{M}}_n(E_1),\sigma \mapsto ({\phi }_{ij}(\sigma ){)}_{n\times n}$$

where $E_1={\mathrm{End}}_R(V_1)$. It is easy to check $\phi$ is a ring homomorphism.

Nest, we show $\phi$ is an isomorphism. If $\phi (\sigma )=0$, then ${\theta }_i^{-1}{\pi }_i\sigma {\theta }_j=0$ for all $i,j$. It deduces that ${\pi }_i(\sigma (V_j))=0$ for all $i,j$ and so $\sigma (V_j)=0$ for all $j$. Hence $\sigma =0$ and so $\phi$ is injective. To see $\phi$ is an epimorphism, for any given $({\sigma }_{ij})\in {\mathrm{M}}_n(E_1)$ we can define

$$\sigma =\sum _{i,j}{\tau }_i{\theta }_i{\sigma }_{ij}{\theta }_j^{-1}{\pi }_j$$

where ${\tau }_i:V_i\hookrightarrow V$ is an inclusion. Therefore, $\phi$ is an isomorphism and so ${\mathrm{End}}_R(V)\simeq {\mathrm{M}}_n(E_1)$ with $E_1={\mathrm{End}}_R(V_1)$. \end{proof}

Artin–Wedderburn

Let $A$ be a semisimple Artinian ring with identity. Then $A$ is a direct sum of matrix rings over division rings. Specifically, if ${}_{A}A\simeq S_1^{n_1}\oplus \cdots \oplus S_r^{n_r}$ where $S_1,\cdots ,S_r$ are non-isomorphic simple modules occurring with multiplications $n_1,\cdots ,n_r$, then $A\simeq {\mathrm{M}}_{n_1}(D_1)\oplus \cdots \oplus {\mathrm{M}}_{n_r}(D_r)$ where $D_i={\mathrm{End}}_A(S_i{)}^{\mathrm{op}}$.

\begin{proof} Since $A$ is a semisimple Artinian ring, we can assume that ${}_{A}A\simeq S_1^{n_1}\oplus \cdots \oplus S_r^{n_r}$. Then by ^qk0ias we have

$${\mathrm{End}}_A({}_{A}A)\simeq {\oplus }_{i,j}\mathrm{Hom}(S_i^{n_i},S_j^{n_j})={\oplus }_{i=1}^r\mathrm{Hom}(S_i^{n_i},S_i^{n_i})={\oplus }_{i=1}^r{\mathrm{End}}_A(S_r^{n_r}).$$

Furthermore, by ^8omj6z there is ${\mathrm{End}}_A(S_r^{n_r})\simeq {\mathrm{M}}_n({\mathrm{End}}_R(S_1))$. By ^qk0ias we know ${\mathrm{End}}_R(S_1)$ is a division ring and so $A$ is a direct sum of matrix rings over division rings.

Specifically, recall that ${\mathrm{End}}_R({}_{R}R)\simeq R^{\mathrm{op}}$ by ^5830df, then by ^hvg28s one can get

$$A\simeq {\mathrm{End}}_A({}_{A}A{)}^{\mathrm{op}}\simeq {\oplus }_{i=1}^r{\mathrm{End}}_A(S_r^{n_r}{)}^{\mathrm{op}}\simeq {\oplus }_{i=1}^r{\mathrm{M}}_n({\mathrm{End}}_A(S_i){)}^{\mathrm{op}}\simeq {\oplus }_{i=1}^r{\mathrm{M}}_n(D_i)$$

where $D_i={\mathrm{End}}_A(S_i{)}^{\mathrm{op}}$. Now we finish the proof. \end{proof}

Simple Modules

Definition

A ring is called simple if it has no ideal other than $0$ and $R$.

Example. ${\mathrm{M}}_n(D)$ is a simple ring if $D$ is a division ring. Additionally, notice that

$${\mathrm{M}}_n(D)=\left\{\left(\begin{array}{cccc}\ast & 0& \cdots & 0\\ \ast & 0& \cdots & 0\\ ⋮& ⋮& & ⋮\\ \ast & 0& \cdots & 0\end{array}\right)\right\}\oplus \cdots \oplus \left\{\left(\begin{array}{cccc}0& \cdots & 0& \ast \\ 0& \cdots & 0& \ast \\ ⋮& & ⋮& ⋮\\ 0& \cdots & 0& \ast \end{array}\right)\right\}$$

and each part is a simple left ${\mathrm{M}}_n(D)$-module.

Algebras

Definition

Let $R$ be a commutative ring with identity, and let $A$ be a ring with identity. We call $A$ an $R$-algebra if $A$ is an $R$-module such that $a(xy)=(ax)y=x(ay)$ for any $a\in R$ and any $x,y\in A$. Equivalently, there is a ring homomorphism $R\to A$ such that the image of $R$ lies in the center of $A$.

Moreover, if $R$ is a field and $A$ is a finite-dimensional over $R$, then we shall say that the algebra is finite-dimensional, which is both Noetherian and Artinian.

Examples.

• Suppose $R$ is a commutative ring with identity and $G$ is a finite group. Define group algebra/group ring as $RG=\{{\sum }_{g\in G}{a}_{g}g:a_g\in R\}$, which is a free $R$-module with basis $\{g:g\in G\}$. Thus, $RG$ is a $R$-algebra.
• For a field $F$, the polynomial ring $F[x]$ is a $F$-algebra.
• For a commutative ring $R$, ${\mathrm{M}}_n(R)$ is a $R$-algebra, where $r\mapsto \mathrm{diag}(r,\cdots ,r)$.

Schur

The following statements hold.

• Let $A$ be a $k$-algebra, where $k$ is a field. Let $S$ be a simple $A$-module. Then ${\mathrm{End}}_A(S)$ is a division ring, which is a $k$-algebra.
• If $A$ is finite-dimensional over $k$, so is ${\mathrm{End}}_A(S)$.

Furthermore, if $A$ is finite dimensional and $k$ is algebraically closed, then ${\mathrm{End}}_A(S)\simeq k$.

\begin{proof} i) Note that ${\mathrm{End}}_A(S)$ is a division ring by ^qk0ias, and one can easily check that ${\mathrm{End}}_A(S)$ is an $k$-module with $k{\phi }_1{\phi }_2=(k{\phi }_1){\phi }_2={\phi }_1(k{\phi }_2)$ by $k\in A$. Thus ${\mathrm{End}}_A(S)$ is a $k$-algebra.

ii) If $A$ is finite-dimensional over $k$, then by ^ef1283 we know $A$ is Artinian. Since $S$ is a simple $A$-module, there exists a left maximal ideal $M\subseteq A$ such that $A/M\simeq S$ by ^09efb3. Then $S$ is finite-dimensional over $k$, because $S$ is a quotient space of $A$. Suppose ${\dim }_{k}S=d$, then ${\mathrm{End}}_k(S)\simeq {\mathrm{M}}_d(k)$ and it is also finite-dimensional over $k$. Then ${\mathrm{End}}_A(S)$ is finite-dimensional by ${\mathrm{End}}_A(S)\subseteq {\mathrm{End}}_k(S)$.

iii) By i) and ii), ${\mathrm{End}}_A(S)$ is a finite-dimensional division ring over algebraically closed field $k$. Then by ^190f7e, ${\mathrm{End}}_A(S)\simeq k$. \end{proof}

Theorem

A finite-dimensional semisimple $k$-algebra $A$ over a field $k$ with $k=\overline{k}$ is isomorphism to a direct sum of matrix algebras over $k$, that is,

$$A\simeq {\mathrm{M}}_{n_1}(k)\oplus \cdots \oplus {\mathrm{M}}_{n_r}(k)\text{ for some }{n}_i\in {\mathbb{N}}_{+}.$$

\begin{proof} Recall that $A={\mathrm{M}}_{n_1}(D_1)\oplus \cdots \oplus {\mathrm{M}}_{n_r}(D_r)$ by ^f43ojz. Since $k$ is algebraically closed, $D_i=k$ for each $i$ by ^evw7zf. \end{proof}

Corollary

If $A$ is a finite-dimensional $k$-algebra, then any simple $A$-module is of finite-dimensional as $k$-vector space.

\begin{proof} See the argument of ^evw7zf, ii). \end{proof}

Proposition

Let $A$ be a finite-dimensional semisimple $k$-algebra over a field $k$. In any decomposition ${}_{A}A=S_1^{n_1}\oplus \cdots \oplus S_r^{n_r}$ where the $S_i$ are pairwise non-isomorphic simple $A$-modules, we have $S_1,\cdots ,S_r$ is a complete set of representative of the isomorphism of simple $A$-modules.

If moreover $k=\overline{k}$, then $n_i={\dim }_k{S}_i$ and ${\dim }_k{}_{A}A={\sum }_{i=1}^r{n}_i^2$.

\begin{proof} By ^09efb3, each simple $A$-module $M$ is a quotient of ${}_{A}A$. Since ${}_{A}A$ is semisimple, we have that $M$ is a direct summand of ${}_{A}A$ and so $S_1,\cdots ,S_r$ are all representative of the isomorphism of simple $A$-modules.

Furthermore, if $k$ is algebraically closed, by ^88dlwq one can get ${\dim }_k{}_{A}A={\sum }_{i=1}^r{n}_i^2$. \end{proof}