## Two applications of Schur’s lemma

Posted: July 11, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
Tags: , ,

Let $k$ be an algebraically closed field, $A$ a $k$ algebra and $V$ a simple $A$ module with $\dim_k V < \infty.$ We know, by Schur’s lemma, that every element of $D = \text{End}_A(V)$ is in the form $\mu 1_D,$ for some $\mu \in k.$

Application 1. If $A$ is commutative, then $\dim_k V = 1.$

Proof. Let $a \in A$ and $\{0\} \neq W$ be a $k$ subspace of $V.$ Define the map $f: V \longrightarrow V$ by $f(v)=av,$ for all $v \in V.$ Clearly $f$ is $k$ linear and, for any $b \in A$ and $v \in V,$ we have

$f(bv)=a(bv)=(ab)v=(ba)v=b(av)=bf(v).$

That means $f \in D$ and hence $f = \mu 1_D$, for some $\mu \in k.$ Thus if $w \in W$, then $aw=f(w)=\mu w \in W,$ which means that $W$ is an $A$ submodule of $V$ and so $W=V,$ because $V$ is simple over $A.$ So every non-zero $k$ subspace of $V$ is equal to $V.$ Hence $\dim_k V = 1.$

Application 2. Let $Z(A)$ be the center of $A.$ For every $a \in Z(A)$ there exists (a unique) $\mu_a \in k$ such that $av=\mu_a v,$ for all $v \in V$ and the map $\chi_V : Z(A) \longrightarrow k$ defined by $\chi_V(a)=\mu_a$ is a $k$ algebra homomorphism.

Proof. Define the map $f_a : V \longrightarrow V$ by $f_a(v)=av,$ for all $v \in V.$ Then $f_a \in D$ because $a \in Z(A).$ Thus $f_a = \mu_a 1_D,$ for some $\mu_a \in k$ and therefore $av=f_a(v)=\mu_a v,$ for all $v \in V.$ The uniqueness of $\mu_a$ is trivial.

To show that $\chi_V$ is a homomorphism, let $\lambda \in k, \ a,b \in Z(A).$ Then

$\mu_{\lambda a + b} v= (\lambda a + b)v=\lambda (av) + bv = \lambda \mu_a v + \mu_b v,$

and so $\mu_{\lambda a + b} = \lambda \mu_a + \mu_b.$ Similarly

$\mu_{ab} v = (ab)v = a(bv)=a (\mu_b v) = \mu_a (\mu_b v)=(\mu_a \mu_b)v$

and so $\mu_{ab}=\mu_a \mu_b.$  $\Box$

Definition. The homomorphism $\chi_V$ is called the central character of $V.$

## Ring of endomorphisms (3)

Posted: June 9, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
Tags: , , , ,

Schur’s lemma states that if $A$ is a simple $R$ module, then $\text{End}_R(A)$ is a division ring. A similar easy argument shows that:

Example 6. For simple $R$-modules $A \ncong B$ we have $\text{Hom}_R(A,B)=\{0\}.$

Let’s generalize Schur’s lemma: let $M$ be a finite direct product of simple $R$-submodules. So $M \cong \bigoplus_{i=1}^k M_i^{n_i},$ where each $M_i$ is a simple $R$-module and $M_i \ncong M_j$ for all $i \neq j.$ Therefore, by Example 6 and Theorem 1, $\text{End}_R(M) \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i),$ where $D_i = \text{End}_R(M_i)$ is a division ring by Schur’s lemma. An important special case is when $R$ is a semisimple ring. (Note that simple submodules of a ring are exactly minimal left ideals of that ring.)

Theorem 2. (Artin-Wedderburn) Let $R$ be a semisimple ring. There exist a positive integer $k$ and division rings $D_i, \ 1 \leq i \leq ,$ such that $R \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i)$.

Proof. Obvious, by Example 1 and the above discussion. $\Box$

Some applications of Theorem 2.

1. A commutative semisimple ring is a finite direct product of fields.

2. A reduced semisimple ring is a finite direct product of division rings.

3. A finite reduced ring is a finite direct product of finite fields.

## Primitive rings; basic facts

Posted: December 18, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
Tags: , , , ,

Fact 1. Let $R$ be a left primitive ring and $M$ a faithful simple left $R$ module. By Schur’s lemma $D=\text{End}_R(M)$ is a division ring and $M$ can be considered as a right vector space over $D$ in the usual way. Let $S=\text{End}_D(M)$ and define $\varphi : R \longrightarrow S$ by $\varphi(r)(x)=rx,$ for all $r \in R$ and $x \in M.$ Then $\varphi$ is a well-defined ring homomorphism. Also $\varphi$ is one-to-one because $M$ is faithful. So $R$ can be viewed as a subring of $S.$

Fact 2. Every left primitive ring $R$ is prime. To see this, suppose $M$ is a faithful simple left $R$ module and $I,J$ be two non-zero ideals of $R$ with $IJ=(0).$  Now $JM$ is a submodule of $M$ and $M$ is simple. Therefore either $JM=(0)$ or $JM=M.$ If $JM=(0),$ then we get $(0) \neq J \subseteq \text{ann}_R M = (0),$ which is nonsense. Finally, if $JM=M,$ then we will have $(0)=(IJ)M=I(JM)=IM.$ Thus $I \subseteq \text{ann}_R M = (0)$ and so $I=(0),$ a contradiction!

Fact 3. A trivial result of Fact 2 is that the center of a left primitive ring is a commutative domain. A non-trivial fact is that every commutative domain is the center of some left primitive ring. For a proof of this see: T. Y. Lam,  A First Course in Noncommutative Ring Theory, page 195.

Fact 4. Let $R$ be a prime ring and $M$ a faithful left $R$ module of finite length. Then $R$ is left primitive. To see this, let $(0)=M_0 \subset M_1 \subset \cdots \subset M_n=M$ be a composition series of $M.$ Therefore $M_k/M_{k-1}$ is a simple left $R$ module for every $1 \leq k \leq n.$ We also let $I_k=\text{ann}_R (M_k/M_{k-1}).$ Then each $I_k$ is an ideal of $R$ and it’s easy to see that $I_1I_2 \cdots I_nM = (0).$ Thus $I_1I_2 \cdots I_n = (0),$ because $M$ is faithful. Hence $I_{\ell} = (0),$ for some $\ell,$ because $R$ is prime. Therefore $M_{\ell}/M_{\ell - 1}$ is a faithful simple left $R$ module.

Fact 5. Every left primitive ring $R$ is semiprimitive. This is easy to see: let $M$ be a faithful simple left $R$ module and $J=J(R)$, as usual, be the Jacobson radical of $R.$ The claim is that $J=(0)$. So suppose that $J \neq \{0\}$  and choose $0 \neq x \in M.$ Then $Rx=M,$ because $M$ is simple, and so $JM=Jx.$ Also either $JM=(0)$, which is impossible because then $J \subseteq \text{ann}_R M=(0)$, or $JM=M.$ If $Jx=JM=M,$ then $ax =x,$ for some $a \in J.$ Thus $(1-a)x=0,$ which gives us the contradiction $x = 0,$ because $1-a$ is invertible in $R.$