Posts Tagged ‘Schur’s lemma’

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


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.

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.

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.