Posts Tagged ‘endomorphism ring’

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.

Example 2. \text{End}_R(R^n) \cong \mathbb{M}_n(R^{op}).

Proof. This is just an obvious result of Example 1 and Theorem 1.

Example 3. Let G be a cyclic group. If |G| = \infty, then \text{End}_{\mathbb{Z}}(G) \cong \mathbb{Z} and if |G|=n, then \text{End}_{\mathbb{Z}}(G) \cong \mathbb{Z}/n \mathbb{Z}.

Proof. The first part is obvious by Example 1. So suppose that |G|=n and let g be a generator of G and f \in \text{End}_{\mathbb{Z}}(G). Let f_i(x)=x^i. Then, since x^{ni}=1, we can choose i anything we like. Now define \varphi : \mathbb{Z} \longrightarrow \text{End}_{\mathbb{Z}}(G) by \varphi(i)=f_i. See that \varphi is an onto ring homomorphism and \ker \varphi = n \mathbb{Z}. (Note that f_0 = 0_{\text{End}_{\mathbb{Z}}(G)}.)

Example 4. Let G_1,G_2, G be cyclic groups of order m,n, \gcd(m,n) respectively. Then \text{Hom}_{\mathbb{Z}}(G_1,G_2) \cong G, as abelian groups.

Proof. Let g_1,g_2 be generators of G_1,G_2 respectively. Let f_i : G_1 \longrightarrow G_2 be defined by f(g_1)=g_2^i. See that f_i \in \text{Hom}_{\mathbb{Z}}(G_1,G_2) if and only if n \mid mi which is equivalent to \frac{n}{\gcd(m,n)} \mid i. So there are \gcd(m,n) possibility for f_i. Let g be a generator of G and define  \varphi : G \longrightarrow \text{Hom}_{\mathbb{Z}}(G_1,G_2) by \varphi(g)=f_1 and see that \varphi is a group isomorphism.

Example 5. Let p be a prime number and G_1 and G_2 be cyclic groups of orders p and p^2 respectively. Then |\text{End}_{\mathbb{Z}}(G_1 \times G_2)| = p^5.

Proof. By Theorem 1:

\text{End}_{\mathbb{Z}}(G_1 \times G_2) \cong \begin{pmatrix} \text{End}_{\mathbb{Z}}(G_1) & \text{Hom}_{\mathbb{Z}}(G_2,G_1) \\ \text{Hom}_{\mathbb{Z}}(G_1,G_2) & \text{End}_{\mathbb{Z}}(G_2) \end{pmatrix}.

By Examplse 3, |\text{End}_{\mathbb{Z}}(G_1)|=|G_1|=p and |\text{End}_{\mathbb{Z}}(G_2)|=|G_2|=p^2. Also, by Example 4


So |\text{End}_{\mathbb{Z}}(G_1 \times G_2)|=p^5.

Throughout R is a ring with 1 and M is a unitary left R module. An R module homomorphism of f: M \longrightarrow M is called an endomorphism of M. The set of endomorphisms of M is denoted by \text{End}_R (M) or  \text{Hom}_R(M,M). See that (\text{End}_R (M),+,\circ) is a ring, where \circ is the function composition.

Example 1. \text{End}_R (R) \cong R^{op}.

Proof. Define \varphi : R^{op} \longrightarrow \text{End}_R (R) by \varphi(r)(s)=sr, for all r,s \in R. It is easy to see that \varphi is a ring homomorphism. It is one-to-one because r \in \ker \varphi if and only if sr=0, for all s \in R. So if we let s=1, we’ll get r=0. It is onto because if \psi \in \text{End}_R (R), then letting r = \psi(1) we’ll have \varphi(r)(s)=sr=s \psi(1)=\psi(s) and thus \varphi(r)=\psi. \Box

Theorem 1. let M=M_1 \oplus M_2 \oplus \cdots \oplus M_n and suppose S is the set of all n \times n matrices A=[a_{ij}] with a_{ij} \in \text{Hom}_R(M_j,M_i). Then \text{End}_R (M) \cong S.

Proof. For every 1 \leq k \leq n define \rho_k : M_k \longrightarrow M and \pi_k : M \longrightarrow M_k by \rho_k(x_k)=x_k and \pi_k(x_1 + \cdots + x_n)=x_k. Now define \varphi : \text{End}_R (M) \longrightarrow S by \varphi(f)=[\pi_i f \rho_j]. Then

1) \varphi is well-defined : \pi_i f \rho_j \in \text{Hom}_R (M_j, M_i), for all i,j and thus \varphi(f) \in S.

2) \varphi is a homomorphism : let f,g \in \text{End}_R(M). Then \varphi(f+g)=\varphi(f) + \varphi(g) clearly holds. Also, since \sum_{k=1}^n \rho_k \pi_k = 1_{\text{End}_R(M)}, we have

 \varphi (f) \varphi (g)=[\sum_{k=1}^n \pi_i f \rho_k \pi_k g \rho_j] = [\pi_i fg \rho_j]=\varphi(fg).

3) \varphi is injective : because f = \sum_{i,j} \pi_i f \rho_j for all f \in \text{End}_R(M).

4) \varphi is onto : for any g =[g_{ij}] \in S let f = \sum_{i,j} \rho_i g_{ij} \pi_j. See that \varphi(f)=g. \ \Box

Remark. If in the above theorem M_1=M_2 = \cdots = M_n=X, then M=X^n. We will also have a_{ij}=\text{End}_R(X), for all i,j and thus S = M_n(\text{End}_R(X)). Therefore we get this important result:

\text{End}_R(X^n) \cong M_n(\text{End}_R(X)).