Ring of endomorphisms (1)

Posted: June 8, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
Tags: ,

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)).

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