Ring of endomorphisms (1)

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

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