## Ring of endomorphisms (3)

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

## Ring of endomorphisms (2)

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

$|\text{Hom}_{\mathbb{Z}}(G_1,G_2|=|\text{Hom}_{\mathbb{Z}}(G_2,G_1)|=p.$

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

## 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:

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