## Ideals of the ring of endomorphisms of a vector space

Posted: October 5, 2011 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
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Notation. Throughout this post we will assume that $k$ is a field, $V$ is a $k$-vector space, $E=\text{End}_k(V)$ and $\mathfrak{I} = \{f \in E : \ \text{rank}(f) < \infty \}.$ Obviously $\mathfrak{I}$ is a two-sided ideal of $E.$

If $\dim_k V = n < \infty,$ then $E \cong M_n(k),$ the ring of $n \times n$ matrices with entries in $k,$ and thus $E$ is a simple ring, i.e. the only two-sided ideals of $E$ are the trivial ones: $(0)$ and $E.$ But what if $\dim_k V = \infty ?$  What can we say about the two-sided ideals of $E$ if $\dim_k V = \infty ?$

Theorem 1. If $\dim_k V$ is countably infinite, then $\mathfrak{I}$ is the only non-trivial two-sided ideal of $E.$

Proof. Let $J$ be a two-sided ideal of $E$ and consider two cases.

Case 1. $J \not \subseteq \mathfrak{I}.$ So there exists $f \in J$ such that $\text{rank}(f)=\infty.$ Let $\{v_1, v_2, \ldots \}$ be a basis for $V$ and let $W$ be a subspace of $V$ such that $V = \ker f \oplus W.$ Note that $W$ is also countably infinite dimensional because $f(V)=f(W).$ Let $\{w_1,w_2, \ldots \}$ be a basis for $W.$ Since $\ker f \cap W = (0),$ the elements $f(w_1), f(w_2), \ldots$ are $k$-linearly independent and so we can choose $g \in E$ such that $gf(w_i)=v_i,$ for all $i.$ Now let $h \in E$ be such that $h(v_i)=w_i,$ for all $i.$ Then $1_E=gfh \in J$ and so $J=E.$

Case 2. $(0) \neq J \subseteq \mathfrak{I}.$ Choose $0 \neq f \in J$ and suppose that $\text{rank}(f)=n \geq 1.$ Let $\{v_1, \ldots , v_n \}$ be a basis for $f(V)$ and extend it to a basis $\{v_1, \ldots , v_n, \ldots \}$ for $V.$ Since $f \neq 0,$ there exists $s \geq 1$ such that $f(v_s) \neq 0.$ Let $f(v_s) = b_1v_1 + \ldots + b_nv_n$ and fix an $1 \leq r \leq n$ such that $b_r \neq 0.$ Now let $g \in \mathfrak{I}$ and suppose that $\text{rank}(g)=m.$ Let $\{w_1, \ldots , w_m \}$ be a basis for $g(V)$ and for every $i \geq 1$ put $g(v_i)=\sum_{j=1}^m a_{ij}w_j.$ For every $1 \leq j \leq m$ define $\mu_j, \eta_j \in E$ as follows: $\mu_j(v_r)=w_j$ and $\mu_j(v_i)=0$ for all $i \neq r,$ and $\eta_j(v_i)=b_r^{-1}a_{ij}v_s$ for all $i.$ See that $g=\sum_{j=1}^m \mu_j f \eta_j \in J$ and so $J=\mathfrak{I}. \ \Box$

Exercise. It should be easy now to guess what the ideals of $E$ are if $\dim_k V$ is uncountable. Prove your guess!

Definition. Let $n \geq 1$ be an integer. A ring with unity $R$ is called $n$-simple if for every $0 \neq a \in R,$ there exist $b_i, c_i \in R$ such that $\sum_{i=1}^n b_iac_i=1.$

Remark 1. Every $n$-simple ring is simple. To see this, let $J \neq (0)$ be a two-sided ideal of $R$ and let $0 \neq a \in J.$ Then, by definition, there exist $b_i,c_i \in R$ such that $\sum_{i=1}^n b_iac_i=1.$ But, since $J$ is a two-sided ideal of $R,$ we have $b_iac_i \in J,$ for all $i,$ and so $1 \in J.$

It is not true however that every simple ring is $n$-simple for some $n \geq 1.$ For example, it can be shown that the first Weyl algebra $A_1(k)$ is not $n$-simple for any $n \geq 1.$

Theorem 2. If $\dim_k V = n < \infty,$ then $E$ is $n$-simple. If $\dim_k V$ is countably infinite, then $E/\mathfrak{I}$ is $1$-simple.

Proof. If $\dim_k V = n,$ then $E \cong M_n(k)$ and so we only need to show that $M_n(k)$ is $n$-simple. So let $0 \neq a =[a_{ij}] \in M_n(k)$ and suppose that $\{e_{ij}: \ 1 \leq i.j \leq n \}$ is the standard basis for $M_n(k).$ Since $a \neq 0,$ there exists $1 \leq r,s \leq n$ such that $a_{rs} \neq 0.$ Using $a = \sum_{i,j}a_{ij}e_{ij}$ it is easy to see that $\sum_{i=1}^n a_{rs}^{-1}e_{ir}ae_{si}=1,$ where $1$ on the right-hand side is the identity matrix.  This proves that $E$ is $n$-simple. If $\dim_k V$ is countably infinite, then, as we proved in Theorem 1, for every $f \notin \mathfrak{I}$ there exist $g,h \in E$ such that $gfh=1_E.$ That means $E/\mathfrak{I}$ is $1$-simple. $\Box$

Remark 2. An $n$-simple ring is not necessarily artinian. For example, if $\dim_k V$ is countably infinite, then the ring $E/\mathfrak{I}$ is $1$-simple but not artinian.

## Centralizers as rings of endomorphisms

Posted: January 31, 2011 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
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Throughout this post $k$ is a commutative ring with identity. We will keep the notation for centralizers in this post.

Remark. Given a $k$-algebra $A$ and a $k$-subalgebra $B$ of $A,$ we can give $A$ a structure of a right $R:=A \otimes_k B^{op}$ module by defining $x(a \otimes_k b)=bxa,$ for all $a,x \in A$ and $b \in B.$ The only thing we need to check is the associativity of product of elements of $R$ by elements  of $A.$ This is easy to see. We have

$(x(a_1 \otimes_k b_1))(a_2 \otimes_k b_2)=(b_1xa_1)(a_2 \otimes_k b_2)=b_2b_1xa_1a_2$

and

$x((a_1 \otimes_k b_1)(a_2 \otimes_k b_2))=x(a_1a_2 \otimes_k b_2b_1)=b_2b_1xa_1a_2.$

Theorem. Let $A$ be a $k$-algebra and let $B$ be a subalgebra of $A.$ Let $R=A \otimes_k B^{op}.$ Consider $A$ as a right $R$-module, as explained in the above remark. Then $C_A(B) \cong \text{End}_R(A).$

Proof. Define the map $f: C_A(B) \longrightarrow \text{End}_R(A)$ by

$f(c)(a)=ca,$

for all $a \in A$ and $c \in C_A(B).$ We are going to prove that $f$ is a $k$-algebra isomorphism.

i) $f$ is well-defined. So we have to show that $f(c) \in \text{End}_R(A)$ for all $c \in C_A(B).$ Let $a \otimes_k b \in R.$ Let $x \in A.$ Then

$f(c)(x(a \otimes_k b))=f(c)(bxa)=cbxa=bcxa. \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The last identity in (1) is true because $c \in C_A(B).$ We also have

$f(c)(x)(a \otimes_k b)=(cx)(a \otimes_k b)=bcxa. \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Now (1) and (2) proves that $f(c)$ is an $R$-endomorphism of $A$ and so $f$ is well-defined.

ii) $f$ is a homomorphism. For, if $c_1,c_2 \in C_A(B)$ and $a \in A,$ then $f(c_1c_2)(a)=c_1c_2a=f(c_1)f(c_2)(a).$

iii) $f$ is injective. For, if $f(c)=0$ for some $c \in C_A(B),$ then $c=f(c)(1)=0.$

iv) $f$ is surjective. To see this, let $\alpha \in \text{End}_R(A).$ Let $c = \alpha(1).$ Then for every $b \in B$ we have

$bc =c(1 \otimes_k b)=\alpha(1) (1 \otimes_k b)= \alpha(1(1 \otimes_k b))=\alpha(b).$

Also

$cb=c(b \otimes_k 1) = \alpha(1)(b \otimes_k 1)=\alpha(1(b \otimes_k 1))=\alpha(b).$

Thus $bc=cb$ and so $c \in C_A(B).$ Finally, for every $a \in A$ we have

$f(c)(a)=ca=c(a \otimes_k 1)=\alpha(1)(a \otimes_k 1)=\alpha(1 (a \otimes_k 1))=\alpha(a).$

Thus $f(c)=\alpha,$ which proves that $f$ is surjective. $\Box$

## Two applications of Schur’s lemma

Posted: July 11, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
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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

$f(bv)=a(bv)=(ab)v=(ba)v=b(av)=bf(v).$

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 $V.$

## 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:

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