## 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. $\Box$