## 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.

## Primitive rings; definition & examples

Posted: December 17, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Let $R$ be a ring and let $M$ be a left $R$-module. Recall the following definitions:

1) $M$ is called faithful if $rM=(0)$ implies $r=0$ for any $r \in R.$ In other words, $M$ is called faithful if $\text{ann}_RM = \{r \in R : \ rM=(0) \}=(0).$
2) $M$ is called simple if $(0)$ and $M$ are the only left $R$ submodules of $M.$

Faithful and simple right $R$-modules are defined analogously.

Definition. A ring $R$ is called left (resp., right) primitive if there exists a left (resp., right) $R$-module $M$ which is both faithful and simple.

Remark. We will show later that left and right primitivity are not equivalent.  From now on, I will only consider left primitive rings. If a statement is true for left but not for right, I will mention that.

Example 1. If $R$ is a ring and $M$ is a left simple $R$-module, then $R_1=R/\text{ann}_RM$ is a left primitive ring.  This is clear because $M$ would be a faithful simple left $R_1$-module.

Example 2. Every simple ring is left primitive. That’s because we can choose a maximal left ideal $\mathbf{m}$ of $R$ and then $M=R/\mathbf{m}$ would be a faithful simple left $R$-module. The reason that $M$ is faithful is that $\text{ann}_R M$ is a two-sided ideal contained in $\mathbf{m}$ and therefore $\text{ann}_R M = (0),$ because $R$ is simple. One special case of this example is $M_n(D),$ the ring of $n \times n$ matrices with entries from a division ring $D.$ If $V$ is an infinite dimensional vector space over a field $F$, then $\text{End}_F V$ is an example of a left primitive ring which is not simple [see Example 4 and this post].

Example 3. If $R$ is a left primitive ring and $0 \neq e \in R$ an idempotent, then $R_1=eRe$ is left primitive: let $M$ be a faithful simple $R$-module. The claim is that $M_1=eM$ is a faithful simple left $R_1$-module. Note that $M_1 \neq (0)$ because $e \neq 0$ and $M$ is faithful. Clearly $M_1$ is a left $R_1$-module.  To see why it’s faithful, let $r_1 = ere \in R_1$ with $r_1M_1=(0).$ Then $(0)=ere^2M=ereM=r_1M.$ So $r_1=0,$ because $M$ is faithful. To prove that $M_1$ is a simple $R_1$-module let $0 \neq x_1 \in M_1.$ We need to show that $R_1x_1=M_1.$ Well, since $x_1 = ex,$ for some $x \in M,$ we have $ex_1=ex=x_1.$ Thus $R_1x_1=eRex_1=eRx_1=eM=M_1.$

Example 4. Let $D$ be a division ring and let $M$ be a right vector space over $D.$ Then $R=\text{End}_D M$ is a left primitive ring. Here is why: $M$ is a left $R$-module if we define $fx = f(x),$ for all $f \in R$ and $x \in M.$ It is clear that $M$ is faithful as a left $R$-module. To see why it is simple, let $x,y \in M$ with $x \neq 0.$ Let $B=\{x_i: \ i \in I \}$ be a basis for $M$ over $D$ such that $x=x_k \in B,$ for some $k \in I.$ Define $f \in R$ by $f(\sum_{i \in I}x_id_i) = yd_k.$ Then $f(x)=f(x_k)=y.$ So, we’ve proved that $Rx = M,$ which shows that $M$ is a simple $R$-module. If $\dim_D M = n < \infty,$ then $R \cong M_n(D),$ which we already showed its primitivity in Example 2. Note that if $M$ was a “left” vector space over $D$ with $\dim_D M = n < \infty,$ then $R$ would be isomorphic to the ring $M_n(D^{op})$ rather than the ring $M_n(D).$