Posts Tagged ‘simple ring’

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.

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).