Posts Tagged ‘ring of endomorphisms’

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.


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


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


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).


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