Archive for the ‘Ring of Endomorphisms’ Category

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.

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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

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.

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.

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.

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)).