Centralizers as rings of endomorphisms

Posted: January 31, 2011 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
Tags: ,

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

Advertisements
Comments
  1. galoistron says:

    There is an extra c after ‘Also’ in the second equality.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s