## 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. $\Box$

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Comments
1. galoistron says:

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

• Yaghoub Sharifi says:

Thanks.