Reduced norm and reduced trace (1)

Posted: October 11, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , , , , ,

Throughout this post, $k$ is a field and $A$ is a finite dimensional central simple $k$-algebra of degree $n.$

Let $a \in M_m(k)$ and suppose that $p(x) = x^m + \alpha_{m-1}x^{m-1} + \ldots + \alpha_1x + \alpha_0 \in k[x]$ is the characteristic polynomial of $a.$ We know from linear algebra that $\text{Tr}(a)= - \alpha_{m-1}$ and $\det(a)=(-1)^m \alpha_0.$ We also know that $\text{Tr}$ is $k$-linear, $\det$ is multiplicative and $\text{Tr}(bc)=\text{Tr}(cb)$ for all $b,c \in M_m(k).$ We’d like to extend the concepts of trace and determinant to any finite dimensional central simple algebra.

Definition. Let $\text{Prd}_A(a,x)=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x]$ be the reduced characteristic polynomial of $a \in A$ (see the definition of reduced characteristic polynomials in here). The reduced trace and the reduced norm of $a$ are defined, respectively, by $\text{Trd}_A(a) = - \alpha_{n-1}$ and $\text{Nrd}_A(a) = (-1)^n \alpha_0.$

Remark. Let $K$ be  a splitting field of $A$ with a $K$-algebra isomorphism $f : A \otimes_k K \longrightarrow M_n(K).$ So $\text{Trd}_A(a)=\text{Tr}(f(a \otimes_k 1))$ and $\text{Nrd}_A(a)=\det(f(a \otimes_k 1))$ because $\text{Prd}_A(a,x)$ is the characteristic polynomial of $f(a \otimes_k 1).$

Theorem. 1) The map $\text{Trd}_A: A \longrightarrow k$ is $k$-linear and the map $\text{Nrd}_A: A \longrightarrow k$ is multiplicative.

2) $\text{Trd}_A(ab)=\text{Trd}_A(ba)$ for all $a,b \in A.$

3) If $a \in k,$ then $\text{Tr}_A(a)=na$ and $\text{Nrd}_A(a)=a^n.$

4) $\text{Nrd}_A(a) \neq 0$ if and only if $a$ is a unit of $A.$ So $\text{Nrd}_A : A^{\times} \longrightarrow k^{\times}$ is a group homomorphism.

5) $\text{Trd}_A$ and $\text{Nrd}_A$ are invariant under isomorphism of algebras and extension of scalars.

Proof. Fix a splitting field $K$ of $A$ and a $K$-algebra isomorphism $f: A \otimes_k K \longrightarrow M_n(K).$

1) We have already proved that the values of $\text{Trd}_A$ and $\text{Nrd}_A$ are in $k.$ Now, let $a_1,a_2 \in A$ and $\alpha \in k.$ Then

$\text{Trd}_A(\alpha a_1 + a_2) = \text{Tr}(f((\alpha a_1 + a_2) \otimes_k 1)) = \text{Tr}(\alpha f(a_1 \otimes_k 1) + f(a_2 \otimes_k 1)) =$

$\alpha \text{Tr}(f(a_1 \otimes_k 1)) + \text{Tr}(f(a_2 \otimes_k 1)) = \alpha \text{Trd}_A(a_1) + \text{Trd}_A(a_2)$

and

$\text{Nrd}_A(a_1a_2) = \det(f(a_1a_2 \otimes_k 1)) = \det(f(a_1 \otimes_k 1)f(a_2 \otimes_k 1))=$

$\det(f(a_1 \otimes_k 1)) \det (f(a_2 \otimes_k 1)) = \text{Nrd}_A(a_1) \text{Nrd}_A(a_2).$

2) This part is easy too:

$\text{Trd}_A(ab)=\text{Tr}(f(ab \otimes_k 1))=\text{Tr}(f((a \otimes_k 1)(b \otimes_k 1)))=\text{Tr}(f(a \otimes_k 1)f(b \otimes_k 1))=$

$\text{Tr}(f(b \otimes_k 1)f(a \otimes_k 1))= \text{Tr}(f(ba \otimes_k 1))=\text{Trd}_A(ba).$

3) Let $I$ be the identity element of $M_n(K).$ Then

$\text{Trd}_A(a)=a \text{Trd}_A(1) = a \text{Trd}_A(f(1 \otimes_k 1)) = a \text{Tr}(I)=na$

and $\text{Nrd}_A(a)=\det(f(a \otimes_k 1))=\det(a f(1 \otimes_k 1)) = \det(aI) \det(I) = a^n.$

4) If $ab=1$ for some $b \in A,$ then $1 = \text{Nrd}_A(ab)=\text{Nrd}_A(a) \text{Nrd}_A(b)$ and thus $\text{Nrd}_A(a) \neq 0.$ Conversely, if $\text{Nrd}_A(a) \neq 0,$ then $\det(f(a \otimes_k 1)) \neq 0$ and so $f(a \otimes_k 1)$ is invertible in $M_n(K).$ Let $U$ be the inverse of $f(a \otimes_k 1).$ Then $U = f(u),$ for some $u \in A \otimes_k K$ because $f$ is surjective. Since $f$ is injective, it follows that $(a \otimes_k 1)u=1.$ Now if $a$ is not a unit of $A,$ then it is a zero divisor because $A$ is artinian. So $ba = 0$ for some $0 \neq b \in A.$ But then $b \otimes_k 1 = (b \otimes_k 1)(a \otimes_k 1)u = (ba \otimes_k 1)u=0,$ contradiction!

5) By Prp 3 and Prp 4 in this post, reduced characteristic polynomials are invariant under those things. $\Box$

Reduced characteristic polynomials (2)

Posted: October 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

See part (1)  here. We will assume that $k$ is a field and $A$ is a finite dimensional central simple $k$-algebra of degree $n.$

Lemma 3. Let $K/k$ be a Galois splitting field of $A$ with a $K$-algebra isomorphism $f: A \otimes_k K \longrightarrow M_n(K).$ Then $\det(xI - f(a \otimes_k 1)) \in k[x]$ for all $a \in A.$

Proof. Recall that $A$ always have a Galois splitting field (see the corollary in this post). Let $a \in A$ and put $q(x) = \det(xI-f(a \otimes_k 1))=\sum_{i=0}^n c_i x^i.$ Clearly $q(x) \in K[x]$ but we want to prove that $p(x) \in k[x].$ Let $\phi \in \text{Gal}(K/k).$ By Lemma 1 in part (1), there exists a $K$-algebra isomorphism $g : A \otimes_k K \longrightarrow M_n(K)$ such that $\det(xI - g(a \otimes_k 1))=\phi_p(q(x)).$ By Lemma 2 in part (1), $\det(xI - g(a \otimes_k 1))=q(x)$ and so $q(x)=\phi_p(q(x)),$ i.e. $\sum_{i=0}^n c_i x^i = \sum_{i=0}^n \phi(c_i)x^i.$ Thus $\phi(c_i)=c_i$ for all $i$ and all $\phi \in \text{Gal}(K/k).$ So $c_i \in k,$ because $K/k$ is Galois, and thus $q(x) \in k[x]. \ \Box$

Lemma 4. Let $F/k$ and $E/k$ be splitting fields of $A$ with algebra isomorphisms $g : A \otimes_k F \longrightarrow M_n(F)$ and $h : A \otimes_k E \longrightarrow M_n(E).$ Then $\det(xI - g(a \otimes_k 1))= \det(xI - h(a \otimes_k 1)) \in k[x]$ for all $a \in A.$

Proof. Fix a Galois splitting field $K/k$ of $A$ and a $K$-algebra isomorphism $f : A \otimes_k K \longrightarrow M_n(K).$ Let $a \in A$ and put

$q(x)=\det(xI - f(a \otimes_k 1)).$

By Lemma 3, $q(x) \in k[x].$ Let

$s(x) = \det(xI - h(a \otimes_k 1)).$

Then, in order to prove the lemma, we only need to show that $q(x) =h(x).$ Let $L = (K \otimes_k E)/\mathfrak{M},$ where $\mathfrak{M}$ is a maximal ideal of $K \otimes_k E.$ So $L$ is a field. Define the $k$-algebra homomorphisms $\phi : K \longrightarrow L$ and $\psi : E \longrightarrow L$ by $\phi(u) = u \otimes_k 1 + \mathfrak{M}$ and $\psi(v) = 1 \otimes_k v + \mathfrak{M}$ for all $u \in K$ and $v \in E.$ By Lemma 1 in part (1), there exist $L$-algebra isomorphisms

$g_1,g_2: A \otimes_k L \longrightarrow M_n(L)$

such that $\det(xI - g_1(a \otimes_k 1))=\phi_p(q(x))$ and $\det(xI - g_2(a \otimes_k 1))=\psi_p(s(x)).$ By Lemma 2, $\det(xI - g_1(a \otimes_k 1))=\det(xI - g_2(a \otimes_k 1))$ and so $\phi_p(q(x))=\psi_p(s(x)).$ We also have $\phi_p(q(x)) = \psi_p(q(x))=q(x)$ because $q(x) \in k[x].$ Hence $\psi_p(q(x))=\psi_p(s(x))$ and therefore $q(x)=s(x)$ because $\psi_p$ is injective. $\Box$

Definition. Let $K/k$ be a splitting field of $A$ with a $K$-algebra isomorphism $f: A \otimes_k K \longrightarrow M_n(K).$ For $a \in A$ let $\text{Prd}_A(a,x)=\det(xI - f(a \otimes_k 1)).$ The monic polynomial $\text{Prd}_A(a,x)$ is called the reduced characteristic polynomial of $a.$

Theorem. Let $a \in A.$ Then $\text{Prd}_A(a,x) \in k[x]$ and $\text{Prd}_A(a,x)$ does not depend on $K$ or $f.$

Proof. By Lemma 2 in part (1), $\text{Prd}_A(a,x)$ does not depend on $f(x).$ By Lemma 4, $\text{Prd}_A(a,x) \in k[x]$ and $\text{Prd}_A(a,x)$ does not depend on $K. \ \Box$

Splitting fields of central simple algebras

Posted: October 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , , ,

Throughout this post, $k$ is a field and $A$ is a finite dimensional central simple $k$-algebra.

Definition. A field $L$ is called a splitting field of $A$ if $k \subseteq L$ and $A \otimes_k L \cong M_n(L),$ as $L$-algebras, for some integer $n \geq 1.$

Remark. By this lemma, the algebraic closure of $k$ is a splitting field of any finite dimensional central simple $k$-algebra. Also, if f $L$ is a splitting field of $A,$ then $\dim_k A = \dim_L A \otimes_k L = \dim_L M_n(L)=n^2$ and so $n = \deg A.$

Theorem 1. Let $A = M_n(D),$ where $D$ is some finite dimensional central division $k$-algebra and let $L$ be a field containing $k.$ Then $L$ is a splitting field of $A$ if and only if $L$ is a splitting field of $D.$

Proof. Let $\dim_k D = m^2.$ Then $\dim_k A = (mn)^2$ and so $\deg A = mn.$ If $L$ is any field containg $k,$ then

$A \otimes_k L \cong M_n(D) \otimes_k L \cong M_n(D \otimes_k L). \ \ \ \ \ \ \ \ \ (*)$

Now, suppose that $L$ is a splitting field of $A.$ Then $A \otimes_k L \cong M_{mn}(L).$ Also, since $D \otimes_k L$ is a finite dimensional central simple $L$-algebra, $D \otimes_k L \cong M_r(D')$ for some division ring $D'$ and some integer $r.$ Therefore, by $(*),$ we have $M_{mn}(L) \cong A \otimes_k L \cong M_{rn}(D')$ and thus $m=r$ and $D' \cong L.$ Hence $D \otimes_k L \cong M_m(L).$ Conversely, if $L$ is a splitting field of $D,$ then $D \otimes_k L \cong M_m(L)$ and $(*)$ gives us $A \otimes_k L \cong M_{mn}(L). \ \Box$

Corollary. There exists a finite Galois extension $L/k$ such that $L$ is a splitting field of $A.$

Proof. Trivial by Theorem 1 and the theorem in this post. $\Box$

Notation. Let $F/k$ and $E/k$ be field extensions and suppose that $\phi : F \longrightarrow E$ is a $k$-algebra homomorphism. Note that, since $F$ is a field, $\phi$ is injective. Now, given an integer $n \geq 1,$ we define the map $\phi_m: M_n(F) \longrightarrow M_n(E)$ by $\phi_m([u_{ij}]) = [\phi(u_{ij})] \in M_n(E).$ Clearly $\phi_m$ is a $k$-algebra injective homomorphisms.

Theorem 2. Let $F/k$ and $E/k$ be field extensions and suppose that $\phi : F \longrightarrow E$ is a $k$-algebra homomorphism. Suppose also that $F$ is a spiltting field of $A$ with an $F$-algebra isomorphism $f : A \otimes_k F \longrightarrow M_n(F).$ Then $E$ is a splitting field of $A$ and there exists an $E$-algebra isomorphism $g : A \otimes_k E \longrightarrow M_n(E)$ such that $g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1))$ for all $a \in A.$

Proof. The map $\phi_m: M_n(F) \longrightarrow M_n(\phi(F))$ is an isomorphism and so

$A \otimes_k \phi(F) \cong A \otimes_k F \cong M_n(F) \cong M_n(\phi(F)).$

Therefore

$A \otimes_k E \cong A \otimes_k (\phi(F) \otimes_{\phi(F)} E) \cong (A \otimes_k \phi(F)) \otimes_{\phi(F)} E \cong M_n(\phi(F)) \otimes_{\phi(F)} E \cong$

$M_n(\phi(F) \otimes_{\phi(F)} E) \cong M_n(E).$

So we have an isomorphism $g: A \otimes_k E \longrightarrow M_n(E).$ It is easy to see that $g(a \otimes_k 1)=\phi_m(f(a \otimes_k 1))$ for all $a \in A. \ \Box$

Reduced characteristic polynomials (1)

Posted: October 1, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , , ,

Throughout, $k$ is a field and $A$ is a finite dimensional central simple $k$-algebra of degree $n.$

If $K$ is a splitting field of $A,$ then, by definition of splitting fields, there exists a $K$-algebra isomorphism $f: A \otimes_k K \longrightarrow M_n(K).$ Now let $a \in A$ and put $q(x) = \det(xI - f(a \otimes_k 1)) \in K[x],$ i.e. the characteristic polynomial of $f(a \otimes_k 1)$ in $M_n(K).$ The goal is to prove that $q(x) \in k[x]$ and $q(x)$ does not depend on $K$ or $f.$ We will then call $q(x)$ the reduced characteristic polynomial of $a.$

Notation. Let $F/k$ and $E/k$ be field extensions and suppose that $\phi : F \longrightarrow E$ is a $k$-algebra homomorphism. Note that, since $F$ is a field, $\phi$ is injective. We now define the map $\phi_p : F[x] \longrightarrow E[x]$ by $\phi_p(\sum_{i=0}^ra_ix^i)=\sum_{i=0}^r \phi (a_i)x^i.$ Clearly $\phi_p$ is a $k$-algebra injective homomorphism.

Lemma 1. Let $F/k$ and $E/k$ be splitting fields of $A$ with a $k$-algebra homomorphism $\phi : F \longrightarrow E.$ Suppose that $f : A \otimes_k F \longrightarrow M_n(F)$ is an $F$-algebra isomorphism. There exists an $E$-algebra isomorphism $g : A \otimes_k E \longrightarrow M_n(E)$ such that $\det(xI - g(a \otimes_k 1))=\phi_p(\det(xI-f(a \otimes_k 1))).$

Proof. We will apply the notation and Theorem 2 in this post. By that theorem there exists an $E$-algebra isomorphism$g : A \otimes_k E \longrightarrow M_n(E)$ such that $g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1))$ for all $a \in A.$ Thus

$\det(xI - g(a \otimes_k 1)) = \det(xI - \phi_m(f(a \otimes_k 1)))=\phi_p(\det(xI - f(a \otimes_k 1))). \ \Box$

Lemma 2. Let $K$ be a splitting field of $A$ and $a \in A.$ Let $f : A \otimes_k K \longrightarrow M_n(K)$ be a $K$-algebra isomorphism and let $g: A \otimes_k K \longrightarrow M_m(K)$ be any $K$-algebra homomorphism. Then $m=rn$ for some integer $r \geq 1$ and $\det(xI - g(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r$ for all $a \in A.$

Proof. Since $A \otimes_k K$ is simple, $g$ is injective and so $g(A \otimes_k K) \cong A \otimes_k K$ is a central simple $K$-subalgebra of $M_m(K).$ Thus, by the theorem in this post, there exists a central simple $K$-algebra $C$ such that

$M_m(K) \cong g(A \otimes_k K) \otimes_K C.$

Let $\dim_K C = r^2.$ Then, since $\dim_K g(A \otimes_k K) = \dim_K A \otimes_k K = \dim_k A = n^2,$ we’ll get from the above isomorphism that $m^2 = \dim_K M_m(K) = n^2r^2$ and so $m=nr.$ Now, let’s define a map $h : A \otimes_k K \longrightarrow M_m(K)$ by

$h(s) = \begin{pmatrix} f(s) & 0 & 0 & \ldots & 0 \\ 0 & f(s) & 0 & \ldots & 0 \\ 0 & 0 & f(s) & \ldots & 0 \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ 0 & 0 & 0 & \ldots & f(s) \end{pmatrix}, \ \ \ \ \ \ \ \ (*)$

for all $s \in A \otimes_k K.$ Note that $f(s)$ is repeated $r$ times in $h(s)$ because $m=nr.$ Clearly $h$ is an $K$-algebra homomorphism because $f$ is so.  Thus, by the Skolem-Noether theorem (see Corollary 2), there exists an invertible matrix $u \in M_m(K)$ such that $g(t)=uh(t)u^{-1},$ for all $t \in A \otimes_k K.$ Thus if $a \in A,$ then $g(a \otimes_k 1)$ and $h(a \otimes_k 1)$ are similar and so their characteristic polynomials are equal. It now follows from $(*)$ that $\det(xI - g(a \otimes_k 1)) = \det(xI - h(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r. \ \Box$

Corollary. Let $K$ be a splitting field of $A$ and $a \in A.$ If $f,g : A \otimes_k K \longrightarrow M_n(K)$ are $K$-algebra isomorphisms, then $\det(xI - f(a \otimes_k 1))=\det(xI - g(a \otimes_k 1)). \ \Box$

To be continued in part (2).

Derivations of Weyl algebras are inner

Posted: September 23, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , , ,

Let $k$ be a field. We proved here that every derivation of a finite dimensional central simple $k$-algebra is inner. In this post I will give an example of an infinite dimensional central simple $k$-algebra all of whose derivations are inner. As usual, we will denote by $A_n(k)$ the $n$-th Weyl algebra over $k.$ Recall that $A_n(k)$ is the $k$-algebra generated by $x_1, \ldots , x_n, y_1, \ldots , y_n$ with the relations $x_ix_j-x_jx_i=y_iy_j-y_jy_i=0, \ y_ix_j-x_jy_i= \delta_{ij},$ for all $i,j.$ When $n = 1,$ we just write $x,y$ instead of $x_1,y_1.$ If $\text{char}(k)=0,$ then $A_n(k)$ is an infinite dimensional central simple $k$-algebra and we can formally differentiate and integrate an element of $A_n(k)$ with respect to $x_i$ or $y_i$ exactly the way we do in calculus.  Let me clarify “integration” in $A_1(k).$ For every $u \in A_1(k)$ we denote by $u_x$ and $u_y$ the derivations of $u$ with respect to $x$ and $y$ respectively. Let $f, g, h \in A_1(k)$ be such that $g_x=h_x=f.$ Then $[y,g-h]=0$ and so $g-h$ lies in the centralizer of $y$ which is $k[y].$ So $g-h \in k[y].$ For example, if $f = y + (2x+1)y^2,$ then $g_x=f$ if and only if $g= xy + (x^2+x)y^2 + h(y)$ for some $h(y) \in k[y].$ We will write $\int f \ dx = xy+(x^2+x)y^2.$

Theorem. If $\text{char}(k)=0,$ then every derivation of $A_n(k)$ is inner.

Proof. I will prove the theorem for $n=1,$ the idea of the proof for the general case is similar. Suppose that $\delta$ is a derivation of $A_1(k).$ Since $\delta$ is $k$-linear and the $k$-vector space $A_1(k)$ is generated by the set $\{x^iy^j: \ i,j \geq 0 \},$ an easy induction over $i+j$ shows that $\delta$ is inner if and only if there exists some $g \in A_1(k)$ such that $\delta(x)=gx-xg$ and $\delta(y)=gy-yg.$ But $gx-xg=g_y$ and $gy-yg=-g_x.$ Thus $\delta$ is inner if and only if there exists some $g \in A_1(k)$ which satisfies the following conditions

$g_y=\delta(x), \ \ g_x = -\delta(y). \ \ \ \ \ \ \ (1)$

Also, taking $\delta$ of both sides of the relation $yx=xy+1$ will give us

$\delta(x)_x = - \delta(y)_y. \ \ \ \ \ \ \ \ (2)$

From $(1)$ we have $\delta(x) = - \int \delta(y)_y \ dx + h(y)$ for some $h(y) \in k[y].$ It is now easy to see that

$g = - \int \delta(y) \ dx + \int h(y) \ dy$

will satisfy both conditions in $(1). \ \Box$

Derivations of central simple algebras

Posted: February 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , , ,

We will assume again that $k$ is a field. We will denote by $Z(A)$ the center of a ring $A.$ The following result is one of many nice applications of the Skolem-Noether theorem (see the lemma in this post!). For the definition of derivations and inner derivations of a $k$-algebra see Definition 2 in this post.

Theorem. Every derivation of a finite dimensional central simple $k$-algebra $A$ is inner.

Proof. First note that, since $A$ is simple, $M_2(A)$ is simple. We also have $Z(M_2(A)) \cong Z(A) = k.$ Finally $\dim_k M_2(A) = 4 \dim_k A < \infty.$ Thus $M_2(A)$ is also a finite dimensional central simple $k$-algebra. Now, let $\delta$ be a derivation of $A.$ Define the map $f: A \longrightarrow M_2(A)$ by

$f(a) = \begin{pmatrix} a & \delta(a) \\ 0 & a \end{pmatrix},$

for all $a \in A.$ Obviously $f$ is $k$-linear and for all $a,a' \in A$ we have

$f(a)f(a') = \begin{pmatrix} aa' & \delta(a)a'+a \delta(a') \\ 0 & aa' \end{pmatrix} = \begin{pmatrix} aa' & \delta(aa') \\ 0 & aa' \end{pmatrix} = f(aa').$

So $f$ is a $k$-algebra homomorphism and hence, by the lemma in this post, there exists $v \in M_2(A)$ such that $v$ is invertible and $f(a)=vav^{-1},$ for all $a \in A.$ Let

$v = \begin{pmatrix} x & y \\ z & t \end{pmatrix}.$

So $f(a)v=va$ gives us

$\begin{cases} ax+\delta(a)z=xa \\ ay+\delta(a)t=ya \\ az=za \\ at=ta. \end{cases} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

Since $(*)$ holds for all $a \in A,$ we will get from the last two equations that $z,t \in Z(A)=k.$ Since we cannot have $z=t=0,$ because then $v$ wouldn’t be invertible, one of $z$ or $t$ has to be invertible in $k$ because $k$ is a field. We will assume that $z$ is invertible because the argument is similar for $t.$ It now  follows, from the first equation in $(*)$ and the fact that $z^{-1} \in Z(A),$ that

$\delta(a)=(xz^{-1})a - a(xz^{-1}). \ \Box$

The double centralizer theorem (2)

Posted: January 31, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

As in part (1), $k$ is a field, $A$ is a finite dimensional central simple $k$-algebra and $B$ is a simple $k$-subalgebra of $A.$ We will also be using notations and the result in the lemma in part (1), i.e. $R = A \otimes_k B^{op},$ $M$ is the unique simple $R$-module,  $A \cong M^n,$ $D = \text{End}_R(M)$ and $C_A(B) \cong M_n(D).$ Finally, as usual, we will denote the center of any algebra $S$ by $Z(S).$ We now prove a nice relationship between dimensions.

Lemma. $\dim_k C_A(B) \cdot \dim_k B = \dim_k A.$

Proof. We have $R \cong M_m(D)$ and $M \cong D^m,$ for some integer $m.$ Thus $A \cong D^{mn},$ as $k$-modules, and hence

$\dim_k A = mn \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

We also have

$\dim_k A \cdot \dim_k B = \dim_k R = \dim_k M_m(D)=m^2 \dim_k D \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

and

$\dim_k C_A(B)=\dim_k M_n(D)=n^2 \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

Eliminating $\dim_k D$ in these three identities will give us the result. $\Box$

Theorem. If $B$ is a central simple $k$-algebra, then $C_A(B)$ is a central simple $k$-algebra too and we have $A =BC_A(B) \cong B \otimes_k C_A(B).$

Proof. By the lemma in part (1), $C_A(B)$ is simple and thus, since $B$ is central simple, $B \otimes_k C_A(B)$ is a simple algebra. Thus the map $\phi : B \otimes_k C_A(B) \longrightarrow BC_A(B)$ defined by $\phi(b \otimes_k c)=bc$ is a $k$-algebra isomorphism. Hence $B \otimes_k C_A(B) \cong BC_A(B)$ and so

$\dim_k BC_A(B)=\dim_k B \otimes_k C_A(B)=\dim_k A,$

by the above lemma. Therefore $BC_A(B)=A.$ Clearly if $c$ is in the center of $C_A(B),$ then $c$ commutes with every element of both $B$ and $C_A(B).$ Hence  $c$ commutes with every element of $A$ and thus $c \in k.$ So $Z(C_A(B))=k,$ i.e. $C_A(B)$ is a central simple $k$-algebra. $\Box$

The Double Centralizer Theorem. $C_A(C_A(B))=B.$

Proof. Obviously $B \subseteq C_A(C_A(B)).$ So, to prove that $B=C_A(C_A(B)),$ we only need to show that $\dim_k B = \dim_k C_A(C_A(B)).$  Applying the above lemma to $C_A(B)$ gives us

$\dim_k C_A(C_A(B)) \cdot \dim_k C_A(B) = \dim_k A. \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

Plugging $\dim_k C_A(B)=\frac{\dim_k A}{\dim_k B},$ which is true by the above lemma, into $(*)$ finishes the proof. $\Box$

Corollary. If $B$ is a subfield of $A,$ then $\deg A = (\deg C_A(B))(\dim_k B).$

Proof. We first need to notice a couple of things. First, since $B$ is commutative, $B \subseteq C_A(B).$ Since $B$ is commutative, $C_A(C_A(B))=Z(C_A(B)).$ Thus, by the above theorem, $Z(C_A(B))=B$ and so $C_A(B)$ is a central simple $B$-algebra. Now, by the lemma

$(\deg A)^2=\dim_k A = (\dim_B C_A(B))(\dim_k B)^2=(\deg C_A(B))^2 (\dim_k B)^2. \ \Box$