## Reduced norm and reduced trace (2)

Posted: October 16, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , , ,

Here you can see part (1).

Introduction. Let’s take a look at a couple of properties of the trace map in matrix rings. Let $k$ be a field and $B=M_m(k).$ Let $\{e_{ij}: \ 1 \leq i,j \leq m\}$ be the standard basis for $B.$ Let $b = \sum_{i,j}\beta_{ij}e_{ij} \in B,$ where $\beta_{ij} \in k.$ Now, $\sum_{r,s}e_{rs}be_{sr}=\sum_{i,j,r,s} \beta_{ij}e_{rs}e_{ij}e_{sr}=\sum_{i,j} \beta_{ii}e_{jj} = \text{Tr}(b)I,$ where $I$ is the identity element of $B.$ Now let’s define $\nu_B : = \sum_{ij} e_{ij} \otimes_k e_{ji} \in B \otimes_k B.$ See that $\nu_B^2=I \otimes_k I = 1_{B \otimes_k B}$ and $(b_1 \otimes_k b_2)\nu_B=\nu_B(b_2 \otimes_k b_1)$ for all $b_1,b_2 \in B.$ We are going to extend these facts to any finite dimensional central simple algebra.

Notation. I will assume that $A$ is a finite dimensional central simple $k$-algebra.

Theorem. There exists a unique element $\nu_A = \sum b_i \otimes_k c_i \in A \otimes_k A$ such that $\text{Trd}_A(a)=\sum b_iac_i$ for all $a \in A.$ Moreover, $\nu_A^2=1$ and $(a_1 \otimes_k a_2) \nu_A = \nu_A(a_2 \otimes_k a_1)$ for all $a_1, a_2 \in A.$

Proof. As we saw in this theorem, the map $g : A \otimes_k A^{op} \longrightarrow \text{End}_k(A) \cong M_n(k)$ defined by $g(a,a')(b)=aba', \ a,a',b \in A,$ is a $k$-algebra isomorphism. Let’s forget about the ring structure of $A^{op}$ for now and look at it just as a $k$-vector space.  Then $A \cong A^{op}$ and so we have a $k$-vector space isomorphism $g: A \otimes_k A \longrightarrow \text{End}_k(A)$ defined by $g(a \otimes_k a')(b)=aba', \ a,a',b \in A.$ Since $\text{Trd}_A \in \text{End}_k(A),$ there exists a unique element

$\nu_A = \sum b_i \otimes_k c_i \in A \otimes_k A$

such that $g(\nu_A) = \text{Trd}_A.$ Then $\text{Trd}_A(a)=g(\nu_A)(a) = \sum g(b_i \otimes_k c_i)(a) = \sum b_iac_i.$

To prove $\nu_A^2=1,$ we choose a splitting field of $K$ of $A.$ Then $B=A \otimes_k K \cong M_n(K)$ for some integer $n,$ which is the degree of $A.$ Let’s identify $A$ and $K$ with $A \otimes_k 1$ and $1 \otimes_k K$ respectively. Then $A$ and $K$ become subalgebras of $B$ and $B=AK.$ Let $a \in A$ and $\gamma \in K.$ Recall that, by the last part of the theorem in part (1), $\text{Trd}_B(a) = \text{Trd}_A(a),$ for all $a \in A.$ Let $a \in A$ and $\gamma \in K.$ Then, since $K$ is the center of $B,$ we have

$\sum b_i(\gamma a) c_i = \gamma \sum b_i a c_i= \gamma \text{Trd}_A(a)= \gamma \text{Trd}_B(a)= \text{Trd}_B(\gamma a).$

Thus $\nu_B = \sum b_i \otimes_k c_i = \nu_A,$ by the uniqueness of $\nu_B.$ Hence $\nu_B^2 = \nu_A^2.$ But $B \cong M_n(K)$ is a matrix ring and so, as we mentioned in the introduction, $\nu_B^2=1.$ So $\nu_A^2=1.$

To prove the last part of the theorem, let $s,t,a \in A.$ Then

$g((a_1 \otimes_k a_2) \nu_A)(a)=g(\sum a_1b_i \otimes_k a_2c_i)(a) = \sum g(a_1b_i \otimes_k a_2c_i)(a)$

$= \sum a_1b_iaa_2c_i = a_1 \text{Trd}_A(aa_2)= \text{Trd}_A(a_2a)a_1.$

The last equality holds by the second part of the theorem in part (1). Also, the image of $\text{Trd}_A$ is in $k,$ the center of $A,$ and so $\text{Trd}_A(a_2a)$ commutes with $a_1.$ Now, we also have

$g(\nu_A(a_2 \otimes_k a_1))(a) = g(\sum b_i a_2 \otimes_k c_ia_1)(a) = \sum g(b_ia_2 \otimes_k c_i a_1)(a)$

$= \sum b_ia_2ac_ia_1 = \text{Trd}_A(a_2a)a_1.$

Thus $g( (a_1 \otimes_k a_2) \nu_A) = g(\nu_A( a_2 \otimes_k a_1))$ and so $(a_1 \otimes_k a_2) \nu_A= \nu_A(a_2 \otimes_k a_1). \ \Box$

Definition. The element $\nu_A \in A \otimes_k A$ in the theorem is called the Goldman element for $A.$

Remark. David Saltman in a short paper used the properties of $\nu_A$ to give a proof of Remark 2 in this post.