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.


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