## Reduced norm and reduced trace (2)

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

## Reduced trace and reduced norm in division algebras

Posted: October 13, 2011 in Division Rings, Noncommutative Ring Theory Notes
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Definition. Let $D$ be a division algebra with the center $k.$ We denote by $[D,D]$ the additive subgroup of $D$ generated by the set $\{ab-ba : \ a,b \in D \}.$ Also, we denote by $D'$ the subgroup of $D^{\times}$ generated by the set $\{aba^{-1}b^{-1}: \ a,b \in D^{\times} \}.$

Lemma. Let $r, s$ be integers, $a_1 , \ldots , a_r \in D$ and $b_1, \ldots , b_r \in D'.$ There exists $v \in D'$ such that $(b_1a_1b_2a_2 \ldots b_ra_r)^s=(a_1a_2 \ldots a_r)^s v.$

Proof. Apply the following repeatedly: if $a \in D$ and $b \in D',$ then $ba = ac,$ where $c = b(b^{-1}a^{-1}ba) \in D'. \ \Box$

Theorem. Let $D$ be a finite dimensional central division $k$-algebra of degree $n.$ For every $a \in D$ there exist $u \in [D,D]$ and $v \in D'$ such that $\text{Trd}_D(a)=na + u$ and $\text{Nrd}_D(a)=a^nv.$

Proof. Let $f(x)$ be the minimal polynomial of $a$ over $k$ and let $m = \deg f(x).$ Let $r = \frac{n}{m}.$ Then, by the theorem in this post, $\text{Prd}_D(a,x)=(f(x))^r$ and by Wedderburn’s factorization theorem there exist non-zero elements $c_1, c_2, \ldots , c_m$ such that

$f(x) = (x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_m a c_m^{-1}) .$

Let $\alpha = \sum_{i=1}^m c_i a c_i^{-1}$ and $\beta = \prod_{i=1}^m c_i a c_i^{-1}.$ Then

$f(x)=x^m - \alpha x^{m-1} + \ldots + (-1)^m \beta$

and so

$\text{Prd}_D(a,x) = (x^m - \alpha x^{m-1} + \ldots + (-1)^m \beta)^r= x^n - r \alpha x^{n-1} + \ldots +$ $(-1)^n \beta^r.$

Therefore $\text{Trd}_D(a) = r \alpha$ and $\text{Nrd}_D(a) = \beta^r.$ Hence

$\text{Trd}_D(a)=r \alpha = r \sum_{i=1}^m c_i ac_i^{-1} = na + r\sum_{i=1}^m(c_iac_i^{-1} - ac_i^{-1}c_i).$

Let $u =r\sum_{i=1}^m (c_iac_i^{-1} - ac_ic_i^{-1}).$ Then $u \in [D,D]$ and $\text{Trd}_D(a)=na+u.$ Now let $b_i=c_iac_i^{-1}a^{-1}.$ Then the lemma gives us

$\text{Nrd}_D(a)=\beta^r = (b_1ab_2a \ldots b_m a)^r = a^n v,$

for some $v \in D'. \ \Box$

## Wedderburn’s factorization theorem (2)

Posted: October 13, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , , ,

You can see part (1) in here.

Wedderburn’s Factorization Theorem. (Wedderburn, 1920) Let $D$ be a division algebra with the center $k.$ Suppose that $a \in D$ is algebraic over $k$ and let $f(x) \in k[x]$ be the minimal polynomial of $a$ over $k.$ There exist non-zero elements $c_1, c_2 , \ldots , c_n \in D$ such that

$f(x)=(x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_nac_n^{-1}).$

Proof. By Remark 2 in part (1), there exists $g(x) \in D[x]$ such that $f(x)=g(x)(x-a).$ Now let $m$ be the largest integer for which there exist non-zero elements $c_1, c_2 , \ldots , c_m \in D$ and $p(x) \in D[x]$ such that

$f(x)=p(x)(x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_m a c_m^{-1}).$

Let

$h(x) = (x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_mac_m^{-1}).$

Claim. $h(cac^{-1})=0$ for all $0 \neq c \in D.$

Proof of the claim. Suppose, to the contrary, that there exists $0 \neq c \in D$ such that $h(cac^{-1}) \neq 0.$ Then, since $f(x)=p(x)h(x)$ and $f(cac^{-1})=0,$ there exists $0 \neq b \in D$ such that $p(bcac^{-1}b^{-1})=0,$ by Lemma 1 in part (1). So, by Remark 2 in part (1), there exists $q(x) \in D[x]$ such that $p(x)=q(x)(x - bcac^{-1}b^{-1}).$ Hence $f(x)=p(x)h(x)=q(x)(x - bca (bc)^{-1})(x-c_1ac_1^{-1}) \ldots (x - c_m a c_m^{-1}),$ contradicting the maximality of $m. \ \Box$

Therefore, by Lemma 2 in part (1), $\deg h(x) \geq \deg f(x)$ and so $h(x)=f(x)$ because $f(x)=p(x)h(x)$ and both $f(x)$ and $h(x)$ are monic. $\Box$

In the next post, I will use Wedderburn’s factorization theorem to find an expression for the reduced trace and the reduced norm of an element in a finite dimensional central division algebra.

## 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. $\Box$