## 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 $v \in D'. \ \Box$