## 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.

Let $A$ be a finite dimensional central simple $k$-algebra. We proved in this theorem that $\text{Nrd}_A: A^{\times} \longrightarrow k^{\times}$ is a group homomorphism. The image of $\text{Nrd}_A$ is an abelian group because it lies in $k^{\times}.$ So if $a,b \in A^{\times},$ then $\text{Nrd}_A(aba^{-1}b^{-1})=\text{Nrd}_A(aa^{-1}) \text{Nrd}_A(bb^{-1})=1.$ Therefore $(A^{\times})',$ the commutator subgroup of $A^{\times},$ is contained in $\ker \text{Nrd}_A=\{a \in A: \ \text{Nrd}_A(a) = 1\}$ and so the following definition makes sense.

Definition. Let $A$ be a finite dimensional central simple algebra. The reduced Whitehead group of $A$ is the factor group $\text{SK}_1(A) = (\ker \text{Nrd}_A)/(A^{\times})'.$

Remark. Let $D$ be a finite dimensional central division $k$-algebra of degree $n.$ By the theorem in this post, for every $a \in D$ there exists $v \in (D^{\times})'$ such that $\text{Nrd}_D(a)=a^nv.$ So if $\text{Nrd}_D(a)=1,$ then $a^n \in (D^{\times})'.$ Therefore $g^n=1$ for all $g \in \text{SK}_1(D).$

Example 1. $\text{SK}_1(M_n(k))=\{1\}$ if $k$ is a field and either $n > 2$ or $n=2$ and $|k| > 3.$

Proof. Let $A=M_n(k).$ Then $A^{\times} = \text{GL}(n,k)$ and

$\ker \text{Nrd}_A = \{a \in M_n(k): \ \det(a) = 1 \}=\text{SL}(n,k).$

We proved in here that $(\text{GL}(n,k))' = \text{SL}(n,k).$ Thus $\text{SK}_1(A)=\{1\}. \Box$

Example 2. $\text{SK}_1(\mathbb{H})=\{1\},$ where $\mathbb{H}$ is the division algebra of quaternions over $\mathbb{R}.$

Proof. Let $z = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H}$ and suppose that $\text{Nrd}_{\mathbb{H}}(z)=\alpha^2+ \beta^2 + \gamma^2 + \delta^2=1.$ We need to prove that $z$ is in the commutator subgroup of $\mathbb{H}^{\times}.$ We are going to prove a stronger result, i.e. $z = aba^{-1}b^{-1}$ for some $a,b \in \mathbb{H}^{\times}.$ If $z \in \mathbb{R},$ then $\beta = \gamma = \delta = 0$ and $z = \alpha = \pm 1.$ If $z = 1,$ then we can choose $a=b=1$ and if $z=-1,$ we can choose $a=i, \ b=j.$ So we may assume that $z \notin \mathbb{R}.$ We also have $z^2 - 2 \alpha z + 1 = 0$ (you may either directly check this or use this fact that every element of a finite dimensional central simple algebra satisfies its reduced characteristic polynomial).  Thus $(z-\alpha)^2 = \alpha^2 - 1.$ Note that since $\text{Nrd}_{\mathbb{H}}(z)=1$ and $z \notin \mathbb{R}$ we have $\alpha^2 < 1.$ So $1 - \alpha^2 = \alpha_0^2$ for some $0 \neq \alpha_0 \in \mathbb{R}.$ Let

$w = \alpha_0^{-1}(z - \alpha). \ \ \ \ \ \ \ \ \ (1)$

So, since the real part of $z$ is $\alpha,$ the real part of $w$ is zero and hence

$w^2 = -1, \ \ \overline{w} = -w. \ \ \ \ \ \ \ \ \ \ (2).$

Since the center of $\mathbb{H}$ is $\mathbb{R}$ and $w \notin \mathbb{R},$ there exists $c \in \mathbb{H}$ such that $u = cw-wc \neq 0.$ Therefore

$uw = -wu = \overline{w}u. \ \ \ \ \ \ \ \ \ \ (3)$

Now, by $(1),$ we have $z = \alpha + \alpha_0 w$ and $\alpha^2+\alpha_0^2=1.$ So we can write $\alpha = \cos \theta$ and $\alpha_0 = \sin \theta.$ Let $v = \cos(\theta/2) + \sin(\theta/2)w.$ Then $(2)$ gives us

$v^2=z, \ \ \ \overline{v}=\cos(\theta/2) + \sin(\theta/2)\overline{w} = \cos(\theta/2) - \sin(\theta/2)w=v^{-1}. \ \ \ \ \ \ \ \ (4)$

We also have $u v^{-1} = \overline{v^{-1}}u=vu,$ by $(3)$ and $(4).$ Thus, by $(4)$ again, $uv^{-1}u^{-1}v = v^2=z. \ \Box$

To be continued …

We will assume that $k$ is a field.

Definition. Let $A$ be a $k$-algebra. A representation of $A$ is a $k$-algebra homomorphism $A \longrightarrow M_m(k),$ for some integer $m \geq 1.$ If $\ker \phi =(0),$ then $\phi$ is called faithful.

If $\phi$ is a representation of a $k$-algebra $A$ and $a \in A,$ then $\phi(a)$ is a matrix and so it has a characteristic polynomial. Things become interesting when $A$ is a finite dimensional central simple $k$-algebra. In this case, the characteristic polynomial of $\phi(a)$ is always a power of the reduced characteristic polynomial of $a.$ This fact justifies the name “reduced characteristic polynomial”!

Theorem. Let $A$ be a finite dimensional central simple $k$-algebra of degree $n$ and let $a \in A.$ Let $\phi : A \longrightarrow M_m(k)$ be a representation of $A$ and suppose that $p(x)$ is the characteristic polynomial of $\phi(a).$ Then $m =rn$ for some integer $r \geq 1$ and $p(x) = (\text{Prd}_A(a,x))^r.$

Proof. Let $K$ be a splitting field of $A.$ We now define $\psi : A \otimes_k K \longrightarrow M_m(K)$ by $\psi(b \otimes_k \alpha) = \alpha \phi(b)$ for all $b \in A$ and $\alpha \in K$ and then extend it linearly to all $A \otimes_k K.$ We now show that $\psi$ is a $K$-algebra homomorphism. Clearly $\psi$ is additive because $\phi$ is so. Now let $\alpha, \beta \in K$ and $b,c \in A.$ Then

$\psi(\beta (b \otimes_k \alpha)) = \psi(b \otimes_k \alpha \beta)= \beta \alpha \phi(b)=\beta \psi(b \otimes_k \alpha)$

and

$\psi((b \otimes_k \alpha)(c \otimes_k \beta))=\psi(bc \otimes_k \alpha \beta)=\alpha \beta \phi(bc) = \alpha \phi(b) \beta \phi(c)$

$=\psi(b \otimes_k \alpha) \psi(c \otimes_k \beta).$

So, by Lemma 2, $m=rn$ for some integer $r \geq 1$ and

$p(x)=\det(xI - \phi(a))=\det(xI - \psi(a \otimes_k 1))=(\text{Prd}_A(a,x))^r. \ \Box$

Example. Let $A$ be a finite dimensional central simple $k$-algebra of degree $n.$ So $\dim_k A = n^2$ and  we have a faithful representation $\phi : A \longrightarrow \text{End}_k(A) \cong M_{n^2}(k)$ defined by $\phi(a)(b)=ab,$ for all $a,b \in A.$ Let $a \in A$ and suppose that $p(x)$ is the characteristic polynomial of $\phi(a).$ Then $p(x) = (\text{Prd}_A(a,x))^n,$ by the above theorem.

## 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. $\Box$

## Basic properties of reduced characteristic polynomials

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

Prp1. Let $K$ be a field and $S=M_n(K).$ The characteristic polynomial and the reduced characteristic polynomial of an element of $S$ are equal.

Proof. Well, $f: S \otimes_K K \longrightarrow S = M_n(K)$ defined by $f(a \otimes_K \alpha) = \alpha a, \ a \in S, \alpha \in K,$ is a $K$-algebra isomorphism. Thus $\text{Prd}_S(a,x) = \det(xI - f(a \otimes_K 1)) = \det(xI - a). \ \Box$

Prp2. (Cayley-Hamilton) Let $A$ be a finite dimensional central simple $k$-algebra and $a \in A.$ Then $\text{Prd}_A(a,a)=0.$

Proof. Let $K$ be a splitting field of $A$ with a $K$-algebra isomorphism $f: A \otimes_k K \longrightarrow M_n(K).$ We have $\text{Prd}_A(a,x)=\det(xI-f(a \otimes_k 1))=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x].$ Since $\text{Prd}_A(a,x)$ is just the characteristic polynomial of $f(a \otimes_k 1) \in M_n(K),$ we may apply the Cayley-Hamilton theorem from linear algebra to get $(f(a \otimes_k 1))^n + \alpha_{n-1}(f(a \otimes_k 1))^{n-1} + \ldots + \alpha_1 f(a \otimes_k 1) + \alpha_0=0.$ Thus, since $f$ is a $K$-algebra homomorphism, we get

$f((a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1)=0.$

Hence, since $f$ is injective, we must have $(a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1 = 0$ which implies

$\text{Prd}_A(a,a)=a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0 = 0. \ \Box$

Prp3. Reduced characteristic polynomials are invariant under extension of scalars.

Proof. First let’s understand the question! We have a finite dimensional central simple $k$-algebra $A$ with $a \in A.$ We are asked to prove that if $K/k$ is a field extension and $B= A \otimes_k K,$ then $\text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x).$ Note that $B$ is a central simple $K$-algebra and $\deg A = \deg B = n.$ Now, let $L$ be a splitting field of $B$ with an $L$-algebra isomorphism $f : B \otimes_K L \longrightarrow M_n(L).$ But

$B \otimes_K L =(A \otimes_k K) \otimes_K L \cong A \otimes_k (K \otimes_K L) \cong A \otimes_k L.$

So we also have an isomorphism $g : A \otimes_k L \longrightarrow M_n(L).$ Clearly $f((a \otimes_k 1) \otimes_K 1)=g(a \otimes_k 1)$ and hence $\text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x). \ \Box$

Prp4. Reduced characteristic polynomials are invariant under isomorphism of algebras.

Proof. So $A_1,A_2$ are finite dimensional central simple $k$-algebras and $f : A_1 \longrightarrow A_2$ is a $k$-algebra isomorphism. We want to prove that $\text{Prd}_{A_1}(a_1,x) = \text{Prd}_{A_2}(f(a_1),x)$ for all $a_1 \in A_1.$ Let $K$ be a splitting field of $A_2$ with a $K$-algebra isomorphism $g : A_2 \otimes_k K \longrightarrow M_n(K).$ The map $f \otimes \text{id}_K: A_1 \otimes_k K \longrightarrow A_2 \otimes_k K$ is also a $K$-algebra isomorphism. Thus we have a $K$-algebra isomorphism $h=g \circ (f \otimes \text{id}_K) : A_1 \otimes_k K \longrightarrow M_n(K).$ So if $a_1 \in A_1,$ then

$\text{Prd}_{A_2}(f(a_1),x)=\det (xI - g(f(a_1) \otimes_k 1))=\det (xI - h(a_1 \otimes_k 1)) = \text{Prd}_{A_1}(a_1,x). \ \Box$

## Reduced characteristic polynomials; an example

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

Recall that $\mathbb{H},$ the real quaternion algebra, as a vector space over $\mathbb{R}$ has a basis $\{1,i,j,k \}$ and multiplication of two elements in $\mathbb{H}$ is done using distributive law and the relations $i^2=j^2=k^2=ijk=-1.$ It is easy to see that $\mathbb{H}$ is a division algebra and its center is $\mathbb{R}.$ So, by the lemma in this post, $\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} \cong M_2(\mathbb{C}).$ This isomorphism is also a result of this fact that $\mathbb{C}$ is a maximal subfield of $\mathbb{H}$ (see corollary 3 in this post!). In the following example, we are going to find the reduced characteristic polynomial of an element of $\mathbb{H}$ by giving a $\mathbb{C}$-algebra isomorphism $\mathbb{H} \otimes_\mathbb{R} \mathbb{C} \longrightarrow M_2(\mathbb{C})$ explicitely. We will learn later how to find the reduced characteristic polynomial of an element of any finite dimensional central division algebra.

Example. If $a = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H}$, then $\text{Prd}_{\mathbb{H}}(a,x)=x^2 - 2 \alpha x + \alpha^2+\beta^2+\gamma^2 + \delta^2.$

Solution. We define $f : \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} \longrightarrow M_2(\mathbb{C})$ as follows: for every $a_1 = \alpha_1 + \beta_1 i + \gamma_1 j + \delta_1 k \in \mathbb{H}$ and $z \in \mathbb{C}$ we define

$f(a_1 \otimes_{\mathbb{R}} z) = \begin{pmatrix} (\alpha_1 + \beta_1 i)z & (\gamma_1 + \delta_1 i)z \\ (-\gamma_1 + \delta_1 i)z & (\alpha_1 - \beta_1 i)z \end{pmatrix}.$

Then of course we extend $f$ linearly to all elements of $\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}.$ See that $f$ is a $\mathbb{C}$-algebra homomorphism. Since $\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}$ is simple, $f$ is injective. Therefore $f$ is an isomorphism because $4=\dim_{\mathbb{C}} M_2(\mathbb{C}) = \dim_{\mathbb{C}} \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}.$ It is easy now to find $\text{Prd}_{\mathbb{H}}(a)$:

$\text{Prd}_{\mathbb{H}}(a,x)=\det(xI - f(a \otimes_{\mathbb{R}} 1)) = \det \left ( \begin{pmatrix} x & 0 \\ 0 & x \end{pmatrix} - \begin{pmatrix} \alpha + \beta i & \gamma + \delta i \\ -\gamma + \delta i & \alpha - \beta i \end{pmatrix} \right).$

The rest of the proof is straightforward. $\Box$

In the above example, the coefficients of $\text{Prd}_{\mathbb{H}}(a,x)$ are all in $\mathbb{R}.$ This is not an accident. We have already proved that the reduced characteristic polynomial of an element of a finite dimensional central simple $k$-algebra is always in $k[x]$ (see the theorem in this post!).

## 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 $K. \ \Box$