Posts Tagged ‘reduced trace’

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.

Advertisements

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

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.

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