Posts Tagged ‘central simple 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

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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

Throughout this post, k is a field and A is a finite dimensional central simple k-algebra.

Definition. A field L is called a splitting field of A if k \subseteq L and A \otimes_k L \cong M_n(L), as L-algebras, for some integer n \geq 1.

Remark. By this lemma, the algebraic closure of k is a splitting field of any finite dimensional central simple k-algebra. Also, if f L is a splitting field of A, then \dim_k A = \dim_L A \otimes_k L = \dim_L M_n(L)=n^2 and so n = \deg A.

Theorem 1. Let A = M_n(D), where D is some finite dimensional central division k-algebra and let L be a field containing k. Then L is a splitting field of A if and only if L is a splitting field of D.

Proof. Let \dim_k D = m^2. Then \dim_k A = (mn)^2 and so \deg A = mn. If L is any field containg k, then

A \otimes_k L \cong M_n(D) \otimes_k L \cong M_n(D \otimes_k L). \ \ \ \ \ \ \ \ \ (*)

Now, suppose that L is a splitting field of A. Then A \otimes_k L \cong M_{mn}(L). Also, since D \otimes_k L is a finite dimensional central simple L-algebra, D \otimes_k L \cong M_r(D') for some division ring D' and some integer r. Therefore, by (*), we have M_{mn}(L) \cong A \otimes_k L \cong M_{rn}(D') and thus m=r and D' \cong L. Hence D \otimes_k L \cong M_m(L). Conversely, if L is a splitting field of D, then D \otimes_k L \cong M_m(L) and (*) gives us A \otimes_k L \cong M_{mn}(L). \ \Box

Corollary. There exists a finite Galois extension L/k such that L is a splitting field of A.

Proof. Trivial by Theorem 1 and the theorem in this post. \Box

Notation. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Note that, since F is a field, \phi is injective. Now, given an integer n \geq 1, we define the map \phi_m: M_n(F) \longrightarrow M_n(E) by \phi_m([u_{ij}]) = [\phi(u_{ij})] \in M_n(E). Clearly \phi_m is a k-algebra injective homomorphisms.

Theorem 2. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Suppose also that F is a spiltting field of A with an F-algebra isomorphism f : A \otimes_k F \longrightarrow M_n(F). Then E is a splitting field of A and there exists an E-algebra isomorphism g : A \otimes_k E \longrightarrow M_n(E) such that g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1)) for all a \in A.

Proof. The map \phi_m: M_n(F) \longrightarrow M_n(\phi(F)) is an isomorphism and so

A \otimes_k \phi(F) \cong A \otimes_k F \cong M_n(F) \cong M_n(\phi(F)).

Therefore

A \otimes_k E \cong A \otimes_k (\phi(F) \otimes_{\phi(F)} E) \cong (A \otimes_k \phi(F)) \otimes_{\phi(F)} E \cong M_n(\phi(F)) \otimes_{\phi(F)} E \cong

M_n(\phi(F) \otimes_{\phi(F)} E) \cong M_n(E).

So we have an isomorphism g: A \otimes_k E \longrightarrow M_n(E). It is easy to see that g(a \otimes_k 1)=\phi_m(f(a \otimes_k 1)) for all a \in A. \ \Box

Throughout, k is a field and A is a finite dimensional central simple k-algebra of degree n.

If K is a splitting field of A, then, by definition of splitting fields, there exists a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). Now let a \in A and put q(x) = \det(xI - f(a \otimes_k 1)) \in K[x], i.e. the characteristic polynomial of f(a \otimes_k 1) in M_n(K). The goal is to prove that q(x) \in k[x] and q(x) does not depend on K or f. We will then call q(x) the reduced characteristic polynomial of a.

Notation. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Note that, since F is a field, \phi is injective. We now define the map \phi_p : F[x] \longrightarrow E[x] by \phi_p(\sum_{i=0}^ra_ix^i)=\sum_{i=0}^r \phi (a_i)x^i. Clearly \phi_p is a k-algebra injective homomorphism.

Lemma 1. Let F/k and E/k be splitting fields of A with a k-algebra homomorphism \phi : F \longrightarrow E. Suppose that f : A \otimes_k F \longrightarrow M_n(F) is an F-algebra isomorphism. There exists an E-algebra isomorphism g : A \otimes_k E \longrightarrow M_n(E) such that \det(xI - g(a \otimes_k 1))=\phi_p(\det(xI-f(a \otimes_k 1))).

Proof. We will apply the notation and Theorem 2 in this post. By that theorem there exists an E-algebra isomorphismg : A \otimes_k E \longrightarrow M_n(E) such that g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1)) for all a \in A. Thus

\det(xI - g(a \otimes_k 1)) = \det(xI - \phi_m(f(a \otimes_k 1)))=\phi_p(\det(xI - f(a \otimes_k 1))). \ \Box

Lemma 2. Let K be a splitting field of A and a \in A. Let f : A \otimes_k K \longrightarrow M_n(K) be a K-algebra isomorphism and let g: A \otimes_k K \longrightarrow M_m(K) be any K-algebra homomorphism. Then m=rn for some integer r \geq 1 and \det(xI - g(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r for all a \in A.

Proof. Since A \otimes_k K is simple, g is injective and so g(A \otimes_k K) \cong A \otimes_k K is a central simple K-subalgebra of M_m(K). Thus, by the theorem in this post, there exists a central simple K-algebra C such that

M_m(K) \cong g(A \otimes_k K) \otimes_K C.

Let \dim_K C = r^2. Then, since \dim_K g(A \otimes_k K) = \dim_K A \otimes_k K = \dim_k A = n^2, we’ll get from the above isomorphism that m^2 = \dim_K M_m(K) = n^2r^2 and so m=nr. Now, let’s define a map h : A \otimes_k K \longrightarrow M_m(K) by

h(s) = \begin{pmatrix} f(s) & 0 & 0 & \ldots & 0 \\ 0 & f(s) & 0 & \ldots & 0 \\ 0 & 0 & f(s) & \ldots & 0 \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ 0 & 0 & 0 & \ldots & f(s) \end{pmatrix}, \ \ \ \ \ \ \ \ (*)

for all s \in A \otimes_k K. Note that f(s) is repeated r times in h(s) because m=nr. Clearly h is an K-algebra homomorphism because f is so.  Thus, by the Skolem-Noether theorem (see Corollary 2), there exists an invertible matrix u \in M_m(K) such that g(t)=uh(t)u^{-1}, for all t \in A \otimes_k K. Thus if a \in A, then g(a \otimes_k 1) and h(a \otimes_k 1) are similar and so their characteristic polynomials are equal. It now follows from (*) that \det(xI - g(a \otimes_k 1)) = \det(xI - h(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r. \ \Box

Corollary. Let K be a splitting field of A and a \in A. If f,g : A \otimes_k K \longrightarrow M_n(K) are K-algebra isomorphisms, then \det(xI - f(a \otimes_k 1))=\det(xI - g(a \otimes_k 1)). \ \Box

To be continued in part (2).

Let k be a field. We proved here that every derivation of a finite dimensional central simple k-algebra is inner. In this post I will give an example of an infinite dimensional central simple k-algebra all of whose derivations are inner. As usual, we will denote by A_n(k) the n-th Weyl algebra over k. Recall that A_n(k) is the k-algebra generated by x_1, \ldots , x_n, y_1, \ldots , y_n with the relations x_ix_j-x_jx_i=y_iy_j-y_jy_i=0, \ y_ix_j-x_jy_i= \delta_{ij}, for all i,j. When n = 1, we just write x,y instead of x_1,y_1. If \text{char}(k)=0, then A_n(k) is an infinite dimensional central simple k-algebra and we can formally differentiate and integrate an element of A_n(k) with respect to x_i or y_i exactly the way we do in calculus.  Let me clarify “integration” in A_1(k). For every u \in A_1(k) we denote by u_x and u_y the derivations of u with respect to x and y respectively. Let f, g, h \in A_1(k) be such that g_x=h_x=f. Then [y,g-h]=0 and so g-h lies in the centralizer of y which is k[y]. So g-h \in k[y]. For example, if f = y + (2x+1)y^2, then g_x=f if and only if g= xy + (x^2+x)y^2 + h(y) for some h(y) \in k[y]. We will write \int f \ dx = xy+(x^2+x)y^2.

Theorem. If \text{char}(k)=0, then every derivation of A_n(k) is inner.

Proof. I will prove the theorem for n=1, the idea of the proof for the general case is similar. Suppose that \delta is a derivation of A_1(k). Since \delta is k-linear and the k-vector space A_1(k) is generated by the set \{x^iy^j: \ i,j \geq 0 \}, an easy induction over i+j shows that \delta is inner if and only if there exists some g \in A_1(k) such that \delta(x)=gx-xg and \delta(y)=gy-yg. But gx-xg=g_y and gy-yg=-g_x. Thus \delta is inner if and only if there exists some g \in A_1(k) which satisfies the following conditions

g_y=\delta(x), \ \ g_x = -\delta(y). \ \ \ \ \ \ \ (1)

Also, taking \delta of both sides of the relation yx=xy+1 will give us

\delta(x)_x = - \delta(y)_y. \ \ \ \ \ \ \ \ (2)

From (1) we have \delta(x) = - \int \delta(y)_y \ dx + h(y) for some h(y) \in k[y]. It is now easy to see that

g = - \int \delta(y) \ dx + \int h(y) \ dy

will satisfy both conditions in (1). \ \Box

We will assume again that k is a field. We will denote by Z(A) the center of a ring A. The following result is one of many nice applications of the Skolem-Noether theorem (see the lemma in this post!). For the definition of derivations and inner derivations of a k-algebra see Definition 2 in this post.

Theorem. Every derivation of a finite dimensional central simple k-algebra A is inner.

Proof. First note that, since A is simple, M_2(A) is simple. We also have Z(M_2(A)) \cong Z(A) = k. Finally \dim_k M_2(A) = 4 \dim_k A < \infty. Thus M_2(A) is also a finite dimensional central simple k-algebra. Now, let \delta be a derivation of A. Define the map f: A \longrightarrow M_2(A) by

f(a) = \begin{pmatrix} a & \delta(a) \\ 0 & a \end{pmatrix},

for all a \in A. Obviously f is k-linear and for all a,a' \in A we have

f(a)f(a') = \begin{pmatrix} aa' & \delta(a)a'+a \delta(a') \\ 0 & aa' \end{pmatrix} = \begin{pmatrix} aa' & \delta(aa') \\ 0 & aa' \end{pmatrix} = f(aa').

So f is a k-algebra homomorphism and hence, by the lemma in this post, there exists v \in M_2(A) such that v is invertible and f(a)=vav^{-1}, for all a \in A. Let

v = \begin{pmatrix} x & y \\ z & t \end{pmatrix}.

So f(a)v=va gives us

\begin{cases} ax+\delta(a)z=xa \\ ay+\delta(a)t=ya \\ az=za \\ at=ta. \end{cases} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

Since (*) holds for all a \in A, we will get from the last two equations that z,t \in Z(A)=k. Since we cannot have z=t=0, because then v wouldn’t be invertible, one of z or t has to be invertible in k because k is a field. We will assume that z is invertible because the argument is similar for t. It now  follows, from the first equation in (*) and the fact that z^{-1} \in Z(A), that

\delta(a)=(xz^{-1})a - a(xz^{-1}). \ \Box

As in part (1), k is a field, A is a finite dimensional central simple k-algebra and B is a simple k-subalgebra of A. We will also be using notations and the result in the lemma in part (1), i.e. R = A \otimes_k B^{op}, M is the unique simple R-module,  A \cong M^n, D = \text{End}_R(M) and C_A(B) \cong M_n(D). Finally, as usual, we will denote the center of any algebra S by Z(S). We now prove a nice relationship between dimensions.

Lemma. \dim_k C_A(B) \cdot \dim_k B = \dim_k A.

Proof. We have R \cong M_m(D) and M \cong D^m, for some integer m. Thus A \cong D^{mn}, as k-modules, and hence

\dim_k A = mn \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

We also have

\dim_k A \cdot \dim_k B = \dim_k R = \dim_k M_m(D)=m^2 \dim_k D \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

and

\dim_k C_A(B)=\dim_k M_n(D)=n^2 \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

Eliminating \dim_k D in these three identities will give us the result. \Box

Theorem. If B is a central simple k-algebra, then C_A(B) is a central simple k-algebra too and we have A =BC_A(B) \cong B \otimes_k C_A(B).

Proof. By the lemma in part (1), C_A(B) is simple and thus, since B is central simple, B \otimes_k C_A(B) is a simple algebra. Thus the map \phi : B \otimes_k C_A(B) \longrightarrow BC_A(B) defined by \phi(b \otimes_k c)=bc is a k-algebra isomorphism. Hence B \otimes_k C_A(B) \cong BC_A(B) and so \dim_k BC_A(B)=\dim_k B \otimes_k C_A(B)=\dim_k A, by the above lemma. Therefore BC_A(B)=A. Clearly if c is in the center of C_A(B), then c commutes with every element of both B and C_A(B). Hence  c commutes with every element of A and thus c \in k. So Z(C_A(B))=k, i.e. C_A(B) is a central simple k-algebra. \Box

The Double Centralizer Theorem. C_A(C_A(B))=B.

Proof. Obviously B \subseteq C_A(C_A(B)). So, to prove that B=C_A(C_A(B)), we only need to show that \dim_k B = \dim_k C_A(C_A(B)).  Applying the above lemma to C_A(B) gives us

\dim_k C_A(C_A(B)) \cdot \dim_k C_A(B) = \dim_k A. \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

Plugging \dim_k C_A(B)=\frac{\dim_k A}{\dim_k B}, which is true by the above lemma, into (*) finishes the proof. \Box

Corollary. If B is a subfield of A, then \deg A = (\deg C_A(B))(\dim_k B).

Proof. We first need to notice a couple of things. First, since B is commutative, B \subseteq C_A(B). Since B is commutative, C_A(C_A(B))=Z(C_A(B)). Thus, by the above theorem, Z(C_A(B))=B and so C_A(B) is a central simple B-algebra. Now, by the lemma

(\deg A)^2=\dim_k A = (\dim_B C_A(B))(\dim_k B)^2=(\deg C_A(B))^2 (\dim_k B)^2. \ \Box