Archive for the ‘Simple Rings’ Category

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.

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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.

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

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

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!).

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