## Archive for the ‘Division Rings’ Category

Throughout this post, $R$ is a ring with $1.$

Theorem (Jacobson). If $x^n=x$ for some integer $n > 1$ and all $x \in R,$ then $R$ is commutative.

In fact $n,$ in Jacobson’s theorem, doesn’t have to be fixed and could depend on $x,$ i.e. Jacobson’s theorem states that if for every $x \in R$ there exists an integer $n > 1$ such that $x^n=x,$ then $R$ is commutative. But we are not going to discuss that here.
In this post, we’re going to prove Jacobson’s theorem. Note that we have already proved the theorem for $n=3, 4$ (see here and here) and we didn’t need $R$ to have $1,$ we didn’t need that much ring theory either. But to prove the theorem for any $n > 1,$ we need a little bit more ring theory.

Lemma. If Jacobson’s theorem holds for division rings, then it holds for all rings with $1.$

Proof. Let $R$ be a ring with $1$ such that $x^n=x$ for some integer $n > 1$ and all $x \in R.$ Then clearly $R$ is reduced, i.e. $R$ has no non-zero nilpotent element. Let $\{P_i: \ i \in I\}$ be the set of minimal prime ideals of $R.$
By the structure theorem for reduced rings, $R$ is a subring of the ring $\prod_{i\in I}D_i,$ where $D_i=R/P_i$ is a domain. Clearly $x^n=x$ for all $x \in D_i$ and all $i \in I.$ But then, since each $D_i$ is a domain, we get $x=0$ or $x^{n-1}=1,$ i.e. each $D_i$ is a division ring. Therefore, by our hypothesis, each $D_i$ is commutative and hence $R,$ which is a subring of $\prod_{i\in I}D_i,$ is commutative too. $\Box$

Example. Show that if $x^5=x$ for all $x \in R,$ then $R$ is commutative.

Solution. By the lemma, we may assume that $R$ is a division ring.
Then $0=x^5-x=x(x-1)(x+1)(x^2+1)$ gives $x=0,1,-1$ or $x^2=-1.$ Suppose that $R$ is not commutative and choose a non-central element $x \in R.$ Then $x+1,x-1$ are also non-central and so $x^2=(x+1)^2=(x-1)^2=-1$ which gives $1=0,$ contradiction! $\Box$

Remark 1. Let $D$ be a division ring with the center $F.$ If there exist an integer $n \ge 1$ and $a_i \in F$ such that $x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0=0$ for all $x \in D,$ then $F$ is a finite field. This is obvious because the polynomial $x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0 \in F[x]$ has only a finite number of roots in $F$ and we have assumed that every element of $F$ is a root of that polynomial.

Remark 2. Let $D$ be a domain and suppose that $D$ is algebraic over some central subfield $F.$ Then $D$ is a division ring and if $0 \ne d \in D,$ then $F[d]$ is a finite dimensional division $F$-algebra.

Proof. Let $0 \ne d \in D.$ So $d^m +a_{m-1}d^{m-1}+ \cdots + a_1d+a_0=0$ for some integer $m \ge 1$ and $a_i \in F.$ We may assume that $a_0 \ne 0.$ Then $d(d^{m-1} + a_{m-1}d^{m-2}+ \cdots + a_1)(-a_0^{-1})=1$ and so $d$ is invertible, i.e. $D$ is a division ring.
Since $F[d]$ is a subring of $D,$ it is a domain and algebraic over $F$ and so it is a division ring by what we just proved. Also, since $d^m \in \sum_{i=0}^{m-1} Fd^i$ for some integer $m \ge 1,$ we have $F[d]=\sum_{i=0}^{m-1} Fd^i$ and so $\dim_F F[d] \le m. \ \Box$

Proof of the Theorem. By the above lemma, we may assume that $R$ is a division ring.
Let $F$ be the center of $R.$ By Remark 1, $F$ is finite. Since $R$ is a division ring, it is left primitive. Since every element of $R$ is a root of the non-zero polynomial $x^n-x \in F[x], \ R$ is a polynomial identity ring.
Hence, by the Kaplansky-Amtsur theorem, $\dim_F R < \infty$ and so $R$ is finite because $F$ is finite. Thus, by the Wedderburn’s little theorem, $R$ is a field. $\Box$

## Tensor product of division algebras (3)

Posted: October 29, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , ,

Here you can see part (2). We are now ready to prove a nice result.

Theorem. Let $k$ be an algebraically closed field. Let $A$ be a commutative $k$-domain and let $B$ be a $k$-domain. Then $A \otimes_k B$ is a $k$-domain.

Proof. Suppose that $A \otimes_k B$ is not a domain. So there exist non-zero elements $u, v \in A \otimes_k B$ such that $uv=0.$ Let $u = \sum_{i=1}^m a_i \otimes_k b_i$ and $v = \sum_{i=1}^n a_i' \otimes_k b_i',$ where $a_i,a_i' \in A, \ b_i,b_i' \in B$ and both sets $\{b_1, \ldots , b_m\}$ and $\{b_1', \ldots , b_n'\}$ are $k$-linearly independent. Let $C=k[a_1, \ldots , a_m, a_1', \ldots , a_n'],$ which is a commutative domain because $C \subseteq A.$ Also, note that $u,v \in C \otimes_k B \subseteq A \otimes_k B.$ Now,  since $u, v \neq0,$ there exist integers $r,s$ such that $a_r \neq 0$ and $a_s' \neq 0.$ Therefore, by the third part of the corollary in part (2), there exists a ring homomorphism $\varphi : C \longrightarrow k$ such that $\varphi(a_r) \neq 0$ and $\varphi(a_s') \neq 0.$ Let $\psi = \varphi \otimes \text{id}_B.$ Then $\psi : C \otimes_k B \longrightarrow k \otimes_k B \cong B$ is a ring homomorphism and hence $\psi(u)\psi(v)=\psi(uv)=0.$ Therefore either $\psi(u)=0$ or $\psi(v)=0,$ because $B$ is a domain. But $\psi(u)=\sum_{i=1}^m \varphi(a_i) \otimes_k b_i \neq 0,$ because $\{b_1, \ldots , b_m \}$ is $k$-linearly independent and $\varphi(a_r) \neq 0.$ Also, $\psi(v)=\sum_{i=1}^n \varphi(a_i') \otimes_k b_i' \neq 0,$ because $\{b_1', \ldots , b_n' \}$ is $k$-linearly independent and $\varphi(a_s') \neq 0.$ This contradiction proves that $A \otimes_k B$ is  a domain. $\Box$

Let $k$ be an algebraically closed field. A trivial corollary of the theorem is a well-known result in field theory: if $F_1,F_2$ are two fields which contain $k,$ then $F_1 \otimes_k F_2$ is a commutative domain. Another trivial result is this: if $k$ is contained in both a field $F$ and the center of a division algebra $D,$ then $F \otimes_k D$ is a domain.

Question. Let $k$ be an algebraically closed field and let $D_1,D_2$ be two finite dimensional division $k$-algebras. Will $D_1 \otimes_k D_2$ always be a domain?

Answer. No! See the recent paper of Louis Rowen and David Saltman for an example.

## Tensor product of division algebras (2)

Posted: October 29, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , ,

Here you can see part (1). We are now going to prove a more interesting result than the one we proved in part (1). But we need to get prepared first. The following important result is known as Zariski’s lemma.

Lemma. (Zariski, 1946) Let $k$ be a field and let $A=k[a_1, \ldots , a_n]$ be a finitely generated commutative algebra. If $A$ is a field, then $A$ is algebraic over $k$  and thus $\dim_k A < \infty.$

Proof. The proof is by induction over $n.$ If $n=1,$ then $A=k[a_1]$ and since $A$ is a field, $a_1^{-1} \in A.$ Thus $a_1^{-1}=\sum_{i=0}^m \gamma_i a_1^i$ for some integer $m \geq 0$ and $\gamma_i \in k.$ Then $\sum_{i=0}^m \gamma_i a_1^{i+1}=0$ and so $a_1,$ and hence $A,$ is algebraic over $k.$ Now suppose that $n \geq 2.$ If all $a_i$ are algebraic over $k,$ then $A=k[a_1, \ldots , a_n]$ is algebraic over $k$ and we are done. So we may assume that $a_1$ is transcendental over $k.$ Since $A$ is a field, $K =k(a_1) \subseteq A$ and thus $A=K[a_2, \ldots , a_n].$ By the induction hypothesis, $A$ is algebraic over $K.$ So every $a_i$ satisfies some monic polynomial $f_i(x) \in K[x]$ of degree $m_i.$ Let $v_i \in k[a_1]$ be the product of the denominators of the coefficients of $f_i(x)$ and put $v=\prod_{i=1}^n v_i.$ Let $m$ be the maximum of $m_i.$ Then multiplying $f(a_i)=0$ through by $v^m$ shows that each $va_i$ is integral over $k[a_1].$ Note that since $a_1$ is transcendental over $k,$ $k[a_1] \cong k[x],$ the polynomial algebra over $x.$ Thus I can choose an irreducible polynomial $p(a_1) \in k[a_1]$ such that

$\gcd(p(a_1), v)=1. \ \ \ \ \ \ \ \ \ (*)$

Now $(p(a_1))^{-1} \in A=k[a_1, \ldots , a_n],$ because $A$ is a field. Thus for a large enough integer $r,$ we have $v^r(p(a_1))^{-1} \in k[va_1, \ldots , va_n]$ and hence $v^r(p(a_1))^{-1}$ is integral over $k[a_1].$ But $k[a_1]$ is a UFD and we know that every UFD is integerally closed (in its field of fraction). Therefore $v^r(p(a_1))^{-1} \in k[a_1],$ which is absurd because then $p(a_1) \mid v,$ contradicting $(*). \ \Box$

Corollary. Let $k$ be a field and let $A=k[a_1, \ldots , a_n]$ be a finitely generated commutative algebra.

1) If $\mathfrak{m}$ is a maximal ideal of $A,$ then $\dim_k A/\mathfrak{m} < \infty.$

2) If $A$ is a field and $k$ is algebraically closed, then $A=k.$

3) If $A$ is a domain, $k$ is algebraically closed and $b_1, \ldots , b_m \in A \setminus \{0\},$ then there exists a ring homomorphism $\varphi : A \longrightarrow k$ such that $\varphi(b_i) \neq 0$ for all $i=1, \ldots , m.$

Proof. 1) Let $\overline{a_i}=a_i + \mathfrak{m}, \ i=1, \ldots , n.$ Then $A/\mathfrak{m} = k[\overline{a_1}, \ldots , \overline{a_n}]$ and we are done by the lemma.

2) By the lemma, $A$ is algebraic over $k$ and thus, since $k$ is algebraically closed, $A=k.$

3) Let $S$ be the set of all monomials in $b_1, \ldots, b_m.$ Clearly $S$ is multiplicatively closed and $0 \notin S$ because $A$ is a domain. Consider $B=S^{-1}A,$ the localization of $A$ at $S.$ Clearly $B=k[a_1, \ldots , a_n, b_1^{-1}, \ldots , b_m^{-1}]$ and so $B$ is finitely generated. Let $\mathfrak{m}$ be a maximal ideal of $B.$ Note that $b_i \notin \mathfrak{m},$ for all $i,$ because each $b_i$ is a unit in $B.$ Now, we have $B/\mathfrak{m} \cong k,$ by 2). Let $f: B/\mathfrak{m} \longrightarrow k$ be a ring isomorphism. We also have the natural ring homomorphism $g: B \longrightarrow B/\mathfrak{m}$ and an inejctive ring homomorphism $h: A \longrightarrow B=S^{-1}A$ defined by $f(a)=a/1.$ Let $\varphi = fgh.$ Then $\varphi : A \longrightarrow k$ and $\varphi(b_i) \neq 0$ for all $i,$ because $f$ is an isomorphism and $b_i \notin \mathfrak{m}$ for all $i. \ \Box$

See part (3) here.

## Reduced trace and reduced norm in division algebras

Posted: October 13, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , , , , ,

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 $v \in D'. \ \Box$

## Wedderburn’s factorization theorem (2)

Posted: October 13, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , , ,

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.

## Wedderburn’s factorization theorem (1)

Posted: October 12, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: ,

Let $R$ be a ring and let $R[x]$ be the ring of polynomials in the central indeterminate $x.$ We will always write the coefficients of an element of $R[x]$ on the left. Now, if $R$ is commutative and $f(x)=g(x)h(x)$ in $R[x],$ then $f(a)=g(a)h(a)$ for all $a \in R.$ This is not true if $R$ is noncommutative. For example, let $g(x)=x-a$ and $h(x)=x - b,$ where $a,b \in R.$ Then $f(x)=g(x)h(x)=x^2-(a+b)x+ab$ and hence $f(a)=ab-ba.$ Thus if $ab \neq ba,$ then $f(a) \neq 0$ but $g(a)h(a) = (a-a)(a-b)=0.$

Remark 1. Let $R$ be a ring. If $f(x)=g(x)h(x)$ in $R[x]$ and if every coefficient of $h(x)$ is in the center of $R,$ then $f(a)=g(a)h(a)$ for all $a \in R.$

Proof. Let $g(x)=\sum_{i=0}^n b_ix^i$ and $h(x)=\sum_{j=0}^m c_jx^j.$ Then, since $a$ commutes with every $c_j,$ we have

$f(a)=\sum_{\ell=0}^{m+n}\sum_{i+j=\ell}b_ic_ja^{\ell}=\sum_{\ell=0}^{m+n}\sum_{i+j=\ell}b_ia^ic_ja^j=g(a)h(a).$

Remark 2. Let $R$ be a ring and $f(x) \in R[x].$ Then $f(a)=0$ if and only if $f(x)=g(x)(x-a)$ for some $g(x) \in R[x].$

Proof. Let $f(x)=\sum_{i=0}^n a_ix^i.$ If $f(a)=0,$ then since $x-a$ commutes with $a,$ we have

$f(x)=\sum_{i=0}^na_i(x-a +a)^i = \sum_{i=0}^n a_i \sum_{j=0}^i \binom{i}{j}a^{i-j}(x-a)^j =$

$g(x)(x-a) + f(a)=g(x)(x-a).$

Conversely, if $f(x)=g(x)(x-a)$ for some $g(x) = \sum_{i=0}^m b_ix^i,$ then

$f(x)=\sum_{i=0}^mb_ix^{i+1} - \sum_{i=0}^m b_iax^i$

and thus $f(a)=\sum_{i=0}^mb_ia^{i+1} - \sum_{i=0}^m b_ia^{i+1}=0. \ \Box$

Lemma 1. Let $D$ be a division algebra with the center $k.$ Suppose that $f(x)=g(x)h(x)$ in $D[x].$ If $a \in D$ and $b = h(a) \neq 0,$ then $f(a) = g(bab^{-1})b.$ Thus if $f(a)=0,$ then $g(bab^{-1})=0.$

Proof. Let $g(x)=\sum_{i=0}^n b_ix^i$ and $h(x)=\sum_{j=0}^m c_jx^j.$ Then

$f(a)=\sum_{i,j} b_i c_j a^{i+j} = \sum_{i=0}^n b_i \left (\sum_{j=0}^m c_ja^j \right)a^i=\sum_{i=0}^n b_i h(a)a^i = \sum_{i=0}^n b_iba^i =$

$\sum_{i=0}^n b_i(bab^{-1})^ib=g(bab^{-1})b. \ \Box$

Lemma 2. Let $D$ be a division algebra with the center $k.$ Suppose that $a \in D$ is algebraic over $k$ and $f(x) \in k[x]$ is the minimal polynomial of $a$ over $k.$ If $g(x) \in D[x]$ is non-zero and $g(cac^{-1}) = 0,$ for all $0 \neq c \in D,$ then $\deg g(x) \geq \deg f(x).$

Proof. Let $\deg f(x) = n$ and suppose that the lemma is false. Let $m \geq 1$ be the smallest integer for which there exists a non-zero polynomial $g(x) \in D[x]$ of degree $m$ such that $m < n$ and $g(cac^{-1})=0$ for all $0 \neq c \in D.$ Note that since for every $0 \neq u \in D,$ the polynomial $ug(x)$ has degree $m$ and $ug(cac^{-1})=0,$ for all $0 \neq c \in D,$ we may assume that $g(x)$ is monic. Let $g(x)=x^m + b_{m-1}x^{m-1} + \ldots + b_1x + b_0.$ To get a contradiction, we are going to find a non-zero polynomial $h(x) \in D[x]$ such that $\deg h(x) < m$ and $h(cac^{-1})=0$ for all $0 \neq c \in D.$ We first note that, since $f(x)$ is the minimal polynomial of $a$ over $k, \ m < n$ and $g(a)=0,$ we have $g(x) \notin k[x].$ So there exists $0 \leq j \leq m-1$ such that $b_j \notin k.$ Let $b \in D$ be such that $b_jb - bb_j \neq 0.$ So $b \neq 0.$ Now we let

$h(x) = g(x)b - bg(x).$

Then $h(x)= \sum_{i=0}^{m-1}(b_ib - bb_i)x^i.$ Hence $\deg h(x) < m$ and $h(x) \neq 0,$ because $b_jb - bb_j \neq 0.$ Let $0 \neq c \in D.$ Then, by Lemma 1,

$h(cac^{-1}) = g(bcac^{-1}b^{-1})b - bg(cac^{-1}) = g(bca(bc)^{-1})b = 0. \ \Box$

## Reduced characteristic polynomials in division algebras

Posted: October 12, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , ,

Theorem. Let $D$ be a finite dimensional central division $k$-algebra of degree $n$ and let $a \in D.$ Suppose that $q(x)=x^d + \alpha_{d-1}x^{d-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x]$ is the minimal polynomial of $a$ over $k.$ Then $\text{Prd}_D(a,x)=(q(x))^{n/d}.$

Proof. We have $\dim_k k(a) = \deg q(x) = d$ and hence $\dim_{k(a)} D = n^2/d.$ Now, the set $\{1,a, \ldots , a^{d-1} \}$ is a $k$-basis for $k(a).$ Let $\{u_1, u_2, \ldots , u_{n^2/d} \}$ be a $k(a)$-basis for $D.$ Then the set

$\mathfrak{B}=\{u_1, au_1, \ldots , a^{d-1}u_1, \ldots , u_{n^2/d}, au_{n^2/d} \ldots , a^{d-1}u_{n^2/d} \}$

is a $k$-basis for $D.$ Now define $\phi : D \longrightarrow \text{End}_k(D) \cong M_{n^2}(k)$ by $\phi(b)(c)=bc$ for all $b,c \in D.$ Let $p(x)$ be the characteristic polynomial of $\phi(a).$ By the example in this post, we have

$p(x)=(\text{Prd}_D(a,x))^n. \ \ \ \ \ \ \ \ \ \ (1)$

We are now going to find $p(x)$ in terms of $q(x).$ To do so, we first find $[\phi(a)]_{\mathfrak{B}},$ the matrix of $\phi(a)$ with respect to the ordered basis $\mathfrak{B}.$ Notice that since $q(a)=0$ we have

$a^d = -\alpha_0 - \alpha_1a - \ldots - \alpha_{d-1} a^{d-1}.$

Now, since $\phi(a)(a^iu_j)=a^{i+1}u_j$ for all $i=0, \ldots , d-1$ and $j =1, \ldots , n^2/d,$ we have the following $n^2 \times n^2$ block matrix

$[\phi(a)]_{\mathfrak{B}} = \begin{pmatrix} C & 0 & \ldots & 0 \\ 0 & C & \ldots & 0 \\ . & . & \ldots & . \\ . & . & \ldots & . \\ . & . & \ldots & . \\ 0 & 0 & \ldots & C \end{pmatrix}, \ \ \ \ \ \ \ \ \ \ \ (2)$

where $C = \begin{pmatrix} 0 & 0 & \ldots & 0 & -\alpha_0 \\ 1 & 0 & \ldots & 0 & -\alpha_1 \\ . & . & \ldots & . & . \\ . & . & \ldots & . & . \\ . & . & \ldots & . & . \\ 0 & 0 & \ldots & 1 & -\alpha_{d-1} \end{pmatrix}.$ It is easy to see that

$\det(xI - C)=x^d + \alpha_{d-1}x^{d-1} + \ldots + \alpha_1x + \alpha_0 = q(x).$

Thus $(2)$ gives us

$p(x)=(\det(xI - C))^{n^2/d}=(q(x))^{n^2/d}. \ \ \ \ \ \ \ \ (3)$

Hence $(q(x))^{n^2/d} = (\text{Prd}_D(a,x))^n,$ by $(1)$ and $(3).$ That means $q(x)$ is the unique irreducible factor of $\text{Prd}_D(a,x)$ and hence $\text{Prd}_D(a,x)=(q(x))^r$ for some integer $r \geq 1.$ Now, since $\deg \text{Prd}_D(a,x)=n$ and $\deg q(x)=d,$ we must have $r = n/d. \ \Box$

Example. (You should also see the example in this post!) Let $\mathbb{H}$ be the division algebra of real quaternions and $a = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H}.$ It is not hard to see that $q(x)=x^2 - 2 \alpha x + \alpha^2 + \beta^2+ \gamma^2 + \delta^2$ is the minimal polynomial of $a$ over $\mathbb{R}.$ Therefore, since $\deg \mathbb{H} = \deg q(x)=2,$ we get from the above theorem that $\text{Prd}_{\mathbb{H}}(a,x)=q(x).$