## Finite rings of square-free order

Posted: July 24, 2019 in finite rings, Noncommutative Ring Theory Notes, Uncategorized

Problem. Let $R$ be a finite ring with $1$ and let $n$ be the number of elements of $R.$ Show that if $n$ is square-free, then $R \cong \mathbb{Z}/n\mathbb{Z}.$

Solution. Since $(R,+)$ is an abelian group of order $n$ and $n$ is square-free, $(R,+)$ is cyclic. Let $a \in R$ be a generator of $(R,+).$ Then

$R=\{0,a,2a, \cdots , (n-1)a\}.$

Thus

$pa=1, \ \ \ qa=a^2,$

for some integers $p,q$ and so

$(1-pq)a=a-pqa=a-pa^2=0.$

Thus $n \mid 1-pq$ because $n$ is the order of $a.$ Hence $q,n$ are coprime. Now consider the map

$f: R \to \mathbb{Z}/n\mathbb{Z}, \ \ \ f(ka)=kq+n\mathbb{Z}, \ \ \ \ \ k \in \mathbb{Z}.$

We show that $f$ is a ring isomorphism, which is what we need.

i) $f$ is well-defined. Because if $ka=0$ for some integer $k,$ then $n \mid k$ and so $f(ka)=kq+n\mathbb{Z}=0.$

ii) $f$ is additive. That’s clear.

iii) $f$ is bijective. Suppose that $f(ka)=kq+n\mathbb{Z}=0$ for some integer $k.$ Then $n \mid kq$ and so $n \mid k,$ because $q,n$ are coprime, implying that $ka=0.$ So $f$ is injective and hence bijective because $R$ and $\mathbb{Z}/n\mathbb{Z}$ have the same number of elements.

iv) $f$ is multiplicative. Because for integers $k,k'$ we have

\begin{aligned} f(kak'a)=f(kk'a^2)=f(kk'qa)=kk'q^2+n\mathbb{Z}=(kq+n\mathbb{Z})(k'q+n\mathbb{Z})=f(ka)f(k'a).\end{aligned} \ \Box

## Finite rings in which sum and product of all non-zero elements are non-zero

Posted: March 29, 2019 in finite rings, Noncommutative Ring Theory Notes
Tags:

Problem. Let $R$ be a finite ring with $1$ and suppose that $R$ satisfies the following properties

i) $\displaystyle \sum_{r \in R} r \ne 0$

ii) $\displaystyle \prod_{0 \ne r \in R} r \ne 0.$

Show that either $R \cong \mathbb{Z}_2$ or $R \cong \mathbb{Z}_4.$

Solution. First notice that both $\mathbb{Z}_2$ and $\mathbb{Z}_2$ satisfy i), ii). Since $\mathbb{Z}_2$ is the only ring with two elements, we may assume that $|R| > 2.$ Let $n:=|R|.$ First we show that $n$ is even. Suppose, to the contrary, that $n$ is odd. Then the group $(R,+)$ would have no element of order $2,$ i.e. $r \ne -r$ for all $0 \ne r \in R,$ and thus

$\displaystyle \sum_{r \in R} r = \sum (r+(-r))=0,$

Now let $m:=\text{char}(A).$ We show that $m=p^k$ for some prime $p$ and $k \in \{1,2\}.$ To see that, let $m=\prod_{i=1}^{\ell}p_i^{k_i}$ be the prime factorization of $m.$ If $\ell > 1,$ then

$\displaystyle \prod_{i=1}^{\ell} p_i^{k_i}1_R=m1_R=0,$

contradicting the property ii), because $p_i^{k_i}1_R$ are non-zero distinct elements of $R.$ So $\ell=1,$ i.e. $m=p^k$ for some prime number $p$ and positive integer $k.$ Now if $k > 2,$ then we can write

$0=m1_A=(p1_R)(p^{k-1}1_R)$

and again, by the minimal property of $m,$ the elements $p1_R$ and $p^{k-1}1_R$ are non-zero and distinct and that contradicts the property ii). So $k \le 2.$
We now show that in fact $m=4.$ We just proved that $m=p^k$ for some prime $p$ and $k \in \{1,2\}.$ Suppose that $q \ne p$ is a prime divisor of $n.$ Then there exists $r \in (R,+)$ that has order $q.$ But then $qa=ma=0$ which gives the false result $r=0,$ because $q,m$ are coprime. So $n$ is a power of $p.$ But we have already showed that $n$ is even. So $p=2$ and thus either $m=2$ or $m=4.$ If $m=2,$ then $R$ would be a vector space over $\mathbb{Z}_2,$ i.e. $R=\sum_{i=1}^s \mathbb{Z}_2r_i$ for some integer $s \ge 2,$ because $n > 2,$ and $r_i \in R.$ But then

$\displaystyle \sum_{r \in R}r =2^{s-1}\sum_{i=1}^s r_i=0,$

contradicting the property ii). So $m=4$ and hence $R$ contains a subring $R_0 \cong \mathbb{Z}_4.$
Finally, we show that $R=R_0.$ Suppose, to the contrary, that $R \neq R_0$ and let $r \in R \setminus R_0.$ We have $(21_R)(2r)=0,$ because $\text{char}(R)=4,$ and so, since $21_R \ne 0,$ we must have either $21_R=2r$ or $2r=0,$ by property ii). If $21_R=2r,$ then $(21_R)(r-1_R)=0,$ which gives $r=1_R \in R_0,$ by property ii), and that’s a contradiction. So $2r=0,$ i.e. we have shown that $2r=0$ for any element $r \in R \setminus R_0.$ But if $r \in R \setminus R_0,$ then $r+1_R \in R \setminus R_0$ too, because $1_R \in R_0,$ and so $2r=2(r+1_R)=0,$ which gives $21_R=0,$ contradiction. So $R=R_0 \cong \mathbb{Z}_4. \ \Box$

## In finite commutative rings of odd order, number of idempotents divides number of units

Posted: October 20, 2018 in finite rings, Noncommutative Ring Theory Notes
Tags: , , , , , , ,

All rings in this post are assumed to have the multiplicative identity $1.$

In this post, we showed that the ring $R:= \mathbb{Z}/n\mathbb{Z}$ has $2^m$ idempotents, where $m$ is the number of prime divisors of $n.$ Clearly $m$ is also the number of prime (= maximal) ideals of $R$ (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let $\varphi$ be the Euler’s totient function. The number of units of $R$ is $\varphi(n).$ If $n=\prod_{i=1}^m p_i^{n_i}$ is the prime factorization of $n,$ then $\varphi(n)=\prod_{i=}^m(p_i^{n_i}-p_i^{n_i-1}).$ If $n$ is odd, then $p_i^{n_i}-p_i^{n_i-1}$ is even for all $i$ and hence $2^m$ divides $\varphi(n).$ So if $n$ is odd, then the number of idempotents of $R$ divides the number of units of $R.$ As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring $R$ is called semilocal if the number of maximal ideals of $R$ is finite and it is called local if it has only one maximal ideal.

Example 1. If $p$ is a prime and $k \ge 1$ is an integer, then $\mathbb{Z}/p^k\mathbb{Z}$ is a local ring with the unique maximal ideal $p\mathbb{Z}/p^k\mathbb{Z}.$ If $n \ge 2$ is an integer, then $\mathbb{Z}/n\mathbb{Z}$ is a semilocal ring (what are its maximal ideals?).

Example 2. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let $R$ be an Artinian ring and let $S$ be the set of all finite intersections of maximal ideals of $R.$ Since $R$ is Artinian, $S$ has a minimal element $I:=\bigcap_{i=1}^m M_i.$ Let $M$ be any maximal ideal of $R.$ Since $I \bigcap M \in S$ and $I$ is a minimal element of $S,$ we must have $M_1M_2 \cdots M_m \subseteq I \subseteq M.$ So $M_i \subseteq M$ for some $i$ and hence $M_i=M.$ Therefore $M_1, \cdots , M_m$ are all the maximal ideals of $R.$

Problem 1. Show that if $R$ is a local ring, then $0,1$ are the only idempotents of $R.$

Solution. Let $M$ be the maximal ideal of $R$ and let $e$ be an idempotent of $R.$ Then $e(1-e)=0 \in M.$ Thus either $e \in M$ or $1-e \in M.$ If $e \in M,$ then $1-e \notin M$ because otherwise $1=e+(1-e) \in M,$ which is false. So $1-e$ is a unit because there’s no other maximal ideal to contain $1-e.$ So $(1-e)r=1$ for some $r \in R$ and thus $e=e(1-e)r=0.$ Similarly, if $1-e \in M,$ then $e \notin M$ and thus $e$ is a unit. So $er=1$ for some $r \in R$ implying that $1-e=(1-e)er=0$ and hence $e=1. \ \Box$

Problem 2 Show that the number of idempotents of a commutative Arinian ring $R$ is $2^m,$ where $m$ is the number of maximal ideals of $R.$

Solution. Since $R$ is Artinian, it has only finitely many maximal ideals, say $M_1, \cdots , M_m$ (see Example 2). The Jacobson radical of $R$ is nilpotent, hence there exists an integer $k \ge 1$ such that

$(0)=\left(\bigcap_{i=1}^m M_i \right)^k=\prod_{i=1}^m M_i^k=\bigcap_{i=1}^m M_i^k.$

Thus, by the Chinese remainder theorem for commutative rings, $R \cong \prod_{i=1}^m R/M_i^k.$ Since each $R/M_i^k$ is a local ring, with the unique maximal ideal $M_i/M_i^k,$ it has only two idempotents, by Problem 1, and so $R$ has exactly $2^m$ idempotents. $\Box$

Problem 3. Let $R$ be a finite commutative ring. Show that if $|R|,$ the number of elements of $R,$ is odd, then the number of idempotents of $R$ divides the number of units of $R.$

Solution. Since $R$ is finite, it is Artinian. Let $\{M_1, \cdots , M_m\}$ be the set of maximal ideals of $R.$ By problem 2, the umber of idempotents of $R$ is $2^m$ and $R \cong \prod_{i=1}^m R/M_i^k$ for some integer $k \ge 1.$
Since $|R|$ is odd, each $|M_i|$ is odd too because $(M_i,+)$ is a subgroup of $(R,+)$ and so $|M_i|$ divides $|R|.$ Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each $R/M_i^k$ is $|R/M_i^k|-|M_i/M_i^k|,$ which is an even number because both $|R|$ and $|M_i|$ are odd. So the number of units of $R,$ which is the product of the number of units of $R/M_i^k, \ 1 \le i \le m,$ is divisible by $2^m,$ which is the number of idempotents of $R. \ \Box$

Remark 1. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, $\mathbb{Z}/2\mathbb{Z}$ has one unit and two idempotents. However, the result is true in $\mathbb{Z}/2^n\mathbb{Z}, \ n \ge 2,$ which has $\varphi(2^n)=2^n-2^{n-1}$ units and two idempotents.
Can we find all even integers $n$ for which the result given in Problem 3 is true in $\mathbb{Z}/n\mathbb{Z}$? Probably not because this question is equivalent to finding all integers $n$ such that $2^m \mid \varphi(n),$ where $m$ is the number of prime divisors of $n,$ and that is not an easy thing to do.

Remark 2. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider $R:=M_2(\mathbb{F}_3),$ the ring of $2\times 2$ matrices with entries from the field of order three. Then $R$ has $(3^2-3)(3^2-1)=48$ units (see Problem 3 in this post!) but, according to my calculations, $R$ has $14$ idempotents and $14$ does not divide $48.$

## Fermat’s last theorem for finite rings

Posted: October 3, 2018 in finite rings, Noncommutative Ring Theory Notes
Tags: , , ,

Problem. Let $R$ be a finite ring with $1$ and let $R^*=R \setminus \{0\}.$ Show that for every integer $n \ge 1$ there exist $x,y,z \in R^*$ such that $x^n+y^n=z^n$ if and only if $R$ is not a division ring.

Solution. Suppose first that $R$ is a division ring. Then, since $R$ is a finite ring, $R$ is a finite field, by the Wedderburn’s little theorem. Let $|R|=q.$ Then $x^{q-1}=1$ for all $x \in R^*$ and so $x^{q-1}+y^{q-1}=z^{q-1}$ has no solution in $R^*.$
Conversely, suppose that $R$ is not a division ring (equivalently, a field because $R$ is finite). So $|R| > 2$ and hence the equation $x+y=z$ has solutions in $R^*$ (just choose $y=1, \ z \ne 0,1$ and $x=z-1$).
Let $J(R)$ be the Jacobson radical of $R.$ Since $R$ is finite, it is Artinian and so $J(R)$ is nilpotent.
So if $J(R) \neq (0),$ then there exists $a \in R^*$ such that $a^2=0$ and so $a^n=0$ for all $n \ge 2.$ Therefore the equation $x^n+y^n=z^n, \ n \ge 2,$ has a solution $x=a, y=z=1$ in $R^*.$
If $J(R)=(0),$ then by the Artin-Wedderburn’s theorem,

$\displaystyle R=\prod_{i=1}^k M_{n_i}(F_i),$

for some finite fields $F_i.$ Since $R$ is not a field, we have either $n_i >1$ for some $i$ or $n_i=1$ for all $i$ and $k \ge 2.$ If $n_i > 1$ for some $i,$ then $M_{n_i}(F_i),$ and hence $R,$ will have a non-zero nilpotent element and we are done. If $n_i=1$ for all $i$ and $k \ge 2,$ then

$x=(1,0,0, \cdots ,0), \ y = (0,1,0, \cdots , 0), \ z = (1,1,0, \cdots , 0)$

will satisfy $x^n+y^n=z^n. \ \Box$

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 I’m not going to prove 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. $\Box$

## GK dimension of Weyl algebras

Posted: April 10, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
Tags: ,

We defined the $n$-th Weyl algebra $A_n(R)$ over a ring $R$ in here.  In this post we will find the GK dimension of $A_n(R)$ in terms of the GK dimension of $R.$ The result is similar to what we have already seen in commutative polynomial rings (see corollary 1 in here). We will assume that $k$ is a field and $R$ is a $k$-algebra.

Theorem. ${\rm{GKdim}}(A_1(R))=2 + {\rm{GKdim}}(R).$

Proof. Suppose first that $R$ is finitely generated and let $V$ be a frame of $R.$ Let $U=k+kx+ky.$ Since $yx = xy +1,$ we have

$\dim_k U^n = \frac{(n+1)(n+2)}{2}. \ \ \ \ \ \ \ \ \ (*)$

Let $W=U+V.$ Clearly $W$ is a frame of $A_1(R)$ and

$W^n = \sum_{i+j=n} U^i V^j,$

for all $n,$ because every element of $V$ commutes with every element of $U.$ Therefore, since $V^j \subseteq V^n$ and $U^i \subseteq U^n$ for all $i,j \leq n,$ we have $W^n \subseteq U^nV^n$ and $W^{2n} \supseteq U^nV^n.$ Thus $W^n \subseteq U^nV^n \subseteq W^{2n}$ and hence

$\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n \leq \log_n \dim_k W^{2n}.$

Therefore ${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)),$ by $(*),$ and we are done.

For the general case, let $R_0$ be any finitely generated $k$– subalgebra of $R.$ Then, by what we just proved,

$2 + {\rm{GKdim}}(R_0)={\rm{GKdim}}(A_1(R_0)) \leq {\rm{GKdim}}(A_1(R))$

and hence $2+{\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)).$ Now, let $A_0$ be a $k$-subalgebra of $A_1(R)$ generated by a finite set $\{f_1, \ldots , f_m\}.$ Let $R_0$ be the $k$-subalgebra of $R$ generated by all the coefficients of $f_1, \ldots , f_m.$ Then $A_0 \subseteq A_1(R_0)$ and so

${\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A_1(R_0))=2 + {\rm{GKdim}}(R_0) \leq 2 + {\rm{GKdim}}(R).$

Thus

${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R)$

and the proof is complete. $\Box$

Corollary. ${\rm{GKdim}}(A_n(R))=2n + {\rm{GKdim}}(R)$ for all $n.$ In particular, ${\rm{GKdim}}(A_n(k))=2n.$

Proof. It follows from the theorem and the fact that $A_n(R)=A_1(A_{n-1}(R)). \Box$

## A theorem of Borho and Kraft

Posted: April 2, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
Tags: , , ,

As usual, I’ll assume that $k$ is a field. Recall that if a $k$-algebra $A$ is an Ore domain, then we can localize $A$ at $S:=A \setminus \{0\}$ and get the division algebra $Q(A):=S^{-1}A.$ The algebra $Q(A)$ is called the quotient division algebra of $A.$

Theorem (Borho and Kraft, 1976) Let $A$ be a finitely generated $k$-algebra which is a domain of finite GK dimension. Let $B$ be a $k$-subalgebra of $A$ and suppose that ${\rm{GKdim}}(A) < {\rm{GKdim}}(B) + 1.$ Let $S:=B \setminus \{0\}.$ Then $S$ is an Ore subset of $A$ and $S^{-1}A=Q(A).$ Also, $Q(A)$ is finite dimensional as a (left or right) vector space over $Q(B).$

Proof. First note that, by the corollary in this post, $A$ is an Ore domain and hence both $Q(A)$ and $Q(B)$ exist and they are division algebras. Now, suppose, to the contrary, that $S$ is not (left) Ore. Then there exist $x \in S$ and $y \in A$ such that $Sy \cap Ax = \emptyset.$ This implies that the sum $By + Byx + \ldots + Byx^m$ is direct for any integer $m.$ Let $W$ be a frame of a finitely generated subalgebra $B'$ of $B.$ Let $V=W+kx+ky$ and suppose that $A'$ is the subalgebra of $A$ which is generated by $V.$ For any positive integer $n$ we have

$V^{2n} \supseteq W^n(kx+ky)^n \supseteq W^ny + W^nyx + \ldots + W^nyx^{n-1}$

and thus $\dim_k V^{2n} \geq n \dim_k W^n$ because the sum is direct. So $\log_n \dim_k V^{2n} \geq 1 + \log_n \dim_k W^n$ and hence ${\rm{GKdim}}(A) \geq {\rm{GKdim}}(A') \geq 1 + {\rm{GKdim}}(B').$ Taking supremum of both sides over all finitely generated subalgebras $B'$ of $B$ will give us the contradiction ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B).$ A similar argument shows that $S$ is right Ore. So we have proved that $S$ is an Ore subset of $A.$ Before we show that $S^{-1}A=Q(A),$ we will prove that $Q(B)A=S^{-1}A$ is finite dimensional as a (left) vector space over $Q(B).$ So let $V$ be a frame of $A.$ For any positive ineteger $n,$ let $r(n) = \dim_{Q(B)} Q(B)V^n.$ Clearly $Q(B)V^n \subseteq Q(B)V^{n+1}$ for all $n$ and

$\bigcup_{n=0}^{\infty}Q(B)V^n =Q(B)A$

because $\bigcup_{n=0}^{\infty}V^n=A.$ So we have two possibilities: either $Q(B)V^n=Q(B)A$ for some $n$ or the sequence $\{r(n)\}$ is strictly increasing. If $Q(B)V^n = Q(B)A,$ then we are done because $V^n$ is finite dimensional over $k$ and hence $Q(B)V^n$ is finite dimensional over $Q(B).$ Now suppose that the sequence $\{r(n)\}$ is strictly increasing. Then $r(n) > n$ because $r(0)=\dim_{Q(B)}Q(B)=1.$ Fix an integer $n$ and let $e_1, \ldots , e_{r(n)}$ be a $Q(B)$-basis for $Q(B)V^n.$ Clearly we may assume that $e_i \in V^n$ for all $i.$ Let $W$ be a frame of a finitely generated subalgebra of $B.$ Then

$(V+W)^{2n} \supseteq W^nV^n \supseteq W^ne_1 + \ldots + W^ne_{r(n)},$

which gives us

$\dim_k(V+W)^{2n} \geq r(n) \dim_k W^n > n \dim_k W^n,$

because the sum $W^ne_1 + \ldots + W^ne_{r(n)}$ is direct. Therefore ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B),$ which is a contradiction. So we have proved that the second possibility is in fact impossible and hence $Q(B)A$ is finite dimensional over $Q(B).$ Finally, since, as we just proved, $\dim_{Q(B)}Q(B)A < \infty,$ the domain $Q(B)A$ is algebraic over $Q(B)$ and thus it is a division algebra. Hence $Q(B)A=Q(A)$ because $A \subseteq Q(B)A \subseteq Q(A)$ and $Q(A)$ is the smallest division algebra containing $A. \Box$