## A theorem of Borho and Kraft

Posted: April 2, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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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 $A. \Box$

## The singular submodule

Posted: December 7, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Throughout $R$ is a ring with 1 and all modules are left $R$-modules. In Definition 2 in this post, we defined $Z(M),$ the singular submodule of a module $M.$

Problem 1. Let $M$ be an $R$-module and suppose that $N_1, \cdots, N_k$ are submodules of $M.$ Prove that $\bigcap_{i=1}^k N_i \subseteq_e M$ if and only if $N_i \subseteq_e M$ for all $i.$

Solution. We only need to solve the problem for $k = 2.$ If $N_1 \cap N_2 \subseteq_e M,$ then $N_1 \subseteq_e M$ and $N_2 \subseteq_e M$ because both $N_1$ and $N_2$ contain $N_1 \cap N_2.$ Conversely, let $P$ be a nonzero submodule of $M.$ Then $N_1 \cap P \neq \{0\}$ because $N_1 \subseteq_e M$ and therefore $(N_1 \cap N_2) \cap P = N_2 \cap (N_1 \cap P) \neq \{0\}$ because $N_2 \subseteq_e M. \ \Box$

Problem 2. Prove that if $M$ is an $R$-module, then $Z(M)$ is a submodule of $M$ and $Z(R)$ is a proper two-sided ideal of $R.$ In particular, if $R$ is a simple ring, then $Z(R)=\{0\}.$

Solution. First note that $0 \in Z(M)$ because $\text{ann}(0)=R \subseteq_e R.$ Now suppose that $x_1,x_2 \in Z(M).$ Then $\text{ann}(x_1+x_2) \supseteq \text{ann}(x_1) \cap \text{ann}(x_2) \subseteq_e M,$ by Problem 1. Therefore $\text{ann}(x_1+x_2) \subseteq_e M$ and hence $x_1+x_2 \in Z(M).$ Now let $r \in R$ and $x \in Z(M).$ We need to show that $rx \in Z(M).$ Let $J$ be a nonzero left ideal of $R.$ Then $Jr$ is also a left ideal of $R.$ If $Jr = \{0\},$ then $J \subseteq \text{ann}(rx)$ and thus $\text{ann}(rx) \cap J = J \neq \{0 \}.$ If $Jr \neq \{0\},$ then $\text{ann}(x) \cap Jr \neq \{0\}$ because $x \in Z(M).$ So there exists $s \in J$ such that $sr \neq 0$ and $srx = 0.$ Hence $0 \neq s \in \text{ann}(rx) \cap J.$ So $rx \in Z(M)$ and thus $Z(M)$ is a submodule of $M.$ Now, considering $R$ as a left $R$-module, $Z(R)$ is a left ideal of $R,$ by what we have just proved. To see why $Z(M)$ is a right ideal, let $r \in R$ and $x \in Z(R).$ Then $\text{ann}(xr) \supseteq \text{ann}(x) \subseteq_e R$ and so $\text{ann}(xr) \subseteq_e R,$ i.e. $xr \in Z(R).$ Finally, $Z(R)$ is proper because $\text{ann}(1)=\{0\}$ and so $1 \notin Z(R). \ \Box$

Problem 3. Prove that if $M_i, \ i \in I,$ are $R$-modules, then $Z(\bigoplus_{i \in I} M_i) = \bigoplus_{i \in I} Z(M_i).$ Conclude that if $R$ is a semisimple ring, then $Z(R)=\{0\}.$

Solution. The first part is a trivial result of Problem 1 and this fact that if $x = x_1 + \cdots + x_n,$ where the sum is direct, then $\text{ann}(x) = \bigcap_{i=1}^n \text{ann}(x_i).$ The second now follows trivially from the first part, Problem 2 and the Wedderburn-Artin theorem. $\Box$

Problem 4. Suppose that $R$ is commutative and let $N(R)$ be the nilradical of $R.$ Prove that

1) $N(R) \subseteq Z(R);$

2) it is possible to have $N(R) \neq Z(R);$

3) if $Z(R) \neq \{0\},$ then $N(R) \subseteq_e Z(R),$ as $R$-modules or $Z(R)$-modules.

Solution. 1) Let $a \in N(R).$ Then $a^n = 0$ for some integer $n \geq 1.$ Now suppose that $0 \neq r \in R.$ Then $ra^n=0.$ Let $m \geq 1$ be the smallest integer such that $ra^m = 0.$ Then $0 \neq ra^{m-1} \in \text{ann}(a) \cap Rr$ and hence $a \in Z(R).$

2) Let $R_i = \mathbb{Z}/2^i \mathbb{Z}, \ i \geq 1$ and put $R=\prod_{i=1}^{\infty}R_i.$ For every $i,$ let $a_i = 2 + 2^i \mathbb{Z}$ and consider $a = (a_1,a_2, \cdots ) \in R.$ It is easy to see that $a \in Z(R) \setminus N(R).$

3) Let $a \in Z(R) \setminus N(R).$ Then $\text{ann}(a) \cap Ra \neq \{0\}$ and thus there exists $r \in R$ such that $ra \neq 0$ and $ra^2=0.$ Hence $(ra)^2 = 0$ and so $ra \in N(R).$ Thus $0 \neq ra \in N(R) \cap Ra$ implying that $N(R)$ is an essential $R$-submodule of $Z(R).$ Now, we view $Z(R)$ as a ring and we want to prove that $N(R)$ as an essential ideal of $Z(R).$ Again,  let $a \in Z(R) \setminus N(R).$ Then $\text{ann}(a) \cap Ra^2 \neq \{0\}$ and thus there exists $r \in R$ such that $ra^2 \neq 0$ and $ra^3 = 0.$ Let $s = ra \in Z(R).$ Then $(sa)^2=0$ and thus $0 \neq sa \in N(R) \cap Z(R)a$ implying that $N(R)$ is an essential ideal of $Z(R). \ \Box$

## A singular module is a quotient of a module by an essential submodule

Posted: December 4, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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We will assume that $R$ is a ring (not necessarily commutative) with 1 and all modules are left $R$-modules.

Definition 1. Let $M$ be an $R$-module and $N$ a nonzero submodule of $M.$ We say that $N$ is an essential submodule of $M,$ and we will write $N \subseteq_e M,$ if $N \cap X \neq (0)$ for any nonzero submodule $X$ of $M.$ Clearly, that is equivalent to saying $N \cap Rx \neq (0)$ for any nonzero element $x \in M.$ So, in particular, a nonzero left ideal $I$ of $R$ is an essential left ideal of $R$ if $I \cap J \neq (0)$ for any nonzero left ideal $J$ of $R,$ which is equivalent to the condition $I \cap Rr \neq (0)$ for any nonzero element $r \in R.$

Definition 2. Let $M$ be an $R$-module and $x \in M.$ Recall that the (left) annihilator of $x$ in $R$ is defined by $\text{ann}(x)=\{r \in R: \ rx = 0 \},$ which is obviously a left ideal of $R.$ Now, consider the set $Z(M):=\{x \in M: \ \text{ann}(x) \subseteq_e R \}.$ It is easy to see that $Z(M)$ is a submodule of $M$ (see Problem 2 in this post for the proof!) and we will call it the singular submodule of $M.$ If $Z(M)=M,$ then $M$ is called singular. If $Z(M)=(0),$ then $M$ is called nonsingular. We will not discuss nonsingular modules in this post.

Problem 1. Prove that if $F$ is a free $R$-module, then $Z(F) \neq F,$ i.e. a free module is never singular.

Solution. Let $x \in F$ be any element of an $R$-basis of $F.$ Let $r \in \text{ann}(x).$ Then $rx = 0$ and so $r = 0.$ Thus $\text{ann}(x)=(0)$ and so $x \notin Z(F). \ \Box$

Next problem characterizes singular modules.

Problem 2. Prove that an $R$-module $M$ is singular if and only if $M = A/B$ for some $R$-module $A$ and some submodule $B \subseteq_e A.$

Solution. Suppose first that $M=A/B$ where $A$ is an $R$-module and $B \subseteq_e A.$ Let $x = a + B \in M$ and let $J$ be a nonzero left ideal of $R.$ If $Ja = (0),$ then $Ja \subseteq B$ and so $\text{ann}(x) \cap J = J \neq (0).$ If $Ja \neq (0),$ then $B \cap Ja \neq (0)$ because $B \subseteq_e A.$ So there exists $r \in J$ such that $0 \neq ra \in B.$ That means $0 \neq r \in \text{ann}(x) \cap J.$ So we have proved that $x \in Z(M)$ and hence $Z(M)=M,$ i.e. $M$ is singular. Conversely, suppose that $M$ is singular. We know that every $R$-module is the homomorphic image of some free $R$-module. So there exists a free $R$-module $F$ and a submodule $K$ of $F$ such that $M \cong F/K.$ So we only need to show that $K \subseteq_e F.$ Note that $K \neq (0),$ by Problem 1. Let $\{x_i \}$ be an $R$-basis for $F$ and suppose that $0 \neq x \in F.$ We need to show that there exists $s \in R$ such that $0 \neq sx \in K.$ We can write, after renaming the indices if necessarily,

$x = \sum_{i=1}^n r_ix_i, \ \ \ \ \ \ \ \ \ \ (1)$

where $r_1 \neq 0.$ For any $y \in F,$ let $\overline{y}=y+K \in F/K.$ Now, since $F/K$ is singular, $\text{ann}(\overline{x_1}) \subseteq_e R$ and so $\text{ann}(\overline{x_1}) \cap Rr_1 \neq (0).$ So there exists $s_1 \in R$ such that $s_1r_1 \neq 0$ and $s_1r_1x_1 \in K.$ Hence $(1)$ gives us

$s_1x= s_1r_1x_1 +\sum_{i=2}^n s_1r_ix_i. \ \ \ \ \ \ \ \ \ (2)$

Note that $s_1x \neq 0$ because $s_1r_1 \neq 0.$ Now, if $s_1r_i = 0$ for all $2 \leq i \leq n,$ then $s_1x =s_1r_1x_1 \in K$ and we are done. Otherwise, after renaming the indices in the sum on the right hand side of $(2)$ if necessary, we may assume that $s_1r_2 \neq 0.$ Repeating the above process gives us some $s_2 \in R$ such that $s_2s_1r_2 \neq 0$ and $s_2s_1r_2x_2 \in K.$ Then $(2)$ implies

$s_2s_1x = s_2s_1r_1x_1 + s_2s_1r_2x_2 + \sum_{i=3}^n s_2s_1r_ix_i. \ \ \ \ \ \ \ (3)$

The first two terms on the right hand side of $(3)$ are in $K$ and $s_2s_1x \neq0$ because $s_2s_1r_2 \neq 0.$ If we continue this process, we will eventually have a positive integer $1 \leq m \leq n$ and $s = s_ms_{m-1} \cdots s_1 \in R$ such that $0 \neq sx \in K. \ \Box$

## Degree of an irreducible factor of composition of two polynomials

Posted: November 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Let $F$ be a field and suppose that $f(x),g(x), p(x)$ are three polynomials in $F[x].$ Prove that if both $f(x)$ and $p(x)$ are irreducible and $p(x) \mid f(g(x)),$ then $\deg f(x) \mid \deg p(x).$

Solution. Let $\mathfrak{m}$ and $\mathfrak{n}$ be the ideals of $F[x]$ generated by $f(x)$ and $p(x),$ respectively. Let $E=F[x]/\mathfrak{m}$ and $L = F[x]/\mathfrak{n}.$ Since both $f(x)$ and $p(x)$ are irreducible, $E$ and $L$ are field extensions of $F.$ Now, define the map $\varphi : E \longrightarrow L$ by $\varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n},$ for all $h(x) \in F[x].$ We first show that $\varphi$ is well-defined. To see this, suppose that $h(x) \in \mathfrak{m}.$ Then $h(x)=f(x)u(x)$ for some $u(x) \in F[x]$ and hence $h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n},$ because $p(x) \mid f(g(x)).$ So $\varphi$ is well-defined. Now $\varphi$ is clearly a ring homomorphism and, since $E$ is a field and $\ker \varphi$ is an ideal of $E,$ we must have $\ker \varphi = \{0\}.$ Therefore we may assume that $F \subseteq E \subseteq L$ and hence $\deg p(x) = [L : F] = [L: E][E:F]=(\deg f(x))[L : E]. \ \Box$

## The Krull dimension of a commutative ring; another view

Posted: November 3, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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We will assume that $R$ is a commutative ring with $1.$ We will allow the case $R=(0),$ i.e. where $1_R=0_R.$ The goal is to describe  the Krull dimension of $R$ in terms of elements of $R$ and not prime ideals of $R.$ The reference is this paper: A Short Proof for the Krull Dimension of a Polynomial Ring. Recall that the Krull dimension of $R$ is the largest integer $n \geq 0$ for which there exist prime ideals $P_i, \ 0 \leq i \leq n,$ of $R$ such that $P_0 \subset P_1 \subset \ldots \subset P_n.$ Then we write $\text{K.dim}(R)=n.$ If there is no such integer, then we define $\text{K.dim}(R)=\infty$ and if $R=(0),$ we define $\text{K.dim}(R)=-1.$

Notation. Let $x \in R$ and set $S_x = \{x^n(1+ax): \ n \geq 0, \ a \in R \}.$ Clearly $S_x$ is multiplicatively closed. Let $R_x$ be the localization of $R$ at $S_x.$ If $0 \in S_x,$ then we define $R_x =(0).$

Problem 1. Let $x \in R$ and suppose that $\mathfrak{m}$ is a maximal ideal of $R.$ Then $\mathfrak{m} \cap S_x \neq \emptyset.$ Moreover, if $P \subset \mathfrak{m}$ is a prime ideal and $x \in \mathfrak{m} \setminus P,$ then $P \cap S_x = \emptyset.$

Solution. if $x \in \mathfrak{m},$ then $x \in \mathfrak{m} \cap S_x$ and we are done. Otherwise, $\mathfrak{m} + Rx = R$ and thus $1 + ax \in \mathfrak{m}$ for some $a \in R.$ Then $1+ax \in \mathfrak{m} \cap S_x.$ For the second part, suppose to the contrary that $x^n(1+ax) \in P$ for some integer $n \geq 0$ and $a \in R.$ Then, since $x \notin P,$ we have $1+ax \in P \subset \mathfrak{m}$ and thus $1 \in \mathfrak{m}$ because $x \in \mathfrak{m}. \Box$

Problem 2. $\text{K.dim}(R)=0$ if and only if $R_x=\{0\}$ for all $x \in R.$

Solution. As we already mentioned, $R_x = \{0\}$ means $0 \in S_x.$ Suppose that $0 \notin S_x$ for some $x \in R.$ Then there exists a prime ideal $P$ of $R$ such that $P \cap S_x = \emptyset,$ because $S_x$ is multiplicatively closed. By Problem 1, $P$ is not a maximal ideal and thus $\text{K.dim}(R) > 0.$ Conversely, suppose that $0 \in S_x$ for all $x \in R$ and $P$ is a prime ideal of $R.$ Let $\mathfrak{m}$ be  a maximal ideal of $R$ which contains $P.$ Choose $x \in \mathfrak{m} \setminus P.$ By Problem 1, $P \cap S_x =\emptyset,$ contradicting $0 \in P \cap S_x. \ \Box$

Problem 3. Let $n \geq 0$ be an integer. Then $\text{K.dim}(R) \leq n$ if and only if $\text{K.dim}(R_x) \leq n-1$ for all $x \in R.$

Solution. The case $n=0$ was done in Problem 2. So we’ll assume that $n \geq 1.$ Suppose that $\text{K.dim}(R) \leq n$ and let $x \in R.$ A prime ideal of $R_x$ is in the form $PR_x,$ where $P$ is a prime ideal of $R$ and $P \cap S_x = \emptyset.$ Also, if $PR_x$ and $QR_x$ are two prime ideals of $R_x$ with $PR_x \subset QR_x,$ then $P \subset Q.$ Now, suppose to the contrary that $\text{K.dim}(R_x) \geq n$ and consider the chain $P_0R_x \subset P_1R_x \subset \ldots \subset P_n R_x$ of prime ideals of $R_x.$ By Problem 1, $P_n$ is not a maximal ideal and so there exists a prime ideal $P_{n+1}$ such that $P_n \subset P_{n+1}$ and that will give us the chain $P_0 \subset P_1 \subset \ldots \subset P_{n+1}$ of prime ideals of $R,$ contradicting $\text{K.dim}(R) \leq n.$ Conversely, suppose that $\text{K.dim}(R_x) \leq n-1$ for all $x \in R.$ Suppose also, to the contrary, that there exists a chain $P_0 \subset P_1 \subset \ldots \subset P_{n+1}$ of prime ideals of $R.$ Let $x \in P_{n+1} \setminus P_n.$ By the second part of Problem 1, $P_n \cap S_x = \emptyset$ and thus $P_0R_x \subset P_1R_x \subset \ldots \subset P_nR_x$ is a chain of prime ideals of $R_x,$ contradicting $\text{K.dim}(R_x) \leq n-1. \ \Box$

Problem 4. Let $n \geq 0$ be an integer. Then $\text{K.dim}(R) \leq n$ if and only if for every $x_0, x_1, \ldots , x_n \in R$ there exist integers $k_0, k_1, \ldots , k_n \geq 0$ and $a_0, a_1, \ldots ,a_n \in R$ such that

$x_n^{k_n}( \ldots (x_2^{k_2}(x_1^{k_1}(x_0^{k_0}(1+a_0x_0)+a_1x_1) + a_2x_2) + \ldots) + a_nx_n)=0.$

Solution. By Problem 3, $\text{K.dim}(R) \leq n$ if and only if $\text{K.dim}((\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n})=-1.$ Therefore $\text{K.dim}(R) \leq n$ if and only if $(\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n}=\{0\}. \ \Box$

Throughout $G$ is a group with the center $Z(G)$ and the commutator subgroup $G'.$ The goal is to prove that if $G/Z(G)$ is finite, then $G'$ is finite too. We will also find an upper bound for $|G'|$ in terms of $|G/Z(G)|.$

Notation. For any $a,b \in G,$ we define $[a,b]=aba^{-1}b^{-1}$ and $a^b = bab^{-1}.$

Problem 1. Let $a,b,c \in G.$

1) $[a,b]=b^ab^{-1}, \ a^a=a, \ ba=a^bb$ and $[a,b]^c=[a^c,b^c].$

2) $c[a,b]=[a^c,b^c]c.$

3) If $[G:Z(G)]=n,$ then $[a,b]^{n+1}=[a,b^2][a^b,b]^{n-1}.$

Proof. 1) The first three identities are trivial and the fourth one is true because the map $f:G \longrightarrow G$ defined by $f(g)=cgc^{-1}=g^c$ is a group homomorphism.

2) By 1) we have $[a^c,b^c]c =[a,b]^cc=c[a,b].$

3) So $g^n \in Z(G)$ for all $g \in G,$ because $[G:Z(G)]=n.$ So $[a,b]^nb^{-1}=b^{-1}[a,b]^n$ with 1) give us

$[a,b]^{n+1}=[a,b][a,b]^n =b^ab^{-1}[a,b]^n = b^a[a,b]^n b^{-1}=b^a[a,b][a,b]^{n-1}b^{-1}$

$=b^ab^ab^{-1}[a,b]^{n-1}b^{-1}= (b^2)^ab^{-2}b[a,b]^{n-1}b^{-1}=[a,b^2](b[a,b]b^{-1})^{n-1}=[a,b^2]([a,b]^b)^{n-1}$

$=[a,b^2][a^b,b^b]^{n-1}=[a,b^2][a^b,b]^{n-1}. \ \Box$

Problem 2. Let $C=\{[a,b]: \ a,b \in G\}.$ If $[G:Z(G)]=n,$ then $|C| \leq n^2.$

Proof. Define the map $\varphi : C \longrightarrow G/Z(G) \times G/Z(G)$ by $\varphi([a,b])=(aZ(G),bZ(G)).$ If we prove that $\varphi$ is one-to-one, we are done because then $|C| \leq |G/Z(G) \times G/Z(G)|=n^2.$ So suppose that $\varphi([a,b])=\varphi([c,d]).$ Then $aZ(G)=cZ(G)$ and $bZ(G)=dZ(G)$ and hence $a^{-1}c \in Z(G)$ and $b^{-1}d \in Z(G).$ Therefore

$[a,b]=aba^{-1}b^{-1}=ab(a^{-1}c)c^{-1}b^{-1} =a(a^{-1}c)bc^{-1}b^{-1}=cbc^{-1}b^{-1} =cbc^{-1}(b^{-1}d)d^{-1}$

$=cb(b^{-1}d)c^{-1}d^{-1}=cdc^{-1}d^{-1}=[c,d]. \ \Box$

Schur’s theorem. If $[G:Z(G)]=n,$ then $\displaystyle |G'| \leq n^{2n^3}.$

Solution. Let $C=\{[a,b]: \ a,b \in G\}$ and $c \in G'.$ Then $c = c_1c_2 \ldots c_m,$ where $c_i \in C.$ We will choose the integer $m$ to be as small as possible, i.e. if $c = c_1'c_2' \ldots c_k',$ with $c_i' \in C,$ then $k \geq m.$ Now, we know from Problem 2, that the number of elements of $C$ is at most $n^2.$ So if we prove that each $c_i$ can occur at most $n$ times in $c = c_1c_2 \cdots c_m,$ then $m \leq n|C| \leq n^3$ and thus there will be at most $(n^2)^m \leq n^{2n^3}$ possible values for $c$ and the problem is solved. To prove that each $c_i$ can occur at most $n$ times in $c = c_1c_2 \ldots c_m,$ suppose, to the contrary, that, say $c_j,$ occurs $r \geq n+1$ times in the product. Then by part 2) of Problem 1, we can move each $c_j$ to the right hand side of the product to get $c = c_1'c_2' \ldots c_s' c_j^r,$ where $c_i' \in C$ and $r + s = m.$ But by part 3) of Problem 1, $c_j^{n+1}$ is a product of $n$ elements of $C$ and hence, since $c_j^r=c_j^{r-(n+1)}c_j^{n+1},$ we see that $c_j^r$ is a product of $r-1$ elements of $C.$ Therefore $c$ is a product of $s+r-1=m-1$ elements of $C,$ which contradicts the minimality of $m. \ \Box$

If $G$ is finitely generated, then the converse of Schur’s theorem is also true, i.e. if $G'$ is finite, then $G/Z(G)$ is finite too. It’s not hard, try to prove it!

## 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.