Tensor product of simple algebras (2)

Posted: January 22, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Throughout this post $k$ is a field. Recall that the center of a simple ring is a field (see Remark 1 in here).

Definition. A ring $A$ is called a central simple $k$-algebra if $A$ is simple and $Z(A)=k.$ If the dimension of $A,$ as a vector space over $k,$ is finite, then $A$ is called a finite dimensional central simple $k$-algebra.

Remark 1. Let $A$ be a finite dimensional central simple $k$-algebra and let $I_1 \supset I_2 \supset \cdots$ be a descending chain of left ideals of $A.$ Each $I_j$ is clearly a $k$-vector subspace of $A$ and therefore $\dim_k I_j < \infty.$ Thus, since $\dim_k I_1 > \dim_k I_2 > \cdots ,$ the chain must stop at some point. This shows that every finite dimensional central simple $k$-algebra $A$ is (left) Artinian. Hence $A$ is simple and Artinian and thus, by the Wedderburn-Artin theorem, $A = M_n(D),$ for some positive integer $n$ and some division ring $D.$ Note that

$k = Z(A)=Z(M_n(D)) \cong Z(D).$

Let $Z=Z(D).$ Then, since $\dim_k A < \infty$ and

$\dim_k A = \dim_Z M_n(D)=(\dim_Z D)(\dim_D M_n(D))=n^2 \dim_Z D,$

we have $\dim_Z D < \infty.$

Note. Since $Z \cong k,$ we may assume that $k$ is the center of $D$ and write $\dim_k D$ instead of $\dim_Z D.$

Remark 2. If $k$ is algebraically closed and $A$ is a finite dimensional central simple $k$-algebra, then $A \cong M_n(k),$  for some $n.$ To see this, we first apply Remark 1 to get a division ring $D$ with $Z=Z(D) \cong k$ and $\dim_Z D < \infty$ such that $A = M_n(D).$ So $Z$ is algebraically closed. Now if $d \in D,$ then $Z[d]$ is an algebraic field extension of $Z$ because $\dim_Z D < \infty.$ Thus $Z[d]=Z,$ because $Z$ is algebraically closed. So $D =Z \cong k$ and hence $A \cong M_n(k).$

Lemma. Let $A$ and $B$ be two $k$-algebras and let $C:=A \otimes_k B.$ Then

1) $Z(C)=k$ if and only if $Z(A)=Z(B)=k.$

2) If $A$ is central simple, then the centralizer of $A$ in $C$ is $B.$

Proof. 1) If $Z(A)=Z(B)=k,$ then by Lemma 2

$Z(C)=Z(A) \otimes_k Z(B)=k \otimes_k k = k.$

Conversely, if $Z(C)=k,$ then

$1 = \dim_k Z(C) =\dim_k Z(A) \otimes_k Z(B) = (\dim_k Z(A))(\dim_k Z(B)),$

and hence $\dim_k Z(A)=\dim_k Z(B)=1.$ Therefore $Z(A)=Z(B)=k.$

2) Just to avoid any possible confusion, I mention that we have identified $A$ and $B$ with $A \otimes_ k 1$ and $1 \otimes_k B$ here. It is obvious that every element of $1 \otimes_k B$ commutes with every element of $A \otimes_k 1.$ So the centralizer of $A \otimes_k 1$ in $C$ contains $1 \otimes_k B.$ Conversely, let $c \in C$ be in the centralizer of $A \otimes_k 1$ in $C.$ Let $\{b_i\}$ be a $k$-basis for $B.$ Then there exist $a_i \in A$ such that $c = \sum a_i \otimes_k b_i.$ Since $c$ is in the centralizer of $A \otimes_k 1,$ we must have $(a \otimes_k 1)c = c(a \otimes_k 1),$ for all $a \in A.$
Thus $\sum (aa_i-a_ia) \otimes_k b_i=0$ and so by Lemma 1, $aa_i=a_ia,$ for all $i.$ Hence $a_i \in Z(A)=k$ and so $c = \sum a_i \otimes_k b_i = \sum 1 \otimes_k a_ib_i = 1 \otimes_k \sum a_ib_i \in 1 \otimes_k B. \ \Box$

Definition. A ring $R$ is called von Nemann regular, or just regular, if for every $a \in R$ there exists $x \in R$ such that $a=axa.$

Remark 1. Regular rings are semiprimitive. To see this, let $R$ be a regular ring. Let $a \in J(R),$ the Jacobson radical of $R,$ and choose $x \in R$ such that $a=axa.$ Then $a(1-xa)=0$ and, since $1-xa$ is invertible because $a$ is in the Jacobson radical of $R,$ we get $a=0.$

Examples 1. Every division ring is obviously regular because if $a = 0,$ then $a=axa$ for all $x$ and if $a \neq 0,$ then $a=axa$ for $x = a^{-1}.$

Example 2. Every direct product of regular rings is clearly a regular ring.

Example 3. If $V$ is a vector space over a division ring $D,$ then ${\rm End}_D V$ is regular.

Proof. Let $R={\rm End}_D V$ and $f \in R.$ There exist vector subspaces $V_1, V_2$ of $V$ such that $\ker f \oplus V_1 = {\rm im}(f) \oplus V_2 = V.$  So if $u \in V,$ then $u=u_1+u_2$ for some unique elements $u_1 \in {\rm im}(f)$ and $u_2 \in V_2.$ We also have $u_1 = v_1 + v$ for some unique elements $v_1 \in \ker f$ and $v \in V_1.$ Now define $g: V \longrightarrow V$ by $g(u)=v.$ It is obvious that $g$ is well-defined and easy to see that $g \in R$ and $fgf=f. \ \Box$

Example 4. Every semisimple ring is regular.

Proof. For a division ring $D$ the ring $M_n(D) \cong End_D D^n$ is regular by Example 3. Now apply Example 2 and the Wedderburn-Artin theorem.

Theorem. A ring $R$ is regular if and only if every finitely generated left ideal of $R$ is generated by an idempotent.

Proof. Suppose first that every finitely generated left ideal of $R$ can be generated by an idempotent. Let $x \in R.$ Then $I=Rx = Re$ for some idempotent $e.$ That is $x = re$ and $e=sx$ for some $r,s \in R.$ But then $xsx=xe=re^2=re=x.$ Conversely, suppose that $R$ is regular. We first show that every cyclic left ideal $I=Rx$ can be generated by an idempotent. This is quite easy to see: let $y \in R$ be such that $xyx=x$ and let $yx=e.$ Clearly $e$ is an idempotent and $xe=x.$ Thus $x \in Re$ and so $I \subseteq Re.$ Also $e=yx \in I$ and hence $Re \subseteq I.$ So $I=Re$ and we’re done for this part. To complete the proof of the theorem we only need to show that if $J=Rx_1 + Rx_2,$ then there exists some idempotent $e \in R$ such that $J=Re.$ To see this, choose an idempotent $e_1$ such that $Rx_1=Re_1.$ Thus $J=Re_1 + Rx_2(1-e_1).$  Now choose an idempotent $e_2$ such that $Rx_2(1-e_1)=Re_2$ and put $e_3=(1-e_1)e_2.$ See that $e_3$ is an idempotent, $e_1e_3=e_3e_1=0$ and $Re_2=Re_3.$ Thus $J=Re_1 + Re_3.$ Let $e=e_1+e_3.$ Then $e$ is an idempotent and $J=Re. \Box$

Corollary. If the number of idempotents of a regular ring $R$ is finite, then $R$ is semisimple.

Proof. By the theorem, $R$ has only a finite number of left principal ideals. Since every left ideal is a sum of left principal ideals, it follows that $R$ has only a finite number of left ideals and hence it is left Artinian. Thus $R$ is semisimple because $R$ is semiprimitive by Remark 1. $\Box$

Remark 2. The theorem is also true for finitely generated right ideals. The proof is similar.

Remark 3. Since, by the Wedderburn-Artin theorem, a commutative ring is semisimple if and only if it is a finite direct product of fields, it follows from the Corollary that if the number of idempotents of a commutative von Neumann regular ring $R$ is finite, then $R$ is a finite direct product of fields.

Schur’s lemma states that if $A$ is a simple $R$ module, then $\text{End}_R(A)$ is a division ring. A similar easy argument shows that:

Example 6. For simple $R$-modules $A \ncong B$ we have $\text{Hom}_R(A,B)=\{0\}.$

Let’s generalize Schur’s lemma: let $M$ be a finite direct product of simple $R$-submodules. So $M \cong \bigoplus_{i=1}^k M_i^{n_i},$ where each $M_i$ is a simple $R$-module and $M_i \ncong M_j$ for all $i \neq j.$ Therefore, by Example 6 and Theorem 1, $\text{End}_R(M) \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i),$ where $D_i = \text{End}_R(M_i)$ is a division ring by Schur’s lemma. An important special case is when $R$ is a semisimple ring. (Note that simple submodules of a ring are exactly minimal left ideals of that ring.)

Theorem 2. (Artin-Wedderburn) Let $R$ be a semisimple ring. There exist a positive integer $k$ and division rings $D_i, \ 1 \leq i \leq ,$ such that $R \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i)$.

Proof. Obvious, by Example 1 and the above discussion. $\Box$

Some applications of Theorem 2.

1. A commutative semisimple ring is a finite direct product of fields.

2. A reduced semisimple ring is a finite direct product of division rings.

3. A finite reduced ring is a finite direct product of finite fields.

The group ring isomorphism problem

Posted: June 6, 2010 in Group Algebras, Noncommutative Ring Theory Notes
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Let $k$ be a field or $\mathbb{Z}$ and $G_1,G_2$ two groups. It is clear that if $G_1 \cong G_2,$ then $k[G_1] \cong k[G_2],$ as $k$ algebras. The group ring isomorphism problem is this question that whether or not $k[G_1] \cong k[G_2],$ as $k$ algebras, implies $G_1 \cong G_2$. Another version of the group ring isomorphism problem is this: given a group $G_1$ and a field $k$ find all groups $G_2$ such that $k[G_1] \cong k[G_2],$ as $k$ algebras. These questions have been answered in special cases only. For example, an old result due to Perlis and Walker states that if $G_1,G_2$ are finite abelian groups and $\mathbb{Q}[G_1] \cong \mathbb{Q}[G_2],$ as $\mathbb{Q}$ algebras, then $G_1 \cong G_2.$ In 2001 Hertweek found two non-isomorphic groups $G_1, G_2$ of order $2^{21}97^{28}$ such that $\mathbb{Z}[G_1] \cong \mathbb{Z}[G_2].$

For now, we’ll only show that it is possible to have $k[G_1] \cong k[G_2]$ and $G_1 \ncong G_2$:

Theorem. If $G_1, G_2$ are two finite abelian groups of order $n,$ then $\mathbb{C}[G_1] \cong \mathbb{C}[G_2] \cong \mathbb{C}^n$ as $\mathbb{C}$-algebras.

Proof. We have already seen in this post that $J(\mathbb{C}[G])=(0)$ for any group $G.$ So if $G$ is a finite abelian group of order $n,$ then $\mathbb{C}[G]$ is a commutative semisimple algebra. Thus, since $\mathbb{C}$ is algebraically closed, the Wedderburn-Artin theorem gives us $\mathbb{C}[G] \cong \mathbb{C}^n$ as $\mathbb{C}$-algebras. $\Box$

Example. Let $G_1$ be the Klein four-group and let $G_2$ be the cyclic group of order four. Then $G_1 \ncong G_2$ but, by the theorem, $\mathbb{C}[G_1] \cong \mathbb{C}[G_2] \cong \mathbb{C}^4$ as $\mathbb{C}$-algebras.