Posts Tagged ‘Wedderburn-Artin theorem’

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.

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.