Posts Tagged ‘semisimple’

Throughout k is a field, K is the algebraic closure of k and A is a finite dimensional central simple k-algebra.

Lemma. A \otimes_k K \cong M_n(K), for some integer n.

Proof. Let S:=A \otimes_k K. By the first part of the corollary in this post we know that S is simple. We also have

Z(S) = Z(A) \otimes_k K = k \otimes_k K \cong K.

It is easy to see that if \{a_i \} is a k-basis for A, then \{a_i \otimes_k 1 \} is an K-basis for S. Thus \dim_K S = \dim_k A. So S is a finite dimensional central simple K-algebra and hence, since K is algebraically closed, S \cong M_n(K), for some n, by Remark 2 in this post. \Box

Theorem. If A is a finite dimensional central simple k-algebra, then \dim_k A is a perfect square.

Proof.  By the lemma, there exists an integer n such that A \otimes_k K \cong M_n(K). Thus

\dim_k A = \dim_K A \otimes_k K = \dim_K M_n(K) = n^2. \ \Box

Definition. The degree of A is defined by \deg A = \sqrt{\dim_k A}.

Remark. Let R be a finite dimensional k-algebra. Then R is reduced if and only if R is a finite direct product of finite dimensional division k-algebras. In this case, \dim_k R=n^2\dim_k Z(R) for some integer n \geq 1.

Proof. obviously R is (left) Artinian because \dim_k R < \infty and so J(R) is nilpotent. Thus J(R)=(0) because R is reduced and so R is semisimple. The result now follows from the Artin-Wedderburn theorem. The converse is trivial. Finally, the fact that, by the above theorem, \dim_{Z(D)}D is a perfect square for any finite dimensional division algebra D, proves the last part of the remark.  \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.

For the first part see here.

6) If R is simple, then Z(R)=Z(Q).

Proof. Let x=s^{-1}a \in Z(Q). Then from xs=sx we get sa=as and thus s^{-1}a=as^{-1}. Hence for every b \in R we’ll have s^{-1}ab=bs^{-1}a=bas^{-1}, which gives us abs=sba. Also, since R is simple, RsR=R, which means \sum_{i=1}^n b_isc_i = 1, for some b_i, \ c_i \in R. Thus \sum_{i=1}^n sb_iac_i = \sum_{i=1}^n ab_isc_i = a=sx and so x=\sum_{i=1}^n b_iac_i \in R. Therefore x \in Z(R) which proves Z(Q) \subseteq Z(R). Conversely, let b \in Z(R) and x=s^{-1}a \in Q. Since bs=sb, we have s^{-1}b=bs^{-1} and thus bx=bs^{-1}a=s^{-1}ba=s^{-1}ab=xb and so b \in Z(Q). \Box

7) The left uniform dimension of R and Q are equal.

Proof. We saw in the previous section that the left ideals of Q are exactly in the form QI, where I is a left ideal of R. Clearly \sum QI_i is direct iff \sum I_i is direct.

8) Let N be a nilpotent ideal of R and let I be the right annihilator of N in R. Then I is an essential left ideal of R and hence QI  is an essential left ideal of Q.

Proof. Let I be the right annihilator of N in R. For an essential left ideal J of R the left ideal QJ of Q is essential in Q because for every non-zero left ideal K of R : (0) \neq \ Q(J \cap K) \subseteq QJ \cap QK. So we only need to prove the first part of the claim. Let J be any non-zero left ideal of R and put n=\min \{k \geq 0 : \ N^k J \neq (0) \}. Then (0) \neq N^n J \subseteq I \cap J.

9) If Q is semisimple, then R is semiprime.

Proof. So we need to prove that R has no non-zero nilpotent ideal. Suppose that N is a nilpotent ideal of R and let I be the right annihilator of N in R. Since Q is semisimple, QI \oplus A = Q, for some left ideal A of Q. But, from the previous fact, we know that QI is essential in Q and thus A=(0), i.e. QI=Q. Thus s^{-1}a=1, for some a \in I=\text{r.ann}_R N. So s=a and Ns=Na=(0). Thus N=Nss^{-1}=(0).

We proved, in the previous section, that if R is prime, then Q is prime too.

10) If Q is simple, then R is prime.

Proof. Let I,J be two non-zero ideals of R. We need to show that IJ \neq (0). We have QIQ=Q, because I \neq (0) and Q is simple. Therefore 1=\sum_{i=1}^n x_ia_iy_i, for some x_i,y_i \in Q and a_i \in I. We can write x_i = s^{-1}b_i, for some b_i \in R. Then s=\sum_{i=1}^n b_ia_iy_i \in IQ. So IQ is a right ideal of Q which contains a unit. Thus IQ=Q. Similarly JQ=Q and hence IJQ=Q. As a result, IJ \neq (0). \Box

Notation. For a ring R let J(R) be the Jacobson radical of R.

Definition. Recall that if k is a field and G is a group, then the group algebra k[G] has two structures. Firstly, as a vector space over k, it has G as a basis, i.e. every element of k[G] is uniquely written as \sum_{g \in G} x_g g, where x_g \in k. In particular, \dim_k k[G]=|G|, as cardinal numbers. Secondly, multiplication is also defined in k[G]. If x = \sum_{g \in G} x_g g and y = \sum_{g \in G} y_g g are two elements of k[G], then we just multiply xy in the ordinary fashion using distribution law. To be more precise, we define xy = \sum_{g \in G} z_g g, where z_g = \sum_{rs=g} x_r y_s.

We are going to prove that J(\mathbb{C}[G])=0, for every group G.

Lemma. J(\mathbb{C}[G]) is nil, i.e. every element of J(\mathbb{C}[G]) is nilpotent.

Proof. If G is countable, we are done by this theorem. For the general case, let \alpha \in J(\mathbb{C}[G]). So \alpha =\sum_{i=1}^n c_ig_i, for some c_i \in \mathbb{C}, \ g_i \in G. Let H=\langle g_1,g_2, \cdots , g_n \rangle. Clearly \alpha \in H and H is countable. So to complete the proof, we only need to show that \alpha \in J(\mathbb{C}[H]). Write G = \bigcup_i x_iH, where x_iH are the distinct cosets of H in G. Then \mathbb{C}[G]=\bigoplus_i x_i \mathbb{C}[H], which means \mathbb{C}[G]=\mathbb{C}[H] \oplus K, for some right \mathbb{C}[H] module K. Now let \beta \in \mathbb{C}[H]. Since  \alpha \in J(\mathbb{C}[G]), there exists \gamma \in \mathbb{C}[G] such that \gamma (1 - \beta \alpha ) = 1. We also have \gamma = \gamma_1 + \gamma_2, for some \gamma_1 \in \mathbb{C}[H], \ \gamma_2 \in K. That now gives us \gamma_1(1 - \beta \alpha)=1. \ \Box

Theorem. J(\mathbb{C}[G])=0, for any group G.

Proof. For any x =\sum_{i=1}^n c_i g_i\in \mathbb{C}[G] define

x^* = \sum_{i=1}^n \overline{c_i} g_i^{-1}.

It’s easy to see that xx^*=0 if and only if x=0 and for all x,y \in \mathbb{C}[G]: \ (xy)^*=y^*x^*. Now suppose that J(\mathbb{C}[G]) \neq 0 and let 0 \neq \alpha \in J(\mathbb{C}[G]). Put \beta = \alpha \alpha^* \in J(\mathbb{C}[G]). By what I just mentioned \beta \neq 0 and (\beta^m)^* = (\beta^*)^m=\beta^m, for all positive integers m. By the lemma, there exists k \geq 2 such that \beta^k = 0 and \beta^{k-1} \neq 0. Thus \beta^{k-1} (\beta^{k-1})^* = \beta^{2k-2} = 0, which implies that \beta^{k-1} = 0. Contradiction! \Box

Corollary. If G is finite, then \mathbb{C}[G] is semisimple.

Proof. We just proved that J(\mathbb{C}[G])=(0). So we just need to show that \mathbb{C}[G] is Artinian. Let

I_1 \supset I_2 \supset \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

be a descending chain of left ideals of \mathbb{C}[G]. Obviously each I_j is a \mathbb{C}-subspace of \mathbb{C}[G]. Thus each I_j is finite dimensional because \dim_{\mathbb{C}} \mathbb{C}[G]=|G| < \infty. Hence (*) will stablize at some point because \dim_{\mathbb{C}} I_1 < \infty and \dim_{\mathbb{C}}I_1 > \dim_{\mathbb{C}} I_2 > \cdots . Thus \mathbb{C}[G] is (left) Artinian and the proof is complete because we know a ring is semisimple if and only if it is (left) Artinian and its Jacobson radical is zero. \Box